PRESENTED TO 
 
 THB CORXELL UNITERSITT, 1870, 
 
 The Hon. William Kelly 
 
 Of Rhinebeck. 
 
arV19311 
 
 Ray's algebra part second 
 
 3 1924 031 223 302 
 olin,anx 
 
Cornell University 
 Library 
 
 The original of tliis book is in 
 tlie Cornell University Library. 
 
 There are no known copyright restrictions in 
 the United States on the use of the text. 
 
 http://www.archive.org/details/cu31924031223302 
 
BCLBCTIO EDUCATIONAL BERIES. 
 
 RAY'S 
 
 ALGEBRA 
 
 PART SECOND: 
 
 A WALYTI GAL TREATISE, 
 
 DESIGNED rOK 
 
 HIGH SCHOOLS AND COLLEGES. 
 
 By JOSEPH RAY, M. D. 
 
 FEOrEESOK or MATHEMATIOB IW WOODWAKB COllESB. 
 
 Stcreotspe Htiftton 
 
 CINCINNATI: 
 WILSON, HINKLE & GO. 
 
 PHII'A: CIAXTON, EEMSEN & HAFFELI'INGER. 
 NEW YORK: CLABK & MAYNAED. 
 
THE BEST AND CHEAPEST 
 
 MATHEMATICAL COURSE. 
 
 ARITHMETIC. 
 RAY'S ARITHMETIC, PART FIRST j very simple leseons for little 
 
 learners. 
 RAT'S ARITHMETIC, PART SECOND ; a complete text-book in 
 
 Mental Arithmetic, by induction and analysis. 
 RAY'S ARITHMETIC, PART THIRD ; for schools and academies i 
 
 a full and complete treatise, on the inductive and analytic methods. 
 KEY TO RAY'S ARITHMETIC, containing solutions to the questions. 
 
 With an appendix embracing Slate and Black Board exercises. 
 RAY'S HIGHER ARITHMETIC ; designed for advanced pupils — 
 
 embracing the ^igTlpy/|^J^pll<^^Hl^t^a n! T.WnBffiVfl of numbers, "svith 
 
 an extensive cour^(i^f^Oiijlfflerj*jlwal<!nia|tio|s/^Prcparin^. 
 
 JTSHHice of ; 
 
 RAT'S ALGEBRll Sill'l flHsfci idt^hrAo 
 
 mies ; a simple, wogre^ive,^d-^crci^^«kn 
 RAY'S ALGEBRA^^|^»Ti^!OirA^lJl^n^JJ^cal Treatise, era 
 
 bracing Sturm's T'h X^ am, nt.H HTji-nar'Q M»ti;i^ /.f Approximation- 
 
 KEY TO RAY'S ALGEBRA, PARTS FIRST and SECOND j Com- 
 
 iscjl }ols and acade- 
 y treatise. 
 
 plete in one volume, 12 mo. 
 
 PREPARING FOR PUBLICATION, 
 
 ELEMENTS OF GEOMETRY ; embracing Plane and Solid Geome- 
 try, with numerous interesting and practical exercises. 
 
 TRIGONOMETRY and MENSURATION ; containing Logarithmic 
 Computations, Plane and Spherical Trigonometry, with their appli- 
 cations, and Mensuration of Planes and Solids, with Logarithmic 
 and other Tables. 
 
 SURVEYING AND NAVIGATION ; containing Surveying and Level- 
 ing, Navigation, Barometric higbts, <tc. 
 The above will be followed by works on Analytical Geometry : — 
 
 Mechanical Philosophy: — Differential and Integral Calculus: — and 
 
 Mathematical Astronomy. 
 Professor Ray, during a long period devoted to actual instruction in 
 
 these several branches, has prepared much of the material requisite 
 
 for the respective volumes. They will appear as rapidly as a due 
 
 regard to their careful publication will permit. 
 
 Entered according to act of Congress, in the year Eighteen Uundred and Fifty- 
 Two, by WlXTHROP B. SMiTn, in the Clerk's Offise of the District Court of the United 
 Btates, for the District of Ohio. 
 
 Stebeotyped by a. James. 
 
PREFACE. 
 
 Algeeka is justly regarded one of the most interesting and 
 useful branches of education, and an acquaintance with it is now 
 sought by all who progress beyond the more common elements. 
 To those who would know Mathematics, a knowledge not merely 
 of its elementary prLnciples, but also of its higher parts, is essen- 
 tial ; while no one cSh lay claim to that discipline of mind which 
 education confers, who is not familiar with the logic of Algebra. 
 
 It is both a demonstrative and a practical science — a system 
 of truths and reasoning, from which is derived a collection of 
 Rules that may be used in the solution of an endless variety of 
 problems, not only interesting to the student, but many of which 
 are of the highest possible utility in the arts of life. 
 
 The object of the present treatise is to present an outline of 
 this science in a brief, clear, and practical form. The aim 
 throughout has been to demonstrate every principle, and to fur- 
 nish the student the means of understanding clearly the rationale of 
 every process he is required to perform. No effort has been made 
 to simplify subjects by omitting that which is difficult, but rather 
 to present them in such a light as to render their acquisition 
 within the reach of all who will take the pains to study. 
 
 To fix the principles in the mind of the student, and to show 
 their bearing and utility, great attention has been paid to the 
 preparation of practical exercises. These are intended rather to 
 illustrate the principles of the science, than as difficult problems 
 to torture the ingenuity of the learner, or amuse the already skill- 
 6il Algebraist. 
 
 An effort has been made throughout the work, to observe a 
 natural and strictly logical connection between tlie different parts, 
 »o that the learner may not be required to rely on a principle, or 
 
IV PREFACE. 
 
 employ a process, with the rationale of which he is not alreadj 
 acquainted. The reference by Articles will always enable him 
 to trace any subject back to its first principles. 
 
 The limits of a preface will not permit a statement of the pecu 
 liarities of the work, nor is it necessary, as those who are in- 
 terested to know, will examine it for themselves. It is, however, 
 proper to remark, that Equations of the Second Degree have re- 
 ceived more than usual attention. The same may be said of 
 Radicals, of the Binomial Theorem, and of Logarithms, al! of 
 which are so useful in other branches of mathematics. 
 
 On some subjects it was necessary to be brief, to bring the 
 work within suitable limits. For example, what is here given of 
 the Theory of Equations, is to be regarded merely as an outline 
 of the more practical and interesting parts of the subject, which 
 alone is sufficient for a distinct treatise, as may be seen by refer- 
 ring to the works of Young or Hymers in English, or of DeFourcy 
 or Reynaud in French. 
 
 Some topics and exercises deemed both useful and interesting 
 will be found here, not hitherto presented to the notice of stu- 
 dents. But these, as well as the general manner of treating the 
 subject, are now submitted, with deference, to the intelligent 
 educational public, to whom the author is already greatly in- 
 debted for the favor with which his previous works have been 
 received. 
 
 WooDWABD School, May, 1852. 
 
 NOTICE. 
 
 (fir" A Ket to this work, containing solutions to the more dif- 
 ficult problems, with remarks and suggestions, intended princi 
 pally for private students, is now published. 
 
 The Key also embraces the Diophantine and Indeterminate 
 Analysis, with the Notation of Numbers, &c., subjects not 
 usually contained in the ordinary course of instruction in Higher 
 Algebra. 
 
CONTENTS. 
 
 CHAPTER I. 
 
 FUNDAMENTAL EULES. 
 
 AETICLES 
 
 Definitions and Notation ; . . . . 1 — 36 
 
 Exercises on the Definitions and Notation 36 
 
 Examples to be written in Algebraic Symbols 3t) 
 
 Addition — General Rule 37 — 41 
 
 Subtraction — Rule 42 — 45 
 
 Bracket, or Vinculum 46 
 
 Observations on Addition and Subtraction 47-^ 51 
 
 MuLTiPLiCikTioN — Preliminary principle 52 — 53 
 
 Rule of the CoeiEcients 55 
 
 Rule of the Exponents 56 
 
 Rule of the Signs 60 
 
 General Rule for Multiplication 61 
 
 Multiplication by Detached Coefficients. ... 62 
 
 Remarks on Algebraic Multiplication 63 — 66 
 
 Division — Rule of the Signs 67 — 69 
 
 Rule of the Coefficients — Exponents 70 
 
 Division of a Monomial by a Monomial 71 
 
 Division of Polynomials by Monomials 74 
 
 Division of one Polynomial by another 75 — 76 
 
 Division by Detached Coefficients 77 
 
 CHAPTER II. 
 
 THEOEEMS — FAOTORINS — GREATEST OOMMOK 
 
 DIVtSOE — LEAST COMMON MULTIPLE. 
 
 Algebraic Theorkms — Square of the sum of two quantities . 78 
 
 Square of the difTerence of two quantities 79 
 
 Product of the Sum and Difference of two quan- 
 tities 80 
 
 What the reciprocal of a quantity is equal to 81 
 
 Any quantity whose exponent is is ^1 g2 
 
 a*" — 6'" is divisible by a — 6 g3 
 
 a2i" — b^r" is divisible by a-\-b 85 
 
 a!m+r_[-42m+i is divisible by a-[-b 88 
 
 FAOTORiisa — Principles of factoring numbers 87 
 
 factoring of Algebraic quantities 88 95 
 
 Greatest Common Divisor 96 108 
 
 Least Common Moltiple X09 1 1 3 
 
VI CONTENTS 
 
 CHAPTER III. 
 
 ALGEBRAIC FRACTIONS. 
 
 AETIOLEa 
 
 Definitions — Proposition — Lowest terms 114 — 113 
 
 To reduce a Fraction to an Entire or Mixed quantity 121 
 
 To reduce a Mixed quantity to tlie form of a Fraction 122 
 
 Signs of Fractions 12? 
 
 To reduce Fractions to a Common Denominator 126 
 
 To reduce a quantity to a Fraction with a given Denominator. 127 — 128 
 
 Addition and Subtraction of Fractions 129 — 130 
 
 Multiplication and Division of Fractions 131 — 132 
 
 Reduction of Complex Fractions 133 
 
 Resolution of Fractions into Series 134 
 
 Miscellaneous Propositions in Fractions — _=0 135 
 
 -= 00 ; _ is indeterminate 1 36 — 137 
 
 Theorems in Fractions 138—139 
 
 • Miscellaneous exercises in Fractions 139 
 
 CHAPTER IV. 
 
 EQUATIONS OF THE FIRST DEGREE. 
 
 Definitions and Elementary principles 140 — 149 
 
 Transposition 1 50 
 
 To clear an Equation of Fractions 151 
 
 Solution of Equations of the First Degree — Rule 152 — 153 
 
 Questions involving Equations of the First Degree 154 
 
 Simultaneous Equations of the First Degree containing two 
 
 unknown quantities 155 
 
 Elimination by Substitution 156 
 
 by Comparison 157 
 
 by Addition and Subtraction '. 158 
 
 Questions involving Simultaneous Equations of the First De- 
 gree containing two unknown quantities 159 
 
 Simultaneous Equations of the First Degree involving three 
 
 or more unknown quantities 160 
 
 Questions producing Simultaneous Equations of the First De- 
 gree containing thrre or more unknown quantities 161 
 
 CHAPTER V. 
 
 SUPPLEMENT TO EQUATIONS OF THE FIRST DEGREE, 
 
 Generalization — ForMation of Rules — Examples 162 — 163 
 
 Negative Solutions 164 
 
 Discussion of Problems 165 
 
 Problem of the Couriers , 166 
 
 Cases of Indetermination and Impossible Problems .167- -169 
 
 An Equation of the First Degree has but One Root 170 
 
CONTENTS. 
 
 CHAPTER Vr. 
 
 FORMATION OF POWERS — EXTKACTION OF BOOTS — 
 
 RADICALS — IMEQUALITIES. AMIOIES 
 
 Involution or Formation of Powers 172 
 
 Square Root of Numbers — Of Fractions 173 — 177 
 
 Perfect and Imperfect Squares — Theorem 178 — 179 
 
 Approximate Square Roots , 180 
 
 Square Root of Monomials — Of Polynomials 182 — 184 
 
 Cube Root of Numbers 185 — 188 
 
 Approximate Cube Roots 189 
 
 Cube Root of Monomials — Of Polynomials 190 — 191 
 
 Fourth Root — Sixth Root — Nth Root, &c 192 
 
 Signs of the Roots — Nth Root of Monomials 193 — 104 
 
 Radical Quantities — Definitions 195 — 198 
 
 Reduction of Radicals 199—203 
 
 Addition and Subtraction of Radicals , 204 
 
 Multiplication and Division of Radicals 205 
 
 To render Rational the Denominator of a Fraction containing 
 
 Radicals , 206 
 
 Powers and Roots of Radicals 207—208 
 
 Imaginary or Impossible Quantities 209 — 210 
 
 Theory of Fractional Exponents 211 
 
 Multiplication and Division of quantities with Fractional Ex- 
 ponents 212—213 
 
 Powers and Roots of quantities with Fractional Exponents. . . . 214 
 
 Equations of the First Degree containing Radicals 216 
 
 LvEQUALiTiES — Propositions I to V 217 — 224 
 
 Examples involving the principles of Inequalities S:24 
 
 CHAPTER VII. 
 
 EQUATIONS OF THE SECOND DEGREE. 
 
 Definitions and Forms 224 — 22S 
 
 Incomplete Equations 227 — 228 
 
 Questions producing Incomplete Equations 229 
 
 Complete Equations of the Second Degree 230 
 
 Completing the Square — General Rule 231 
 
 Hindoo Method of Solving Equations of the Second Degree. . . 232 
 
 Questions producing Complete Equations of the Second Degree 233 
 
 Discussion of the General Equation of the Second Degree 234 — 238 
 
 Problem of the Lights 23? 
 
 Trinomial Equations — Definitions 240 
 
 Binomial Surds — Theorems 241 
 
 Extraction of the Square Root of the Binomial A±^B 241 
 
 Trinomial Equations — Varieties of 242 
 
 Equations to be reduced to Trinomial Equations 243 
 
 Simultaneous Equations of the Second Degree 244 
 
CONTENTS. 
 
 AKTICLi fi 
 
 Pure Equations 245 
 
 Adfected Equations 246—250 
 
 Questions producing Simultaneous Equations of the Second 
 
 Degree 251 
 
 FormulsB — General Solutions 252 
 
 Special Solutions and Examples 253 
 
 CHAPTER VIII. 
 
 RATIO — PKOPORTION — PKOGEESSIONS. 
 
 Ratio — Arithmetical — Geometrical 254 
 
 Antecedent and Consequent defined 256 
 
 Ratio — Multiplication and Division of 259 
 
 Ratio of Equality — Of greater and less Inequality 260 
 
 Ratio — Compound — Duplicate Triplicate 261 
 
 Ratios — Comparison of 262 
 
 PaoroKTioN — Definitions 263 — 266 
 
 Proanct of Means equal to Product of Extremes 267 
 
 Proportion from two Equal Products 268 
 
 Product of the Extremes equal to the Square of the Mean. . . . 2G9 
 
 Propoilion by Alternation 270 
 
 Proportion by Inversion 271 
 
 Propoition from equality of Antecedents and Consequents. . . . 272 
 
 Propoilion by Composition 273 
 
 Proportion by Division 274 
 
 Proportion by Sum and Difference of Antecedent and Conse- 
 
 quem 275 
 
 Like Powers or Roots of Proportionals are in Proportion 276 
 
 Products of Proportionals are in Proportion 277 
 
 Continued Proportion. 278 
 
 Exercises in Proportion 278 
 
 Problems involving Ratio and Proportion 279 
 
 Haemonioal Proportion 280 
 
 Variation — Propositions — Exercises ,281 — 290 
 
 Arithmetical Progression — Increasing and Decreasing 291 
 
 Value of nth term — Rule 292 
 
 Sum of Series — Rule 293 
 
 Table of General Formute 294 
 
 To insert m Arithmetical Means between two numbers, Ex. 16 294 
 
 Geometrical Progression — Increasing and Decreasing 295 
 
 Value of Last Term — How to find it 296 
 
 Sum of the Series — Rule 297 
 
 Bum of Decreasing Infinite Geometrical Series 299 
 
 Table of General Formute 300 
 
 To find a Geometric Mean between two numbers, Ex. 23 300 
 
 To insert m Geometrical Means between Two numbers, Ex. 24 300 
 
 Circulating Decimals — To find the value of 301 
 
CONTENTS. 
 
 RETICLES. 
 
 Haruonioal Progression — Proposition 302 — 303 
 
 Problems in Arithmetical and Geometrical Frogression 304 
 
 CHAPTER IX. 
 
 pekmutations — combinations — 
 binomial theokem. 
 
 Permutations 305 — 30? 
 
 Combinations 308 — 309 
 
 Binomial Theorem when the Exponent is a Positive Integer . . 310 
 
 Binomial Theorem applied to Polynomials 3U 
 
 CHAPTER X. 
 
 INDETEEMINATE COEFFICIENTS — SEEIES. 
 
 Indeterminate Coefficients — Theorem 314 
 
 Evolution by Indeterminate CoefBoients 317 
 
 Decomposition of Rational Fractions 318 
 
 Binomial Theorem for any Exponent 319 — 320 
 
 Application of Binomial Theorem 321 
 
 Extraction of Roots by the Binomial Theorem 322 
 
 Limit of Error in a Converging Series 323 
 
 Differential Method of Series — Orders of Differences. . . . 324 
 
 To find the nth term of a Series 396 
 
 To find the Sum of n terms of a Series 327 
 
 Piling of Cannon Balls and Shells 328 — 331 
 
 Interpolation of Series 333 — 335 
 
 Summation of Infinite Series 336 — 338 
 
 Recurring Series 339 — 343 
 
 Reversion of Series ., 344 — 346 
 
 CHAPTER XI. 
 
 continued fractions — logab items — ex- 
 ponential equations — inter- 
 est and annuities. 
 
 Continued Fractions 347 — 356 
 
 Logarithms — Definitions — Characteristic 357 — 358 
 
 Logarithms of numbers from 1 to 100 359 
 
 General Pi operties of Logarithms — Multiplication 360 
 
 Division 361 
 
 Formation of Powers — Extraction of Roots 362 — 363 
 
 Logarithms of Decimals 364 — 365 
 
 Logarithm of Base — Of 367—368 
 
 Computation of Logarithms 370 — 372 
 
 Logarithmic Series 373 
 
 Naperian Logarithms — Computation of 375 
 
CONTENTS. 
 
 ARTIOXES 
 
 Common Logarithms — Computation of by Series 371 
 
 Single and Docble Position 380 — 381 
 
 Exponential Equations 382 — 383 
 
 Interest and Annuities — Simple Interest 384 — 3&3 
 
 Compound Interest — Increase of Population 386 — 387 
 
 Compound Discount — FormnlsB 388 
 
 Annuities Certain 389—391 
 
 CHAPTER XII. 
 
 GEKEKAL THEOET OF EQUATIONS. 
 
 Definitions — General Form of Equations 393 — 394 
 
 An Equation whose root is o is divisible by x — a 395 
 
 An Equation of the nth Degree has n roots 396 
 
 An Equation of the nth Degree has only n roots 397 
 
 Relations of the Roots and CoSfEcients of an Equation 398 
 
 What Equations have no Fractional Roots 399 
 
 To change the Signs of the Roots of an Equation 400 
 
 Number of Imaginary Roots of an equation must be even. . . , 401 
 
 Descartes' Rule of the Signs 402 
 
 Limits of a Root — a method of finding 403 
 
 Tbansformation of EaoAT'ONS — To Multiply or Divide the 
 
 Roots 404 — 405 
 
 To Increase or Diminish the Roots 406 — 408 
 
 Synthetic Division 409 
 
 Transformation of Equations by Synthetic Division 410 
 
 Derived Polynomials — Lav? of 41 1 — 412 
 
 Transformation of Equations by Derived Polynomials 41 3 
 
 Equal Roots 414 
 
 Limits of the Roots of Equations 415 
 
 Limit of the greatest root 416 — 417 
 
 Limit of the negative roots 418 
 
 Sturm's Theorem 420 — 427 
 
 CHAPTER XIII. 
 
 RESOLUTION OF NUMERICAL EQUATIONS. 
 
 Rational Roots — Rule for finding 429 
 
 Horner's IVIethod (t^ Approximation — Rule 430—434 
 
 Approximation by Double Position 436 
 
 Newton's Method of Approximation 437 
 
 Cardan's Rule for Solving Cubic Equations 438—441 
 
 Reciprocal or Recurring Equations 442 
 
 Binomial Equations 443—444 
 
RAY'S 
 
 ALGEBRA. 
 
 PART SECOND. 
 CHAPTER I. 
 
 DEFINITIONS AND NOTATION. 
 
 Article 1. Quantity is anything capable of being increased 
 or diminished'; such as numbers, lines, space, time, motion, &c. 
 
 Remark. — If the pupil does not already kno'w, let the instructor here 
 explain to him what is meant by unit of quantity, the nwmerical value 
 of quantity, &c. See Ray's Algebra, Part First, Arts. 9 to 10. 
 
 Art. 2. Mathematics is the science of quantity, and of the 
 symbols by which quantity is represented. 
 
 Art. 3. Algebra is that branch of Mathematics which relates 
 to the solution of problems and the demonstration of theorems, 
 when any of the quantities employed are designated by letters. 
 
 Art. 4. A ■problem is a question proposed" for solution; a theo- 
 rem is a proposition to be proved by a chain of reasoning. 
 
 Art. 5. The operations of A Igebra are conducted by means of 
 figures, letters, and signs. The letters and signs are often called 
 symlols. 
 
 Art. 6. Known quantities are generally represented by the 
 first letters of the alphabet, as a, b, c, &c. ; and unknown quanti- 
 ties by the last letters, as t, v, x, y, z. 
 
 Art. 7. The principal signs used in Algebra are the following* 
 
 Each sign is the representative of certain words ; and is used for 
 the purpose of expressing the various operations in the most cleai 
 and brief manner. 
 
12 RAY'S ALGEBRA, PART SECOND. 
 
 Akt. 8. The sign =, is termed the sign of equality. It is read 
 e^al to, or equals, and denotes that the quantities between which 
 it is placed are equal to each other. Thus, x=^5, denotes that the 
 quantity represented by x is equal to 5. 
 
 Aet. 9. The sign -j-, is termed the sign of addition. It is read 
 plus, and denotes that the quantity to which it is prefixed is to be 
 added. 
 
 Thus, a-]-b denotes that b is to be added to a. If a=3, and 
 6=5, then a-|-J=3-|-5, which are =8. 
 
 Aet. lO. The sign — , is termed the sign of subtraction. It ia 
 read minus, and denotes that the quantity to which it is prefixed 
 is to be subtracted. Thus, a — b denotes that b is to be subtracted 
 from a. If a=7 and 6=4, then 7 — 4=3. 
 
 Aet. 11. The signs + ■ind — ^■'e called tlte signs ; the former 
 is called the positive, and the latter the negative sign ; they are 
 said to be contrary, or opposite. 
 
 Aet. 12. Every quantity is supposed to have either the positive 
 or negative sign. Quantities having the positive sign are called 
 positive; those having the negative sign are called wgalive. 
 When a quantity has no sign prefixed to it, the sign -j- is under- 
 stood, and it is to be considered positive. 
 
 Aet. 13. Quantities having the same sign are said to have like 
 signs ; those having different signs are said to have unlike signs. 
 Thus, -\-a and -{-b, or — a and — b have like signs ; while -\-c and 
 — d have unlike signs. 
 
 Art. 14. The sign X> is termed the sign of multiplication. It 
 is read into, or mvMplied by, and denotes that the quantities 
 between which it is placed are to be multiplied together. 
 
 A dot, or period, is sometimes used to denote multiplication. 
 Thus, ayj), and a.b, each denote that a and b are to be multiplied 
 together. The dot if) not often employed to denote the multipli- 
 cation of figures, since it is used, by some authors, to separate 
 whole numbers and decimals. 
 
 The product of two or more letters is generally denoted oy 
 writing them in close succession. Thus, db denotes the same aa 
 aY,b, or a.b ; and abc means the same as oX^Xc, or a.b.c ; each 
 signifying the continued product of the numbers designated by 
 I/, b, and c. 
 
 Aett 15. When two or more quantities are to be multiplied 
 together, each of them is called a, factor. Thus, in the product 
 ab there are two factors, a and i; in the product Zy^byft there 
 are three factors, 3, 5, and 7. 
 
DEFINITIONS AND NOTATION. 13 
 
 Akt. 16. The sign -i-, is termed tlie sign of division. It ia 
 read divided by, and denotes that the quantity preceding it is to be 
 • divided by that following it. The most common method of repre- 
 senting the division of two quantities is to place the dividend as 
 the numerator, and the divisor as the denominator of a fraction. 
 
 Thus, a-i-h, or -, signifies that a is to be divided by b. 
 b 
 Division is also represented thus, a\b, where u, denotes 'Jis 
 
 dividend, and b the divisor. 
 
 Art. 17. The sign >, is termed the sign of inequality. It 
 denotes that one of the two quantities between which it is placed 
 is greater than the other, the opening of the sign being turned 
 toward the greater quantity. 
 
 Thus, a^b, denotes that u is greater than b. It is read, a 
 greater than b. If o=7 and J=2, then 7>2. 
 
 Also c<Cid, denotes that c is less than d. It is read, c'less than 
 d. If c=3 and d=5, then 3<5. 
 
 Akt. 18. A coefficient is a number or letter prefixed to a quan- 
 tity, to show how often it is taken. 
 
 Thus, if a is to be taken 4 times, instead of writing a-\-a-\-a 
 -\-a, we write 4a; in like manner, az-\-az-\-az=Zaz. 
 
 The coefficient is called numeral, or literal, according as it is a 
 number or a letter. Thus, in the quantities 5x and mx, 5 is a 
 numeral and m a literal coefficient. 
 
 In 3az, 3 may be considered as the coefficient of az, or 3a may 
 be considered as the coefficient of z. 
 
 When no numeral coefficient is expressed, 1 is always under- 
 stood. Thus, a is the same as la, and ax is the same as lax. 
 
 Art. 19. A power of a quantity, is the product arising from 
 multiplying the quantity by itself one or more times. When the 
 quantity is taken twice as a factor, the product is called the second 
 power ; when taken three times, the third power, and so "on. Thus, 
 2x2=4, is the second power of 2. 
 2X2X2=8, is the third power of 2. 
 2X2X2X2=16, is the fourth power of 2. 
 Also, aY_a=aa, is the second power of a. 
 
 a'/.ay,a=aaa, is the third power of a; and so on. 
 The second power is often termed the square, and the third 
 power, the cube. To avoid repeating the same quantity as a fac- 
 tor, a small figure, termed an exponent, is placed on the right, and 
 a little above it, to denote the number of times the quantity ia 
 taken as n factor, Thus, aa is written a' ; aaa is Written a' ; 
 aabbb is written a'6'. 
 
i4 RAY'S ALtJEBRA, PART SECOND. 
 
 When no exponent is expressed, 1 is always understood. Thus, 
 a IS the same as a', each signifying the ^rst power of a. 
 
 Akt. 20. A root of a quantity is a factor, which multiplied by' 
 itself a certain number of times, will produce the given quantity. 
 
 The root is called the square or second root, the cube or tJiird 
 root, the fourth root, &s.c., according to the number of times it 
 must be taken as a factor to produce the given quantity. Thus, 
 2 is the second or square root of 4, since 2x2=4. In like 
 manner, a is the fourth root of a'', since a'X.ay(,a'X,a=a*. 
 
 Art. 21. The sign ^, or J, is called the radical sign. 
 When prefixed to a quantity, it denotes that its root is to be 
 extracted. Thus, 
 
 ^a, or ^a, denotes the square root of a. 
 
 %/8, or ^8, denotes the cube root of 8, which is 2. 
 
 ija, or jja, denotes the fourth root of a. 
 
 The number placed over the radical sign is called the index of 
 the root. Thus, 2 is the index of the square root, 3 of the cube 
 root, 4 of the fourth root, and. so on. When the radical sign has 
 no index over it, 2 is understood; thus ^^^9 and 2/9 signify the 
 same thing. 
 
 Art. 22. An algebraic quantity, or an algebraic expression, is 
 any quantity written in algebraic language, that is, by means of 
 symbols. Thus, 
 
 5a, is the algebraic expression of 5 times the number a ; 
 
 36-|-4c, is the algebraic expression for 3 times the number b 
 increased by 4 times the number c; 
 
 3o' — lab, is the algebraic expression for 3 times the square of 
 a, diminished by 7 times the product of the number a 
 by the number b. 
 
 Art. 23. a monomial is a quantity not united to any other by 
 I3ie sign of addition or subtraction ; as 4o, a^ic, — Axy, &c. 
 
 A monomial is often called a simple quantity, or term. 
 
 Art. 24. A polynomial, or compound quantity, is an algebraic 
 expression composed of two or more terms ; as a-\-b, c^x-\-y, dz.c. 
 
 Art. 25. A binomial is a quantity having two terms ; bm 
 a-\-b, x^-\-y, &c. 
 
 A binomial, of which the second term is negative, as a — b, is 
 sometimes called a residical. 
 
 Art. 26. A trinomial is a quantity consisting of three terma j 
 e£, a-\-b — c. 
 
 Binomials and trinomials are polynomials. 
 
DEFINITIONS AND NOTATION. 15 
 
 Art. SV. The numerical value of an algebraic expression is the 
 number obtained by giving a particular value to each letter, and 
 then performing the operations indicated. Thus, in the algebraic 
 expression 4a — 3c, if o=5 and c=6, the numerical value is 
 4X5—3X6=20—18=2. 
 
 Art. 2§. The value of a polynomial is not afTected by chang- 
 ing the order of the terms, provided each term retains its respect- 
 ive sign. Thus b^ — 2ab-\-o is the same as i'-j-c — 2ab. This is 
 self-evident. 
 
 Art. 29. Each of the literal factors of any term is called a 
 dimension of that term. The degree of any term is equal to the 
 number of literal factors which it contains. Thus, 
 
 5a is of \h& first degree ; it contains one literal factor. 
 
 ax is of the second degree ; it contains two literal factors. 
 
 3o'J^c^3aaaiic is of the sixth degree ; it contains six literal 
 factors. 
 
 It is obvious, that the degree of any term is equal to the sum of the 
 exponents of the letters which compose that term. 
 
 Thus, 5o'ic' is of the seventh degree, since 2-\-\-\-4c=1 . 
 
 Art. 30. A polynomial is said to be homogeneous, when each 
 of its terms is of the same degree. Thus, 
 
 a-\-b — 3c is of the first degree ; it is homogeneous. 
 
 a;' — txy^ is of the second degree ; it is homogeneous. 
 
 x' — 3sy' is not homogeneous. 
 
 Art. 31. An algebraic quantity is said to be arranged accord- 
 ing to the dimensions of any letter it contains, when the expo- 
 nents of that letter occur in the order of their magnitudes, either 
 increasing or decreasing. Thus, 
 
 ax''-\-a''x — a'a;', is arranged according to the ascending powers 
 of a. bx^ — b'x^-\-b'x, is arranged according to the descending 
 powers of x. 
 
 Art. S2. A parenthesis, ( ), is used to show that all the terms 
 of a polynomial are to be considered together, as a single term. 
 Thus, 
 
 10 — (a — 6) means that a — b is to be subtracted from 10 ; 
 
 6{a-\-b — c) means that a-\-b — c is to be multiplied by 5 ; while 
 
 5a+(ft — c)meansthat?i — cisto be added to 5a. 
 
 When the parenthesis is used, the sign of multiplication is gen- 
 erally omitted. Thus, (a — i)X(.a-\-b) is the same as (a — b)(a^b). 
 
 A vinculum, , is sometimes used instead of a parentliesis. 
 
 Thus, 0-1-6x5 means the same as 5(a-|-J). Sometimes tb" ', 
 lum is placed vertically ; it is then called a bar- 
 
16 RAY'S ALGEBRA, PART SECOND. 
 
 a 
 
 —b 
 
 x^, is the same as (a — i-\-c)x'. 
 
 Akt. 33. Similar, or like quantities, are such as differ only in 
 their signs, or numerical coefficients, or both. Thus, 2ah and 
 — 2ab, 3a'J and Sa^ft, Za'b and — bd^b, are respectively similar. 
 
 Unlike quantities are different combinations of letters. Thus, 
 3a62, and 3a'i, are unlike or dissimilar. 
 
 Remahk. — An exception, however, must be made in those cases 
 where letters are taken to represent coefficients. Thus, as? and 6z' are 
 like quantities, when a and i are talien as coefficients of i* 
 
 Art. 34. The reciprocal of a quantity is unity divided by that 
 quantity. Thus, 
 
 1 1 
 
 The reciprocal of a-\-h is ~^rKt and of 3 , is o- 
 
 Art. 35. The same letter, accented, is often used to denote 
 quantities which occupy similar positions in different equations 
 or investigations. Thus, a,' a', a", a'", represent four different 
 quantities, of which a' is read, a prime ; a" is read, a second ; a'" 
 is read, a third ; and so on. 
 
 Art. 36. The following signs are also used, for the sake of 
 brevity : 
 
 00 , a quantity indefinitely great, or infinity 
 
 . • . , signifies therefore, or consequently. 
 
 • . • , signifies since, or because. 
 
 /~^, is used to represent the difference between two quantities, 
 as c\jd, when it is not known which is the greater. 
 
 EXERCISES 
 ON THE DEFINITIONS AND NOTATION. 
 
 Each example is intended to furnish to the pupil two distinct 
 exercises. First, to be copied on the slate, or blackboard ; and 
 then read, that is, expressed in common language. Second, the 
 numerical value to be found, supposing a=2, 6=3 , c=:4, i=5, 
 
 1. h-\-c — X. Ans. 2. ex — ay 
 
 3. ib+x^y. Ans. 20. "• "^^Zy ■*"*• *■ 
 
 3. abx — cy. Ans. 6. Jc ay 
 
 4. a'by—3xK Am. —3. 9- y+— Ans. 5. 
 
 5. c+aXo-a. Ans. 10. iq. 2c'— a(x+y)(y— i) 
 
 6. (c+a)(c— a.) Ans. 12. Ans. 10 
 o'4-ft+c — V cibCc — a) 
 
 7. ^3 — Ans. 2i. 11. -^zi^—Jabv Ans. 
 
DEFINITIONS AND NOTATION. H 
 
 12. Find the difference between abx, and a-\-h-\-x, wlien a=4 
 h=l, x=Z ; and when a=5, J=7, x=12. Atw. 1^ and 39fi 
 
 13. Required the values of a^-\-2ah-^b^, and a' — 2ah-\-b^ 
 when a=7 and i=4. Ans. 121 and 9. 
 
 14. What is n(n — 1) (n — 2) (n — 3), when n=4, and when 
 1=10 1 Ans. 24, and 5040. 
 
 15. Find the difference between Gabc — 2ab, and 6abc-i-2ah, 
 iivhena,J,c, are 3, 5, and 6 respectively. Ans. 492. 
 
 16. When a=3, what is the difference between a' and 2a; 
 «' and 3a; a* and 4a; a' and 5o ? .Ans. 3, 18, 69, and 228. 
 
 a'—b' 
 
 17. Find the value of rj-ru, when a=5 and b=3. Ans. 2 
 
 18. Find the difference between each pair of the following 
 expressions, when a=6 and b=8 : (a-^by and o'-j-i' ; 5(a4-t) 
 ind 5a+i; a-\-b and V(o'+*'); («+*)' and a+¥ ; ^(a2_[.j2) 
 and Ja^+^b^. Ans. 96, 32, 4, 2226, and 4. 
 
 19. What is SJc+2a^(2a-lrb—x) when a=6, 6=5, c=4, 
 ind:r=l! Ans. 54. 
 
 20. What is ra(7i— 1) (ti— 2) (7^—3) (n— 4), if n be 1, 2, 3, 
 "l, 5, and 6, in succession 1 Ans. 0, 0, 0, 0, 120, and 720. 
 
 The pupil may verify the following expressions, by giving to each 
 etter any value whatever. 
 
 21. a(m-\-n)(m — n)=am^ — an'. 
 
 23. (x'+a;2+l)(a;2— l)=a:6— 1. 
 a-\-l a — 6 4ab 
 
 24 
 
 -b~a+b~a^—b' 
 
 EXAMPLE S 
 TO BE EXPRESSED IN ALGEBKAIC SYMBOLS. 
 
 1. Five times a, plus the second power of 6. 
 
 2. X, plus y divided by 3z. 
 
 3. X plus y, divided by 32. 
 
 4. 3 into X minus n times y, divided by m minus n. 
 
 5. m into x squared, minus m plus r squared, plus c into the 
 cabe of x. 
 
 6. a third power minus x third power, divided by a second 
 power minus x second power. 
 
 7. a minus x third power into a plus a; second power, dirided by 
 a second power plus x second power. 
 
 2 
 
18 
 
 RAY'S ALGEBRA, PART SECOND. 
 
 8. m squared plus n squared, into the square of m plus n. 
 
 9. The square root of m minus the square root of n. 
 
 10. The square root of m minus n. 
 
 11. The square root of m, minus n. 
 
 12. The square root of h squared, minus m plus % into e 
 
 
 
 ANSWERS. 
 
 1. 
 
 5a+6'. 
 
 
 ^- a'+a;' " 
 
 8. (OT=-f7l=')(jJl+»)'. 
 
 9. ^771 — ^n. 
 
 2. 
 3. 
 
 3« • 
 
 
 4 
 
 3 a; — ny 
 
 
 10. <J{m-^). 
 
 5. 
 
 m — n ' 
 ma:'— (m+a:)'+car». 
 
 11. ^m — n. 
 
 6. 
 
 a?—x? 
 
 
 12. V {&»-(m+n)c^ 
 
 ADDITION. 
 
 Aet. SV. Addition in Algebra, is the process of finding the 
 simplest expression for the sum of two or more algebraic 
 quantities. 
 
 There are three cases of algebraic addition : 
 
 1st. When the quantities are all similar, and have the same 
 sign, either positive or negative. 
 
 2d. When the quantities are all similar, but part positive, and 
 part negative. 
 
 3d. When the quantities are dissimilar, or part similar and part 
 dissimilar. 
 
 AnT. 38. First Case. — Let it be required to find the sum of 
 Sar'j/, 5a;'!/, and Ix'y. 
 
 Here x'y is taken, in the first term, 3 times ; in opekation 
 the second, 5 times ; and in the third, 7 times ; + 3a;'y 
 hence, in all, it is taken 15 times. Since adding -j- 5x'y 
 the quantities together cannot change their charac- -j- '7x''y 
 ter, and since each term is positive, therefore their -f-15a;'j 
 sum is positive. 
 
ADDITION. 19 
 
 Next, let it be required to find the sum of — Zxh/, — bx^y, and 
 -Ix'y. 
 
 Here x^y is taken, in tlie first term, — 3 times ; in opekation 
 the second, — 5 times ; and in the third, — 7 times ; — Zx'y 
 hence, in all, it is taken — 15 times. — 5a^y 
 
 Therefore, To aid together quantities having the — Ix^y 
 same sign ; Jind tJie sum of their coefficients, and prefix — Idx^y 
 it, with tlie common sign, to the literal part. 
 
 Akt. 39. Second Case. — Let it be required to find the sum of 
 -(-9a, — 5ffi, -\-ia, and — 2o. 
 
 Before solving this example, the pupil must understand the fol- 
 lowing principle. Since the sign plus denotes that the quantity 
 before which it is placed is to be added, and the sign minus, that 
 the quantity before which it is placed is to be subtracted ; there- 
 fore, the sum of two equal quantities, of which one is positive and tlie 
 other negative, is zero, or 0. Thus, -\-a — a=0 ; -\-5a^ — 5o^=0; 
 and so on. 
 
 Here -\-9a-\-i.a is 13a ; and — 5a — 2a is — la. operation 
 Now, — 7a will cancel -|-7a in the quantity -\-lda, -\-9a 
 and leave -\-6a for the aggregate, or result of the — 5a 
 four quantities. -|-4a 
 
 In like manner, if it be required to obtain the — 2a 
 sum of — 9a, -|-5a, t— 4a, and -\-2a, we find the sum -|-6o 
 of — 9a and — 4a is — 13a, and the sum of -f-Sa 
 and -(-2a is -\-7a. Now, -(-7a will cancel — 7a in operation, 
 the quantity — 13a ; which leaves — 6a for the — 9a 
 aggregate, or sum of the quantities. -(-5a 
 
 Therefore; To add similar quantities having dif- 4a 
 
 ferent signs ;Jind the sum of the positive quantities and -(-2a 
 
 the sum of the negative quantities, then take the dif- 6a 
 
 ference of their coefficients, and prefix it, with the sign of 
 the greater quantity, to the literal part. 
 
 Am. 40. Thikd Case. — Let it be required to find the sum of 
 5a2— 86-(-c, -1-6— aS and 5i-(-3a2. 
 
 In writing the quantities, we place those which operation. 
 
 are similai under each other, for the sake of con^ 5a^ 8b-\-e 
 
 venience in performing the operation. We then a'-(- i 
 
 find, as in the preceding case, that the sum of 3a=-f 56 
 
 the quantHies in the first column is -(-7a', and 7a=— 26-(-c 
 
 in the second, — 2i; and there being no term 
 
 similar to c, it is connected to the other quantities by its f//cpei 
 
 Bign. 
 
20 
 
 RAY'S ALGEBRA, PART SECOND. 
 
 Art. 41. From the pre?.eding, we derive the following 
 
 GeNEEAL rule for the addition of algebraic QUAHTITIES.— 
 
 Write the quantities to he added, placing those that are similar 
 under each other ; then reduce each set of similar terms, hy taking 
 the difference of the positive and negative coefficients, and prefix- 
 ing it, with the sign of the greater, to the literal part ; after this, 
 annex the other terms with their proper signs. 
 
 Reuares. — 1. It is immaterial in wliat order tlie quantities are se 
 down, if care is talten to prefix to eacli its proper sign. 
 
 2. It will often happen tiiat the sum of two or more quantities is less 
 than either. See Observations on Addition and Subtraction, page S4. 
 
 XXAM FI.B S. 
 
 1. Add together 4aa;-|-3Jj/, 5ax-\-Shy, Qax-\-Gby, and iiOax-\~hy. 
 
 Ans. Srrax+I8by. 
 
 2. Add together lOcz— 2ax', 15cz—^ax', 2icz—9ax^, and 
 '6cz—8ax'. Ans. 52cz—22ax'. 
 
 3. Find the sum of Si'y'— lOy", —x'y^-\-5y\ 8x'y'—6y\ and 
 4xY+2y''. Ans. 14a;2y=— 9s^ 
 
 4. Add together a+h-\-c-\-d, a-\-b-\-c — d, a+6 — c-\-d, a — b-\-c 
 i-d, and —a-{-b+c+d. Ans. 3o+3J+3c+3d. 
 
 5. Add together ^x^—y^), 8^x'—y^), and —5 (a;^— y'). 
 
 Ans. 6(a;2— y=). 
 
 6. Required the sum of lOa'J— 12a'6c— ISi'c^+lO, — 40=^4 
 +8a5ic— lOJV— 4, — 3a=i— 3a3Jc+20J2c<— 3, and 2a^+12a^bc 
 +5i'c<+2. Ans. 5a'b+5a^bc+5. 
 
 7. Add together a'+i'+c'+d^, ab—2a^-{-ac—2c'+adr—2d^, 
 a»— 3ai+i3— 3ac+c'— 3ad, and 2ab—a+2ac—b+2<fd—c. 
 
 Ans. a'+J34-c3+J=— a2— c=— <i'— fl— 5— c. 
 
 8. Add together o"— 6"+3*', 20'"— 36"-^'', and a'"+4i''— a?. 
 
 Ans. 4a"'-\-2x'—iifl. 
 
 SUBTRACTION 
 
 Art. 42. Subtraction, in Algebra, is the process of finding the 
 simplest expression for the difference between two algebraic 
 quantities. 
 
 The quantity to be subtracted is called the Subtrahend. 
 
 The quantity from which the subtraction is to be made is called 
 the Minvjmd. 
 
SUBTRACTION. 21 
 
 The quantity left, after the subtraction is performed, is called 
 the Difference, or Remainder. 
 
 The explanation of the principles on which the operations 
 depend, may be divided into two cases. 
 
 1st. Where all the terms of the quantity to be subtracted are 
 positive. 
 
 2d. Where the quantity to be subtracted is either partly or 
 wholly negative. 
 
 Art. 43. To explain the first case, let it be required to subtract 
 4 a from 7a. 
 
 It is evident that 7 times any quantity, operation. 
 
 less 4 times that quantity, is equal to 3 7a Minuend 
 times the quantity ; therefore, 7a less ia is 4a Subtrahend 
 equal to 3a. Hence, to find the difference 3a Remainder 
 between two similar quantities, tee take the 
 difference between their coefficients, and prefix it to the common Utter 
 or letters. 
 
 If it be required to subtract h from a, operation. 
 
 unless we know the number of units repre- a Minuend 
 
 sented by each, we can only indicate the ope- t Subtrahend 
 
 ration, which is done by placing the sign a — ft Remainder 
 mimis before the quantity to be subtracted. 
 
 Art. 44. To explain the second case, let it be required to 
 subtract h — c from a. 
 
 If we subtract b from a, the result, a — h, operation. 
 
 is obviously too little, for the quantity b, a Minuend 
 
 taken from o, ought to be diminished by c b — c Subtrahend 
 
 before the subtraction is effected. We a — b-\-c Remainder 
 have, in fact, subtracted a quantity too 
 
 great by c, and, therefore, to obtain a true result, the difference 
 a — b must be increased by c; this gives, for the true remainder 
 o — b-\-c. 
 
 This operation may be explained by figures, thus: 
 
 Let a=9, t=5, and c=3 ; and let it be required to subtract 
 5—3 from 9. 
 
 If we subtract 5 from 9, the remainder is 9 — 5; but the quan- 
 tity to be subtracted is 3 less than 5, therefore we have subtracted 
 3 too much ; hence, we must add 3 to 9 — 5, which gives 9 — 5-1-3, 
 or 7, for the true remainder. 
 
 The operation and illustration may be compared, thus : 
 
 From a From 9 =9 
 
 Take b—c Take 5— 3 =2 
 
 Rem. «— 6-fc Rem. 9-^+3 ^ 
 
22 RAY'S ALtiEIiRA, PART SECOND. 
 
 The same principle may be further illustrated by the following 
 examples : 
 
 a — (c — ii)=a — c+a=2o — c. 
 
 a — (a — c)=a — a-j-c^c. 
 
 a-\-c — (ffl — c)=a-\-c — a-|-c=2c. 
 
 In all these cases, we see that the same remainder would have 
 been obtained, by changing the signs of the quantity to be 
 subtracted, and then adding it. 
 
 Akt. 45. Hence we have the following 
 Rule for the subtraction of algebraic quantities. — Writt 
 
 the quantity to be subtracted under that from which it is to be taken, 
 
 placing similar terms under each other. 
 Conceive the signs of all the terms of the subtrahend to he changed, 
 
 from -\- to — ,or from — to -{-, and then reduce the result to its 
 
 simplest form. 
 
 Remarks,—] . Beginners may solve a few examples by actually chang- 
 ing the signs of the subtrahend. After tliis, it is better merely to con- 
 ceive the signs to be changed ; that if it becomes necessary to refer to 
 the operation, we may be in no doubt with regard to the signs of the 
 terms originally. 
 
 2. Subtraction in Algebra may be proved in the same manner as in 
 Arithmetic, by adding together the remainder and the subtrahend; the 
 sum should equal tile minuend. 
 
 (1) (1) 
 From 8a=i— 3ca;— zH The same, with the f 8«'6— 3cx— z^ 
 Take Sa^b+Acx—Sz' Ligns of the Bab-.< — 2a^ — icx+Sz" 
 Rem. 5a'b—7a:+2z'J trahend ch^ged. l^i^_ 5a'b—7^x+2T' 
 
 (2) . (3) 
 From 5a' — Bmz-{-5y* From ax' — Scy' — t' 
 T ake — 2'a'+3mz-i-6y' ' Take hx'-^cy'-\-y' 
 Rem. la' — 6mz — y* Rem. (a — b)x'— y z' 
 
 EXAMPLES FOR PRACTICE. 
 
 4. From 4a — 2i-|-3c take 3a+4i — c. Ans. a — 6i4-4c 
 
 5. From 9a;2 — iy+Q take Tx'-\-by — 14. Ans. 2a;' — 9y-4-23. 
 
 6. From 22xy'—ly+llx' take nxy'—5y—9x\ 
 
 Ans. 120^2— 2y+20a;' 
 
 7. From 12a:+18 take 12a;— 18+y. Ans. 26— y 
 
 8. Prom aP — y' take — 4 — y^-\-ix'. Am. 4 3a;' 
 
 9. Prom 4aa;'-|-ia;-fc take 3a;'— 2a:+5. 
 
 Ans. (4a— 3>'-l-(24-J)x-f-o— S 
 
THE BRACKET, OR VINCaLUM. 23 
 
 10. FTom—l'7x'-\-9ax'—Ta'x-Jrl5aHake—19x>+9ax'—9a^a 
 -fl7a3. Ans. 2x'-Jt-2a^x—2aK 
 
 11. Froma;5+3a;'+3a;+l takea;'— 3a:=+3a;— 1. Ams. 6x2+2. 
 
 12. From 9arx^—13-{-20a¥x—ib'"cx'' take 3b'"cx^+9a"'x-'—6 
 +Zah^x. Ans. IT aVx—lb'"cx^—'I . 
 
 13. From a — x — (x— 2a)+2o — x take a — 2x — (2a — x) + 
 (x — 2a). Ans. 8a — 3*. 
 
 14. From 4a'"+2a:''— ^' take a"*— 6»+3a* and aa"— 3J»— iu'. 
 
 Ans. a"'+4i"- 
 
 Reuark. — The number of exercises in both Addition and Subtrac- 
 tion is purposely small, as ample practice of the best kind will be found 
 In the operations of Multiplication and Division. 
 
 THE BRACKET, OR VINCULUM. 
 
 As the Bracket, or Vinculum, is frequently employed, it is 
 proper that the pupil should become acquainted with the rules 
 which govern its use in relation to Addition and Subtraction. 
 
 Akt. 46. 1st. Where the sign plus precedes a vinculum, it may 
 be omiited withcmt affecting the expression. This principle is self- 
 evident. 
 
 Thus, a-\-{b — c) is the same as a-\-b — c. The first shows that 
 6 is to he diminished by the number of units in c, and the 
 remainder added to a ; the second shows that a is to be increased 
 by the number of units in b, and the result diminished by the 
 number of units in c. Or, if o=6, i^5, and c=3. 
 
 Then 6+(5— 3)= 6+2=8; 
 
 And 6+5—3=11—3=8. 
 
 From this it follows, that any number of terms of an algebraic 
 expression may be included within a vinculum, if it be preceded 
 by the sign plus. 
 
 Thus, x-\-y — z=:a;+(y — z). 
 
 2d. Where the sign minus precedes a vinculum, it may be omitted 
 if the signs of all the terms within it he changed. This is evidenti 
 because th& sign minus indicates subtraction, which is effected 
 by changing the signs of all the terms of the quantity to be 
 subtracted. Thus, 
 
 a — (b — c)^=a — 6+c. 
 
 a — {x — y-{-z)=a — x-\-y — z. 
 
 Sometimes several brackets, or vinculums, are employed in the 
 same expression ; by this principle they may all be removed. 
 Thus, 
 
24 RAY'S ALGEBRA, PART SECOND. 
 
 a — f a-f-J — [a+* — '^ — (" — *4"'^)] I • 
 =a — \ a-\-h — {a^-h — c — a-f-ft — c] \ . 
 =a — \ a-\-b — a — b-{-c-\-a — b-\-c \ . 
 =a — a — i-j-a+i — c — o-j-i — c=h — 2c. 
 3d. Any quantity may he inclosed in a mnoAum, and preceded hy 
 the sign minus, provided the signs of all the terms in the vinculum be 
 changed. This is evident from the preceding principle. Thus, 
 a — b-{-c=a — (b — c) =c — (b — a) . 
 This principle often enables us to express the same quantity 
 under several different forms. Thus, 
 
 a — b-\-c-\-d=a — \ b — c — d ] . 
 =a — \b — {c-\-d)\. 
 
 EXAMPLES FOR PRACTICE. 
 
 Simplify, as much as possible, the following expressions. 
 
 1. (l_2a;+3a;2)+(3+2a;— a;2). Ans. 4+2a;'. 
 
 2. 5a— 46+3c-j-(— 3a+22>— c). Am. 2a—'2b-\-2c. 
 
 3. (o_J— c)+(64-c— <f)+((Z— e+/)+(e— /— ^). Ans. Or-g. 
 
 4. 2(,x^+y^)—\(x^+2xy+y')—(2xy—x^—y')]. Ans. x'+yK 
 
 5. a — (x — a) — I a: — (a — x)]. Ans. 9a— 3x. 
 
 6. 1— fl— [1— (1— a;)]|. Ans.x. 
 
 7. a — (b — c) — (a — c)4-c— (a — i). Ans. 3c— a. 
 
 8. a— ja+J— [a+ft+c— (a+i+c+<0]r Ans. —b—d. 
 
 OBSERVATIONS ON ADDITION AND SUBTRACTION. 
 
 In order that the pupil may have clear and precise ideas con- 
 cerning the various operations in Algebra, it is important to 
 understand the meaning of the signs plus (-J-) and minus ( — ), 
 and their relation to each other. 
 
 Art. 47. All quantities are to be regarded as positive, unless, 
 for some special reason, they are otherwise designated. Negative 
 quantities embrace those that are, in some particular respect, the 
 opposite of positive quantities. 
 
 Thus : If the sums of money put into a drawer be considered 
 positive, those taken out would be negative ; if a merchant's gains 
 are positive, his losses are negative ; if latitude north of the equa- 
 tor is reckoned -|-, then latitude soiith is — ; if distance to the 
 right of a certain line be reckoned -\-, then distance to the left 
 would be — ; if elevation above a certain point, or plane, be 
 regarded as -|-, then distance below would be — ; if time after a 
 certain hour is -|-, then time before that hour is — ; if motion 
 in any given direction be -|-, then motion in the opposite direction 
 wouhl be — ; and so on. 
 
OBSERVATIONS. 25 
 
 Akt. 48. This relation of the signs ^ives rise to some im- 
 portant particulars. 
 
 1st. The addition, to any quantity, of a negative number, pro- 
 duces a less result than adding eJiO. Thus, if we take any num- 
 ber, for example, 10, and add to it the numbers 3, 2, 1, 0, — 1 
 —2, — 3, we have 
 
 10 10 10 10 10 10 10 
 3 _2 J. 0—1—3—3 
 
 13 12 11 10 9 8 7 
 
 We see from this, that adding a negative number produces the 
 same result as subtracting an equal positive number. Thus, 
 adding — 3 to 10 gives the same result as subtracting -1-3 
 from 10. 
 
 2d. The subtraction of a negative quantity produces a greater 
 result than subtracting fcero. Thus, take any number, for exam- 
 ple, 10, and subtract from it ihe numbers 3, 2, 1, 0, — 1, — 2, 
 —3, we have 
 
 10 10 10 10 10 10 10 
 
 _3 _2 _1_ —1 —2 —3 
 
 "7 8 9 10 11 12 13 
 
 We see, also, from this, that subtracting a negative number 
 produces the same result as adding an equal positive number. 
 
 Art. 49. When two negative quantities are considered alge- 
 braically, that is called the least which contains the greatest num- 
 ber of units. Thus, — 3 is said to be less than — 2. But, of two 
 negative quantities, that which contains the greatest number of 
 units is said to be numerically the greatest ; thus, — 3 is numer- 
 ically greater than — 2. 
 
 Akt. 50. The sum of two positive quantities is always greater 
 than either of them ; and the sum of two negative quantities, 
 algebraically considered, is less than either of them. But the sum 
 of a positive and negative quantity is always less than thei 
 positive quantity. 
 
 Ex. A merchant gains 3o (-1-30) dollars, but soon after loses 
 2ffi ( — 20) dollars ; how much will his property be increased by 
 the two operations 1 
 
 3a-l-(— 2a)=-l-a; 
 Or, +30-l-(— 20)=:-l-10. 
 That is, his property will be increased a (10) dollars. 
 
 Had he gained So, and lost ba dollars, then the sum would have 
 been — 20 dollars, and his property would, evidently, have beeB 
 diminished. 
 
28 
 
 RAY'! 'V.LGEBRA, PART SECOND. 
 
 Aet. 51. The difference of two positive quantities, as in 
 Arithmetic, is always less than the greater quantity. Thus, 5a 
 — (-|-2a)=+3a. 
 
 The difference of two negative quantities is always greater, 
 algehraically considered, than the minuend. Thus, — 5a — ( — 2a) 
 =— 3 a. 
 
 The difference between a positive and a negative quantityj 
 found by subtracting the latter from the former, is always greatei 
 than either of them. Thus, 2a — ( — a)=3a. 
 
 Examples : 1. The latitude of A is 10° N. (+); the latitude 
 of B is 5° S. (— ); what is their difference of latitude 1 Ans. 15°. 
 
 2. At 7 A. M. of a certain day, the thermometer stood at — 9° 
 that is, 9 degrees below zero ; at 2 P. M., at 4-15°. that is, 16 
 degrees above zero ; what was the change of temperature betweeh 
 these hours 1 , Am. 24°. 
 
 MULTIPLICATION. 
 
 Akt. 52. Multiplication, in Algebra, is the process of taking 
 one algebraic quantity as often as there are units in another. 
 
 The quantity to be multiplied is called the Multiplicand. , 
 
 The quantity by which we multiply is called the Multiplier. 
 
 The result of the operation is called the Product. 
 
 The multiplicand and multiplier are generally called factors. 
 
 Art. 53. To understand the subject of algebraic multiplica- 
 tion, it is necessary that the pupil should be made acquainted with 
 the following preliminary principle : 
 
 The pritduct of two factors is the same, whichever he made the 
 multiplie' . 
 
 To prove this, suppose we have a sash containing a vertical 
 and h horizontal rows. It is evident that the whole number of 
 panes in the sash will be equal to the number in one row, taken 
 as many times as there are rows. 
 
 Since there are a vertical rows and b panes in each row, the 
 whole number of panes will be represented by h taken a times ; 
 that is, by ab. 
 
 Again, since there are h horizontal rows, and a panes in each 
 row, the whole number of panes will be represented by a taken i 
 times ; that is, by ha. Hence, ab is equal to ba ; that is, the pro- 
 duct of two factors is the same, whichever be made the multiplier. 
 
MULTIPLICATION. 
 
 If we have three factors, a, b, and c, by the preceding principle, 
 the product of two of them, as a and b, will be either ai or ba; 
 f, then, we regard this product as a single factor, and multiply it 
 jy c, the product may be written either abc, cab, bac, or cba, all of 
 which, by the preceding principle, are equal to each other. From 
 this, it is evident that the product of three factors is ike same, in 
 whatever order they are taken. Thus, 2x3x4=4X2x3=3x2 
 X4=4x3x2; the product in each case being 24. 
 
 In a similar manner, it may be shown that the proauct of anj 
 number of factors is the same, in whatever order they are taken 
 
 It follows from this principle, that (zcX6=6ac, zyx'^y.^~-=i>x'h)z 
 and so on. 
 
 It also follows -from this principle, that when either of the facton 
 of a product is multiplied, the product itself is multiplied. Thus, if 
 we take the product of two factors, as 2x3, and multiply it by 5, 
 the product may be written 5x2x3, or 5x3x2; that is, 10x3, 
 or 15x2, either of which is equal to 30. 
 
 Remark. — In the multiplication of numbers, since each figure of the 
 multiplicand is multiplied by the multiplier, pupils sometimes suppose 
 that in multiplying the product of two or more factors, as ab, by a third 
 factor, that each of the factors ought to be multiplied. That this would 
 be erroneous is evident from the preceding principle. 
 
 Art. 54. In multiplication there are four things to be con- 
 eidered in relation to each term, viz : 
 The sign ; 
 The coefficient ; 
 The exponent ; 
 The literal part. 
 
 Remark. — In writing a monomial product, we generally write, first 
 the sign, then the coefficient, and then the literal part ; but, in explain- 
 ing the principles, it is most convenient to consider the sign last. 
 
 Art. 55. Of the Coefficient. — To determine the rule of the 
 coefficients, let it be required to find the product of 2a by 3i. 
 
 To indicate the multiplication, we may write operation 
 the product thus, 2ax3i. But, by Art. 53, this 2a 
 is the same as 2x3Xa6, and 2x3=6, therefore 36 
 the product is Gab. Hence, the coefficient of the Gab product. 
 product is obtained by multiplying together the coeffi- 
 cients of the factors. This is termed, the rule of the coefficients. 
 
 From this example we see, also, that the literal part of the pro- 
 duct is obtained by annexing to the coefficient all the Utters in the two 
 factors. 
 
28 RAr'S ALGEBRA, PART SECOND. 
 
 
 
 E X AM P L E S . 
 
 2. 
 3. 
 4. 
 
 x^yXxi/= 
 a?x^zXa^2-= 
 
 a'J. 
 x^y^. 
 
 5. a'"Xfl"= 
 
 6. c"+'X<:"~'= 
 
 7. a;'"+''Xa!"~''= 
 
 EXAMPLES. 
 
 2. 3acX5i=^ ISoic. I 4. 5aX4ffia^ 20aar. 
 
 3. 2amy.cn= 2acmn. I 5. 7ci/X32/z= 21c3/y2:. 
 Art. 56. Of the Exponent. — To determine the rule of the 
 
 exponents, let it be required to find the product 
 of 2a' by 3a'. operation 
 
 Since 2a' is the same as 2affi, and 3a' the 2a==2aa 
 eame as 3aao, the product will be 2aaX3aaa, 3a'=3 ca<z 
 or Qaaaaa, which, for the sake of brevity, is Qa^=Qaaaaa 
 written 6o'. Hence, the exponent of a Utter in 
 Ike product is equal to the sum of its exponents in the two faciors. 
 This is termed the rule of the exponents. 
 
 0""+". 
 
 Art. 57. From the two preceding articles we derive the 
 following 
 
 Rule fok mdltiplting one positive monomial by anothee. — 
 Multiply the coefficients of the two terms together, and to the product 
 annex all the letters in both quantities, giving to each letter an 
 exponent equal to the sum of its exponents in the tvio factors. 
 
 Note. — Although the product is the same, in whatever order the let- 
 tars are placed (Art. 53), yet, for the sake of convenience, they are 
 generally written alphabetically. 
 
 E XAMPLE S. 
 
 1. Multiply ic by z. Ans. ba 
 
 2. Multiply Sax by hy. Ans. 3dbxy. 
 
 3. Multiply 4am by 2bn. Ans. lHabmn. 
 
 4. Multiply 5a'a; by lax^y. Ans. Z5a'x*y. 
 
 5. Multiply Sa^x" hy 9a"a?». Ans. 27a'"+"a;"^». 
 Akt. 58. Let it be required to find the product o' a-{-b by m 
 Here, the sum of the units in a and 6 is to be 
 
 taken m times. The units in a taken m times operation 
 
 =rma, and the units in 6 taken m times =:mb; a-{-b 
 
 hence, both together taken m times =ma-^mb. m 
 
 Hence, when the sign of each term is positive, ma+mft 
 we have the following 
 
 Rule foe multiplying a polynomial by a monomial. — Multi- 
 ply each term of the multiplicand hy the multiplier. 
 
MULTIPLICATION, 29 
 
 E XAM PL E S. 
 
 2. Multiply x-\-y by n. Atis. nx-\-ny. 
 
 3. Multiply ax'-^-cz by 3ac. Ans. 3a'cx'-l-3ac'2. 
 
 4. Multiply 2ffi=+3i' by 5a6. A?w. \Qa^b-\-lf)ab' 
 
 5. Multiply mx-\-ny-\-vz by ro'». Ans. m'na;-(-m'?i^!/-}"'"''i''* 
 Abt. 59. Let it be required to find the product of a-\-h by 
 
 m-\-n. Here, a-\-b is to be taken as many times as there are 
 units in m-\-n, which is evidently as many times as there are unltn 
 in m, plus as many times as there are units in n. Thus 
 a+b 
 m-\-n 
 7na-)-m6=the multiplicand taken m times. 
 
 na-\-nb=the multiplicand taken n times. 
 ma-\-mb-\-na-\-nb=:=th.e multiplicand taTten (m-f-ra) times. 
 Hence, when all the terms in each are positive, we have the 
 following 
 
 RtlLE FOR MTJLTIPLTINO ONE POLYNOMIAL BY ANOTHER. — Multi- 
 ply each term of the multiplicand by each term, of the multiplier, and 
 add the products together. 
 
 EXAMPLES 
 
 2. Multiply x-\-y by a-\-c. Ans. ax~\-ay-\-cx-\-cy, 
 
 3. Multiply 2x+3z by 3a;+2z. Ans. QxJ'-^-lSxz+Gz^. 
 
 4. Multiply 2a+c by a+2c. Ans. 2a^-\-5ac-{-2c'. 
 
 5. Multiply x'^-\-xy\-y^ by x+y. Ans. x^-{-2x^y-\-2xy^-\-f 
 
 6. Multiply a2-i-2aJ+6' by a-\-b. Ans. a^-\-'ia'^b-\-Zab''-\-b'. 
 
 Art. 60. Of the Signs. — In the preceding article it was as- 
 sumed that the product of two positive quantities is also positive. 
 It may, however, be shown, as follows : 
 
 1st. Let it be required to find the product of -|-i by a. 
 
 The quantity b, taken once, is -{-b; taken twice, is evidently 
 ■\-Zb; taken 3 limes, is -|-34, and so on. Therefore, taken a 
 times, it is -\-ab. Hence, the product of two positive quantities 
 is positive ; or, as it may be more briefly expressed, plus multiplied 
 by plus gives plus. 
 
 2d. Let it be required to find the product of — b by a. 
 
 The quantity — b, taken once, is — J ; taken twice is — 26 ; 
 taken 3 times, is — 3J ; and hence, taken a times, is — ab ; that 
 is, a negative quantity, multiplied by a positive quantity, gives a 
 negative product. This is generally expressed by saying, that 
 minus, multiplied by plus, gives minus. 
 
80 RAT'S ALUEBRA, PART SECOND. 
 
 3d. Let it be required to multiply h by — a. 
 
 Since, when two quantities are to be multiplied together, eithei 
 may be made the multiplier (Art. 53), this is the same as to mul- 
 tiply —a by b, which gives —ab. That is, a positive quantity 
 multiplied by a negative quantity, gives a negative product ; or 
 more briefly, plus multiplied by minus, gives minus. 
 
 4th. Let it be required to multiply — h by — a. 
 
 The negative multiplier signifies that the multiplicand is tc be 
 taken positively as many times as there are units in the multi- 
 plier, and then subtracted. 
 
 The product of — b by a, is — ab, and then changing the sign to 
 subtract, it becomes -\-ab. Hence, the product of two negative 
 quantities is positive ; or, more briefly, minus multiplied by 
 minus, gives plus. 
 
 Note. — Tho following proof of the last principle, that the product 
 of two negative quantities is positive, is generally regarded by mathe- 
 maticians as more satisfactory than the preceding, though it is not quite 
 80 simple. Either method may be used. 
 
 5th. To find the product of two negative quantities. 
 
 To do this, let it be required to find the product of c — d by 
 a — i. Here it is required to take c — d as many times as there 
 are units in a — b. It is obvious that this will be done by taking 
 c — d as many times as there are units in a, and then subtracting, 
 from this product, c — d taken as many times as there are units 
 in b. 
 
 Since plus, multiplied by plus, gives plus, and minus, multiplied 
 by plus, gives minus, the product of c — d by a is ac — ad. In 
 the same manner, the product of c — d by h is he — bd ; changing 
 the signs of the last product to subtract it, it becomes — bc-\-bd ; 
 hence, the product of c — d by a — i, is ac — ad — bc-\-bd. But the 
 last term, -\-bd, is the product of — d by — b; hence, the product 
 of two negative quantities is positive ; or, more briefly, minus 
 multiplied by minus, produces plus. 
 
 The multiplication of c — d by a — 6 may be written thus : 
 
 c — d 
 
 a — 6 
 ac — ad=c — d taken a times. 
 
 — ic-)-M=c — d taken b times, and then subtracted. 
 ac — ad — bc-\-bd 
 
 To illustrate the operation by figures, let it be required to find 
 ihe product of 9 — 4 by 5 — 3. 
 
MULTIPLICATION. 31 
 
 OPERATION We first take 5 times 9 — 4; this gives a 
 
 9 — i product too great by 3 times 9 — 4, or 27 
 
 5 — 3 — 12; subtracting this from the first product, 
 
 45—20 we have, for the true result, 45 — 47-|-12, 
 
 — 27+12 which reduces to -|-10. This is evidently 
 
 45:i47-j_12 correct, for 9 — 4=5, and 5—3=2, and the 
 
 product of 5 by 2 is 10 . 
 Prom the preceding illustrations, we derive the following 
 
 General rule for the sisns. — Plus multiplied hy plus, or minus 
 multiplied hy minus, gives plus. Plus multiplied by minus, or 
 minus multiplied by plus, gives minus. 
 
 Or, the product of like signs gives plus, and of unlike signs gives 
 minus. 
 
 Art. 61. From the preceding, we derive the following 
 
 General rule for the multiplication of algebraic quan- 
 tities. — Multiply every term of the mtdtiplicand by each term of 
 the multiplier, observing, 
 
 1st. That the coefficient of any term is equal to the product of the 
 coefficients of its factors. 
 
 2d. That the exponent of any letter in the product is equal to the sum 
 of its exponents in the two factors. 
 
 3 d , That the product of like signs gives plus in the product, and unlike 
 signs gives minus. Then, add the several partial products together. 
 
 NUMERICAL EXAMPLES TO VERIFY THE RULE OF THE SIGNS. 
 
 1. Multiply 7—4 by 5. Ans. 35—20=15=3x5 
 
 2. Multiply 8+3 by 6— 4. jlns. 48— 14— 12=22=11x2. 
 
 3. Multiply 5—8 by 4—9. 
 
 Ans. 20— 77+72=+15=— 3x— 5 
 
 4. Multiply 7—3 by 8—2. Ans. 56—38+6=24=6x4 
 
 GENERAL EXAMPLES. 
 
 1. Multiply 4a'— 3ac+2 by 5ax. Ans. 2Qa^x—\5a''cx-\-lQax. 
 
 2. Multiply 5ffl— 2fflt+10 by —Qab. 
 
 Ans. — 45a2i+18o2J'— 90ffl6, 
 
 3. Multiply 2a;+32 by 2a;— 3z. Ans. ix^—Qz'. 
 
 4. Multiply 4a=— 6a+9 by 2a+3. Ans. 8a3+27. 
 
 5 . Multiply a — J+c — d by a-\-b — c — d. 
 
 Ans. a=— 6=— c2+(Z2— 2a(i+2Ac 
 
 6. Multiply x^+y'+z' by x'+y'. 
 
 Ans. a:=+a;'y'+a;'22+a;'y=+v*+y'i' 
 
32 RAY'S ALGEBRA, PAllT SECOND. 
 
 7. Multiply a5-l-3a'i+3ai=+/>' by a'—3a^b+Sab<^b'>. 
 
 Ans. a^—2a'>¥+Za'b*—h' 
 
 8. Multiply Ux^—Sx-y+loxy'^—lOyS by 3a:+2j^. 
 
 Ans. Z6x''-\-^9xY—20i/, 
 
 9. Multiply a^-\-ax-\-x^ by a^ — ax -{-x^. Ans. a*-\-aV-{-x*. 
 10. Multiply a^-\-2al+2h' by a^—2ab-}-2b^. Ans. ai+U*. 
 1 1 Multiply x^-\-2xy—Sy^ by a;'— 5iy+4j(2. 
 
 Anj. a'— 3a;')/— 9a:=y=+23a^'— 12i/4. 
 
 12. Multiply l+x+x^+x'+x* by l^c. ylns. 1— i*. 
 
 13. Multiply 37a;'-j-9j;=3/+3a:i/=-|-y' by 3x—y. Ans. 81x*—y'. 
 
 14. Multiply x*+2x'y+ixY+Sxy^+lGy'' by a:— 2y. 
 
 Atjs. x»— 32i/». 
 15 Multiply a^—2a^+ia^l>^—Sa¥+16b'' by a+2i. 
 
 Ans. o'+326'. 
 
 16. Multiply a^+2a^b+2ab'+¥ by o=— 2a=i+2ai=— i'. 
 
 Ans. o^ — 6*. 
 
 17. Multiply a'+J'+c^ — ab — ac — be by a-\-b-\-c. 
 
 Ans. a^-\-h'-\-c' — 2abc. 
 
 18. Multiply x'—x'-l-x^—x-l-l byx=+a:— 1. 
 
 Ans. x^ — a;''-|-x' — x''-\-2x — 1 . 
 
 19. Multiply 1+^4"^'+^' '^y 1 — x-^x' — x'. Ans. 1 — a:*. 
 
 20. Multiply 1— 2a;+3a;' — ix'+Sa:"— 0x'+7a:«— 8j;M)y 1 -f2a 
 -y-x^ Ans. 1— 9a;«— 8a;'. 
 
 21. Multiply together x — 3, a;+4, x — 5, and .c-f-O. 
 
 Ans. a;''+2.c'— 41a;2— 42a;+360. 
 
 22. Multiply together a^+db+b^, a^—a^+¥, and a—b. 
 
 Ans. a^ — a''b-\-a^¥ — b'. 
 
 23. Multiply together a-j-fc, a — 6, a'-{-ab-\-b', and a' — ab-\-b'. 
 
 Ans. a* — i«. 
 
 24. Prove that 
 
 x(a;+l)(a;+2)+a;(a;— l)(a;— 2)+4(av-l)a;(x+l)=6x'. 
 
 25. Find the value of the expression 
 (_x+a)ix+b%x+c)—(_a+b-\-c)(,x+a){x+l)+{a^-\-ab+V)(^x+a). 
 
 Ans. x'-\-a' 
 
 JIULTIPLICATION BY DETACHED COEFFICIENTS. 
 
 Art. 62. In the multiplication of polynomials, it is evident 
 that the coefficients of the product depend on the coefficients of 
 the factors, and not upon the literal parts of the terms. 
 
 Hence, by detaching the coefficients of the factors from the 
 literal parts, and multiplying them together, we shall obtain tho 
 coefficients of the product. If to these coefficients the propei 
 
MULTIPLICATION. 
 
 33 
 
 letters are then annexed, the whole product will be obtained. 
 
 This method is applicable where the powers of the same letter 
 increase or decrease regularly. 
 
 1. Multiply a^ — 2ab-]-i^ hy a-{-h. operation. 
 After finding the coefficients, it is obvious 1 — 2-|-l 
 
 that a^ will be the first term and 6' the last l+l 
 term ; hence, the entire product is o' — a'b — aJ' 
 
 2. Multiply a'— 3a'6+6' by a'— i'. 
 
 In this example, supposing the powers operation. 
 
 of a to decrease regularly toward the 1 — 3+0-|-l 
 left, it is obvious that there is a term l-(-0 — 1 
 wanting in each factor. In such cases 1 — 3-|-0-|-l 
 the coefficient of each absent term must — 1+3 — — 1 
 
 be considered zero, and supplied before 1 — 3 — 1+4 — — 1 
 commencing the operation. 
 
 This method, termed multiplication by detached coefficients, is 
 useful in leading the pupil to consider the properties of coefficients 
 by themselves. 
 
 £ X AUPLES . 
 
 3. Multiply m'+m'ra+mn'+n' by m — n Atis. m*—n^. 
 
 4. Multiply l+22;+3z=+42'+5z< by 1—z. 
 
 AnS. 1 +2+22+25+2"— 5z-\ 
 
 The pupil may also solve, by this method, the general examples, 
 Art. 61, from 7 to 20 inclusive, except example 17. 
 
 REMARKS ON ALGEBRAIC MULTIl'LICATION. 
 
 Akt. 63. The degree of the product of any two monomials, is 
 equai to the sum of the degrees of the multiplicand and multi- 
 ylier. This is evident, since all the factors of both quantities 
 appear in the product. Thus, 2a'J, which is of the 3d degree, 
 multiplied by Zalfl which is of the 4th degree, gives Ga'i'', which 
 is of the 7th degree. Hence, if two polynomials are homogeneous, 
 their product will be homogeneous. Thus, in example 7, Art 
 61, both multiplicand and multiplier are homogeneous, each term 
 being of the 3d degree, and the product is homogeneous, each 
 term being of the 6 th degree. 
 
 Art. 64. In the multiplication of two polynomials, when the 
 partial products do not contain similar terms, the whole number of 
 terms in the final prod' ct will be equal to the number of terms in 
 the multiplicand, multi^ilied by the number of terms in the multi- 
 
34 RAY'S ALGEBRA, PART SECOND. 
 
 plier. Thus, if there be m terms in the multiplicand, and n terms 
 in the multiplier, the number of terms in the product will be 
 mXra- Thus, in example 6, Art. 01,, there are 3 terms in the 
 multiplicand, 2 in the multiplier, and 3x2=6 in the product. 
 
 Art. 65. If the partial products contain similar terms, the 
 number of terms in the reduced product will evidently be less 
 than mX»; see examples 7 to 21 inclusive, Art. 61. It is 
 Important to note that th;;re are two terms which can never be 
 reduced with any others ; these are, 
 
 1. That term which is the product of the two terms In the 
 factors which contain the highest power of the same letter. 
 
 2. That term which is the product of the two terms in the 
 factors which contain the Vywest power of the same letter.- 
 
 Akt. 66. The multiplication of two polynomials is indicated 
 by inclosing each in a parenthesis, and writing them in succes- 
 sion. Thus, the multiplication of the polynomials m-\-n and p — q, 
 is indicated by (m-|-?j)(p — q.) 
 
 When the operation is actually performed, the expression is 
 said to be expanded, or developed. 
 
 DIVISION. 
 
 Art. 67. Division, in Algebra, is the process of finding how 
 often one algebraic quantity is contained in another. Or, having 
 the product of two factors, and one of them given, division teaches 
 the method of finding the other. 
 
 As in Arithmetic, the quantity, by which we divide is called the 
 divisor; the quantity to be divided, the dividend; the result of the 
 operation, the quotient. 
 
 Art. 68. In division, as in multiplication', there are four thing 
 to be considered, viz : 
 
 The sign ; 
 The coefficient ; 
 The exponent ; 
 The literal part. 
 Art. 69. To ascertain the rule of the signs. 
 Since +aX+i=+a5'~l f+al- — f-o=-|-a 
 
 r ^+^=-1 y therefore ^ -«f-+J=-« 
 +oX — b=—ah r j — db-i i=-|-a 
 
 — aX— 4=+aiJ XA-ah-. h=—a 
 
DIVISION. 35 
 
 From which we derive the following 
 Rule of the signs. — When both divisor and, dividend have lite 
 same sign, the quotient will have the sign -\- ; when they have 
 different signs, the quotient will have the sign — . 
 
 Art. 70. To ascertain the rule of the co'ejjlcients, the rule of 
 the exponents, and the rule of the literal part. These may all be 
 derived from the solution of a single example. 
 
 Let it be required to find how often 2a' is contained in 6a'6. 
 
 6a=J 6 
 -=-a^-^t=2a^b. 
 
 Since division is the reverse of multiplication, the quotient, 
 multiplied by the divisor, must produce the dividend ; hence, to 
 obtain this quotient, it is obvious, 
 
 1st. That the coefficient of the quotient must be such a num- 
 ber, that when multiplied by 2 the product shall be 6 ; therefore, 
 to obtain it, we divide 6 by 2. Hence, we have the following 
 
 Rule op the coefficients. — To obtain the coefficient of the quo- 
 tient, divide the coefficient of the dividend by the coefficient of the 
 divisor. 
 
 2d. The exponent of a in the quotient must be such a number, 
 that when 2,' the exponent of a in the divisor, is added to it, ihe 
 sum shall be 5; hence, to obtain it, we must subtract 2 from b; 
 that is 5^2=3, is the exponent of a in the quotient. This 
 gives the following 
 
 Rule of the exponents. — From the exponent of any letter in the 
 dividend siibtract its exponent in the divisor, the remainder will be 
 its exponent in the quotient. 
 
 3d. The letter b, which is a factor of the dividend, but not of 
 the divisor, must be found in the quotient, in order that the pro- 
 duct of the divisor and quotient may equal the dividend. Hence, 
 mery letter fmmd in the dividend, and not in the divisor, must be 
 found in the quotient, with the same exponent as in the dividend. 
 This, in connection with the rule of the exponents, furnishes the 
 rule of the literal parts. 
 
 Art. Tl. The preceding rules taken together, give the following 
 Rule for dividing one monomial bt another. — Divide the 
 coefficient of the dividend by that of ihe divisor; observing, that like 
 signs give plus and unlike signs give minus. 
 After the coefficient, write the letters common to both divisor and divi- 
 dend, giving to each an exponent, equal to the excess of the exponent 
 of the same letter in the dividend, over that in ihe divisor- 
 
86 RAY'S ALGEBRA, PART SECOND. 
 
 In the quotient, tcrite the tetters, with their respective expimenis, that 
 are foani in the dividend, hat not in the divisor. 
 
 EXAMPLES. 
 
 1. Divide 4a' by 2a^ and by —2a^. Am. 2a? and —2a' 
 
 2. Divide 30a«A' by Sa'i. Ans. Oa^ft 
 
 3. Divide — aSaiyz" by —Ixy'^z. Ans. Axh/^z^ 
 
 4. Divide —ZbaWc by 5a5'. Ans. —Tdbc 
 
 5. Divide Z^iayz by —Sxy. Ans. — 4«. 
 
 6. Divide 42c'm'ra by — 3cnm. Ans. — 14c'm, 
 
 7. Divide a;"H-" and a™"" each by 3f. Ans. a;"" and a;'"-^. 
 
 8. Divide t)"M-» by ■u'M-r. Ans. if^r. 
 
 Note. — In solving the following examples, the pupil must recollect, 
 that the quantities included within the vinculum are to be considered 
 together, as a single quantity. 
 
 9. Divide Ca+by by (a+t)'. Am. {a+h) 
 
 10. Divide (rra — ra)' by (m — ny. Ans. (m — n)' 
 
 11. Divide 8(a~ t)'a;^2/ by 2(a — b)xy. Atw. 4(a — hyx 
 
 12. Divide e(ix+zyia-^by by 3 (a:+z)(ffl— ft)'. Am. 2(a:+2)=, 
 
 13. Divide aV(x — y)(_y — zy by aft'(y — zy. Am. ai(x — y) 
 
 14. Divide (a-\-bx^y+^ by la+bx^)'^K Am. (a+bx^y. 
 Art. 72. It is evident that one monomial cannot be divided by 
 
 another in the following cases : 
 
 1st. When the coefficient of the dividend is not exactly divisi- 
 ble by the coefficient of the divisor. 
 
 2d. When the same literal factor has a greater exponent in the 
 divisor than in the dividend. 
 
 3d. When the divisor contains one or more literal factors not 
 found in the dividend. 
 
 In each of these cases the division is to be indicated by writing 
 the divisor under the dividend, in the form of a fraction. This 
 fraction may often be reduced to lower terms. See Art. 119. 
 
 Art. 73. It has been shown, in Art. .53, that any product is 
 
 multiplied by multiplying either of its factors ; hence, converselv, 
 
 any dividend will be divided by dividing either of its factors. 
 
 6X9 
 Thus, -g— =2x9=18, by dividing the factor 6. 
 
 6X9 
 Or, —g— =6x3=18, by dividing the factor 9 
 
 DIVISION OF POLYNOMIALS BY MONOMIALS. 
 
 Art. 74:. In multiplying a polynomial by a monomial, we 
 multiply each term of the multiplicand by the multiplier. Thus, 
 
DIVISION. 37 
 
 (oJ— i')Xa=ffi'i — ab^; hence, (a'i — db'')-i-a=.-ab — h^ ; therefore, 
 we have the following 
 
 RitLE FOR DIVIDING A POLTKOMIAL BY A MONOMIAL. — Divide each 
 
 term of Che dividend by the divisor, according to the rule for thf. 
 division of monomials. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Divide a^+ai'by a. Ans. a-\-b 
 
 2. Divide 3xy-\-2x^y by — xy. Ans. — 3 — 2x. 
 
 3. Divide 10a'z—15z^—25z by bz. Ans. 2a^—Zz—5. 
 
 4. Divide Zab-\-V2abx — 9a^ by — 2ab. Ans. — 1 — ix-\-'3a. 
 .'5. Divide 5x'y^ — i0a^x''y''-\-25a''xy by — 5xy. 
 
 Ans. — x^y'-\-8a'xi/—5a* , 
 
 6. Divide 4aic— 24a&=— S^aM by —iaD. Ans. —c+6b+8d. 
 
 7. Divide a'"b'+a"^V+a"'''b by ab. Ans. a'"-'b^-\-a'"b-\-a"-K 
 
 8. Divide 4a'(B+a;)+6a(a;+y) by 2a. 
 
 Ans. 2a(^a-\-x)-\-3(x-\-y). 
 
 9. Divide Sa{x-\-y)-{-c\x-\-yy by x-\-y. Ans. 3a-\-c\x-\-y). 
 lO. Divide b^c(m-{-n) — bc'(m-{-n) by bc(m-\-n). Ans. b — c. 
 tl. Divide (b+c){b—cy—(b—c)(b+cy by (J-]-c)(?;— c). 
 
 Ans. (6— c)— (i+c)=— 2c 
 12. Divide (m — ny(,x-^zy — (m — ny(x-{-zy by (m — ny(x-{-zy. 
 
 Ans. (.x-\-z) — (m — n). 
 
 DIVISION OF ONE POLYNOMIAL BY ANOTHER. 
 
 Art. 75. To explain the method of dividing one polynomial 
 by another, we may regard the dividend as a product, of which 
 the divisor and the quotient are the two factors. We shall first 
 form a product, and then, by a reverse operation, explain the 
 process of division. 
 
 Multiplication, or forma- Division, or decomposition of a product. 
 
 lion of a product. 
 
 a^-\-2a b—V 
 a>—6a*b 
 
 —a^b'-{-5a^¥ 
 
 a'— 3 a'b—l 1 a^J^-f-S a=6' 
 
 a^—^a''b—lla^b^-\-5a^^ a'—Sa'b 
 a'—5a^b 
 
 istr. +2a^b—Ua^b'+5a^^ 
 
 -]-2a''b—10a^b' 
 2d remainder, — a^6^-l-5a^6^ 
 — a'J2-f-5a=i' 
 
 a^+2ab—b' 
 Quotient 
 
 3d rcDiainder, 
 
 By comparing the product with the two factors, each being 
 arranged according to the decreasing powers of the letter a, we 
 see that the 1st term a' of the total product, is the product of the 
 !st term a' of the multiplicand, by a' the 1st term of the divisor • 
 
38 RAY'S ALGEBRA, PART SECOND 
 
 that the last term -\-5aV of the total product, is the result of the 
 product of — 5a'b the last term of the multiplicand, by — b' the 
 last term of the multiplier ; and that the other terms of the total 
 product ai-e the result of the reduction of the similar terms of the 
 partial products. See Arts. 64 and 65 
 
 Consequently, the division of a', the first term of the dividend 
 by a', the first term of the divisor, will give a', the first term of 
 the quotient. 
 
 The dividend expresses the sum of the partial products of the 
 divisor by the different terms of the quotient ; therefore, if we 
 subtract from the dividend a^ — 5a'b, which is the product of the 
 divisor a' — 5a'J by a' the first term of the quotient, the remainder 
 -\-2a*b — lla^b''-\-5aV, will be the product of the divisor by the 
 other terms of the quotient. 
 
 Knowing the 1st term a^ of the quotient, and the 1st remainder, 
 it is now required to find the other terms of the quotient. We 
 remark, that the 1st remainder expresses the product of the divi- 
 sor by the unknown terms of the quotient, and that, consequently, 
 the 1st term -j-Sa^Jof the 1st remainder, is the product of the 1st 
 term a' of the divisor by the 1st of the unknown terms of the 
 quotient ; therefore we shall obtain the 1st of these terms, that is< 
 the 2d term of the required quotient, by dividing the 1st term 
 -\-2a*b of the 1 st remainder, by the 1st term o' of the divisor; this 
 gives -\-2ab the 2d term of the required quotient. 
 
 Lastly, to find the 3d term of the quotient, we subtract from 
 the tst remainder, the product of the divisor by -\-2ab, the 2d 
 term of the quotient ; the 2d remainder is the product of the 
 divisor by the 3d term of the quotient ; hence, the division of the 
 1st term — a'b^ of this 2d remainder, by the 1st term a^ of the 
 divisor, must give the 3d term of the quotient, which is thus found 
 to be — 6'. 
 
 Subtracting from the 2d remainder, the product of the divisor 
 by — fc', the remainder zero, shows that the quotient a^-\-2ab — i' 
 is exact ; for we have arrived at this remainder by subtracting 
 from the dividend and the several remainders, the partial products 
 of the divisor by the terms a^, -\-2ab, — i' of the quotient. 
 
 Since there is no remainder when we subtract from the divi- 
 dend the product of the divisor by a^-\-2ab — b^, therefore the divi- 
 dend is the exact product of the divisor by a^-\-2ab — b^, which is, 
 therefore, the required quotient. 
 
 Since each term of the quotient is found, by dividing that term 
 of the dividend containing the highest power of a particular let- 
 ter, by the term of the divisor c-sntaining the highest power of the 
 
DIVISION. 
 
 39 
 
 same letter, the divisor ani dividend should always be arranged 
 (Art. 31) with reference to a certain letter. 
 
 The situation of the divisor in regard to the dividend, is a mat- 
 ter of arbitrary arrangement ; by placing it on the right it is more 
 easily multiplied by the respective terms of the quotient. 
 
 Am. YG. From the preceding we derive the following 
 
 Rule fob the division o? one poltnomial by another.— 
 Arrange the dividend and divisor with reference to a certain letter, 
 and place the divisor on the right of the dividend. 
 
 Divide the first term of the dividend hy the first term of the divisor ; 
 the result will he the first term of the quotient. Multiply the divisor 
 by this term, and svhtract the product from the dividend. 
 
 IHvide the first term of the remainder by the first term of the divisor; 
 the result will be the second term of the quotient. Multiply the divi- 
 sor by this term, and subtract the product from the last remainder. 
 
 Proceed m the same manner, and if you obtain for a remainder, the 
 division is said to be exact. 
 
 Remarks. — 1st. When there are more than two terms in the quotient, 
 it is not necessary to Ifriag down any more terms of the remainder, at 
 each successive subtraction, than have corresponding terms in the quan- 
 tity to be subtracted. 
 
 2d. It is evident that the exact division of one polynomial by another 
 wiir be impossible, when the first term of the arranged dividend is not 
 exactly divisible by the first term of the arranged dipisor ; or when any 
 remainder is not divisible by the first term of the divisor. 
 
 ] . Divide 1 5x^+1 6x^1— ISy^ by 5x—2y. 
 
 operation. 
 15x^+16xy— 15y' |5x— .3y 
 15x' — 9xy 'ix-\-5y Quotient. 
 
 -15j-= 
 
 -\-25xy- 
 +25xy-l5y' 
 
 2. Divide m^ — n' hjm-\-n. 
 operation. 
 m' — n' \m-\-n 
 m'-\-mn m — n Quotient. 
 — mn — n' 
 — mn — 7i' 
 
 3 . Divide x'-|-i/' by x-\-y. 
 operation. 
 
 X3-|-X^J( 
 
 -xy-{-i/^ Quot 
 
 -X^y-\- y3 
 -x^y — xy^ 
 
 +xy^+y^ 
 
*0 RAY'S ALGEBRA, PART SECOND. 
 
 4. Divide lx^y-\-5m/'+2x^+y^ by 3xy+x^-\-y^. 
 
 OPERATION. 
 
 2x'-\-6x^y-\-2xy' 2x-\-y Quotient. 
 
 xh/+3xy^+y' 
 x'y-^Sxy'-\-y^ 
 
 In this example, neither divisor nor dividend being arranged 
 with reference to either x or y, we arrange them with reference 
 to X, and then proceed to perform the division. 
 
 5. Divide I'+x' — ^7a^+5a;= by x — a'. 
 
 Division performed, by arranging 
 ooth quantities , according to the 
 ascending powers of x. 
 i2+ar'— 7a;<+5a;5 \x—x' 
 x^—x' x+2x'—5x' 
 
 2a:'— Tx* Quotient. 
 
 2x^—2x* 
 
 Division performed, by arranging 
 both quantities according to th» 
 descending powers of x. 
 
 5x^—Tx*+x'-{-x'\—x'-\-x 
 bx^—Sx" —5x'-\-2x'+x 
 
 — 2x*-\- a:' Quotient. 
 
 — 2jcy-2a' 
 
 — x'-f-*' 
 — x'+i' 
 
 The learner will perceive that the two quotients are the same, but 
 iiflerently arranged. 
 
 EXAMPLES FOR PRACTICE. 
 
 6. Divide 6x^-{-53cy — iy^ by 3x-|-4j/. Ans. 2x — y. 
 
 7. Divide x'— 40x— 63 by x— 7. Atis. x=4-7x+9. 
 
 8. Divide 3h^+ieh*k—Sdh%^+14W by h^+lhk. 
 
 Ans. M'—5h'k+2hk'. 
 
 9. Divide o'— 243 by a— 3. Ans. a<+3a'+9o2+27o+81. 
 
 10. Divide x«— 2a'x'4-o^ by x=— 2ax+a=. 
 
 Ans. x<-f2ax5-f3a'x=+2a'x+o<. 
 
 11. Divide 1—6x5+5x5 by 1— 2x+x'. 
 
 Ans. l+2x+3x2+4x'+5x<. 
 
 12. Divide;)'+p5+2j)r— 2g=-)-7gr— 3r= hj p—g-\-2r. 
 
 Ans. p-{-2q—r. 
 
 13. Divide 4x5-|-4x— x' by 3x-|-2x=+2. Ans. 2x'— 3x2-f2ar. 
 
 14. Divide x«— a» by x'+2ax=i-i-2a=x+a3. 
 
 Ans. x' — 2ax'-f-2a'x — a*. 
 
 15. Divide m'-|-2mp — n' — 2nq-\-p^ — 3' by m — n-\-p — q. 
 
 Ans. m-\-7ir^-\-q. 
 
 16. Divide a'+J'+c'— 3aJc by a+h+c. 
 
 Ans. a^+J'-j-c' — ab — ac — be. 
 
DIVISION. 41 
 
 17. Divide af<+''-\-x"y"-{-afy"'-\-y"'+'' by af^-\-y"'. Ans. x'"-{-y". 
 
 18. Divide ax' — {a^-\-b)x'-\-h' by ax — b. Ans. x' — ax — 6, 
 
 19. Divide mpx'-\-(mq — np)x' — (mr-\-nq)x-\-nr by tux — n. 
 
 Ans. px^-\-qx — r, 
 
 20. Divide 0="— 3o'"c"+2c2'' by a^—c". Ans. a""— 2c". 
 
 21. Divide x^-\-x~' — x^ — x~^ by x — i"'. Ans. x' — x '. 
 
 22. Divide a^-\-a^b^-^a^¥-\-a'^¥-\-V by a*-\-a'b-\-a''b^-\-ab'J^b\ 
 
 Ans. a<— o36-|-a2i2— ai3-f-M. 
 23 Divide a=+(a— l)a;=+(a— l)x'+(a— l)a;<— a;' by a—x. 
 
 Ans. a.-{-x-\-x^-\-x'-\-x^. 
 
 24. Divide a;(a^-l)a'+(a:'+2af— 2)a2-)-(3x'— a;')a— x'' by a^x 
 
 +2a— x2. Arts, (x— I)a4-x2. 
 
 25. Divide x'— 83/'+1250'+3Oxyz by x— 2y+5z. 
 
 Am. x=-|-2xT/— 5x2-j-4y2_(-10yi-)-25z2. 
 
 26. Divide 1— 9x«— Sx^ by 14-2x+x2. 
 
 Ans. 1— 2x+3x2— 4x'-t-5x''— 6x5+7x«— 8x'. 
 
 27. Divide l-)-2x by 1 — 3x to -5 terms in the quotient. 
 
 Ans. l+5x+15x=+4.'5x'+135x<4-&c. 
 
 28. Divide 1— 3x— 2x2 by 1 — 4x to 6 terms in the quotient. 
 
 Am. l+x+2x2+8x'+32x<+128x'+&,c. 
 
 DIVISION BY DETACHED COEFFICIENTS. 
 
 Art. 77. From Art. 62, it is evident that division sometimes 
 may be conveniently performed, by operating on the coefficients 
 detached from the letters, and afterwards supplying the letters. 
 Thus, if it be required to divide x'-\-2xy-^y^ by x-\-y, we may 
 perform the operation as follows : 
 1+2+1 [1+^ Hence the coefficients of the quotient 
 
 1+1 1+1 are 1 and 1. Also, x'-hi;=x, and y'-i-y=y; 
 +1+1 therefore the quotient is Ix+lw, or x+v. 
 
 1+1 
 
 2. Divide 12a^—26a'b—8a%^-{-10ab'—8¥hy3a^—2ab-\-b\ 
 12— 26-8+10— 8 |3— 2+1 Hence the coefficie.nts of the 
 
 12— 8+4 4—6—8 quotientare4— 6— 3. Also, 
 
 —18—12+10 a''-r-a^=a^, and ^—-b'^b' ; 
 
 — 184-12 — 6 therefore the quotient is 4a' 
 
 —24+16— a —Qab—8b\ 
 —26+16—8 
 
 When any of the intermediate powers of the letters are want- 
 ing, the coefficients of the corresponding terms must be supplied 
 with zero, as in the following example. 
 
 4 
 
42 RAY'S ALGEBRA, PART SECOND. 
 
 3. Divide a^-\-a^ by a-\-x. 
 
 1+0+0+1 |1+1 
 1+1 1-1+1 
 
 - — 1 a' — ax-\-x^ Quotient. 
 
 —1—1 
 
 +1+1 
 +1+1 
 
 E X AM FLE S. 
 
 4. Divide 6x*+4x>y—9x'y'-Sxy'+2y' by 2x'+2xy—y'. 
 
 Ans. 3x^ — xy — 2y', 
 
 5. Divide m' — SmV+lOmW — lOmV -\-5mn^ — »' by m' 
 
 — 2mn-\-n'. Ans. m? — Zrrfin-\-Zmn? — n'. 
 
 6. Divide o«— Sa^i'+Sa^J'— 6« by d>—Za'b-\-^db^—bK 
 
 Ans. a'+3o2J+3a&2+i'. 
 , Most of the examples in the preceding article may be solved bj 
 this method. 
 
 CHAPTER II. 
 ALGEBRAIC THEOREMS, 
 
 DKRIVED FROM MULTIPLICATION AND DIVISION 
 
 Rkmakk. — One of the chief objects of Algebra is to establish certain 
 general truths. The pupil has now obtained the necessary knowledge 
 to prove the following theorems, which may be regarded as the simplest 
 application of Algebra. 
 
 Art. 'S'S. Theorem I. — The square of the sum of two quantities 
 is equal to the square of the first, plus twice the product of the first by 
 the second, plus the square of the second. 
 
 Let a represent one of the quantities, and b the other ; 
 then a+i=their sum ; 
 
 and (a+i)x(a+6), or (a+6)2=the square of their sum. 
 But (o+i)X(fl!+i)=o'+2ai+i2, which proves the theorem 
 
 AP P L I CAT ION. 
 
 1. (2+5)2=4+20+25=49. 
 
 2. (2OT+3m)s=4OT2+12»n7i+9»=. 
 
ALGEBRAIC THEOREMS. 43 
 
 3. (ax+byy=a'x'+2ahxy-{-bY- 
 
 Art. 79. Theoeem II. — The square of ihe difference of two 
 quantities is equal to the sqruare of the fast, minus timce the product 
 of the first by the second, plus the square of the se^nd. 
 
 Let a represent one of the quantities, and b the other ; 
 then a — i=their difference ; 
 
 and (a — i)X(a — b), or (a — i)'=the square of their difference. 
 But (a — b)y.(a — b)=a' — 2ab-\-b^, which proves the theorem. 
 
 APPLICATION. 
 
 1. (5— 3)2=25— 30+9=4. 
 
 2. (2x-^y='ix^—4xy+y'. 
 
 3. (3a^-5^)'=9x=— 30x2+2522. 
 
 4. {az — 3cx)2:=a22' — 6aca;2+9c'x'. 
 
 Art. 80. Theokem III. — The product of the sum and difference 
 of two quantities, is equal to the difference of their squares. 
 
 Lei a represent one of the quantities, and b the other ; 
 then a+J=their sum, 
 and a — Zi=their difference. 
 And (a+i)(a — b)=a^ — i', which proves the theorem. 
 
 AP PL IC AT ION. 
 
 1. (7+4)C7—4)=49— 16=33=11X3. 
 
 2. (2x+2/)(2x— y)=4x2— y=. 
 
 3. (3a='+4J2)(3a2— 4J2)=9a''— 16M. 
 
 4. (3ax+56y)(3ax— 5i^)=9a'x2— 25J=^'. 
 
 Art. 81. Theorem IV. — The reciprocal of a quantity, is equal 
 to the same quantity with the sign of its exponent changed. 
 
 If we divide a' by a*, the quotient is expressed by^,or by 
 
 a'~^=a~', since the rule for the exponents in division (Art. 70) 
 requires that the exponent of the same letter in the divisor should 
 
 be subtracted from that of the dividend. But ^ is a fraction, and 
 
 if we divide both terms by a', which does not alter its value 
 (Ray's Arith., part 3d, 
 since each is equal to 
 
 (Ray's Arith., part 3d, Art. 147), it becomes L; hence a-2=i. 
 
 a' 
 
44 RAY'S ALliEBRA, PART SECOND. 
 
 In the same manner, ^=0"*-", by subtracting the exponents; 
 
 ml"" 1 
 
 or 1-=L^, by dividing both terms by o"; hence, 0'"-''= -^j^, 
 
 a" 
 
 a" 
 
 which proves the theorem. 
 
 E X AH PL E S. 
 
 1. 
 
 
 2. 
 
 0" 
 
 3. 
 
 " ui""™ 
 
 4. 
 
 ==■-'- 
 
 5. 
 
 r*'- 
 
 6. 
 
 L =a->6 
 
 We see also from this, that any factor may be transferred from 
 one term of a fraction to the other, if at the same time the sign of 
 its exponent be changed. Thus, 
 
 z X"' X 'z V a~V o~' 
 
 Art. 82. Theokem V. — Any quantity, whose exponent is 0,is 
 iqual to unity. 
 
 If we divide a' by a', and apply the rule for the exponents 
 
 (Art. 70), we find _=o' '=0° ; but, since any quantity is con- 
 
 tained in itself once, — =1 . 
 
 Similarly, ^=a"'-'"=a'>. But, —=1. 
 
 a™ o"" „ 
 
 a 
 Hence, 0"=!, since each is equal to — ; which proves the 
 
 theorem. 
 
 This notation is used, when we wish to preserve the trace of a 
 
 letter, which has disappeared in the operation of division. Thus, 
 
 if we divide a^b by ab, the quotient is ^i=a2~'i'~'=a'i''=a. 
 
 ab 
 Now the quotient is expressed correctly, either by o'i", or by a, 
 since both have the same value. The first form is used when we 
 wish to show that the letter b was originally a factor, both of the 
 dividend and divisor. 
 
 Akt. 83. TijEOKEM VI. — The difference of the same power of 
 boo quantities is always divisible by the difference of the quantities. 
 
 1. If we divide o' — b^ by a — b, the quotient is a-\-b. 
 
 2. If we divide a' — h^ by a — b, the quotient is a^-\-ab-\-b^. 
 
 In the same manner, we would find by trial, that the difierenoe 
 of the same power of any two quantities is divisible by the 
 
ALGEBRAIC THEOREMS. 45 
 
 difference of the quantities. The general and direct proof of this 
 theorem is as follows : 
 
 Let us divide a™ — i»> by a — b. 
 a"' — b'"\a — 6 
 
 6(a"'-'— fi"'-') „ 
 " ^ZZt Q.uotient. 
 
 a"— a^-'ft 
 
 a'"- 1 J — J" 
 =6(a"'"' — J"-'). 
 
 In performing this division, we see that the first term of the 
 quotient is a'"-', and the first remainder, 6(0'""' — i""'). 
 
 The remainder consists of two factors, b and a™""' — 6"""'. Now 
 it is evident, that if the second of these factors is divisible by a — b, 
 then will the quantity a"' — J"" be divisible by a — b. Thus, if a — b 
 is contained c times in o""' — b"'~', the entire quotient of a" — J", 
 divided by a — 6, would be a'"~'-|-ic. 
 
 This proves, that IF o'""' — 6"""' is divisible by a — J, then- will 
 o"— ft^jbe divisible by a — b. That is, ip the difference of the same 
 ■powers of two quantities is divisible by the difference of the quantities 
 themselves, tlien will the difference of the next higher powers of the 
 same quantities, be divisible by the difference of the quantities. 
 
 But we have seen, already, that a' — i' is divisible by a — 5 ; 
 hence, it follbws, that a' — V is also divisible by a — b. Again, 
 since a' — b' is divisible by a — b, it follows that a'' — b* is divisible 
 by it. And so on, without limit, which proves the truth of the 
 theorem generally. 
 
 Note. — In dividing the difference of the same powers of two quan- 
 tities by the difference of those quantities, the quotients follow a simple 
 law. Thus, 
 
 {a^—¥)-L.{a—b)=a'>+db+V ; 
 (a''_J4)^(a_i)=ffi3_|_a2j_j.aj5_|.5s . 
 
 The law is, that the exponent of the first letter decreases by unity, 
 while that of the second increases by unity. 
 
 Aet. 84. Lemma. — In proving the next two theorems, it is 
 necessary to remind the student, that the even powers of a nega- 
 tive quantity are positive, and the odd powers negative. Thus, 
 
 — a, the 1st power of — a, is negative. 
 
 — aX — a=:a^, the 2d power, is positive. 
 
 — aX — oX — a= — a', the 3d power, is negative. 
 
 — oX — aX — aX — a=a^,thp 4th power, is positive ; and so on. 
 
46 RAT'S ALGEBRA PART SECOND. 
 
 Art. 85. Theorem VII. — The difference of the even powers of 
 the same degree of two quantities, is always divisible by the sum of llit. 
 quantities. 
 
 If we take the quantities a — b, and a'" — i", and put — c instead 
 ,of b, a — b will become a — ( — c)=a+c, and, when m is even, 
 fc" will become C", and ffi"— d" will become a""— (4-c"')=a'" — C" ■ 
 but a" — i™ is always divisible by a — b ; therefore, 
 
 C"— <;"■ is always divisible by a-\-c when m is even, which is the 
 theorem 
 
 E X AMFL E S . 
 
 1. (a2— t')-!-(a+5)=a— i. 
 
 2. (a<— 6<)-^(a+i)=o'— o'J+ai'— 63. 
 
 3. (a6— i«)-^(a+i)=o5— a'lJ+o'J^— a=J'+a6<— J^ 
 
 Art. 86. Theorem VIII. — The sum of the odd powers of the 
 same degree of two quantities, is always divisible by the sum of the 
 quantities. 
 
 If we take the quantities a — b, and a" — R", and put — c instead 
 of b, a — b will become a — ( — c)=a-\-c, and when m is odd, i"" 
 will become — c" (Art. 84), and a" — 6"" will become o" — ( — c") 
 =a"'-|-c'" : but o" — 6"" is always divisible by a — b ; therefore, 
 
 gm_|_gm is always divisible by a-\-c when m is odd, which is tho 
 theorem. 
 
 E X AM PL ES. 
 
 1. ia'+V)-i-(a+b)=a'—ab+b'. 
 
 2 . (a'-\-b^)^a+b)=a*—a'b+a'b'—ab'+¥. 
 
 3 . {d'+b^)^{a+■b)=a'—a<■b+a^b'—a^'+a^¥—abi-\-b'. 
 
 FACTORING. 
 
 Note. — Previous to studying the factoring of algebraic quantities, 
 the pupil should be well acquainted with factoring numbers. See 
 Ray's Arith., Part 3d, Arts. 121 to 124. 
 
 Art. B'7. The following is a summary of the principles and the 
 most useful rules employed in factoring numbers. 
 
 1st Principle. A factor of a number is a factor of any multiple 
 of that number. 
 
FACTORING. 4T 
 
 2d Peincifle. a factor of any two numbers is also a factor of 
 
 their sum. 
 Propositions deduced from these principles : 
 
 1. Every number ending with 0, 2, 4, 6, or 8, is divisible by 2. 
 
 2. Every number is divisible by 4, when the number denoted 
 by its two right hand digits is divisible by 4. 
 
 3. Every number is divisible by 5, whose right hand digit 
 is or 5. 
 
 4. Every number whose first digits are 0, 00, &c., is divisible 
 by 10, 100, &c. 
 
 The converse of each of the preceding propositions is also true. 
 Thus, no number is divisible by 2, unless it ends with 0,2, 4, 6, 
 or 8. 
 
 5. Every composite number is divisible by the product of any 
 two or more of its prime factors. 
 
 6. Every prime number, except 2 or 5, ends with 1,3,7, or 9. 
 
 Rule fok resolving a composite number into its prime fac- 
 tors. — Divide the given number by any prime number that will 
 exactly divide it ; divide the quotient again in the same manner, and 
 so continue to divide until a quotient is obtained which is a prime 
 number ; then the last quotient and the several divisors are the prime 
 factors of the given number. 
 
 FACTORING OF ALGEBRAIC QUANTITIES. 
 
 Art. §8. A divisor, or factor of a quantity, is a quantity that 
 will exactly divide it ; that is, without a remainder. Thus, a is a 
 factor or divisor of ab, and a-\-x is a divisor or factor of a^ — x'. 
 
 Art. §9. A prime quantity is one which is exactly divisible 
 only by itself and by unity. Thus, x, y, and x-\-z, are prime 
 quantities ; while xy, and ax-^az, are not prime. 
 
 Art. 90. Two quantities, like two numbers, are said to be 
 prime to each other, or relatively prime, -when no quantity except 
 unity will exactly divide them both. Thus, ab and cd are prime 
 tc each other. 
 
 Art. 91. A composite quantity, is one which is the product of 
 two or more factors, neither of which is unity. Thus, a^ — a^ is 
 a composite quantity, of which the factors are a-\-x and a — x. 
 
 Art. 93. To separate a monomial into its prime factors. 
 
 Rule. — Resolve the coefficient into its prime factors; then, these with 
 the literal factors of the monomials, will be the prime factors of the 
 given quantity. The reason of this rule is self-evident. 
 
48 RAY'S ALGEBRA, PAKT SECOND. 
 
 Find the prime factors of tlie following monomials. 
 
 1. ISali^. Am.2y.^X'-iy.a-l>-b. 
 
 2. 28 x't/zK Ans.2x2xTXx-x.y.z.z.z. 
 
 3. 36a'4yz. Atw. 2x2x3x3.a.a.i.4.y.2/-y-i. 
 
 4. 21Q(u^yz^. Ans. 2x3x5X7.0 i.xx.y.z.*. 
 Art. 93. To separate a polynomial into its factors, when one 
 
 of them is a monomial. 
 
 Rule. — Divide the given qaantity by the greatest monomial that wiU 
 exactly divide each of its terms. Then the monomial divisor will 
 be erne factor, and the quotient the other. The reason of this rule 
 is self-evident. 
 Separate the following expressions into factors. 
 
 1. a+ax. Ans. a(,l-{-x). 
 
 2. xz-\-yz. Ans. z(.x^y). 
 
 3. x'^y+xy''. Ans. xy{x^). 
 
 4. 6ai'+9a'6c. Ans. 3aJ(2b4-3''^)- 
 ■5. 4a'6c+6ai='c— lOaftc'. Ans. 2alc(2a+Zb—5c). 
 
 6. a^x'y — ab^xy^-\-abcxyz^. Ans. ahxyiax'^ — by-\-cz^). 
 
 7. 3i=J/— 6i3^2+9a:y. Ans. Zx^y0.—2xy+Zy^) . 
 
 8. \2amH — 18am'm'+30am7i'. Ans. 6amn(2m^ — 3mn-\-5n') 
 
 Art. 94. To separate any binomial or trinomial which is the 
 product of two or more polynomials, into its prime factors. 
 
 1st. Any trinomial can be separated into two binomial factors, 
 when the extremes are squares and positive, and the middle term 
 is twice the product of the square roots of the extreme terms. 
 The factors will be the sum or difference of the square roots of 
 the extreme terms, according as the sign of the middle term is 
 plus or minus. (See Arts. 78, 79.) 
 
 Thus, a^-\-2ab-\-b'=ia+bXa+b) ; 
 a^—2ab-\-b^=(,a—b)ia—b). 
 
 2d. Any binomial, which is the difference of two squares, can 
 be separated into factors, one of which is the sum and the other 
 the difference of their roots. (See Art. 80.) 
 Thus, a'— i2=(a+i)(a— i). 
 
 3d. Any binomial which is the difference of the same powers 
 of two quantities, can be separated into at least tvM factors, one 
 of which is the difference of the two quantities. (See Art. 83). 
 
 Thus, a"'— J'"=(a— i)(a'"-i+<i'"-=Z) .... +a6''-2-|-J'»-', where 
 a, b, and m, may be any quantities whatever. 
 
 In this case, one of the factors is the difference of the quanti- 
 ties, and the other may be found by dividing the given expression 
 by this difference. 
 
I'ACTORiNG. 
 
 49 
 
 Thus, to find the other factor of x' — !/', we divide by x — y, and 
 the quotient is x'^-\-xy-\-y'' ; hence, 
 
 x'—if={x—y){x''+xy-\-y'^). 
 
 Similarly, x^ — 2/'=(^ — 2/)(^''+'^^3'+^^i/^+^i/'-l-3/0- 
 
 4th. Any binomial which is the difference of the even powers of 
 
 wo quantities, higher than the second degree, can be separated 
 
 nto at least three factors, one of which is the sum, and another 
 
 lie difference of the quantities. (See Art. 85.) 
 
 Thus, by Art. 84, a'' — h*, is divisible by a-\-b, and, by Art. 85, 
 t is divisible by a — h ; hence it is divisible by both a-\-b and a — ft, 
 md the other factor will be found by dividing by their product. 
 Or, it may be separated into factors, thus, 
 
 o«— 6^=(a2-|-62)(a2— J2)=(a2+i=)(a+ft)((v— ft). 
 
 5th. Ar.y binomial which is the sum of the aid powers of two 
 quantities, can be separated into at least two factors, one of 
 which is the sum of the quantities. (Art. 86.) The other factor 
 will be found by dividing the given binomial by this sum. Thus, 
 
 Separate the following expressions into their simplest factors 
 
 1. c2+2cd+d'. 
 
 2. ffi23.i_|_2aa;'2/-f-y'. 
 
 3. 25a:y4-20a:y2j,_^4i». 
 
 4. 9a;<— 6a;=2:'+«^ 
 
 5. 4m2a;^ — ^mn'x-Y-n*. 
 
 6. x' — z'^. 
 
 7. Oa^x'i— 2.5. 
 
 8. 16— o'ft''2«. 
 
 9. 4m'a;^- 
 
 10. a^—x\ 
 
 11. 2^+1. 
 
 12. y'— 1. 
 
 13. l+c». 
 
 14. a'x' — fty. 
 
 15. x=+ys. 
 
 16. x' — y*. 
 
 %n^z\ 
 
 1. (c+d)(c+d). 
 
 2. {ax'^-\-y){ax^Ary)- 
 
 3. (5a:i/24-22)(5a;2/2+2«). 
 
 4. (3x2_22)(3x=— z2). 
 
 5. {2mx — re^)(2mx — ra'). 
 
 6. \x-\-z)(x—z). 
 
 7. (3ffir2+5)(3aa;2_5). 
 
 8. (4+aft2z3)(4._aft2z3). 
 
 9. (2ma!+3n2=) (27na?— 3nz2). 
 iO. (a2+x2)(ffl2— vc^i) 
 
 =(a'+x^)(a-f-*)(a — x). 
 
 11. (2+l)(22-^+l). 
 
 12. (j/_l)(y2+3,^_l). 
 
 13. (l+c)(l-c+c=). 
 
 14. {ax — lry){a''x''^abxy-\-Vy^) 
 
 15. (a;-j-y)(x'' — x^y-\-x^y'^ — xy^ 
 
 16. (x'+y3) (x3— ^3^ = (x'+j/') 
 (x— y)(x2+xj/+2/2)=(x+y) 
 
 (a;2_^y4-!/2) {x—y) (X^+X!/ 
 
 4.i/2)=(x+y)(a^-2/)(x»— aj/ 
 +y2)(x=-|-xj/+y2)= 
 
 (x'-j,=)(x<+xy+3/^) 
 
60 RiSY'S ALGEBRA, PART SECOND. 
 
 Aar. 94i To separate a quadratic trinomial into its factors. 
 
 A quadratic trinomial is of the form, a;'-|-aa;-|-6, in which the 
 sign of the second term may be either plus or minus. 
 
 To explain the method of performing this operation, let us 
 examine the relation that exists between two binomial factors and 
 their product. 
 
 1. (,x-\-a)(,x+i)=x^+(a+h)x+ab. 
 
 2. (_z—a)(_x—b)=x^—(,a+b)x+ab. 
 
 3. (x+a)(x—b)=x'+^a—b)x—ab. 
 
 4. (jx — a)(a;+J)=a;'+(J — a)x — ab. 
 
 From this we see that any trinomial may be resolved into two 
 factors, when the first term is a square, and the coefficient of the 
 second term equal to the sum of any two quantities, whose product 
 is equal to the third term. 
 
 Remark. — In Equations of the Second Degree (Art. 234), it will be 
 shown how to perform this operation by a direct method ; it is, however, 
 a useful exercise for the pupil to do it by inspection, the only difficulty 
 being to find two quantities whose sum is equal to the coefficient of the 
 second term, and product equal to the third term. 
 
 Trinomials to be decomposed into binomial factors. 
 
 1. a;=+3a;+2. Ans. (x+l)(x-\-'2). 
 
 2. a^-\-5a+6. Am. (a-(-2)(a-l-3), 
 
 3. a;=— 7a;+12. Arts, (i— 3)(a;— 4) 
 
 4. a;=— ex+15. Ans. (aj— 3)(a;— 5> 
 
 5. a;'— X— 2. Ans. (x+l)(a:— 2), 
 
 6. a;'+a:^12. Ans. (x— 3)(a;+4) 
 
 7. x2— X— 12. Atis. (x+3)(x— 4), 
 
 8. x2— 5x+6. Ans. (x— 2)(x— 3), 
 
 9. x2+2x— 35. Ans. (x— 5Xx+7) 
 10. x^-i-x— 56. Ans. (x— 7)(x+8), 
 
 Art. 95. Examples of binomials and trinomials that may be 
 aeparated into factors, by first separating the monomial factor, 
 ■nd then applying the principles in Art. 93. 
 
 Ex. 1. ax'y — axy^=axy{x'^ — y^)=axy{x-^){x — y). 
 
 2. 3ax'-j-6axy-|-3ay^. Ans. 3a(x+i/)(x-j-y}. 
 
 3. 2cx2— 12cx4-18c. Ans. 2c(x— 3)(x— 3). 
 
 4. 27a— 18ax-4-3ax'. Ans. 3a(3— x)(3— x). 
 
 5. Zm?n — 3m7i'. Ans. 3m.n(m-\-n)(m — n), 
 6- 82>—2z^. Ans. 2z(2+2)(2— z). 
 
 7. 2x'j; — 2xy^. Ans. 2xy{a?-\-y''-)(x-\-y){x — y). 
 
 8. 2x'+6x— 8. Ans. 2(x-l 4)(x'— 1). 
 
GREATEST COMMON DIVISOR. 51 
 
 9. 2a3+4a;'— 70a;. Ans. 2a;(a;4-7)(a;— 5). 
 
 10. 3a'J— 3a'6— 60a6. Am. 3<zi(a— 5Xa+4). 
 Solve the following questions by first indicating the operations 
 
 to be performed, and then canceling the factors common to the 
 dividend and divisor. 
 
 11. Multiply 4a; — 12 by 1 — x', and divide the product by 2-|-2a. 
 (4a;-12)(l-a;^ ) 4(a^-3)(l+a;)(l-cQ 
 
 2+2^ = 2(1+^) =2(^-3)(l-a;)= 
 
 8a;— 6— 2a;2. 
 
 12. Multiply x^-^-'ixy-^-y^ by x — y, and divide the product by 
 «' — yK Ans. x-\-y. 
 
 13. Multiply 6om' — 6<z7i' by m-f-re, and divide the product by 
 2m'-]-4:mn-\-2ji'. Ans. 2a(m — n). 
 
 14. Multiply together 1 — c, 1 — c', and 1+c', and divide the 
 product by 1 — 2c-\-c'. Ans. l+c+c'+c'. 
 
 15. Multiply together x'-\-x — 2 and a:' — x — 6, and divide the 
 product by x^-["4a;-|-4. Ans. x^ — 4x-|-3- 
 
 16. Multiply together x^—x^ — 30a; and a;=-f-lla;+30, and di- 
 vide the product by the product of x' — 36 and a;2-|-10a;-|-25, 
 
 Ans. X. 
 
 GREATEST COMMON DIVISOR. 
 
 Art. 96. Any quantity that will exactly divide two or more 
 quantities, is called a common divisor, or common measure, of those 
 quantities. Thus, db is a common divisor of ab' and abx. 
 
 Remark. — Two quantities, lilie two numbers, often have more than 
 one common divisor. Thus, a?ex and ahdx have three common divisors, 
 If, j:, and tix. 
 
 Aet. 97. That common divisor of two quantities which is the 
 greatest, both with regard to the coefficients and exponents, ia 
 called their greatest common divisor, or greatest common measure. 
 Thus, the greatest common divisor of 6a^x^ and 9a'cxz is 3a'a 
 
 Art. 98. Quantities that have a common divisor are said to b» 
 commensuraMe, and those that have no common divisor are said to 
 be incommensurable. 
 
 Art. 99. To find the greatest common divisor of two or mors 
 monomials. 
 
52 
 
 KAY'S ALGEBRA, PART SECOND. 
 
 1 . Let it be required to find the greatest common divisor of the 
 two monomials, lAa'cx and 21a'Jx. 
 
 By separating each quantity into its prime factors, we have 
 14:a^cx=l X^Xaaacx, and 21a^bx='7 X^Xaabx. 
 
 By examining these quantities we find that 7, aa or a', and x, 
 are the only factors common to both ; hence, both the quantities 
 can be exactly divided by either of these factors, or by their pro- 
 duct, 7a'x, and by no other quantity whatever ; therefore, Ta'x is 
 their greatest common divisor. This gives the following 
 
 Rule for finding the greatest common divisor of two ob 
 MORE Monomials. — Resolve the quantities into their prime fac- 
 tors ; then the product of those factors that are common to all the 
 terms, will form their greatest common divisor. 
 
 Note. — The greatest common divisor of the literal parts of the quan- 
 tities may generally be found most easily by inspection, by taking each 
 letter tliat is common to two or more of the quantities, with its hast 
 exponent. 
 
 2. Find the greatest common divisor of Qa'xy, 9a'a;', and 
 15oVj/'. 
 
 Here we find that 3 is the only nu- 
 
 oPERATioN. merical factor, and a and x the only 
 
 Qa'^xy =3x2a'x}/ letters common to all the quantities. The 
 
 Qa^x^ =3 XSa'x' least powers of a and x, in either of the 
 
 l5o'a;y=3x5a'a;'2/' quantities, are a' and x; hence, the 
 
 greatest common divisor is Sa^a. 
 
 Find the greatest common divisor of the following quantities. 
 
 3. 15aJc^ and21i2crf. Ans.Zbc. 
 
 4. 4o'J, lOa'c, and 14a'&c. Ans. 2a'. 
 
 5. 15aa^3/,18x'3/', and 2\x^y^. Ans. 2x^y. 
 
 6. 4:axy, 20a:<2/^^. and l^x'fz'. Ans. Ai^y^. 
 
 7. 2a^¥cx, Za^h^'c^^z, and 5a%'>cy. Ans. aWe. 
 
 8. 12a'x2z', \Qax^z\ 30aVz, and Qai^z^. Ans. 6aaH. 
 
 Art. 100. Previous to investigating the rule for finding the 
 greatest common divisor of two polynomials, it is necessary to 
 demonstrate the following 
 
 Proposition. — Any common divisor of two quantities, vnlt 
 always exactly divide their remainder after division. _ 
 
 Let AD and BD be either two monomials, or polynomials, of 
 which D is a common divisor, and let AD be greater than BD. 
 
GREATEST COMMON DIVISOR. 53 
 
 Divide AD by BD, and if BD is net con- 
 tained an exact number of times in AD, BD)AD(Q 
 suppose it is contained Q, times with a BDQ. 
 
 remainder, which may be called R. Then, AD — BDCi=R 
 
 since the remainder is found, by subtract- 
 ing the product of the divisor by the quotient from the dividend 
 we have, R=AD — BDQ,. Dividing both sides by D, we get 
 
 ~ =A — BQ, ; but A and BQ are each entire quantities, therefore 
 
 their difference, vr, must be an entire quantity. Hence, any com- 
 mon divisor of two quantities (and of course the greatest common 
 divisor), will always exactly divide their remainder after division. 
 
 Remark. — In the preceding demonstration it is assumed that the 
 pupil understands the following axioms : 
 
 First. // tioo equal quantities be divided by the same quantity the quotients 
 teitl be equal. 
 
 Second. TVie difference of two entire quantities is also an entire quantity 
 
 Aet. 101. Let it be required to find the greatest common 
 divisor of two polynomials, A and B, of which A is the greater. 
 
 If we divide A by B, and there is 
 
 no remainder, B is, evidently, the B)A(Q, 
 
 greatest common divisor, since it can BQ 
 
 have no divisor greater than itself. A — BQ=R, 1st Rem. 
 
 Divide A by B, and call the quo- 
 tient Q, then if there is a remainder R)B(Q' 
 R, it is evidently less than either of RQ' 
 the quantities A and B ; and by the B — RQ'^R', 2d Rem. 
 preceding theorem it is also exactly 
 
 divisible by the greatest common A=BQ+R Since the 
 
 divisor ; hence, the greatest common B=RQ'+R' **''"^'='"' " 
 
 divisor must divide A, B, and R, and product of the dmsor by the 
 
 cannot be greater than R. But if R , quotient, plus the remainder 
 
 will exactly divide B, it will also 
 
 exactly divide A, since A^BQ-|-R, and therefore will be the 
 
 greatest common divisor sought. 
 
 Suppose, however, that when we divide R into B, to ascertain 
 if it will exactly divide it, we find that the quotient is Q', with a 
 remainder R'. Now, it has been shown that whatever exactly 
 divides two quantities, will divide their remainder after division 
 (Art. 100) ; and since tlie greatest common divisor of A and E, has 
 been shown to diviile B ard R, it must also divide their remainder 
 
64 
 
 HA IS ALlrfcBKA, PART SECOND. 
 
 R', and therefore cannot be greater than R'. And, if R' exactly 
 divides R, it will also divide B, since B=RQ,'-|-R'; and whatever 
 exactly divides B and R, will also exactly divide A, since A=BQ 
 4-R ; therefore, if R' exactly divides R, it will exactly divide both 
 A and B, and will be their greatest common divisor. 
 
 In the same manner, by continuing to divide the last divisor by 
 the last remainder, it may always be shown, that the greatest 
 common divisor of A and B will exactly divide every new 
 remainder, and, of course, cannot be greater than either of them. 
 It may also always be shown, as above, in the case of R', that any 
 remainder, which exactly divides the preceding divisor, will also 
 exactly divide A and B. Then, since the greatest common divisor 
 of A and B cannot be greater than this remainder, and as this 
 remainder is a common divisor of A and B, it will be their 
 greatest common divisor sought. 
 
 The same principle may be illustrated by numbers, by calling 
 A, 55, and B, 15, and proceeding to find their greatest common 
 divisor. 
 
 Aht. i02. When the remainders decrease to unity, or when 
 we arrive at a remainder which does not contain the letter of ar- 
 rangement, it is evident that there is no common divisor of the 
 two quantities. 
 
 Aet. 103. If either quantity contains a factor not found in the 
 other, that factor may be canceled without affecting the common 
 divisor. Thus, in the two quantities, x{x' — y') and y{x^-\-2m/-\-y'^) , 
 of which the greatest common divisor is x-\-y, we may cancel x in 
 the first, or y in the second, or both of them, and the greatest 
 common divisor of the resulting quantities will still be x-\-y. 
 
 Art. 104. We may multiply either quantity by a factor not 
 fouTid in the other, without changing the greatest common divi- 
 sor. Thus, in the two quantities, xQc' — y^) and y{x^-\-'2.xy-\-y^, 
 if we multiply the first by m and the second by n, we have 
 mx{x^—y^) and ny(,x^-\-2xy-\-y^), of which the greatest common 
 divisor is still x-j-y. 
 
 Art. 105. But if we multiply either quantity by a factor found 
 m the other, we change the greatest common divisor. Thus, in 
 the two quantities, a;(a;'— y'-*) and y(a;2-f Siy+y'), if we multiply 
 the second by x, the two quantities become x{x'' — y'^) and xij{x^ 
 -l-2a;y-|-!/2), of which the greatest common divisor is x(x+y') 
 instead of x-]-y as before. In like manner, if we multiply the 
 first quantity by y, the greatest common divisor of the two result- 
 ing quantities will be y{x-\-y') 
 
GB.EATEST COMMON DIVISOR. 
 
 &5 
 
 Art. 1©6. From Art. 101 it is evident that the greatest CDtn- 
 tnon divisor of two quantities will exactly divide each of the suc- 
 cessive remainders ; therefore, the principles of the three pre- 
 ceding articles apply to the successive remainders that arise in 
 hnding the greatest common divisor. 
 
 Aet. 107. It is evident that any common factor of two quan- 
 tities, must also be a factor of their greatest common divisor. 
 Where such common factor is easily seen, as when it is a mono- 
 mial, it simplifies the operation to set it aside, and find the great- 
 est common divisor of the remaining quantities. 
 
 We shall now show the application of these principles. 
 
 1. Find the greatest common divisor of *' — z' and x* — xV. 
 
 Here the second quantity contains x^ as a fac- 
 tor, but it is not a factor of the first ; we may 
 therefore cancel it (Art. 103), and the second 
 quantity becomes x^ — z'. Then divide the first 
 by it. After dividing, we find that z' is a factor 
 of the remainder, but not of x' — z', the next 
 dividend. We therefore cancel it (Art. 103), 
 and the second divisor becomes x — z- Then, 
 dividing by this, we find there is no remainder ; 
 therefore x — z is the greatest common divisor. 
 
 OPERATION. 
 
 x^ — z^\x' — z' 
 x^ — xz^ \x 
 xz' — z^ 
 or (x — z)z' 
 
 x' — z^ \x — z 
 x' — xz \x-\-ii 
 
 xz- 
 xz- 
 
 2. Find the greatest common divisor of a' 
 -\-xV and x' — x^z'. 
 
 The factor x- is common to both quantities ; 
 it IS, therefore, a factor of the greatest divisor 
 (Art. 107), and may be taken out and reserved. 
 Doing this, the quantities become x'-^z' and 
 a:' — xz'. The second quantity still contains a 
 common factor, x, which the first does not ; 
 canceling this, it becomes x' — z'. Then pro- 
 ceeding as in the first example, we find that 
 x-\-z divides without a remainder ; therefore, 
 x\i:-\-z) is the required greatest common divisor. 
 
 3. Find the greatest common divisor of lOa'x'- 
 and 5bx' — lUx-{-db. 
 
 By separating the monomial factors, we find 
 
 10a'x=— 4a=a:— 8a2=2a2(5x2— 2a^-3), 
 and 5bx'~Ubx-{-6b=b(,5x'—}lx-\-&). 
 
 OPERATION. 
 
 x^+z' \x'—z' 
 x' — xz' Ix 
 
 xz'-^z^ 
 or (_x-]-z)z^ 
 
 x' — z' \x-\-z 
 
 x'-\-xz \x — z 
 
 — xz — z' 
 
 — xz — z' 
 
 ia'x — 6a' 
 
 [07EB.J 
 
56 RAY'S ALGEBRA, PART SECOND. 
 
 The factors 2a and 4 have no com- operation. 
 
 moi> measure, and therefore are not 5x' — 11 j+6| 5j:'— 2a>— 3 
 factors of the common divisor. We Sj' — 2j — 3 |1 
 may therefore suppress them (Art. — 9a;-(-9 
 
 103), and proceed to find the great- or — 9(a: — 1) 
 
 est common divisor of the remain- 
 ing quantities, virhich is found to be 5x' — 2x — 3 \ x — 1 
 t_-l . Sa"— 6a |.5a+3 
 
 3a— 3 
 3a— 3 
 
 4. Find the greatest common divisor of 4a' — 5m/-\-y', and 
 3a3— 3a'y+a^'— j/^ 
 
 In solving this exam- operation. 
 
 pie, there are two in- 2a' — Za^y-\-ay'' — y'lia^ — 5ay-\-y' 
 
 stances in which it is ne- 4 
 
 cessary to multiply the 12a' — 12a'y+4a]/- — 4j/' |3a-l-3j 
 
 dividend, in order that 12a' — 15a^y-\-^ay'' 
 
 the coefficient of the Ba^y-\-ay^ — 4?/' 
 
 first term may be exactly 4 
 
 divisible by the first 12a^y-\- iay^ — ICi/' 
 
 termof the divisor (Art. 12a'y — 15ay^-\- 3 y' 
 
 104). 19ay'— 19!/= 
 
 We find 19y' is a or lOy'-'ia—y) 
 factor of the first re- 
 mainder, but it is not a 4a' — fjay-\-y'\a — y greatest «)m.di-riK)r 
 factor of the first divisor, 4(i' — iay |4a — y 
 and, therefore, cannot — <'y-\-y' 
 be a factor of the great- — ay-f-j' 
 est common divisor ; it 
 must, therefore, be suppressed. 
 
 Art. l€i§. From the preceding demonstrations and examples, 
 «re derive the following 
 
 Rule for finding the greatest commoh divisor of two 
 POLTKOMIALS. — 1 . Divide the greater polynomial hy the less, and 
 if there is no remainder, the less quantity will he the divisor 
 sought. 
 
 2. If there he a remainder, divide the first divisor hy it, and continut 
 to divide the last divisor by the last remainder, until a divisor is 
 obtained lohich leaves no remainder ; this will he the greatest com' 
 mon. divisor of the two given polynomials. 
 
GREATEST COMMON DIVISOR. 5T 
 
 Notes. — 1. When the highest power of the leading letter is the sams 
 in botli, it is lininaterial wliicli of the quantities is made the dividend. 
 
 2. If both quantities contain a common factor, let it be set aside, aa 
 forming a factor of tlie common divisor, and proceed to find the greatest 
 common divisor of the remaining factors, as in Example 2. 
 
 3. If either quantity contains a factor not found in the other, it may- 
 be canceled before commencing the operation, as in Example 3. See 
 Art. 103. 
 
 4. Whenever it is necessary, the dividend may be multiplied by any 
 quantity which will render the first term exactly divisible by the first 
 term of the divisor. See Art. 104. 
 
 5. If, in any case, the remainder does not contain the leading letter, 
 there is no common divisor. 
 
 6. To find the greatest common divisor of three or more quantities, 
 first find the greatest common divisor of two of them ; then of that divi- 
 sor and one of the other quantities, and so on. The last divisor thus 
 found will be the greatest common divisor sought. 
 
 7. Since the greatest common divisor of two quantities contains all the 
 factors common to both, it may be found most easily by separating the 
 quantities into factors, where this can bo done by the rullbs for factoring. 
 Arts. 92 to 95. 
 
 Find the greatest common divisor of the quantities in each of 
 the follovifing 
 
 E XAMPI.E S. 
 
 1. 5a:2— -2a:— 3 and 5a;2— lla;+6. Am. x—l. 
 
 2. dx^—i and Qx^— 15a;— 14. Ans. 3x+2. 
 
 3. (j2_a6_l2i2 and a'+5ab-]-Gb'. Ans. a+3i. 
 
 4. 4o2— i2 and 4a'+2a6— 21'. Ans. 2a— b. 
 
 5. a'' — a;'' and a^-\-a'x — ax' — x'. Ans. a? — x'. 
 
 6. x^— 5x2+13x— 9 and x'— 2x2-(-4x— 3 . Atis. x— 1. 
 
 7. x'— 5x--fl6x— 12 and x'- 2x2_l5a;_|-l6. Ans. x— 1. 
 
 8. 21a;3— 26x=+8x and Gx'— x— 2. Ans. 3x— 2. 
 
 9. 2x''+llx'— 13x2— 99x— 45 and 2x'— Tx^— 46x— 21. 
 
 Arai. 2x2+7x+3. 
 
 10. x<+2x2+9 and7x3— 11x2+1 5x+9. Ans. x'- 2x4-3. 
 
 11. 48x2+16x— 15 and 24x3— 22x=+17x— 5. ^„s_ 12x— 5. 
 
 12. x2+5x+4, x2+2x— 8, and x'+7x+12. Ans. x+4, 
 
 13. x'-j-o^x'-l-Bi and x-i+ax'— n'x— o". Ans. x=+Bx+a'. 
 
 14. 2J'— 10ai2-f8a=6 and 9a-'— 3ai'+3a2j3— 9a3j. A.ns. a~-b. 
 
 15. x-"— yx'-j-Cj — l)x2-f-px — q and x* — qp^A^i^f — V)x'^-\-qx-^, 
 
 Ans. X- — 1. 
 
68 RAY'S ALGEBRA, PART SECOND. 
 
 LEAST COMMON MULTIPLE. 
 
 Art. 109. A multiple of a quantity is any quantity that con- 
 tains it exactly. Thus, 6 is a multiple of 2 or of 3 ; and o6 is a 
 multiple of a or of 6 ; also, a(h — c) is a multiple of a or of (b — c). 
 
 Art. no. A common mvltvple of two or more quantities, is a 
 quantity that contains either of them exactly. Thus, 12 is a 
 common multiple of 2 and 3 ; and 20m/ is a common multiple of 
 2a; and 5y. 
 
 Art. 111. The least common multiple of two or more quanti- 
 ties, is the least quantity that will contain them exactly. Thus, 
 6 is the least common multiple of 2 and 3; and XQxy is the least 
 common multiple of 2x and by. 
 
 Remark. — Two or more quantities can have but one least common 
 multiple, while they may have an unlimited number of common mul- 
 tiples. 
 
 Art. 112. To find the least common multiple of two or more 
 quantities. 
 
 From the nature of the least common multiple of two or more 
 quantities, it is evident that it contains all the prime factors ol 
 each of the quantities once, and does not contain any prime fac- 
 tor besides ; for, if it did not contain all the prime factors of any 
 quantity, it would not be divisible by that quantity ; and if it con- 
 tained any prime factor not found in either of the quantities, it 
 would not be the least common multiple. Thus, the least common 
 multiple of ab and he must contain the factors a, h, c, and no other 
 factor. Hence, 
 
 Tlie least common multiple of two or more quantities, contains aU 
 the prime factors of those quaniiiies once, and does not contain any 
 other factor. 
 
 With this principle let us find the least common multiple of 
 mx, nx, and m'^nz. 
 
 Arranging the quantities as in the 
 OPERATION. margin, we see that m is a prime factor 
 
 common to two of them. It must, 
 therefore, even if found in only one of 
 the quantities, be a factor of the least 
 common multiple, and we place it on 
 myCn'X.xy.mz=m'7Uicz the left of the quantities. Then, since 
 
 m 
 
 mx 
 
 nx 
 
 mHz 
 
 n 
 
 X 
 
 vx 
 
 mnz 
 
 X 
 
 X 
 
 X 
 
 mz 
 
 
 1 
 
 1 
 
 mz 
 
LEAST COMMON MULTIPLE. 59 
 
 the same factor can occur but once in the least common multiple, 
 we cancel m in each of the quantities in which it is found, which 
 is done by dividing by it. 
 
 We next observe that n is a factor common to two of the re- 
 maining quantities ; we therefore place it on the left, as another 
 factor of the least common multiple, and cancel it in each of the 
 terms in which it is found. 
 
 By examining the remaining quantities, we find that a; is a fac- ' 
 lor common to two of them. We then place it on the left, aa 
 another factor of the least common multiple, and cancel it in each 
 of the terms in which it is found. 
 
 We thus find that the least common multiple must contain the 
 factors m, n, and x ; it must also contain the factor mz, otherwise 
 it would not contain all the prime factors found in one of the 
 quantities. Hence the products, mXraX^X'n2^»"'na;2, contains 
 all the prime factors of the quantities once, and does not contain 
 any other factor ; it is, therefore, the required least common 
 multiple. Hence we have the following 
 
 Rule for finding the least common multiple of two ok 
 MORE quantities. — 1 . Arrange the quantities in a horizontal 
 line, and divide them by any prime factor that will divide two or 
 more of -them without a remainder, and set the quotients and the 
 undivided quantities in a line beneath. 
 
 2. Continue dividing as before, until no prime factor, except unity, 
 will divide two or more of the quantities without a remainder. 
 
 3. Multiply the divisors and the quantities in the last line togethe-^, 
 and the product will be the least common multiple required. 
 
 Or, Separate ike given quantities into their prime factors, and then 
 multiply together such of these factors as are necessary to form a 
 product that will contain all tlie prime factors in each quantity : 
 this product will be the least common multiple required. 
 
 Art. 113. Since the greatest common divisor of two quanti- 
 ties contains all the factors common to both, it follows, that if u>6 
 divide the product of two quantities by their greatest common divisor, 
 the quotient will be their least common multiple. 
 
 Find the least common multiple of the quantities in each of 
 the following 
 
 E X AMPLE S. 
 
 1. 6a', 9ax\ and 24a:'. Ans. 72a V. 
 
 2. B2xY, 40ot;*3/, and 5a'x(,x-^). Ans. ieOa^x'y''(x—^), 
 
60 RAY'S ALGEBRA, PART SECOND. 
 
 3. 3x+6^ and 2a:'— Sy'. Ans. 6a:'— 24^' 
 
 4. a?-\-x^ and a? — x^. Ans. a* — a'ai+oa;^ — x* 
 
 5. 4(a2+aa:), 12(aa;=— x'),andl8(a'-^=). 
 
 Ans. 36ax'(a=— a;'). 
 
 6 . 2a;— 1 , 4a;'— 1 , and 4a:'+l . Ans. 1 6x<— 1 . 
 
 7. a>-l, a:'- 1, a>— 2, and a;'— 4. Ans. a;^— 5a;'4-4. 
 
 8. a;'— 1, a;2+l, (a;—!)', (x+1)', x»— 1, and a;'+l. 
 
 Ans. a;'" — a* — x*-\-l 
 
 9. 4(1— a;)», 8(1 ^r), 8(1 +a;), and 4(1 +a;'). 
 
 Atis. 8(1— a;)(l— a;<). 
 
 10. 3x'— lla;+6, 2a:2— 7a:+3, and 6x'— 7a;+2. (See Art. 
 
 113.) Ans. 6a;'— 25a;'+23a;— 6 
 
 CHAPTER III. 
 ALGEBRAIC FKACTIONS. 
 
 DEFINITIONS. 
 
 Art. 114. Algebraic fractions are represented in the same 
 manner as common fractions in Arithmetic. The quantity below 
 the line is called the denominator, because it denominates, or shows 
 the number of parts into which the unit is divided ; and the quantity 
 above the line is called the numerator, because it numbers, or shows 
 
 how many parts are taken. Thus, in the fraction, ° , if a=5 
 
 c-\-d 
 6=3, c=2, and d=\, the denominator c-\-d shows that a unit is 
 divided into 3 equal parts, and a — b shows that 2 of those parts 
 are taken. 
 
 Art. 115. The terms proper, improper, simple, compound, and 
 complex, have the same meaning when applied to algebraic frac- 
 tions, as to common numerical fractions. 
 
 Art. 116. Every quantity not expressed under the form of a 
 fraction, is called an entire algebraic quantity. Thus, ex — d is an 
 entire quantity. 
 
 Art. 117. Every quantity composed partly of an entire quan- 
 tity and partly of a fraction, is called a mixed quantity. Thus, 
 
 a-|--) is a mixed quantity. 
 h 
 
ALGEBRAIC FRACTIONS. 61 
 
 Note. — The same principles and rules are applicable to algebraic and 
 to common numerical fractions. However, as a good knowledge of frac- 
 tions is of great importance to the student, we shall present a concise 
 demonstration of the fundamental principles and rules of operation. 
 In these demonstrations the pupil is supposed to be acquainted with this 
 self-evident principle : If we perfonn the same operations on two equal 
 quantities the results will be equal. 
 
 Art. H8. Proposition. — TJie valve of a fraction ts not 
 allied, if we multiply or divide both terms by the same quantity. 
 
 Let _ be a fraction whose value is Q,. 
 B 
 
 A 
 
 Then _=Q, ; but, from the nature of fractions, A represents 
 a dividend, B the divisor, and Q the quotient ; 
 and by the nature of division, 
 A=BQ. 
 If m represents any number, then 
 
 OTA=mBQ, ; dividing these equals by mB, we have 
 
 "L_=Q ; which proves the 1 st part of the proposition. 
 mB 
 
 Again, take the equals 
 
 A=:BQ„ and divide each by m, we have (Art. 73), 
 
 — =:_Q,; divide each of these equals by _, then 
 mm m 
 
 m 
 
 — =0, ; which proves the 2d part of the proposition. 
 
 B 
 
 m 
 
 Case I. — To reduce a fkaction to its lowest terms. 
 
 Art. 119. Since the value of a fraction is not changed by 
 dividing both terms by the same quantity (Art. 118), we have the 
 following 
 
 RtiLE. — Divide both terms by their greatest common divisor. 
 Or, Resolve both terms into their prime factors, and then cancel thosa 
 factors which are common. 
 
 EXAMPLES. 
 
 1 . Reduce to its lowest terms. 
 
 I0acx'_ 2aX5cx^_ 2a ^^ 
 
 15Jcjc3 3bxX5cx^ Sbx 
 
82 
 
 RAY'S ALGEBRA, PART SECOND. 
 
 Fractions to be reduced to their lowest terms. 
 
 Ans. 
 
 Sab 
 
 . a4-x 
 Ans. — i- . 
 3i— c 
 
 Ans. 
 
 a+b 
 a — b 
 
 10 
 
 11 
 
 12 
 
 13. 
 
 14. 
 
 15. 
 
 16. 
 
 ax-\-x^ 
 3bx — ex 
 Za^+3ab 
 Sa^—Sab 
 21aWx—9aVx^ 
 15a^b^>x—3a^b*x'' 
 1—x 
 1— j:»" 
 5o'-f-5aa! 
 a^—x^ 
 x^+2x—^ 
 x^+5x-\^' 
 a''— 3 a:— 70 
 aJ_39a;+70* 
 x^ — 4a°-|-5 
 
 x'+l 
 ix'—12ax+9a ' 
 8a;'— 27a' 
 1.53:'+35a'+3a;4-7 
 27a;''+63a3— 1 2a;2— 28 a 
 2x'+8a''y+l 6ay'+l 6y' 
 8x^-{-Axy—2iy' 
 
 5. -!!_ 
 
 (. mnp — m?p 
 
 2ax — iax' 
 
 7. 
 
 6ax 
 
 Ans. 
 a 
 
 n — m 
 
 m^-p 
 
 1— 2a r 
 
 3 
 
 7o'— 3ia 
 
 Ans 
 
 Atis. 
 
 5a — a^i'a 
 Atis. 
 
 Ans. 
 
 Ans. 
 
 1+x 
 
 5a 
 
 a — X 
 
 Ans.'^ 
 
 x+2 
 
 a— 10 
 
 a-— 7a+10 
 Ans. ^'rrSa+S 
 
 Ans. 
 
 x^—x-{-l 
 2x—3a 
 
 4^2+6 aa+9ffl2 
 Ans. ^-'-^1 
 
 9a'— 4a 
 
 Ans. ^^+^^+^3'° 
 2(2a— 3i-> 
 
 Remake. — Instead of finding the greatest common divisor by the -^le, 
 Art 119, it is often preferable to separate the quantities into factor" by 
 the rules for factoring (Arts. 87 to 95), and then cancel those fap*^ra 
 common to both terms. The following examples should be solvec in 
 this manner. 
 
 17. Reduce '^'i"^""t°?'^i"f'^ to its lowest terms. 
 
 x^-{-(b-\-c)x-\-bc 
 
 x'-\-{a-\-c)x-\-ac=x'-{-ax-\-cx-\-ac 
 =x(x-{-a)-\-c(^x-\-a)=(x-{-c)(x-\-a). 
 Also, a2+(j4-c)a+ic=(a+c)(a+J) ; 
 
 .-. the fraction becomes (?±£XH:^)_^a ^^, 
 (a;+c)(a+6)~a+f 
 
 18. <«>i-h/+ay+ic . c+y 
 af+2bx+2ax+bf' ' f+'ix 
 
 jg 6ae+10te+9aa+15ia . So-fSJ 
 
 6c2+9ca— 2c— 3a " "*' 3o— 1 * 
 
A.LGEBRAIC FRACTIONS. 68 
 
 20. '''+^f+fy+y\ Ans. ^!iv 
 
 a;i — y* x' — y^ 
 
 21 _ a'+Ca+&)gx+tj;' ^ ^^^ a-\-x 
 
 22. Ans. 
 
 a^Jx — ft'x^ J(a+ix) 
 
 ffl^x^ — i- ace — h 
 
 24. a^+aP-a'!H-ft' ^^^ a=+6^ 
 
 4o<— 2a=ift2_4a36+2ai> 2a(2a=— J^) 
 
 25. ^^!±^5=*1. An.. 2.^* 
 
 a'+a^J— a— J a^— 1 
 
 Art. 120, Exercises in Division (see Art. 72), in which the 
 quotient is a fraction, and capable of being reduced to lower 
 terms. 
 
 1. Divide 2a V by 5a V6. Ans.—. 
 
 5i 
 
 2. Divide lejc'x^ by 20ac'x'. Am. ^IH 
 
 5 ax 
 
 3. Divide ax-\-x^ by 3ix — ex. Ans. ^p^. 
 
 oh — c 
 
 4. Divide a'— J' by a^—b'. Ans. °°+''^+^'. 
 
 5. Divide a^— i^ by (a— J)». Ans. '^'+°^+^'. 
 
 a — b 
 
 6. Divide n' — 2n2 by n' — 4n4-4. Ans. . 
 
 n — 2 
 
 7. Divide 3x'— Sx'— 63x +135 by 3x'— 2x— 21. 
 
 Ans. 3^^+6x-45 
 3x+7 
 
 Case II. — To EEDtfcE a fraction to an entire or mixed 
 
 QUANTIir. 
 
 Art. 121. Since the numerator of the fraction may be re- 
 garded as a dividend, and the denominator the divisor, this is 
 merely a case of division. Hence we have the following 
 
 Rule. — Divide the numerator by the denominalor, for the entire part, 
 and if tliere be a remainder, place it over the denominator, for tht 
 fractional part. 
 
 Note. — The fractional part should be reduced to its lowest terms. 
 
64 RAY'S ALGEBRA, PART SECONU. 
 
 1 . Reduce " ~' " TT"^' to an entire or mixed quantity. 
 a^ — ax 
 
 t±^LlI^=a+x+-^ =a+x+-±- Am. 
 a' — ax a^ — ax a — x 
 
 Reduce the following fractions to entire or mixed quantities. 
 
 2 ax—o:^ ^^ ^^^^^ 
 
 a a 
 
 3.1:^'. Am.l-<c ?L 
 
 1+x 1+x 
 
 A.'^l^. Am.a+b+1^. 
 
 as2-& a — A 
 
 5. l±?f. Ans.l+6x+l^. 
 
 1—31 1— 3x 
 
 a;2 — fta; a; — b 
 
 ^ax^-ax-x+l_ J^ns.x-l. 
 
 ax — a a 
 
 i'— 1 a;+l 
 
 Case III. — To eeduce a mixed quantity to the foem of a 
 
 FRACTION. 
 
 Aet. 122. Let it be required to reduce a-\-- to the form of a 
 
 fraction. 
 
 It is evident that a is the same as -, and -= ^2Sf = —. Art. 118. 
 
 1 1 IXc c 
 
 Hence, a+^-=^+^=^^\ 
 c c c c 
 
 Similarly, a — .=?? Hence we have the following 
 
 c c 
 
 Rule. — Multiply the entire part by the denominator of the fraction ; 
 then add the numerator to the product, and place the result over the 
 denominator. 
 
 Before proceeding to the application of this rule, it is necessary 
 for the learner to consider 
 
 THE SIGNS OF FRACTIONS. 
 
 Aet. 123. Each of the several terms of the numerator and 
 denominator of a fraction, is preceded by the sign plus or minus, 
 expressed or understood, and the fraction taken as a whole, is also 
 preceded by the sign plus or minus, expressed or understood. 
 
ALGEBRAIC FRACTIONS. 65 
 
 Tims, in the fraction '—"' , the sign of a', the first term of the 
 
 numerator, is plus ; of the second, V, minus ; while the sign of 
 each term of the denominator is plus ; but the sign of the fraction, 
 taken as a whole, is minus. The pupil must always recollect, 
 that the signs of the several terms relate only to those terms to 
 which they are prefixed, while the sign placed before the fraction 
 relates to it as a whole. 
 
 Art. 124. It is often convenient to change the signs of the 
 numerator or denominator of a fraction, or of both. We will now 
 show the law regulating these changes. 
 
 By the rule for the signs, in Division (Art. 69), we have, 
 
 '-— =+i; or, changing the signs of loth terms, =-(-6. 
 
 -\-a -'^a 
 
 But, if we change the sign of the numerator, we have = — b. 
 
 +a 
 
 And, changing the sign of the denominator, we have '11- = — J. 
 
 — a 
 
 Hence, The signs of both terms of a fraction may be changed, 
 without altering its value or changing its sign, as a, whole ; but, if 
 the sign of either term be changed, the sign of the fraction will be 
 changed. 
 
 Hence, also, The signs of either term of a fraction may be changed, 
 without altering its valve, if the sign of the fraction he changed at the 
 tame time. 
 
 Thus, ?!=:f =— Z:?!±^ =— -?!ll^'=— (— a-^)=a+a;. 
 a — X a — X — a-{-x 
 
 And, fl_"_!=^=a+i:^±f =a+ ''llH^-x. 
 a — X a — X — a-\-x 
 
 £ X ADIPLES. 
 
 Reduce the following quantities to a fractional form. 
 
 1 I I a^ — ax . a'4-a!' 
 
 X X 
 
 2 . a^—ax-^x''— ^ . Ans. 
 
 a-\-x a-\-x 
 
 3. 2«-x+^^=^'. Am.^. 
 
 X X 
 
 4. a-^. Ans. ^ . 
 
 o+o a-\-b 
 
 5. x+y+JL. Am. J^. 
 
 X — y X — y 
 
66 RAY'S ALGEBRA, PART SECOND. 
 
 6. a — a; — ~ .. Ans. — 
 
 a-\-x a-\-x 
 
 1 a'+ai2+_^. Ans. "" 
 
 8. l-(^^'. ATt*. -^^. 
 
 x^+f x'+y' 
 
 Case IV. — To reduce fractions of different denominatobi 
 ro equivalent fractions having a common denominator. 
 
 Art. 125. 1 . Let it be required to reduce — , -, and _, to a 
 ... m n r 
 
 common denominiitor. 
 
 It is evident that we may multiply both terms of each fraction 
 
 by the same quantity, since this (Art. 118) will not change its 
 
 value. Now, if we multiply both terms of each fraction by the 
 
 denominators of the other two fractions, the new denominators 
 
 of each will be the same, since, in each case, they will consist of 
 
 the product of the same factors ; that is, of all the denominators. 
 
 Thus, ^X^OCr^am; 
 
 mY.n'yi.r mnr 
 
 i XwX'' Jnnr 
 
 my,m.y,r mnr 
 
 cy.mXn cmn 
 
 ry.m'Xn mnr 
 It is evident that the value of each fraction is not changed, 
 and that they have the same denominator. Hence, we have the 
 following 
 
 Rule for reducing fractions to a common denominator. — 
 Multiply both terms of each fraction by the product of att the 
 denominators, except its own. 
 
 Remark. — Since each denomlaator of the new fractions will consist 
 of the product of all the denominators of the given fractions, it is 
 unnecessary to perform the multiplication more than once. 
 
 EXAMPLES. 
 
 Reduce the fractions in each of the following examples, ta 
 others having a common denominator. 
 
 2. l,?,and?. Ans.yJ., ^, ^. 
 X y z xyz xyz xyz 
 
 3. f and t Ans. ^ and ^1. 
 i a ah ab 
 
 4. ^_ and _?- Ans. t^ and "f^II^. 
 x+o 3p — a' a;' — a' 
 
ALGEBRAIC FRACTIONS. 61 
 
 Art. 126. It frequently happens, that the denominators of the 
 fractions to be reduced, contain a common factor. In such casea 
 the preceding rule does not give the least common denominator. 
 
 1 . Let it be required to reduce — , — . , and _ , to their least 
 common denominator. "* ™" "*■ 
 
 Since both terms of a fraction may be multiplied by the same 
 quantity without altering its value, the first fraction may have any 
 denominator that is a multiple of m ; the second, any denomina- 
 tor that is a multiple of mn ; and the third, any denominator 
 that is a multiple of nr. Hence, any common denominator of the 
 three fractions must be a multiple of m, mn, and nr, and their 
 least common denominator must be the least common multiple of 
 the three given denominators. 
 
 The least common multiple of the three denominators is easily 
 found (Art. 112) to be mnr. It novir remains to reduce each frac- 
 tion to another virhose denominator shall be mnr. 
 a 
 
 The first fraction is ~ 5 in order to change this to another, 
 
 whose denominator shall be mnr, we must multiply both terms by 
 
 the same quantity, and by such a quantity that when multiplied 
 
 by m the product shall be mnr. But this multiplier will evidently 
 
 be obtained by dividing mnr by m ; that is, by dividing the least 
 
 common multiple of the given denominators, by the denominator 
 
 of the first fraction. It is evident that the other fractions may be 
 
 reduced in the same manner ; the operation is as follows : 
 
 J a y.nr anr 
 
 mnr— m=nr, a,na ■ — '^__= — 
 
 m y.nr mnr' 
 
 J b Xr hr 
 mnr-i.'mn=r, and — _ = 
 
 mny.r mnr 
 
 mnr~r'^r^m, and — — = 
 
 nr Xm mnr' 
 
 The process of multiplying the denominators by the quotients 
 may be omitted, since the product in each case will be equal to 
 the least common multiple. This gives the following 
 
 RULB FOR REDUCING FRACTIONS OF DIFFERENT DENOMINATORS TO 
 EQUIVALENT FRACTIONS HAVING THE LEAST COMMON DENOMINA- 
 TOR. — 1 . Find the least common multiple of all the denominators; 
 this will he the common denominator. 
 
 2. Divide the least common multiple by Ihejirst of the given denomi- 
 nators, and multiply the quotient by the first of the given numerators : 
 the product will be the first of the required numerators. 
 
 3. Proceed, in a similar mann£r, to find each of the other n/umerators. 
 
68 RiY'S ALvJEBRA, PART SECOND. 
 
 Note. — Before commencing the operation, each fraction must be in 
 its lowest terms. 
 
 E X AM PL E S, 
 
 Reduce the fractions, in each of the following examples, to 
 
 equivalent fractions having the least common denominator. 
 
 „ a i c . a Uby Sex 
 
 Gxy Zx 2y 6a^ 6a^ Qxy 
 
 ^ X y z . x{a—b) y{a^-b) z 
 
 ■ • ^' <iZrj' J5I^2- ■ a2_5J ' a.-'—b-' a'—b'' 
 
 . m — n m-\-n mHi' •„, (m — n)' {m-\-ny mV 
 
 *' ■ — :— ' > — :; "n. A7W. — , — - — ■ — - , — 
 
 m-{-n m — n w? — w m? — n' m' — n' m' — n' 
 
 Other exercises will be found in tlie Addition of Fractions. 
 
 Note. — The two following articles depend on the principles explained 
 in the preceding article, and are therefore introduced here. They will 
 both be found of frequent use,'especially in completing the square in 
 the solution of equations of the second degree. 
 
 Art. 127. To reduce an entire quantity to the form of a 
 fraction having a given denominator. 
 
 Rule. — Multiply the entire quantity by the given denominator, and 
 write the product over it. 
 
 EXAMPLES. 
 
 1. Reduce a; to a fraction whose denominator is a. Ans. — . 
 
 2. Reduce 2az to a fraction whose denominator is 2'. 
 
 Ans. . 
 
 3. Reduce x-\-y to a fraction whose denominator is x — y. 
 
 Ans. "^ y 
 x—y 
 
 4. Reduce m — n to a fraction whose denominator is a(m — n)'. 
 
 Ans. <^-^y 
 a{m — n) 
 Akt. 128. Ts convert a fraction to an equivalent one, having 
 a given denominator. « 
 
 Rule. — Divide the given denominator by the denominator of tht 
 given fraction, and multiply both terms by the quotient. 
 
 Remark. — This rule is perfectly general, but it is never applied azcept 
 when the required deuomlnator is a multiple of the given one. In 
 other cases it would produce a complex fraction. Thus, if it were 
 required to reduce § to an equivalent fraction with a denominator 5, the 
 numerator of the new fraction would be 3J. 
 
ALGEBRAIC FRACTIONS. 89 
 
 EXAMPLES. 
 
 1 . Convert | to an equivalent fraction, having for its denomi- 
 nator, 49. Am. -^^. 
 
 2. Convert _ and _ to equivalent fractions having the denomi- 
 natrr Oc\ " Am. ^, 1^-= 
 
 3. Convert Jl and to equivalent fractions having the 
 
 , „ "+ («+*)' («-*)' 
 
 denominator a' — b'. Ans. -; — rr > -^ — rr • 
 
 a' — b' a' — b' 
 
 Case V. — Addition and subtraction or fkactions. 
 
 Art. 129. It is self-evident that two algebraic fractions, like 
 two arithmetical fractions, must have a common denominator, 
 before we can find ejther their sum or their difference. 
 
 1 . Let it be required to find the value of ? , _ , and - . 
 „ J. - d d d' 
 
 Let _=m, -=n, and -=r. 
 d d d 
 
 Then a=md, b=nc!, and c=rd ; 
 and a-\-b-]-c=md-\-nd-\-rd ; 
 or, a-\-b-\-c=(m-\-n-{-r)d; 
 
 hence 'tti+?=m+ra+r. 
 d 
 
 This gives the following 
 
 Rule for the addition of fractions. — Beduce the Jractiom, if, 
 necessary, to a common denominator ; add the numerators together 
 and place their sum over the common denominator. 
 
 AaT. 130. 2. Let it be required to subtract _ from " 
 
 T * a A^ d d- 
 
 Let -=:m, and -=:n. 
 d d 
 
 Then a-=md, and J=nd ; 
 
 and a — b=md — nd, =z(m — n)d ; 
 
 hence . =m — n. 
 i 
 
 This gives the following 
 
 Rule for the subtraction of fractions. — Reduce the fractims, 
 if necessary, to a common denominator ; then subtract the numera- 
 tor of the fraction to be subtracted from the mimerator of the other, 
 and place the remainder over the /•ommon, denominator 
 
70 
 
 RAY'S ALGEBRA, PART SECOND. 
 
 EXAMPLES IN ADDITION OF FRACTIONS 
 1. Add 
 
 - and — toeether. 
 b 4i ^ 
 
 ' a. Add - and _ together. 
 b a 
 
 3 Add and together. 
 
 l-|-a; 1 — X 
 
 Find the value 
 
 •c , b , a 
 
 4. Of. 
 
 X x' x^ 
 
 5. Of L_ + _?+? . 
 
 3(1— a;V 3(l+a;-Hr») 
 
 « ^p 5 , ad — he 
 
 d d{c-\-dxy 
 a& ac be 
 
 7. ofZ+i.+!: 
 
 8. Of ?:+_i-+_?-. 
 X x+1 a; — 3 
 
 9. Of- 
 
 10. Of 
 
 1 1 
 
 4(l+a;) M(l— I) ' 2(l+a;')' 
 
 11.0f?=Li_'Z^jL?=r. 
 M ?"■ ?'■ 
 2 3ffl 3o— 2a; 
 
 12. Of 
 
 13. Of 
 
 14. Of 
 
 A To 
 
 4& 
 
 An*. 
 
 2 
 
 ,3' 
 
 ATM. °+ftx'+ex^ 
 a 
 
 Ans. 
 Ans. 
 
 1 
 
 1— a;'' 
 a+Ja; 
 
 4„o ;'e+g&+ra 
 
 A7». 
 
 A7». 
 
 Ans. 
 
 a;+a+(x4-a)2"1"a:2— 2aa^-i-3o2■ 
 4aXa+^) 4o'(o— x)^2a2(a'+a!=) 
 
 1 1 1 + L_ 
 
 a(a— i)((Z — c)~J(6— a)(i— c)^c(c— a)(o— i) 
 
 ahc 
 4a^— ax— 3 
 x'— 2a;2— 3a;' 
 
 Ans. 
 
 1— a;i' 
 
 A?w. 0. 
 
 18(z3 
 
 a!*+4o3a;4-3ffl<' 
 1 
 
 An«. 
 
 abc 
 
 EXAMPLES IN SUBTRACTION OF FRACTIONS. 
 
 In the first ten examples the second fraction is to be subtracted 
 from the first. 
 
 1. ^ and ?y 
 
 la 7 
 
 2. -L and J_. 
 a — 6 a-\-b 
 
 3. ?+?and?=? 
 p—q p-j-9 
 
 Ans. 
 
 Ans. 
 
 Am. 
 
 5x — Say 
 7^" 
 
 2b_ 
 
 a^—i^' 
 
 Apq 
 
ALGEBRAIC FRACTIONS 
 
 71 
 
 4.^=1 and 
 
 n 
 
 5. and - 
 
 1— a; 1 
 
 n 
 2 
 
 and . 
 
 Ans. 
 Ans.f=}:=. 
 
 l—2n 
 rC — n 
 
 1-^2 
 
 l+« 
 
 (a;+l)(a:+2) (x+l)(a;+2)(ar+3)' 
 
 Ans. 
 
 7. 
 
 1 
 
 (a:+l)(a:+3) 
 
 (a:+l)(a;+2) 
 
 and 
 
 
 8. ? and Cf^:±)f. 
 c c(c-)-dj:) * 
 
 g 1 3m+2?t ^^j 1 3m— 2m 
 ' 2'3m— 2re 2'3m+2re" 
 
 10 "Al and ^±1 
 
 (o — 6)(a; — o) (o — J)(a: — 6)' 
 
 Find the value 
 
 11 Of ^"^ — ^'^ ro+3re 2m 
 3 (1 — n)"~3 (1 _ra)+i:^' 
 
 12. Of ^_?=f I ^. 
 aJ oc Je ' 
 
 (a;+l)(a:+2)(a^4-3)- 
 
 Ans. 
 
 12mn 
 9m' — in'' 
 x-\-c 
 
 (a; — a)Qc — b)' 
 
 Ans.Jl- 
 1— n' 
 
 13. Of 
 
 3x 
 
 14. Of ?i3? -^ iE'— «'y 
 
 Ans. 
 
 15. Of 
 
 a+y a;^ — y'' 
 1 a;+3 
 
 Ans. 0. 
 
 X 
 
 1x' — y'' 
 Ans. 1. 
 
 a>- 1 2(a;+l) 2(a;'+l)' 
 
 16. Of . 
 
 x—l 
 
 Ans.^+l 
 
 x^~r 
 
 Ans.^±t^±L 
 a^(a;=4-l)2' 
 
 X' • x' X ' x'+l (x'+iy 
 
 Case VI. — Multiplication of fractions. 
 
 Aet. 131. 1. Let.it be required to find the product of ?by£ 
 
 6 d' 
 
 Let -=m, and i=». 
 6 d 
 
 Then a=d)m, and cz=dn 
 
 . • . ac=bmdn,=hdXmn; or, dividing ty bd, —=mn. 
 
 hd 
 Hence, to find the product of two or more fractions, v^e have 
 the following 
 
 RtTLE. — MvUiply the numerators together for a new rmmeratw, and 
 the denominators together for a new denaminatar. 
 
72 RAY'S ALGEBRA, PART SECOND. 
 
 Remarks. — 1st This rale is general, and embraces all the cases in 
 which a fraction is a factor. Thus, if it be required to multiply a frac- 
 tion by an integral quantity, the latter may be placed under the form of 
 a fraction, by writing unity beneath it. 
 
 2d. If either of the factors is a mixed quantity, it is best to reduce it 
 to an improper fraction, before comtoencing the operation. 
 
 3d. When the numerators and denominators have common factors, 
 the process may be abbreviated by indicating the operation, and then 
 canceling the factors common to both terms. 
 
 Thus 2a' (a+by 2a^y.(.a+b){a+b) a+6 
 a'~b'^ ia'b (a+6)(o— 6)4o2j —26(o— 6)" 
 
 E XAM PLE S . 
 
 Find the products of the fractions in each of the following 
 ixercises, expressed in their simplest forms. 
 
 1- -7- by __ and .^ — by __. Ans. - and — 
 
 4 -^ 3y c 8o» y o' 
 
 2. Iff, "y-lf" Am. ^<'''—y) 
 
 cy Qx^-\-Qxy' ' 3c(x+y) ' 
 
 3. 
 
 X' a X 
 
 — ) — — . 
 
 a X a 
 
 4. 1-^ and 2+-?^-. Ans. >^ 
 
 x-\-y x—y a;2— y2' 
 
 5. l±±t?L and 1=? Ans. 1=?! 
 
 o' — a'x-\-ax'' — a' a-\-x ' a* — x* 
 
 rj_ a:'-9x+20 ^^^ x '-13x+i2 ^^ x'-Ux+29 
 
 x' — 6a; a;' — 5x ' ' x' ' 
 
 8.?!±?f±?and^!+5£±i Ans.^J:^ 
 
 a:'+2a:+l x'+lx+12' x+S' 
 
 9. ^, "'— ^' bc+bx ^^ 4a:(a4-a) 
 
 Sby c' — x^ a? — oa;' ' Zy{_e — xy 
 
 10. f!=±', ^1=£, "' Am. °'(''+^) 
 
 a:+y a—b (a:— y)'" x-^ ' 
 
 11. .'+.+1 by ^1+1. Ans. a;=+l+^,. 
 
 12. x+l-f- by a:— 1+-. Ans. a;'+l+i. 
 
 3J ill j;* 
 
 13. lf+?5 by ?L^. Ans. ?^+24.??! 
 3a;^^ ' SiUi' 9ar»^ ^8ai6' 
 
ALGEBRAIC FKACTIONP. 78 
 
 p — qx i""!"?-* 
 
 Ans. rs-\-(Tt-\-qs)x-\-qtx^. 
 Find the value 
 
 ^^•°^(H)^+(H)f+(H)I- 
 
 Ans. ^l^-i) 
 
 ad bo ' 
 Case VII. — Division of fractions. 
 
 ^RT. 132. 1. Let it be required to find the quotient of - by -_ 
 
 a 
 
 Let --=)n, and-,=». Then, 
 a 
 
 01=6771, and c=^7i. 
 
 Multiplying both terms of the first equality by d, and of the 
 
 second by J, we find 
 
 ad=bdm, ana bc=bd7i. 
 
 ^, f ad bdm m 
 therefore — = — =— ■ 
 be bdn n 
 
 that IS, _=-x-. 
 n c 
 
 Hence, to find the quotient of one fraction divided by another, 
 
 we have the following 
 
 Rule. — Invert the divisor, and proceed as in multiplication of 
 
 fractions. 
 
 Remark. — This rule is general, and embraces not only all the cases 
 in which either divisor or dividend is a fraction, but is also applicable 
 when both are integral quantities, siace any integral quantity may be 
 placed under the form of a fraction, by writing unity beneath it. 
 
 Thusa^i=-Xi=- 
 
 Remarks 2 and 3, Art. 131, apply equally well to division as to multipli' 
 cation of fractions. 
 
 E XAMP LES. 
 
 Required, in their simplest forms, the quotients 
 
 1. Of ^^^J^ Aw. "X 
 
 2. Of ?+^^?z:f. Aru>. ^=^. 
 
 ffl+c ' a — 6 o' — c' 
 
14 RAY'S ALGEBRA, PART SECOND. 
 
 3 
 
 A7w.^-!±^ 
 
 a- 
 
 5. Of a^+y' ■ ^'— ^+y' _ Atw. 1. 
 
 x' — y' ■ a; — y 
 
 6 Of °'-^' • ''''^+^ Ans. ?±5(a"+<w+a:»). 
 
 ll+x^ I / W+x X / 2x'— 1- 
 
 9. Of fj^4-^=y )-i-f^..^-y) An*.^±y-'. 
 
 \ X— y ' x-|-3/ / ■ \ X — y x-\-y I • 2xy 
 
 10. Of ^^ - ^ Ans. ?. 
 
 2x— 2"x— 1" 4 
 
 \ X— 3 / \ X— 3 / X— 5 
 
 12 Of 4a(a'-^°) . a'— gx 4^, 4(a+x) 
 
 3i(c'— x') ■ ic+6x' ' 3(c— x)' 
 
 13. Of f x^— i ) -i- ( x-1 ) . Ans. x'+l+x+l . 
 
 Vx*/\x/ x'x 
 
 Akt. 133. To reduce a complex fraction to a simple one. 
 This is merely a case of division, in which the dividend and 
 divisor are either fractions or mixed quantities. 
 
 C 71 
 
 Thus is the same as to divide a-1 — hy m , 
 
 n ' e ' r 
 
 m — _ 
 
 r 
 
 r acr+6r 
 
 mr — n cmr—cn 
 Let the following examples be solved in the same manner. 
 3x 
 
 >c/\ r / c r c K 
 
 1. ^^ An. 3 
 
 2x ^^- 1' 
 
 -+- 
 
 f h 
 
ALGEBRAIC FRACTIONS. TS 
 
 "+1 I °-l 
 
 a— I a+l 
 
 4. ^ Am. *. 
 
 5. A^ A»,.Hd 
 
 Art. 134. Resolution of fractions into series. 
 
 Def. — An infinite series consists of an unlimited number of 
 terms which observe the same law. 
 
 The law of a series is a relation existing between its terms, 
 BO that, when some of them are known, the succeeding terms 
 may be easily obtained. 
 
 be found by multiplying the preceding terra by — . 
 
 x' 
 
 Any proper algebraic fraction, whose denominator is a polyno- 
 mial, may, by division, be resolved into an infinite series ; for the 
 numerator is a dividend, and the denominator a divisor, so related 
 to each other that the division can never terminate, and the quo- 
 tient will therefore be an infinite series. After finding a few 
 terms of the series, the law of continuation is, in general, easily 
 Been, and the succeeding terms may be found without continuing 
 the division. 
 
 EXAMPLES. 
 1 
 
 1. Convert the fraction into an infinite series. 
 
 l-\-x 
 
 l—x\l-\-x 
 
 l+»_ l—2x-\-2x'—2x>-{- &c. It is evident that the law of 
 
 — 2x this series is, that each term, 
 
 — 2r — 2a!' after the second, is equal to 
 
 +2a:' the preceding term, mnl*i- 
 
 Thus, in the infinite series 1 — - -I 4'^c., any term may 
 
 ■4-23i;'+2j' plied by —x. 
 
 ^2x' 
 
76 RAY'S ALGEBRA, PART SECOND. 
 
 In a Bimilar manner, let the fractions in each of the following 
 examples be resolved into an infinite series. 
 
 2. .^_=1— r'+r"— r6+r»— &c., to infinity. 
 
 3. ^_i__ =14-r— r^— r''+rS+r'— r»— r"'+ &c. 
 1 — r-^^ 
 
 4. 1 — 1_r-[-ri— r<-|-r«— r'-fr»— r'»+ &C. 
 
 5. _?_=!-* 4-^ _*i+&c. 
 
 MISCELLANEOUS PROPOSITIONS DT FRACTIONS. 
 
 Of the forms -, -, and -. 
 6 
 
 When the two terms of a fraction - are finite determinate 
 
 6 
 
 quantities, the fraction has necessarily a finite determinate value, 
 which is, the quotient of o divided by h. 
 
 Let us now examine the cases where the numerator or denom- 
 inator, or both, reduce to zero. 
 
 Akt. 135. To prove that -=0. 
 b 
 
 While the denominator i is a constant number, if the numera- 
 tor a diminishes, the value of the fraction diminishes. Thus, in 
 the fractions ^, |, |, and i, each is less than the preceding. 
 Heiice, as the numerator a diminishes, and approaches to zero, 
 
 the value of - diminishes and approaches to zero; and finally, 
 b 
 
 vhen a=0, the expression _ reduces to zero. 
 6 
 
 Or thus: Since the product of zero, by any number, la zero, 
 therefore the quotient of zero, divided by any number, is zero. 
 
 That is, since X6=0, therefore -=0. 
 
 b 
 
 Aet. 136. To prove that f =qo. 
 
 
 If the numerator a, of a fraction, remains constant, and the 
 
 denominator diminishes, the value of the fraction increases. 
 
 Thus: 1st Suppose the denominator 1 ; then ?=o. 
 
ALGEBRAIC FRACTIONS. T7 
 
 8nd. Suppose the denominator — ; then _=10o. 
 
 3rd. Suppose the denominator : then — ==100a. 
 
 ^^ 100 .01 
 
 4th. Suppose the denominator ; then ___=1000a. 
 
 From this it is evident, that if the denominator is kss than 
 any assignable quantity, that is 0, the value of the fraction is 
 greater than any assignable quantity, that is irifinitely great, or 
 infinity. This is designated by the sign oo ; that is 
 a 
 
 5=00. 
 
 Akt. 137. To prove that - is indeterminate in value. 
 
 When both numerator and denominator are zero, the fraction 
 - becomes _. Now since the divisor zero, multiplied by ant, 
 
 number whatever, produces the dividend zero; therefore the quotient 
 of zero, divided by zero, may be taken any number vifhatever; that 
 
 is, the fraction _ is indeterminate. 
 
 
 It is important, however, for the pupil to know, that the form 
 _ is often the result of a particular supposition, when both terms 
 of a fraction contain a common factor. 
 
 ^2 J2 (j2 ^2 Q 
 
 Thus, if x= , and we make J=a, it becomes =:- ; 
 
 a — J a — a 
 
 but if we cancel the common factor, a — b, and then make J=a, 
 we have x=2a. 
 
 Similarly, the fraction x= becomes - whena=l; but 
 
 a'-\-a — 2 
 
 if we divide both terms by their common factor, a — 1, we havo 
 
 x= Jl-, which reduces to - when a=l . 
 a+2 3 
 
 These examples show, that if the value of any quantity is _ 
 
 before we decide that it is really indeterminate, we must see thai 
 the apparent indetermination has not arisen from the existence 
 of a factor, which, by a particular supposition, became equal to 
 zero. 
 
78 RAY'S ALGEBRA, PART SECOND. 
 
 Art. 138. Theorem. — If the same quantity ie added to both 
 terms of a proper fraction, the new fraction resvUing will ie greater 
 than the first; but if the same quantity be added to both terms of an 
 improper fraction, the Tteto fraction resulting will be less than the first. 
 
 Let _ be a proper fraction, a being less than b. 
 
 Let m represent the quantity to be added to each term, then 
 
 the resulting fraction is — X_ . 
 b-\-m,' 
 
 To determine which of the fractions, - and ~^ , is the great- 
 
 6 b-\-m 
 
 er, we must reduce them to a common denominator; 
 
 ., . . a db-\-am 
 
 this gives _= — ! , 
 
 ^ 6 J2+im 
 
 J aA-m ab-\-bm 
 
 and -^! — = — ! . 
 
 b-\-m b^-\-bm 
 
 Since the denominators are the same, that fraction is the 
 
 greatest which has the greatest numerator. 
 
 When _ is a proper fraction, a is less than 6; 
 b 
 
 therefore am is less than bm, 
 and ab-\-am<^ah-\-bm; 
 that is, the resulting fraction is greater than the first. 
 
 But if _ is an improper fraction, it is evident that 
 
 ah-\-airC^ab-\-im ; 
 that is, the resulting fraction is less than the first. 
 
 Art. 139. Theorem. — If the same quantity be subtracted Jrom 
 both terms of a proper fraction, the new fraction resulting will be less 
 than the first; but if the same quantity be subtracted from both terms 
 of an improper fraction, the new fraction resulting vrill be greater 
 than the first. 
 
 Let - be a proper fraction, a being less than 6. Let m repre- 
 b 
 
 sent the quantity to be subtracted from each term, then the re- 
 sulting fraction is "' "^ . To determine which fraction is the 
 
 b — m 
 greater, we reduce them to a common denominator, and compare 
 
 their numerators: , 
 
 4. . . a ab — am 
 this gives - =_ 
 
 b b^ — bm 
 a — m ab — 
 i — m b' — bm ' 
 
 J a — m ab — bm 
 
ALGEBRAIC FRACTIONS. 79 
 
 If a<^6, then am<^bm; and if am<CJ)m, then 
 ah — arn^ab — hm ; 
 Ihat is, the resulting fraction is less than the first. 
 But if a^b then arri^hm; and if ani^hm, then 
 ah — am<^aii — bm ; 
 that is, the resulting fraction is greater than the first. 
 
 MISCELLANEOUS EXERCISES IN FRACTIONS. 
 
 1. Prove that -^ ^?=?+_^ ^±?= JL. 
 
 a;— 3 X ^x-j-3 x x'—9 
 
 i. Prove that a'+'^+l ^ ^'+^+1 ^ c'+c+l ^^^ 
 (o— 6)(a— c)^(6— o)(J— c)~(c— a)(c— J) 
 
 3. Find the value of ( x+-^ ) -j- ( x—^ ) , when x=5 
 
 Ans. 9. 
 
 4. Find the value of _ _P''~^— ^'^~^ l^^ , when x=ii . 
 2 13 4 J • 2 ^ 
 
 Ans. 2 1. 
 
 .5. Find the value of ax-\-h/, when x=?S~JL and y=°'" '^, 
 
 05 — ftp c^ — 6p' 
 
 Atis. c. 
 
 6. Find the value of^+^^^+i^ when x=if^. Ans. 2. 
 
 a: — 2o a: — 2 J a-[-i 
 
 7. Find the value of "" -1 - , when a:=?!±^ 
 
 2no"— 27W ' 2?ii"— 2m 2 " 
 
 re' 
 
 8. Prove that the sum or difference of any two quantities divi- 
 ded by their product, is equal to the sum or difference of their 
 reciprocals. 
 
 9. If two fractions are together equal to 1, prove that their 
 difference is the same as the difference of their squares. 
 
 10. If the difference of two fractions is equal to t, show that 
 
 1 
 f tirties tneir sum is equal to q times the difference of their 
 
 ■quares. 
 
 11. Prove that a'+^^ , i^+^' 0'+^ _^. 
 
 (_a—b)(_a—cy{b—a)(b—cy(,o-a)(c—b) 
 that when the terms are multiplied respectively by b-\-c, a-^c, and 
 B-f-i, the sum =0 : and that when multiplied respectively by fto, 
 
 ac, and ab, it is ^=A^. 
 
80 RAY'S ALGKBRA, PART SECOND. 
 
 CHAPTER IV. 
 EQUATIONS OF THE FIRST DEGEEE. 
 
 DEFINITIONS AND ELEMENTARY PRINCIPLES. 
 
 Akt. 140. An equation is an algebraic expression, stating 
 tlie equality between two quantities. Thus 
 
 X — 5=3, 
 is an equation stating that if 5 be subtracted from x, the remain- 
 der will be 3. 
 
 Art. 141. Every equation is composed of two parts, separated 
 from each other by the sign of equality. The quantity on the 
 left of the sign of equality, is called the _first memher or side of 
 the equation. The quantity on the right, is called the second 
 member or side. The members or quantities are composed of one 
 or more terms. 
 
 Art. 143. There are generally two classes of quantities in 
 an equation, the hnovm and the unknown. The known quantities 
 are represented either by numbers, or the first letters of the 
 alphabet, as a, b, c, &c. ; and the unknown quantities by the last 
 letters of the alphabet, as x, y, z, &c. 
 
 Art. 14S. Equations are divided into Jegrees, called Jlrst, 
 second, third, and so on. The degree of an equation depends on 
 the highest power of the unknown quantity which it contains. 
 Thus, an equation which contains no power of the unknown 
 quantity higher than the first, is termed a?i equation of the first 
 degree, or a simple equation. 
 
 An equation in which the highest power of the unknown 
 quantity is of the second degree, is called an equation of the secona 
 degree, or a quadratic equation. 
 
 Similarly, we have equations of the third degree, fourth degree 
 and so on; those of the third degree are generally called cubi^ 
 equations, and those of the fourth degree, biquadratic equations. 
 
 Thus, 
 ax — t=c, is an equation of the 1st degree. 
 x^-\-2px=q, " " " 2d " or Quadratic eouatiof 
 
 a;3 — jpx=q, " " " 3d " or cubic equation. 
 
 3s*-\-ax^-\-px=!!:q, " 4 th " or biquadratic eq. 
 
 i"+aa;"' '-\-bx"~'=:c, " reth degree. 
 
EQtAllONS OF THE FIRST DEGREE. 81 
 
 When any equation contains more than one unknown quantity; 
 its degree is equal to the greatest sum of the exponents of thfl 
 unknown quantity, in any of its terms. Thus, 
 
 xy-\-ax — hy:=c, is -an equation of the 2nd degree. 
 
 x''y-\-x' — cx=a, is an equation of tlie 3rd degree. 
 
 Art. 144. An equation of any degree is said to be complete, 
 when it contains all the powers of the unknown quantity, from 
 up to the given degree. When one or more terms are wanting, 
 ihe equation is said to be incomplete. 
 
 Thus, a:'-|-^j:-j-j=0, is' a complete equation of the second 
 degree, the term q being equivalent to qx", since a;''=l . (Art. 82.) 
 
 x^-\-px^-\-qx-{-r=0, is a complete equation of the third degree. 
 
 ax^=q, is an incomplete equation of the second degree. 
 
 x^-\-px=q, is an incomplete equation of the third degree. 
 
 Art. 145. An identical equation, is one in which the twc 
 members are identical; or, one in which one of the members is 
 the result of the operations indicated in the other. 
 Thus, ax — 5=ax — b, 
 8x — 2x:=5x, 
 (a:+3)(x — Z)=x^ — 9, are identical equations. 
 
 Equations are also distinguished as numerical and literal. 
 
 A numerical equation is one in which all the known quantities 
 are expressed by numbers. 
 
 Thus 2a;'-|-3a;=10a;-|-15, is a numerical equation. 
 
 A literal equation is one in which the known quantities are 
 represented by letters, or by letters and numbers. 
 
 Thus, ax-\-h^cx-\-d, 
 
 and ax-\-b=2x — 5, are literal equations. 
 
 Art. 146. Every equation may be regarded as the statement, 
 in algebraic language, of a particular question. 
 
 Thus, x — 5=9, may be regarded as the statement of the fol- 
 lowing question: — To find a number from which, if 5 be sub- 
 tracted, the remainder shall be 9. 
 
 If we add 5 to each member, we shall have 
 ar— 5-|-5=9+5, or a:=14. 
 
 To solve an equation, is to find the value of Che unknown quantity; 
 or, to find a number or expression, which, being substituted for 
 the unknown quantity, will render the two members identical. 
 
 Remark. — The solution of equations is the moat useful and interest- 
 ing part of algebra. 
 
 An equation is said to.be verified, when the value of the 
 unknown quantity being substituted for it, the two members are 
 rendered equal to each other. 
 
82 RAT'S ALGEBRA, P ART SECOND. 
 
 Thus, in the equation x— 5=9, if 14, the value of x, be sub- 
 Btituted instead of it, we have 
 
 14—5=9; 
 or, 9=9. 
 Art. 147. The value of the unknown quantity, in any equa 
 tion, is called the root of that equation. 
 
 EQUATIONS OF THE FIRST DEGREE, CONTAINING BUT ONE 
 UNKNOWN aUANTITY. 
 
 Art. 148. The operations employed to find the value of the 
 unknown quantity in any equation, are founded on this evident 
 principle: 
 
 If we perform the same operation on two equal quantities, the re- 
 sults will be equal. 
 
 This principle or axiom may be otherwise stated, as follows: 
 
 \.If,to two equal quantities, the same quantity be added, the sums 
 will be equal. 
 
 2. If , from two equal quantities, the same quantity be subtracted, 
 the remainders will be equal. 
 
 3 . If two equal quantities be multiplied by the seme quantity, the 
 products will be equal. 
 
 4. If two equal quantities be divided by the same quantity, the 
 quitients will be equal. 
 
 5. If two equal, quantities be raised to the same power, the results 
 will be equal. 
 
 6. If the same root of two equal quantities be extracted, the results 
 will be equal. 
 
 Remark. — An axiom is a Belf-evident truth. The preceding axioma 
 are the foundation of a large part of the reasoning in mathematics. 
 
 Art. 149. There are two operations of frequent use in the 
 solution of equations. These are, first, to clear an equation of 
 fractions; and second, to transpose the terms in order to find the 
 value of the unknoion quantity. 
 
 These are named in the order in which they are used in the 
 solution of an equation; we shall, however, first consider the 
 subject of 
 
 TRANSPOSITION. 
 
 Art. 1 50. Suppose we have the equation 
 
 ax-\-b=e — dx- 
 Since, by the preceding principle, the equa.^ty will not be 
 tffectpd by adding the same quantity to both members ; or, b> 
 
EQUATIONS OF THE FIRST DEGREE. 83 
 
 tubtracting the same quantity from both members ; if we add dx 
 tt each side, we have 
 
 ax-\-b-\-dx=c — dx-\-dx. 
 
 If we cubtract b from each member, we have 
 ax-\-h — h-{-dx=c — dx-\-<ix — b. 
 
 But -\-b — b cancel each other, so do — dx-\-dj:; omitting these, 
 we have ax-\-dx=^c — b. 
 
 But thkls result is the same as if we had removed the terms -{-b 
 &nd — dx to the opposite members of the equation, and at the 
 same time changed their signs. Hence, 
 
 Any quantihj may be transposed from one side of an equation to 
 the other, if, at the same time, its sign be changed. 
 
 This is termed the Rule of Transposition. 
 
 TO CLEAR AN EQUATION OF FRACTIONS. 
 
 Art. 151. 1. Let it be required to clear the following equa- 
 tion of fractions. 
 
 X X , 
 
 ab be 
 
 Since the first term is divided by ab, if we multiply it by ab, the 
 divisor will be removed; but if we multiply the first term by ab, 
 we must multiply all the other terms by ab, in order to preserve 
 the equality of the members. Again, since the second term is 
 divided by be, if we multiply it by be, the divisor will be removed; 
 but if we multiply the second term by be, we must multiply all 
 the other terms by be, in order to preserve the equality of the 
 members. Hence, if we multiply all the terms on both sides by 
 ahy,bc, the equation will be cleared of fractions. 
 
 Instead, however, of multiplying every term by abycbc, it is 
 evident, that if each term be multiplied by such a quantity as 
 will contain the denominators, without a remainder, that all the 
 denominators will be removed. This quantity is evidently the 
 least common multiple of the denominators, which, in this case, ia 
 abc; then, multiplying both sides of the equation by abc, we 
 have 
 
 ex — ax^^abcd. 
 
 From which we derive the following 
 
 Rule for clearing an equation of tkactions. — Find the least 
 common multiple of all the denominators, and multiply each term oj 
 the equation hij it. 
 
^* RAY'S ALGEBRA, PART SECOND. 
 
 EXAMPLES FOR PRACTICE, 
 
 In clearing equations of fractions. 
 X x_ 
 3~4^ 
 
 2- |— j=l. A>M. 4z— 3i=12 
 
 3- 4+f=^- A.ns. 3a:-f-2as=60 
 
 4- |— |+^=3i- ^'W- 6a— 3a4-2*=84 
 
 5- ^'^+-5-=-2 ■ ■^««- 20i+2a— 6=5a;-(-45. 
 6. 2a— ?ZL=?ZL, A;ji. 20a^— 2i+6=5a!— 15 
 '''• '*~^-^=i^— -g-- ^'"- ISa— 3a;+6=60— 2a;— 4. 
 
 9' <" ^~'^'^ • ^^- <"o:i:—bz-\-cz=cb:z+6x—cx. 
 
 in X — a X — a 2nb 
 
 a-\-b a — b a' — 4'" 
 
 Ans. ax—a^—bx-\-ab — ax-\-a^—bx-\-ab='^nb 
 
 SOLUTION OF EaUATIONS OF THE FIRST DEGREE, CONTAIN- 
 ING ONLY ONE UNKNOWN QUANTITY. 
 AitT. 152. The unknown quantity in an equation may bs 
 combined with the known quantities, either by addition, subtrac 
 tion, multiplication, or division; or by two or more of these dif- 
 ferent methods. 
 
 1. Let it be required to find the value of x, in the equation 
 
 a-^x=b, 
 where the unknown quantity is connected by addition. 
 By subtracting a from each side (Art. 148), we have 
 a;=J — a. 
 
 2. Let it be required to find the value of x, in the equation 
 
 X — a=b, 
 where the unknown quantity is connected by subtraction. 
 By adding a to each side (Art. 148), we have 
 x=zb-\-a. 
 
 3. Let it be required to find the value of x, in the equation 
 
 ax=b, 
 where the unknown quantity is connected by multiplication. 
 
EQUATIONS OF THE FIKST DEGREE. 88 
 
 By dividing each side by a, we have 
 b 
 a 
 
 4. Let it be required to find the value of x, in the equation 
 
 a 
 i»heie tlie unlinown quantity is connected by divtston. 
 By multiplying each side by a, we have 
 
 x=J)y.a=ab 
 From the solution of these examples, we see that 
 
 When the unknown quantity is connected by addition, it is to be 
 separated by subtraction. When it is connected by subtraction, it is 
 to be separated by addition. When it is connected by multiplication, 
 it is to be separated by division. And, when it is connected hy divi- 
 tion, it is to be separated by multiplication. 
 
 5. Let it be required to find the value of x, in the equation 
 
 Clearing the equation of fractions, we have 
 2 1 a:— (24— 2a;)=7a;+56 , 
 or 21a;— 24+2a;=7x4-56. 
 Transposing the terms 7x and — 24, we have 
 2 1 x+2a;— 7a;=56+24 ; 
 reducing, IGa^SO; 
 
 dividing by 16, a;=f §^5. 
 
 It will be readily seen that this solution consists of three 
 steps, viz.: 
 
 1st. Clearing the equation of fractions. 
 
 2nd. Transposition. 
 
 3rd. Reducing like terms, and dividing by the coefficient ot a:. 
 
 Let this value of x be substituted instead of x in the original 
 equation, and, if it is the true value, the two members wiU he 
 equal to each other. 
 
 Original equation, 3a;— ?lz^=a;4-8. 
 
 Substituting 5 in place of x, it becomes 
 3 j^ 5_24--^X5_5_|_g^ 
 
 or 15—2=5+8, 
 or 13=13. 
 
86 RAY'S ALGEBRA, PART SECOND. 
 
 The operation of substituting the value of the unlcnown quan- 
 tity instead of itself, in the original equation, to see if it wiU 
 render the two members equal to each other, is called verification 
 
 6. Find the value of x, in the equation 
 
 X — ! ="+ — . 
 
 ab be' 
 
 Ist step . . . obex — dc — ac=dbcd+ax. 
 
 2nd step . . . obex — ex — ax=:abed-\-ac. 
 
 Factoring . . . {abc — c — a)x=ac(bd-\-l). 
 
 „ , , ac(bd+l) 
 
 abc — c — a 
 
 Aet. 15S. From the solution of the preceding examples, we 
 
 derive the following 
 
 Rule fok the solution of an equation of the y'KST degree. — 
 
 1. If Tiecessary, dear the equati-m of fractions; and pe'form. all the 
 operations indicated. 
 
 2 . Transpose all the terms containing the unknovm quantity to one 
 side, and the knmim quantities to the other. 
 
 3. Reduce each member to its simplest form, and divide htth sides by 
 the coefficieTit of the unkTumm quantity. 
 
 Remake. — This rule gives the method of proceeding most generally 
 advantageous, but in some cases it is best to perform the cperationa 
 indicated, and transpose the necessary terms, before clearing of fractions. 
 Experience can alone determine the best method in particular cases. 
 
 EXAMPLES FOR PRACTICE. 
 Note. — Let the pupil verify the value of the unknown quo'itity is 
 each example. 
 
 Find the value of the unknown quantity in each of the foUoW' 
 log examples. 
 
 ■ 14 21 "^ * ~i~' 
 
 1st. step . . . 181+42— 8a^+28+231=21a^-84; 
 2nd step . . . 18a;— 8a;— 211=— 231— 42— 28— 84 
 3rd sfep . . . — lliE=— 385, 
 x=35. 
 Verification. 3X35+7 2x35-7 ..^35^ 
 14 21 ^ * 4 
 
 8_3+2|=7|, 
 71=71. 
 8. 5(1+1)— 2=3(a^+5). Ans. as=6. 
 
 ». 3(a^2)+4=4(3-a;). Ans. x=2. 
 
EQUATIONS OF THE FIRST DEGREE. 81 
 
 10. 5— 3(4^r)+4(3— 2x)=0. Atis. x=1 
 
 1 1 . 3(a;— 3)— 2(a;— 2)+x— l=a;+3+2(a;+2)+3(a;-f 1). 
 
 • Ans. x=: — 4. 
 
 12. 5(5a:— 6)— 4(4x— 5)+3(3a!— 2)— 2a!— 16=0. 
 
 Ans. x=2. 
 
 13. ? + ?=?+7 Am. *=12. 
 2^3 4^ 
 
 14. ?4.?_?.+?=75. Ans. »=10. 
 2^3 4^5 5 
 
 16. ?^_?f=10+??=l Ans. x=U. 
 
 2 3 ^ 6 • 
 
 5^-^i_3_^ 27-5X ^^_ ^„ , 
 
 2" 
 
 17. 
 
 2 4 
 
 T7' 
 
 18. 5j!—?^ +1=31+^+7. Anj. a;=8. 
 
 3 ' 2 ' 
 
 19. ?^=l+5=?_?^=5=2-^t? A7«. x=6. 
 
 7 ,^ 4 12 28 • 
 
 20. 7^+9 3x+l_ 9a:-13 249-9^ ^^_ ^_9_ 
 
 8 7 4 14 ■ 
 
 21. |(2a>— 10)— TV(3a;— 40)=15— i(57— x). Ans. x=n. 
 
 22. 4(«-5)-5'^f— ^)=l-3§- ^n*- «=4f- 
 
 23. |C4+ix)-4(2a^i)=i|._ * Ani. x=|. 
 
 24. ^(|j;+4)— Zii:5=f (-— 1 ) Atw. x='3. 
 
 25. 3ix{28- (?+24 ) |=3^x{2-|+^}. Ans. x=4. 
 
 27. ^(^||)-^=^3(l-3x)=x-,>^ ( 5a:- 1=^ ) . 
 
 Ans. a=ll. 
 
 28. bx+2x—a=:Sx—2c Ans. x="llHf 
 
 . i— 1 ' 
 
 29. a2a^f J'=6»x+a5. Ans. x=''- +°^+^^ . 
 
 a-\-b 
 
 30. ax-\-h''=a'-\-bx. Ans. x=ci-\-b. 
 
 SI. ^_^=?_^. Ans. x=^, 
 
 a c h d be' 
 
88 RAY'S ALGEBRA, PART SECOND. 
 
 32. l—-=a->—b\ A.ns. x=- 
 
 hx ax "* 
 
 33 "'~^= ''+^ A.ns. x=UZa—b). 
 
 x—L a;+2c" 26 
 
 34. f— 1— _4-3oi=0. A.ns. a;=_i _'' 
 
 a c^ c— <"« 
 
 35. ?^4i?+^=5^+i(/i-«c) Av^. r- .YSr^L • 
 
 6 T^^^ i -I yv./ af-^2bc—bfg 
 
 36 .^=34—^ ^»w. a:=2ffi— 56+?^. 
 
 ■ 0—26 2o— 6 a 
 
 37. 1(1— o)— I (2a:-36)— K«-«)=10«+l !*• 
 
 A»M. a;=25o4-246. 
 
 Of, 3a— a , j:4-2& _7a_ a ^^^ ,y_ 8i'— 4ge+aie 
 
 ~6 ' 'c c 4' ■ 12(2J— c.) ■ 
 
 39. _J_+_2_=_J_. An.. ^*i^=±h=). 
 
 ab — ax bo—bx ac — ax a 
 
 aUBSTIONS PRODUCING EttUATIONS OF THE FIRST DEGREE, 
 CONTAINING ONLY ONE UNKNOWN aUANTITY. 
 
 Art. 154. The solution of a problem by algebra, consists o( 
 two distinct parts. 
 
 1st. To express the conditions of the problem in algebraic lan- 
 guage; that is, to form the equation. 
 
 2nd. To solve the equation; that is, to find the value of the un- 
 known quantity. 
 
 Sometimes the statement of the question proposed, furnishes 
 tlie equation directly; and sometimes it is necessary, from the 
 conditions given, to deduce others, from which to form the equa- 
 tion. When the conditions furnish the equation direc.tly, they 
 are called explicit conditions. When the conditions are deduced 
 from those given in the question, they are called implied conditions. 
 
 It is impossible to give a precise rule by means of which every 
 question may be readily stated in the form of an equation. The 
 first step is, to understand fully the nature of the question, so as 
 to be able to prove the correctness or incorrectness of any pro- 
 posed answer. After this, the equation, by the solution of which 
 the value of the unknown quantity is to be found, may generally 
 be formed by the following 
 
 Rule. — Denote the required quantity by one of the final letters of 
 the alphabet; then, by means of signs, indicate the same operations 
 that it would he necessary to make on the answer, to verify it. 
 
EQUATIONS OF THE FIKST DEGREE. 89 
 
 E X AM FLE S. 
 
 1 . Find two numbers such, that their sum shall be 50, and 
 their difFerencfe 12. » 
 
 Let X denote the least of the two required numbers. 
 
 Then will . . a;-|-12= the greater, 
 
 And a;+a;+12=50, by the question. 
 
 Transposing, . a;-|-x=50 — 12. 
 
 Reducing, . . 2x=38. 
 
 Dividing, . . . a;=19, the less number; 
 
 And a;+12=19-(-12=31, the greater number. 
 
 Verification. 31+19=50, and 31—19=12. 
 
 2. What number is that whose ^ part exceeds its 5 part by 6! 
 Let x=: the required number. 
 
 Then will its J- part be denoted by - , and its ^ part, by - . 
 
 3 5 
 
 Therefore, . . . ?_?=6. 
 3 5 
 
 Clearing, . . . 5x — 3a;=:90. 
 
 Reducing, . . . 2k=90. 
 
 Dividing, . . . a;=45, the number required. 
 
 Verification. | of 45=15, ^ of 45=9; 15—9=6. 
 
 3. Divide $500 among A, B, and C, so that B shall have $20 
 more than A, and C $75 more, than A. 
 
 Let . . . a;=A's share. 
 Then . . a:+20=B's share. 
 And . . . a;-|-75^C's share. 
 
 Then x+i+20+k+75=500, by the question. 
 Reducing, . . . 3a:-|-95=500. 
 Transposing, . . 3a;=500 — 95=405. 
 Dividing, .... 1^135, A's share. 
 
 a;+20=155, B's share. 
 a;4-75=210, C's share. 
 Verification. 1354-155+210=500. 
 
 4. Out of a cask of wine which had leaked away 5, 35 
 gallons were drawn, and then, being guaged it was | full; 
 now much did it hold! 
 
 Let x= the number of gallons it held; 
 
 then ?= " « " leaked out. 
 
 5 
 
 There had been taken away - +35 gallone, 
 .5 
 
 8 
 
00 RAY'S ALGEBRA, PART SECOND, 
 
 and there remained a — ( - -j-35 J gallons. 
 
 Clearing, .... 15a; — 3x— 35Xl5=5a;; 
 Transposing, . . 15» — 'Ax — 5a;=35xl5> 
 Reducing, . . 7x==35xl5; 
 . . 1=5X15=75. 
 
 5. A laborer was engaged for 20 days. For each day that he 
 worked, he received 50 cents and his boarding; and for each day 
 that he was idle, he paid 25 cents for his boarding. At the expi- 
 ration of the time, he received $4 ; how many days did he work, 
 and how many days was he idle ? 
 
 Let . . x= the number of days he worked; 
 Then, . 20— a= " " " " was idle. 
 Also, . 50a;= wages due for work. 
 
 And . 25(20 — a;)= the amount to be deducted for boarding, 
 .-. 50av- 25(20— a;)=400; 
 50a;— 500+25x=400 ; 
 75a;=400-l-500=900; 
 
 a;=12=: the number of days he worked. 
 
 20-^=8= " ' " was idle. 
 
 Pkoof. 50x12=600 cents = wages; 
 25 X 8=200 " = boarding. 
 Diff. =400 " or $4. 
 In solving this example, we reduce the $4 to cents, in order 
 that all the quantities on both sides of the equation, may be of 
 the same denomination, it being regarded as a self-evident prin. 
 ciple, that we can only compare quantities of the same name. 
 Hence, all the quantities in both members of an equation, must be of 
 the same denomination. 
 
 6. What two numbers are as 3 to 5, to each of which, if 9 be 
 added, the sums shall be to each other as 6 to 7. 
 
 If we put X to represent the first number, the second will be 
 
 5a: 
 
 — . But we may avoid fractions by putting 3a; for the first num- 
 
 o 
 
 ber, and 5a; for the second, which fulfills the first condition. 
 
 Then, 3a:+9 : 5a;+9 : : 6 -7. 
 
 But in every proportion, the product of the means is equal to 
 
 the product of the extremes. (Arith. Part 3rd, Art. 209.) 
 
 Hence, 6(5x+9)=7(3a;+9). 
 
 30a;-|-54=21a;+63, 
 
 30a;— 21x=63— 54, 
 
EQUATIONS OF THE FIRST DEGREE. 91 
 
 9ar=9, 
 1=1, 
 .•. 3x=S, and 5x=:5. 
 The method of representing the quantities by Sx and 5x, so as 
 to avoid fractions, is of general application, and may be expressed 
 thus: — When two or more unknown quantities, in any problem, havt 
 to each other a given ratio, it is best to assume each of them a muU 
 tiple of some other unknown quantity, so that they shall have to each 
 other the given ratio. 
 
 7. A courier who traveled at the rate of SlA miles in 5 hours, 
 was dispatched from a certain city; 8 hours after his departure, 
 another courier was sent to overtake him. The second courier 
 traveled at the rate of 22\ miles in 3 hours. In what time did 
 he overtake the first, and at what distance from the place of 
 departureT 
 
 Let x=: the number of hours that the second courier travels. 
 Then, since the first courier travels at the rate of Sl^ miles in fy 
 hours, that is, -£1 miles in 1 hour, he will travel . — _ miles in x 
 
 hours, and since he started 8 hours before the second courier, the 
 whole distance traveled by him will be (8-4-k)4^. 
 
 Again, since the second courier travels at the rate of 22^ 
 miles in 3 hours, that is, -Y" miles in 1 hour, he will travel ^^x 
 miles in x hours. 
 
 But the couriers are supposed to be together at the end of the 
 time X, and, therefore, the distance traveled by each must be the 
 same; hence 
 
 450a;=(8-t-x)378; 
 .-. 72a;=378x8; divide each side by 8; 
 9a;=378; 
 a;=42. 
 Hence the second courier will overtake the first in 4S hours, 
 and the whole distance traveled by each is ^if-x42=315 miles. 
 
 8. A smuggler had a quantity of brandy, which he expected 
 would sell for 198 shillings; after he had sold 10 gallons, a 
 revenue ofiicer seized one third of the remainder, in consequence 
 of which he sells the whole for only 162 shillings. Required the 
 number of gallons he had, and the price per gallon. 
 
 Let x= the number of gallons; 
 
92 RAY'S ALtrEBRA, PART SECOND. 
 
 1 PR 
 then — is the price per gallon, in shillings; 
 
 X 
 
 ''~^" is the quantity seized, the value of which is 
 198—162=36 shillings. 
 
 . . ?=Hx^=36. 
 3 X 
 
 (x— 10)66=36a;, by clearing of fractions; 
 66a:— 660=36a;; 
 30a^=660; 
 .•. 1=22, the number of gallons; 
 
 and 1^=W=9 shillings, the price per gallon. 
 
 X 
 
 9. There are three numbers whose sum is 133; the second ii 
 twice the first, and the third twice the second. Required the 
 numbers. Ans. 19, 38, and 76. 
 
 10. There are three numbers whose sum is 187; the second is 3 
 times, and the third 4^ times, the first. Required the numbers. 
 
 Am. 22, 66, and 99. 
 
 11. There are two numbers, of which the first is 3^ times the 
 second, and their difference is 100. Required the numbers. 
 
 Ans. 40 and 140. 
 
 12. There are three numbers, whose sum is 156; the second 
 is 3 J times the first, and the third is equal to the remainder left, 
 after subtracting the difference of the first and second from 100. 
 Required the numbers. Ans. 28, 98, and 30. 
 
 13. What number is that, whose half, third, and fourth parte, 
 taken together, are equal to .52'! Ans. 48. 
 
 14. What number is that, which being increased by its six' 
 sevenths, and diminished by 20, shall be equal to 45'! Atis. 35. 
 
 15. What number is that, to which if its third and fourth parts 
 be added, the sum will exceed its sixth part by 51"! Ans. 36. 
 
 16. Find a number which, being multiplied by 4, becomes as 
 much above 40 as it is now below it. Ans. 16. 
 
 17. What number is that, to which if 16 be added, 4 times 
 the sum will be equal to 10 time3 the number increased by IT 
 
 Ans. 9. 
 
 18. The sum of two numbers is 30; and if the less be sub- 
 tracted from the greater, one-fourth of the remainder will be 3. 
 Required the numbers. Ans. 9 and 21. 
 
 19. A laborer was engaged for 28 days, upon the condition 
 that for every day he worked he was to receive 75 cents, and for 
 
EQUATIONS OF THE FIRST DEGREE. 93 
 
 every day he was absent, to forfeit 25 cents. At the end of hi» 
 time he received $12. How many days did he workT Ans. 19. 
 
 20. A has three times as much money as B, but if B give A 
 $50, then A will have four times as much as B. Find the money 
 of each. Ans. A, $750; B, $250. 
 
 21. From a bag of money which contained a certain sum, 
 there was taken $20 more than its half; from the remainder, 
 $30 more than its third part; and from the remainder, $40 more 
 than its fourth part, and then there was nothing left. What sum 
 did it contain"! Ans. $290. 
 
 22. A merchant gains the first year, 15 per cent, on his cap- 
 ital; the second year, 20 per cent, on the capital at the close of 
 the first; and the third year, 25 per cent, on the capital at the 
 close of the second; when he finds that he has cleared $1000.50. 
 Required his capital. Ans. $1380. 
 
 23 . A is twice as old as B ; 22 years ago, he was three times 
 as old. What is A's age? Ans. 88. 
 
 24. A person buys 4 houses; for the second, he gives half as 
 much again as for tiie first; for the third, half as much again as 
 for the second; and for the fourth, as much as for the first and 
 third together: he pays $8000 for them all. Required the cost 
 of each. Ans. $1000, $1500, $2250, and $3250. 
 
 25. A cistern is filled in 24 minutes by 3 pipes, the first of 
 which conveys 8 gallons more, and the second 7 gallons less, 
 than the third every 3 minutes. The cistern holds 1050 gal- 
 lons. How much flows through each pipe in a minute? 
 
 Ans. 17/g, 123^, 141^. 
 
 26. A can do a piece of work in three days, B in 6 days, and 
 C in 9 days. Find the time in which all together can perform it. 
 
 Let x= the required number of days; 
 
 then - = part of the work performed by all in qne day. 
 But A does -5, B g, and C \, in one day. 
 
 ••• i+g+i=^- ^'"^ ^tV days- 
 
 27 If A does a piece of work in 10 days, which A and B can 
 do together in 7 days; how long would B take to do it alone? 
 
 Ans. 23| days. 
 28. A performs f of a piece of work in 4 days; he then re- 
 eeives the assistance of B, and the two together finish it In 6 
 days. -^Required the time in which each can do it alone. 
 
 Avs. A, 14 days; B, 21 days. 
 
94 RAY'S ALGEBRA, PART SECOND. ^ 
 
 29. A person bought an equal number of sheep, cows, and 
 oxen, for $330; each sheep cost $3, each cow $12, and each oJi 
 $18. Required the number of each. Ans. 10. 
 
 30. A sum of money is to be divided among five persons; A, 
 B, C, D, and E. B received $10 less than A; C, $16 more than 
 B; D, $5 less than C; E, $15 more than D; and the shares of 
 the last two are equal to the sum of the shares of the other 
 three. Required the share of each. 
 
 Ans. A, $21; B, $11; C,$27; D,$22; E,.$37. 
 
 31. A bought eggs at 18 cents a dozen, but had he bought 5 
 more for the same money, they would have cost him 2l cents a 
 dozen less. How many eggs did he buyl Ans. 31. 
 
 32. A person bought a certain number of sheep for $94; 
 having lost 7 of them, he sold one-fourth of the remainder at 
 prime cost, for $20. How many sheep had he at first .' 
 
 Ans. 47. 
 
 33. There are two places, 154 miles distant from each other, 
 from which two persons, A and B, set out at the same instant, to 
 meet on the road. A travels at the rate of 3 miles in 2 hours, 
 and B at the rate of 5 miles in 4 hours. How long, and how far, 
 did each travel before they met] 
 
 Atis. 56 hours, and A traveled 84, and B, 70 miles. 
 
 34. Find that number, which, multiplied by 5, and 24 taken 
 from the product, the remainder divided by 6, and 13 added to 
 the quotient, will still give the same number. Ans. 54. 
 
 35. In a bag containing eagles and dollars, there are three 
 times as many eagles as dollars ; but if 8 eagles and as many 
 dollars be taken away, there will be left five times as many 
 eagles as dollars. How many were there of each? 
 
 Ans. 48 eagles, 16 dollars. 
 
 36. If 10 apples cost a. cent, and 25 pears cost 2 cents, and 
 you buy 100 apples and pears for Oj cents, how many of each 
 will you have] Ans. 75 apples and 25 pears. 
 
 37. Suppose that for every 8 sheep a farmer keeps, he should 
 plough an acre of land, and allow one acre of pasture for every 
 5 sheep, how many sheep may he keep on 325 acres'! 
 
 Ans. 1000. 
 
 38. A person has just 2 hours spare time; how far may he ride 
 In a stage which travels 12 miles an hour, so as to return home 
 in time, walking back at the rate of 4 miles an hour. 
 
 Ans. 6 miles. 
 
 39. If 651b of sea-water contain 25b of salt, how much fresh 
 watRr must be added to these 65ft, in order that the quantitv of 
 
EQUATIONS OF THE FIRST DEGREE. 95 
 
 Bait contained in 25fti of the new mixture, shall be reduced to 
 ■1 ounces, or \ of a lb. Ans. 135Sj. 
 
 40. A mass of copper and tin weighs 801b; and for every 7H> 
 of copper, there are 3!b of tin. How much copper must be 
 added to the mass, that for every 111b of copper, there may be 
 41b of tin? Atis. IOBj. 
 
 41. A merchant maintained himself for 3 years, at a cost of 
 $250 a year; and in each of those years, augmented that part of 
 his stock which was not so expended, by -| thereof. At the end 
 of the third year his original stock was doubled. What was that 
 stock? Ans. $3700. 
 
 SIMULTANEOUS EaUATIONS OF THE FIRST DEGREE, CONTAINING 
 TWO UNKNOWN aUANTITIES. 
 
 Akt. 155. From what we have already seen, it is evident that 
 the value of any one of the symbols concerned in an equation, is 
 entirely dependent on the rest, and it can become known, only 
 when the values of the rest are given, or known Thus, in the 
 equation 
 
 x-\-y=a, 
 the value of x depends on the values of y and u, and can only 
 become known when they are known; therefore, to find (he value 
 of any unknown quantity, we must Main a single equation contain- 
 ing it and known quantities. Hence, when we have two or more 
 equations containing two or- more unknown quantities, we must 
 obtain from them a single equation containing only one unknown 
 quantity. The method of doing this is termed elimination, which 
 may be defined briefly, thus: — Elimination is the process of deduc- 
 ing, from two or more equations containing two or more un- 
 known quantities, a single equation containing only one unknown 
 quantity. 
 
 There are three principal methods of elimination: 
 
 1st. Elimination by substitution. 
 
 2nd. Elimination by comparison. 
 
 3rd. Elimination by addition and subtraction. 
 
 ELIMINATION BY SUBSTITUTION. 
 
 Art. 156. Elimination by substitution consists in finding the 
 value of one of the unknown quantities in one of the equations, 
 in terms of the other unknown quantity and known terms, and 
 substituting this, instead of the quantity, in the other equaticn 
 
66 RAY'S ALGEBRA, PART SECOND. 
 
 To explain this method, let it be required to find the values of* 
 ind y, in tlie following equations. 
 
 2a;+3y=33, (1) 
 4j:+5^=59. (2) 
 From eq. (1), by transposing 3y and dividing by 2, vye have 
 ^3— 3y . 
 '^ 2 
 Substituting this value of x, instead of a; in eq. (2), we have 
 
 or, 66 — 6y+5y=59; 
 -y=-7; 
 
 and a;=^^^=6. 
 The following is the general form to which two equations of 
 the first degree, containing two unknown quantities, may always 
 be reduced. The signs of the known quantities, a, b, c, &c., may 
 be either plus or minus. 
 
 ax+bi/=c, (1) 
 a'x-\-h'y=:c'. (2) 
 From eq. (1), by transposing hy, and dividing by a, we have 
 
 a 
 Substituting this value of x in eq. (2), we have 
 
 a'c — a'ly-\-ah'y=ac! ; 
 
 (oJ' — db)y=ac' — a'c; 
 y_ac'—a'c 
 
 ab' — ab 
 But 
 
 ,, ( ad — a'c \ 
 ■=':Z^=z [ ab'—a'b I . J^'o—a'ie—abc'-^-a be 
 - a(ab' — a'6) 
 
 a 
 
 _ b'c—bc' 
 ab' — a'6' 
 Hence, when we have two equations, containing two unknown 
 quantities, we have the following 
 
 Rule fok elimination bt sttbstitction. — Find an expression 
 for the value of one of the uriknown quantities in either equation, 
 and substitute this valve, instead of the same unknown quantity, m 
 the other equation; there wiU thus be formed a new equation, con- 
 taining only ome unJcnoim quantity. 
 
EQUATIONS OF THE FIRST DEGREE. 91 
 
 ELIMINATION BY COMPARISOJJI. 
 
 Art. 157. Elimination by comparison consists in finding the 
 value of the same unlinovvn quantity in two different equations, 
 and then placing these values equal to each other. 
 
 To illustrate this method, we will take the same equations as 
 tn the preceding article. 
 
 2^+3j/=33, (1) 
 ^ 4x+52/=59. (2) 
 
 From eq. (1), by transposing and dividing, we have x= ^. 
 
 lit 
 
 From eq. (2), by transposing and dividing, we have x=' " 
 
 4 
 
 Placing these values of x equal to each other, 
 59— 5y 33— 3y 
 "^~=~2— ' 
 
 59 — 52/=66 — 6y, by clearing of fractions; 
 v=7, by transposition. 
 
 The value of x may be found similarly, by first finding the 
 values of y, ai.d placing them equal to each other. But after 
 finding the value of one of the unknown quantities, that of the 
 other may generally be found most readily by substitution. Thus 
 
 4x+5x'7=59; 
 
 whence »=^^=i^=6. 
 
 4 
 
 General equations, ax-\-by=c, (1) 
 a'x-\-b'y=c'. (2) 
 
 From eq. (1), by transposing and dividing, x=' ^, 
 
 a 
 
 From eq. (2), by transposing and dividing, j=° V; 
 
 a' 
 equating these values of x, 
 
 c — iy^c' — Vy^ 
 
 ~r — * 
 
 a a 
 
 a'c — a'by=ac' — ab'y, by clearing of jractiona; 
 
 (ab' — a'h)y=ac' — a'c, by transposing; 
 
 __ac' — a'c 
 
 ab' — a'b ' 
 
 from eq. (1), y='^; 
 From eq. (2), 3P=?^; 
 
 n 
 
88 RAY'S ALGEBRA, PART SECOND. 
 
 equating these values of y, 
 
 c — a'x c — ax. 
 
 V b ■ 
 
 be' — a'bx=b'c — ab'x ; 
 
 {ah' — a'b)x=b'c- — be' ; 
 
 b'c—bc' 
 
 x= . 
 
 ab' — a'b 
 
 Hence, when we have two equations, containing two unknown 
 quantities, we have the following 
 
 Rule for elimination by compakison. — FtTid an expression 
 for the value of the same unknovm quantity in eaeh of the given 
 equations, and place these values equal to each other; there will thus 
 he formed a neto equation, containing only one unknown quantity. 
 
 ELIMINATION BY ADDITION AND SUBTRACTION. 
 
 Aet. 158. Elimination by addition and subtraction consists 
 in multiplying or dividing two equations, so as to render the co- 
 efficient of one of the unknown quantities, the same in both; 
 and then, by adding or subtracting, to cause the terms Containing 
 it to disappea'. 
 Taking the same equations as in the preceding articles, 
 2x+Zy=33, (1) 
 4x+5y=59. (2) 
 It is evident that if we multiply eq. (1) by 2, that the coeffi 
 cient of x will be the same in the two equations. 
 
 4a;+6y=66 (3), by Xing eq. (1) by 2. 
 4x-\-by=:!)9, eq. (2) brought down. 
 
 Since the coefficients of x have the same sign in these equa- 
 tions, if we subtract, the terms containing x will cancel each other, 
 and the resulting equation will contain only y, the value of which 
 may then be found. It is evident that if the signs of the coeffi- 
 cients of X had been different, that by adding, it would have been 
 canceled. 
 
 Having obtained the value of y, that of x may be obtained by 
 substitution, or similar to that of y, as follows: 
 
 It is evident that if we multiply eq. (1) by 5, and eq. (2) by 3, 
 that the coefficients of ;/ will be the same in both. 
 
 10a;+15y=165, (4) by Xing eq. (1) by 5. 
 12j;+1.5y=:177, (5) by Xing eq. (2) by 3. 
 2as=12, by subtracting eq. (4) from (5). 
 
 a=6. 
 
EQUATIONS OF THE FIRST DEGREE. 09 
 
 General equations, ax-\-hy=^c, (1) 
 
 a'x-f6'y=o'. (2) 
 
 It is evident that we shall render the coefficients of x the same 
 
 in both equations, by multiplying eq. (1) by a, and eq. (2) by a. 
 
 aa'x-\-a'hy^a'c, (3) by Xing eq. (1) by a'; 
 
 aa'x-\-ab'y=ac', (4) by Xing eq. (2) by a; 
 
 (ab' — a'b)y=ac' — a'c, by subtracting; 
 
 ac' — a'c 
 
 y= 
 
 ah' — ab' 
 
 The coefficients of y in the two equations will evidently be- 
 come equal by multiplying eq. (1) by V, and eq. (2), by b. 
 (d>'x-\-bb'y=b'c, (5) by X>ng eq. (1) by b'; 
 a'bx-\-bb'y=bc', (6) by Xing eq. (2) by b; 
 (ab' — a'b)x=b'c — be, by subtracting; 
 
 b'c — be' 
 
 x= . 
 
 ab' — a'b' 
 
 It is evident that after we have rendered the coefficients of the 
 quantity to be eliminated the same in both equations, if the signs 
 are alike we must subtract ; but if they are urdike we must add. 
 
 Hence, when we have two equations containing two unknown 
 quantities, we have the following 
 
 Rule, tor elimination bt addition and subtraction. — Multi- 
 ply, or divide the equalions, if necessary, so that one of the unknown 
 quantities will have the same coefficient in both. Then take the dif- 
 ference, or the sum of the equalions, according as the signs of the 
 equal terms are alike or unlike, and the resulling equation will con- 
 tain only one unknown quantity. 
 
 Remark. — When the coefficients of the quantity to be eliminated are 
 prime to each other, they may be equated by multiplying each equation 
 by the coefficient of the unknown quantity in tne other. When the 
 coefficients are not prime, find their least common multiple, and multiply 
 each equation by the quotient obtained by dividing the least common 
 multiple by the coefficient of the unknown quantity to be eliminated in 
 the other equation. 
 
 If the equations have fractional coefficients, they ought to be cleared, 
 before applj ing the rule. 
 
 EXAMPLES FOR PRACTICE. 
 
 Note. — It is recommended to the pupil to solve several of the fol- 
 lowing examples, by each of the preceding rules. 
 
 I. a!+3j(=10, Anjs. x=\, I 2. 2a;+3jf=18, Ans. 1=3, 
 
 3a:+2y=9. j/=3. I 3r— 2y=l. y=^. 
 
100 
 
 RAY'S ALGEBRA, PART SECOND. 
 
 3. 2a>-9y=ll, 
 3a:— 12y=15. 
 
 4 Zx—ly=T, 
 lla;+5y=87. 
 
 6. 9x— 42/=8, 
 13a;+7y=101. 
 
 Ans. a:=:l, 
 y=-l. 
 
 Ans. x=7, 
 y=2. 
 
 Ans. x=4, 
 y=7. 
 
 fix — 4(y — ^2)=5, Ans. 0=5, 
 
 7.? -1-^=8, Ans. x=:=18, 
 
 3^5 
 
 y=10. 
 
 8. l^±5y=x-y, A»w. x=i, 
 '^+2,= -. y=.. 
 
 51—33^=0. 
 
 Ans. a;=99 
 y=15 
 
 10. 
 
 11. 
 
 4 =_^, Ans. x=2, 
 
 *_1-=1 
 9 10 
 
 12. 2z-J±?=7+?y=^, 
 4 ^6 
 
 4y+^=26i-?d:i. 
 
 13 3x-4-4y+3 2x+'}—y _^-_^J-9 
 10 15 ^ 5 ' 
 
 9y+5x—a_x+y_'Tx+6 
 12 ~4 11 * 
 
 14. 
 
 5+1/ 12+a; 
 2a:+5y=35 
 
 3y+2 7' 
 8a!-4=9y. 
 
 3 I 
 
 Ans. j;^5, 
 
 y=4 
 Ans. a;=5, 
 
 Ans. a:=7 
 
 a;4-y=c. 
 
 15. a^-ay=6, 
 aa; — hy=c. 
 
 16. 3ax— 2iy=c, 
 o'a;+J'y=5fie. 
 
 17. (a — 6)i+(o-}-5)y=c, 
 (a'— 4')(x-4-y)=», 
 
 Ans. . 
 
 y=9 
 be 
 
 Am. i 
 
 a+b 
 
 ac 
 
 ~a+b' 
 _flc-|-4' 
 
 o2+i 
 db — c 
 
 Ans. x=. 
 
 a'+V 
 lite 
 
 18. .^+^=a 
 
 X 
 
 n I 77t 
 
 y J Apply Rule, / 
 
 l Art, 158. J 
 
 2a2+3a6 
 c(15b-a\ 
 ^ b\2a+3bl 
 
 Ans. x=:L ( JL-c] 
 26 V a—b I 
 
 " 26 I a+b I 
 Atw. 
 
 wifl — n6 
 
 nib — nd 
 
EQUATIONS OP THE FIKST DEGREE. 101 
 
 19. f+2?=l-f, j„o ^ abc(.ab+ao-bc) 
 a~b c a^b^+aV—bV 
 ^_L?=14-y abc(ac—ab—bc') 
 a^b c * o2J2_|_a5c2— 6=c=* 
 
 20. (ffl»— i2)(5a:+3y)=(4<t-i)2a6, An*. x=z—, 
 
 a-\-b 
 
 a'j/— i5^+(a+ft-|-c)ia=J=j/+(a+2i)oJ. y=—-. 
 
 a-\-b a — V 
 
 Remark. — Transpose bh) in eq. (2), multiply by 3 and subtract, than 
 will then result an equation involving x. 
 
 aUESTIONS PRODUCING SIMULTANEOUS EQUATIONS CONTAIN- 
 ING TWO UNKNOWN QUANTITIES. 
 
 Art. 159. The questions contained in Art. 154, are all capa- 
 ble' of being solved by using one unknown quantity; although in 
 some of the examples, the number of unknown quantities was 
 two or more. But in those questions where there was more than 
 one unknown quantity, there was such a relation existing between 
 the several quantities, that it was easy to express each one in 
 terms of the other. It frequently happens, however, that in a 
 problem containing more than one unknown quantity, there may 
 be no direct relation existing between them, by means of which 
 either may be found in terms of the other. In such a case it 
 becomes necessary to use a separate symbol for each unknown 
 quantity, and then to find equations containing these symbols, on 
 the same principle as when there was but one unknown quantity; 
 that is, in brief, regard the symbols as the answer to ihejfuestion, 
 and then proceed in the same manner which it would be necessary to 
 do to prove the answer. After the equations are obtained, the 
 values of the unknown quantities may be found, by either of the 
 three different methods of elimination. 
 
 We shall now give an example, to show that the same question 
 may sometimes be solved by using either one or two unknown 
 quantities. 
 
 1. The difference of two numbers is o, and the lest 'a to th» 
 greater as m to n'; required the numbers. 
 
 Solution by using one unknown quantity. 
 
 Let mx= the less number, then nx^ the greater; 
 
 And nx — mx=a. 
 
102 RAY'S ALGEBRA, PART SECOND. 
 
 the less number; 
 
 n — m 
 
 7tx=-— , the greater number. 
 n — m 
 
 Solution by using two unknown quantities. 
 Let x= the less number, and v= the greater. 
 Then, y— 1=0, (1) 
 and X : y : : m : n; or my=nx. (2) 
 
 Since my=nx, we have y=: — ; 
 m 
 
 substituting this value of x in eq. (1), 
 
 nx , ^ 
 
 m, 
 
 nx — mx^ma; 
 
 whence a;= , 
 
 and y= — a; 
 
 n — 771 
 mna 
 
 m m(n — m) n — m 
 
 2. The hour and minute hands of a watch are opposite at 6 
 o'clock; when are they next opposite) 
 
 Let x= minute spaces moved over by the hour hand, and y= 
 minute spaces moved over by the minute hand. Then since the 
 minute hand moves 12 times as fast as the hour hand, 
 x:y::l: 12, or jf=12a:. (1) 
 But the minute hand must evidently pass over 60 minutes more 
 than the hour hand; hence 
 
 y=x+60. (2) 
 • Substituting, 12a;=a;-|-60, 
 11j:=60, 
 x=5^x min. 
 
 y=65/-,min.^lh., 5/ym. 
 Hence, the hands are next opposite at 5-,^jm. past 7. 
 In a similar manner the period of coincidence of the hands 
 may be found. 
 
 3. There is a number consisting of two digits, which divided 
 by the sum of its digits, gives a quotient 7; but if the digits be 
 written in an inverse order, and the number thence arising be 
 divided by the sum of the digits increased by 4, the quotient =3. 
 Required the number. Ans. 84. 
 
 In solving questions of this kind, the pupil must observe that 
 any number consisting of two places of figures, is equal to 10 
 times the tigm-e in the ten's place plus the figure in the unit's 
 
EQUATIONS or THE FIRST DEGREE. 103 
 
 place. Thus, 35 is equal to 10x3-|-5. In a similar manner 
 458 is equal to 100x4+10x5+6. 
 
 Let x= the digit in ten's place, and y= the digit in unit's 
 place. 
 Then lOx+y^the number. 
 
 And 10y+x=the number when the digits are reversed. 
 Also, 10x+y _^ 10y+x _^ 
 
 x+y ' x+y+4: 
 
 From these equations we readily find a;=8, and y=4. 
 
 4. A farmer sells to one man 5 sheep and 7 cows for $111, 
 and to another, at the same rate, 7 sheep and 5 cows for $93. 
 Required the price of a sheep and that of a cow. 
 
 Ans. Sheep, $4; cow, $13. 
 
 5^ If 7Hj of tea and 91b of coffee cost $5.20, and at the same 
 rate 41b of tea and llJb of coffee cost $3.85; it is required to 
 find the price of a pound of each. Ans. Tea, 55c.; coffee, 15c. 
 
 6. A and B are in trade together with different sums; if $50 
 be added to A's money, and $20 be taken from B's, they will have 
 the same sum; but if A's money were 3 times, and B's 5 times 
 as great as each really is, they would have together $2350. 
 How much has each] Ans. A, $250; B, $320. 
 
 7. A and B have together $9800; A invests the sixth part of 
 his money in business, and B the fifth part, and then each has the 
 same sum remaining. How much has each? 
 
 Ans. A, $4800; B, $5000. 
 Let 6a;^A's money, and 52/=B's. 
 
 8. What fraction is that, such that if the numerator and de- 
 nominator be each increased by 1, the value is I; but if each be 
 diminished by 1, the value is ^1 Ans. f. 
 
 9. Find two numbers, such that one-third of the first exceeds 
 one-fourth of the second by 3, and one-fourth of the first and one- 
 fifth of the second are together equal to 10. Ans. 24 and 20. 
 
 10. A grocer knows neither the weight nor the first cost of a 
 box of tea which he had purchased. He only recollects that if 
 he had sold the whole at 30 cts. per Hi, he would have gained $1, 
 but if he had sold it at 22 cts. per lb, he would have lost $3. 
 Required the number of pounds in the box, and the first cost 
 per ib. Ans. 501b at 28 cts. 
 
 11. The rent of a field is a certain fixed number of bushels of 
 wheat, and a fixed number of bushels of corn. When wheat is 
 55 cents, and corn 38 cents per bushel, the portions of rent by 
 wheat and corn are equal; but when wheat is 65 cents and corn 
 
104 RAY'S ALGEBRA, PART SECOND. 
 
 41 cents per bushel, the rent is increased by $1.40. What ie 
 the grain rent? 
 
 Ans. 6 bushels of wheat and 10 bushels of com. 
 
 12. The quantity of water which flows from an orifice is pro- 
 portional to the area of the orifice, and the velocity of the water. 
 Now there are two orifices in a reservoir, the areas being as 5 to 
 13, and the velocities as 8 to 7, and from one there issued in a 
 certain time 561 cubic feet more than from the other. How 
 much water did each orifice discharge in this time? 
 
 Ans. 440 and 1001 cubic feet. 
 
 13. Find two numbers in the ratio of 5 to 7, to which two 
 other required numbers in the ratio of 3 to 5 being respectively 
 added, the sums shall be in the ratio of 9 to 13; and the differ- 
 ence of those sums =16. Ans. 30 and 42, and 6 and 10. 
 
 14. A boy spends 30 cents in apples and pears, buying his 
 apples at 4 and his pears at 5 for a cent; he then finds that half 
 his apples and one-third of his pears cost 13 cents. How many 
 of each did he buy? Ans. 72 apples and 60 pears. 
 
 15. A farmer rents a farm for $245 per year; the tillable land 
 being valued at $2 per acre, and the pasture at $1.40; now the 
 number of acres of tillable, is to half the excess of the tillable 
 above the pasture, as 28 to 9. How many acres are there of 
 each? Ans. 98 acres tillable, and 35 of pasture. 
 
 16. Find that number of 2 figures to which, if the number 
 formed by changing the places of the digits be added, the sum is 
 121; and if the less of the same two numbers be taken from the 
 greater, the remainder is 9. Ans. 65. 
 
 17. To determine three numbers such that if 6 be added to the 
 first and second, the sums will be in the ratio of 2 to 3; if 5 be 
 added to the first and third, the sums will be in the ratio of 7 :11; 
 but if 36 be subtracted from the second and third, the remainders 
 will be as 6 to 7. Ans. 30, 48, 50. 
 
 Suggestion. — Let 2x— 6, 3av— 6, and y be the numbers. 
 
 18. Two persons, A and B, can perform a piece of work in 16 
 days. They work together for 4 days, when A being called ofl', 
 B is left to finish it, which he does in 36 days more. In whit 
 time could each do it separately? Ans. A in 24, B in 48 days. 
 
 19. A and B drink from a cask of beer for 2 hours, after which 
 A falls asleep, and B drinks the remainder in 2 hours and 48 
 minutes; but if B had fallen asleep and A had continued to drinkj 
 it would have taken him 4 hours and 40 minutes to finish the 
 cask. In what time could each singly drink the whole? 
 
 Avs. A in 10 hrs., B in 6 hrs. 
 
EQUATIONS OF THE FIRST DEGREE. 105 
 
 20. Divide the fraction J into two parts, so that the numera- 
 tors of the two parts taken together shall be equal to theii 
 denominators taken together. Ans. ^ and j J. 
 
 21. A purse holds 19 crowns and 6 guineas. Now 4 crowns 
 and .5 guineas fill gj of it. How many of each will it hold! 
 
 Ans. 21 crowns or 63 guineas. 
 
 22. When wheat was 5 shillings a bushel, and rye 3 shillings, 
 a man wanted to fill his sack with a mixture of rye and wheat 
 for the money he had in his purse. If he bought 7 bushels of 
 rye and laid out the rest of his money in wheat, he would want 
 2 bushels to fill his sack; but if he bought 6 bushels of wheat, 
 and filled his sack with rye, he would have 6 shillings left. How 
 must he lay out his money, and fill his sack? 
 
 Ans. He must buy 9 bushels of wheat, and 12 bushels of rye. 
 
 SIMULTANEOUS EOUATIONS OF THE FIRST DEGREE, INVOLVING 
 THREE OR MORE UNKNOWN aUANTITIBS. 
 
 Art. 160. Simultaneous equations of the first degree involv- 
 ing three or more unknown quantities, may be solved by either 
 of the three methods of elimination, explained in Arts. 15.5 to 
 159; but the method most generally applicable, is that of elim- 
 ination by addition and subtraction, which we shall now proceed 
 to apply in the solution of a problem containing three unknown 
 quantities. 
 1. Given 5x— 4j/+22=48, (1) 
 3a;+3y— 4z=24, (2) 
 2a;— 52/+32=19, (3) to find a;, y, and 2. 
 To eliminate 2 from the first two equations, we may multiply 
 eq. (1) by 2, and then add this to eq. (2); thus, 
 
 10x—8!/-f 42=96, by Xing eq. (1) by 2, 
 3x+3y— 4z=2 4, (2) 
 ISx— 5]/ =120, (5) by adding. 
 We may eliminate z from equations (1) and (3), by multiplying 
 eq. (1) by 3, and eq. (3) by 2, and then subtracting; thus, 
 15a;— 12y+62=144, by Xing eq. (1) by 3, 
 4a;— 10y-j-6z= 38, by Xing eq. (3) by 2, 
 llx— 2y =106, (6) by subtracting. 
 We may now eliminate y from equations (5) and (6), by muJi 
 tiplying eq. (5) by 2, and eq. (6) by 5, and then subtracting; thus, 
 26x— 10y=240, by Xing eq. (5) by 2, 
 55x— 10i/=530 , by Xing eq. (6) by 5, 
 
 29x =290; 
 
 x=in. 
 
106 RAY'S ALGEBRA, PART SECOND. 
 
 110—2^=106, by substituting 10 for x in eq. (6); 
 whence i/=2. 
 50 — 8-^2z='k8, by substituting for x and y in eq. (1); 
 whence 2=3. 
 It is evident that the same method may be applied when the 
 number of equations is four or more. Hence we derive the 
 following 
 
 General rule for elimination by addition and subtkao 
 Tios. — 1st. Combine any one of the equations with each of the 
 others, so as to eliminate the same unknoum quantity; there will 
 thus arise a 7i£w class of eqtiations, containing one less unknoum 
 quantity. 
 
 2nd. Combine any one of these new equations with each of the others, 
 so as to eliminate another unknown quantity; there will thus arise 
 another class of equations, containing two less unknown quantities. 
 
 3rd. Continue this series of operations until a single equation is 
 obtained, containing but one unknown quantity, from which its 
 value may be easily found; then by going back, and substituting 
 this value in the derived equatioTis, the values of the other unknown 
 quantities may be readily found. 
 
 Remark. — Although the method of elimination by addition and sub- 
 traction is generally the best when the number of unknown quantities 
 is three or more, yet in some particular instances, solutions may be ob- 
 tained more easily and elegantly by other means, which the pupil must 
 acquire by experience and tact. As a specimen, we present the follow- 
 ing question and solution. 
 
 2. Given — x-\-y-\-z=a, (1) 
 x-y+z=b, (2) 
 
 x-{-y — z=zc, (3) to find ui, y, and z. 
 By adding the three equations together, and calling a-f-J-|-c^», 
 we find 
 
 x-\-y-\-z=s. (4) 
 Then by subtracting eqs. (1), (2), and (3), respectively from 
 (4), and dividing by 2, we find 
 
 x=l(s—a), 
 
 2=i(s— c). 
 
 EXAMPLES, 
 
 To be solved by either of the different methods of elimination. 
 Z. x-\-y-\-z=Q, 1 Ans.x=l, 
 
 3i— y+22=7,> y=2, 
 
 Ax+Sy—z=7.} 2=3. ' 
 
EQUATIONS OF THE FIRST DEGREE. 107 
 
 4. 3x-\-4y—5z=32,^ Ajis. a;=10 
 4J^— 5y4-3z=18> y=8, 
 5x—5y—iz= 2.) z=6. 
 
 5. x—9y+2z—l0u=21, \ Ans. x=lOO, 
 2x+7y—2—u=682, [ ~ j/=60, 
 3x-|-y+52+2a=195, [ 2=— 13, 
 4x—6y—2z—9u=516.J u=—50. 
 
 6. x+iy=10—lz, \ Ans. ib=1, 
 \(x-^)=2y-'l .) 2=3. 
 
 ^•^+^=1+5, "j An!.x=5, 
 
 X — 1 y — 2 2+3 
 
 ~4 5 W 
 
 3 ^* 12 
 
 y=7. 
 
 =11-45.) «=-3. 
 
 8. 9i— 22+M=41, \ Ans. x=5, 
 1y—bz—t=l2, I 3/=4, 
 4i/— 3x+2m=5, V 2=3, 
 83/— 4k+3«=7, \. «=2, 
 
 72— 5a=ll.J <=1. 
 Examples, to be solved by special methods. 
 
 9. - + -=a, 1 Ans. x= 
 
 - + -=a, ) 
 
 - + -=o, > 
 x z I 
 
 1,1 \ 
 
 a-\-b — c 
 - + -=*'> y= ^ , 
 
 2 
 
 y 2 ft+c— «• 
 
 Suggestion. — Subtrsct eq. (3) from (2), then combine the 
 resulting equation with (1), to find x and y; z may be found 
 similarly. 
 
 10. __^ ^^-f5' ) Ans.ic^S, 
 
 
 a; V 2 ^ ' 
 
108 RAY'S ALGEBRA, PART SECOND. 
 
 11.? 5 1 
 
 
 Ans. x=6 
 
 r.+J+H-4 
 
 
 y=9. 
 
 5 l,l-.4 3 3.\ 
 
 6x i^z ^« J 
 
 
 z=h 
 
 12. —x-^y+z+v—a,-\ 
 
 Arts. 
 
 ■ x=h{s—a), 
 
 x—y+z+v=b, [ 
 
 
 y=iis-b), 
 
 x+y—z-^-v=c, C 
 
 
 z=i(s—c), 
 
 x-\-y-\-z — v=d. -' 
 
 
 »— a(s-<0. 
 
 
 where s=l{a-\-b-\-c-\-d). 
 
 aUESTIONS PRODUCING SIMULTANEOUS EQUATIONS CONTAIN 
 DfG THREE OR MORE UNKNOWN QUANTITIES. 
 
 Art. 161. When a question contains three or more unknown 
 quantities, equations involving them can be found on the same 
 principle as in questions dontaining OTie or two unknown quanti- 
 ties. (See Arts. 154 and 159.) The values of the unknown 
 quantities may then be found, in the same manner as in the pre- 
 ceding examples. 
 
 1 . The stock of three traders amounts to $760; the shares of 
 the first and second exceed that of the third by $240; and the 
 sum of the second and third exceeds the first by $360: what is 
 the share of each) Ans. $200, $300, and $260. 
 
 2. What three numbers are there, each greater than the pre- 
 ceding, whose sum is 20, and such that the sum of the first and 
 second is to the sum of the second and third, as 4 is to 5; and 
 the difierence of the first and second, is to the difference of the 
 first and third, as 2 to 3 ! Ans. 5, 7, and 8. 
 
 3. Find four numbers, such that the sum of the first, second, and 
 third, shall be 13; the sum of the first, second, and fourth, 15; 
 the sum of the first, third, and fourth, 18; and lastly, the sum of 
 the second, third, and fourth, 20. Ans. 2, 4, 7, 9. 
 
 4. The sum of three digits composing a certain number is 16- 
 the sum of the left and middle digits, is to the sum of the middle 
 and right ones as 3 to Sj; and if 198 be added to the number, 
 the digits will be inverted. Required the number. Atis. 547. 
 
 5. At an election where each elector may give two votes to 
 difierent candidates, but only one to the same, it is found on 
 counting the votes, that of the candidates A B, C, A had 158 
 votes, B had 132, and C 58. Now 26 voted for A onlv, 30 for 
 
EQUATIONS OV THE FLRST DEGREE. 109 
 
 B only, and 28 for C only. How many voted for A and B 
 jointly; how many for A and C; and how many for B and C] 
 Ans. For A and B, 102; A and C, 30; B and C, 0. 
 
 6. It is required to find three numbers such, that l of the first, -3- 
 of the second, and \ of the third, shall together make 46 ; 3 of 
 the first, I of the second, and 5 of the third, shall together 
 make .S.5; and | of the first, 1 of the second, and 5 of the third, 
 shall together make 283. Ans. 12, 60, and 80. 
 
 7. The sum of three numbers, taken two and two, are a, b, 
 and c. What are the numbers'! 
 
 Ans. 2(0+6— c), |(a-|-c — 6), and j(i+c — a). 
 
 8. A person has four casks, the second of which being filled 
 from the first, leaves the first four-sevenths full. The third being 
 filled from the second, leaves it one-fourth full; and when the 
 third is emptied into the fourtK, it is found to fill only nine-six- 
 teenths of it. But the first will fill the third and fourth and have 
 fifteen quarts remaining. How many. quarts does each holdl 
 
 Ans. 140, 60, 45, and 80, respectively. 
 
 9. In the crew of a ship consisting of sailors and soldiers, 
 there were 22 sailors to every 3 guns, and 10 sailors over; also 
 the whole number of hands was 5 times the number of soldiers 
 and guns together; but after an engagement, in which the slain 
 were one-fourth of the survivors, there wanted 5 men to be 13 
 men to every 2 guns. Required the number of guns, soldiers, 
 and sailors. Ans. 90 guns, 55 soldiers, 670 sailors. 
 
 CHAPTER V- 
 
 SUPPLEMENT TO EQUATIONS OF THE 
 FIRST DEGREE. 
 
 I. GENERALIZATION. 
 
 Art. 162. Equations are termed ZiferaZ when the known quan- 
 tities are represented, either entirely or partly, by letters. Quan- 
 tities represented by letters are termed ffenercd values — because 
 the solution of one problem furnishes a general solution which 
 embraces all others, where the letters have specified numerical 
 values. 
 
no 
 
 RAY'S ALGEBRA, HART SECOND. 
 
 The answer to a problem, where the known quantities are 
 represented by letters, is termed a formtUa; and a formula ex- 
 pressed in ordinary language, furnishes a rule. 
 
 By the application of algebra to the solution of general ques- 
 tions, a great number of useful and interesting truths and rules 
 may be established. We shall now illustrate this subject by an 
 example. 
 
 Akt. 163. It is required to divide a given number a into three 
 parts, having to each other the same ratio as the numbers m, n, 
 a.ndp. 
 
 Let mx, nx, and px, represent the required parts, since these 
 are evidently to each other as m, n, and p. 
 Then mx-\-nx-\-px=a, 
 
 and x=: 1 , 
 
 m-\-n-\-p 
 
 rttd 
 
 px= 
 
 m-\-n-\-p 
 
 na 
 m-\-n-\-p 
 
 pa 
 
 This formula translated into ordinary languar^"?. (ri\'es me 
 following 
 
 Rule foe dividing a given number- into parts having to each 
 OTHER A GIVEN RATIO. — Multiply the given number by each term 
 of the ratios respectively, and divide the products hy the sum of the 
 numbers expressing the ratios. The respective quotients will be 
 the required parts. 
 
 The pupil may solve the following examples by this rule, and 
 test its accuracy by verifying the results. 
 
 2. Divide 69 into three parts, having to each other the same 
 ratio as the numbers 5, 7, and 11. Ans. 15, 21, and 33. 
 
 3. Divide ZQh into four parts, having to each other the same 
 ratio as the fractional numbers j, \, \, and y. 
 
 Ans. 15, 10, 7i, and 6. 
 The pupil may now solve the following general examples, ami 
 express the formula in ordinary language, so as to form a general 
 rule. 
 
 4. The sum of two numbers is a, and their difference b. Re- 
 quired the numbers. ^^ ^ a I j a _ b 
 
 2^2' ' 2 3 
 
EQUATIONS OF THE FIRST DEGREE. Ill 
 
 5. The difference of two numbers is a, and the greater is to 
 
 the less as m to n: find the numbers. . ma , va 
 
 Ans. and 
 
 m — n m — n 
 
 6. The sum of two numbers is a, and their sum is to their dif. 
 ference as m to n: required the numbers. 
 
 Ans. Greater =^'"+">, less ^'"-")'' 
 2m 2m 
 
 7. Divide the number a into three such parts, that the second 
 shall exceed the first by b, and the third exceed the second by c. 
 
 Ans "—^^—^ a+b—c, a-{-b-\-2c 
 3 ' 3 ' 3 • 
 
 8. Divide the number a into four such parts, that the first in- 
 creased by m, the second diminished by m, the third multiplied by 
 m, and the fourth divided by m, shall be all equal to each other. 
 
 Suggestion. — The simplest method of solving questions of 
 this kind, is to make such a supposition for the values of the un- 
 known quantities as will fulfill one or more of the conditions. 
 In this question the last four conditions will be fulfilled by repre- 
 senting the four parts by x — m, x-\-m, — , and mx. 
 
 m 
 
 Ans. 
 
 (m+iy (m+l)2 (m+l)2 (m+l)^- 
 
 9. A person has just a hours at his disposal; how far may he 
 
 ride in a coach which travels b miles an hour, so as to return 
 
 home in time, walking back at the rate of c miles an hour? 
 
 . abc ., 
 
 Ans. miles. 
 
 6+c 
 
 10. Given the sum of two numbers =a, and the quotient of 
 the greater divided by the less =i; required the numbers. 
 
 Atis. Less =_?_, greater =-? 
 
 A+l' ^ 6+1- 
 
 This formula gives the following simple rule: — To find the lest 
 number, divide the sum of the numbers by their quotient increased by 
 unity. 
 
 11. A person distributed a cents among n oeggars, giving b 
 
 cents to some, and c to the rest. How many were there of each' 
 
 1 a — nc , I . , nb — a 
 
 Ans. — — at 6 cts.i and at c cts. 
 
 b — c b — c 
 
 12.- Divide the number n into two such parts, that the quolienl 
 of the greater divided by the less shall be 9, with a remainder r. 
 
 Ans. l^+r, '^rl 
 1+9 1+9- 
 
112 RAY'S ALGEBRA, PART SECOND 
 
 13. If A and B together can perform a piece of work in a dayi, 
 
 A and C together the same in b days, and B and C togetner m r 
 
 days: find the time in which each can perform it separately. 
 
 . , . 2aic T, • '2abc ^ . 2a6c , 
 
 Ans. A in , B in , O in days. 
 
 ac-\-bc — ab ab-\-bc — ac ab-\-ac — be 
 
 14. A, B, and C, hold a pasture in common, for which they 
 psy P$ a year. A puts in a oxen for m months; B, b oxen for n 
 months; and C, c oxen for ^months: required each one's share 
 of the rent. 
 
 Ans. A's, "Hf: Pf ; B's, "* 
 
 ma-\-nb-\-pc mar^nb-\-pc 
 
 C's, ?1 P$. 
 
 ma-\-nb-\-pc 
 
 From these formula is derived the rule of Compound Fellowship 
 
 15. A mixture is made of alb of tea at m shillings per lb, ilb 
 at n shillings, and cib at^ shillings: what will be its cost per lb. 
 
 Ans. ""'+"&+?" 
 a-\-b-\-c 
 From this formula is derived the rule termed Alligation Medial. 
 
 16. A waterman rows a given distance a and back again in h 
 liours, and finds that he can row c miles with the current for d mile3 
 against it: required the times of rowing down and up the stream, 
 also the rate of the current and the rate of rowing. 
 
 Ans. Time down, ; time up, — — ; 
 
 c-\-d c-\-d 
 
 rate of current, "("'— ^'). rate of rowing, <^S^+^ 
 2bcd ^ 2bcd ' 
 
 II. NEGATIVE SOLUTIONS. 
 
 Aet. 164. It has been stated already (Art. 12), that when a 
 quantity has no sign prefixed, the sign plus is understood; and 
 also (Art. 47), that all numbers or quantities are regarded as 
 positive, unless they are otherwise designated. Hence, in all 
 problems it is understood that the results are required in positive 
 numbers. It sometimes happens, however, in the solution of a 
 problem, that the result has the minus sign. Such a result is 
 termed a negative solution. We shall now examine a question 
 of this kind. 
 
 1. What number must be subtracted from 3 that the remaindai 
 shall be 71 
 Let x= the number. 
 
 Then 3— a!=7. 
 
EQUATIONS OF THE FIRST DEGREE. 113 
 
 whence — a;=7 — 3, 
 01 a;= — 4 . 
 
 Now — 4 subtracted fro.n 3, according to the rule for algebraic 
 Bubtraction, gives a remainder equal to 7; thus 3 — ( — 4)=7. 
 The result, — 4, is said to satisfy the question in an algebraic 
 smse: but the problem is evidently impossible in an arithmetical 
 tense, since any positive number subtracted from 3 must diminish 
 Instead of increasing it; and this impossibility is shown by the re- 
 sult being negative. But, since subtracting — 4 is the same as 
 adding -)-4 (Art. 48), the result is the answer to the following 
 question: — What number must be added to 8, that the sum shall 
 be equal to Ti 
 
 Let the question now be generalized, thus: 
 
 What number must be subtracted from a, that the remainder 
 shall be i! 
 
 Let x= the number. 
 
 Then a — x=b, 
 whence x=a — b. 
 
 Now, since a — (a — i)=i, this value of x will always satisfy the 
 question in an algebraic sense. 
 
 While b is less than a, the value of x will be positive; and 
 whatever values are given to a and b, the question will be con- 
 sistent, and can be answered in an arithmetical sense. Thus, 
 if ffi=10, and i=4, then a;=:6. 
 
 But if 6 becomes greater than a, the value of a; will he negative; 
 and whatever values are given to a and i, the result obtained will 
 satisfy the question in its algebraic, but not in its arithmetical 
 sense. 
 
 Thus if a— 8 and J=10, then x=—2. Now 8— (— 2)=8+2 
 =10; that is, if we add 2 to 8, the sum will be 10. We thus see 
 that when b becomes greater than a, the question, to be consistent 
 in an arithmetical sense, should read: — What number must be 
 added to a that the sum shall be equal to 6? 
 
 From this we derive the following important general principles: 
 
 1 St. A negative solution indicates some inconsistency or absurdity 
 in the question from which the equation was derived. 
 
 2nd. When u negative solution is obtained, the question, to which 
 it is the answer, may be so modified as to be consistent. 
 
 The pupil may now read carefully the " Obsekvations on Ad- 
 dition AND Subtraction," page 24, and then modify the follow- 
 ing questions so that they shall be consistent, and the results true 
 in an arithmetical sense. 
 10 
 
114 RAY'S ALGEBRA, PART SECOND. 
 
 2. What number must be added to the number 30, that the 
 turn shall be 191 {x=—\ 1). 
 
 3. The sum of two numbers is 9, and their difference 25; 
 required the numbers. Ans. 17 and — 8. 
 
 4. What number is that whose half subtracted from its third 
 leaves a remainder 151 (a;= — 90). 
 
 5. A father's age is 40 years; his son's age is 13 years; in 
 Vow many years will the age of the father be 4 times that of the 
 jonl (.X— — 4). 
 
 6. The triple of a certain number diminished by 100, is equal 
 to 4 times the number increased by 200. Required the number 
 
 (a;=— 300). 
 
 III. DISCUSSION OF PROBLEMS. 
 
 Art. 165, After a question has been generalized and solved, 
 we may inquire what values the results will have, when particular 
 suppositions are made with regard to the known quantities. The 
 determination of these values, and the examination of the various 
 results, to which different suppositions give rise, constitute the 
 discussion of the problem. 
 
 The various forms which the value of the unknown quantity 
 may assume, are shown in the discussion of the following ques- 
 tion. 
 
 1 . After subtracting b from a, what number, multiplied by the 
 remainder, will give a product equal to cl 
 
 Let x= the number. 
 
 Then (o — b)x^c, 
 
 J c 
 and x= .. 
 
 a — 6 
 Now, this result may have five different forms, depending on 
 the different values that may be given to a, b, and c. 
 
 Remark. — In the following forms, A denotes merely some quantity. 
 
 1st. When 5 is less than a. This gives positive values of the 
 form -f-A. 
 
 2nd. When b is greater than a. This gives negative values of 
 the form — A. 
 
 3rd. When b is equal to a. This gives values of the form — . 
 4th. When c is 0, and b either greater or less than a. This 
 
 gives values of the form — 
 ^ A' 
 
EQUATIONS OF THE FIRST DEGREE. 115 
 
 5. When J is equal to a, and c is equal to 0. This gives values 
 of the form §. 
 
 We shall now examine each of these cases. 
 
 I. Values of the form -(-A, or when b is less than a. 
 
 In this case, a — b is positive, and the value of x is positive. 
 To illustrate this form, let a=10, J=3, and c=35, then as=&. 
 
 II. Values of the form — A, or when b is greater than a. 
 
 In this case, a — J is a negative quantity, and the value of x will 
 be negative. This evidently should be so, since minus multiplied 
 by minus, gives plus; that is, if a — b is minus, x must be minus, in 
 order that their product shall be equal to c, a positive quantity. 
 
 To illustrate this form by numbers, let a=5, J=8, and c=12; 
 then a—b=—S, x= — 4, and — 3X — 4=-f-12. 
 
 A 
 
 III. Values of the form — , or when b is equal to a. 
 
 In this case x becomes equal to — But the value of a fraction 
 cf which the numerator is any finite quantity, and the denomina- 
 tor zero (Art. 186), is infinite; that is, _=ao . 
 
 This is interpreted by saying, that no finite value of x will sat- 
 isfy the equation; that is, there is no number, which being multi- 
 plied by 0, will give a product equal to c. 
 
 IV. Values of the form _, that is, when c is and 6 is eithei 
 
 A 
 
 greater or less than a. 
 
 If we put a — b=d, then x=:- =0 , since d X =0 ; that is, when 
 a 
 
 the product is zero, one of the factors must be zero. 
 
 V. Values of the form §, that is, when b=a, and c=0. 
 
 In this case we have a:=~5_=g, or a:XO=0. But g is the 
 
 symbol of indetermination (Art. 137); hence any finite value of a 
 whatever will satisfy this equation; that is, a; is indeterminate. 
 
 The discussion of the following problem, originally proposed 
 by Clairaut, will serve to illustrate further the preceding princi- 
 pies, and show that the results of every correct solution corres 
 pond to the circumstances of the problem. 
 
116 RAY'S ALGEBRA, PART SECOND. 
 
 PROBLEM OF THE COURIERS. 
 
 Art. 166. Two couriers depart at the same time, from two 
 places, A and B, distant a miles from each other; the Ajrmer 
 travels m miles an hour, and the latter n miles: where will they 
 meetl 
 
 There are two cases of this problem, according as the couriers 
 are traveling toward each other, or in the same direction. 
 
 I. When the couriers travel toward each other. 
 
 Let P be the point where they meet, A I 1 B 
 
 and a=:AB, the distance between the 
 two places. 
 
 Let a;=AP, the distance which the first travels. 
 
 Then a — i==BP, the distance which the second travels. 
 
 But the distance each travels, divided by the number of miles 
 traveled per hour, will give the number of hours he was traveling. 
 
 Therefore, — = the number of hours the first travels. 
 m 
 
 And ?IZ?=: « « " « second travels. 
 
 n 
 
 But they both travel the same number of hours, therefore 
 
 X a — x_ 
 
 m n 
 
 ra:=ma — mx, by clearing of fractions; 
 
 whence x= ""^ ; 
 m-\-n 
 
 and a — x= "" 
 
 1st. Suppose m=n, then x=^=-, anda— a;=^; tliat is, if 
 
 the couriers travel at the same rate, each travels precisely half 
 the distance. 
 
 2nd. Suppose n=0, then x=^=a; that is, if the second 
 
 m 
 
 courier remains at rest, the first travels the whole distance from 
 A toB. 
 
 Both these results are evidently true, and correspond to the cir- 
 cumstances of thejiroblem. 
 
 IL When the couriers travel in the same direction. 
 
 As in the first case, let P be the point I i I 
 
 of meeting, each traveling from A toward A B P 
 
 P, and let (Z=AB, the distance between the places; 
 a;=AP, " " tlie first travels; 
 
 then I— o=BP. " " " second travels. 
 
EQUATIONS OF THE FIRST DEGREE. HI 
 
 Then, reasoning as in the first case, we have 
 
 X X — a, 
 
 m n 
 
 nx^=mx — ma; 
 
 , ma 
 
 whence a;=-- ; 
 
 m — n 
 
 and X — a= 
 
 m — n 
 
 1st. If we suppose m greater than re, the value of a; will be 
 positive; that is, the couriers will meet on the right of B. This 
 evidently corresponds to the circumstances of the problem. 
 
 2nd. If we suppose n greater than m, the value of u:, and 
 also that of x — a, will be negative. This value of x being nega- 
 tive, shows that there is some inconsistency in the question 
 (Art. 164). Indeed, where m is less than n, it is evident that 
 the couriers cannot meet, since the forward courier is traveling 
 faster than the hindmost. 
 
 Let us now inquire how the question may be modified, that the 
 value obtained for x shall be consistent. 
 
 If we suppose the direction changed in which the couriers travel, 
 that is, that the first travels from A, and P' I i I 
 
 the second from B, toward P'; and that o=AB, ■^ " 
 
 a;=AP', 
 a-|-a:=BP', we have, 
 
 reasoning as before, -^=?IL^; 
 m n 
 
 whence x= ""^ , and a-\-. 
 
 n — m n — m 
 
 The distances traveled are now both positive, and the question 
 will be consistent, if we regard the couriers, instead of traveling 
 toward P, as traveling in the opposite direction, toward P'. The 
 change of sign thus indicating a change of direction (Art 47'). 
 
 3rd. Let us suppose m equal to n. 
 
 In this case x is equal to ??^, and x — a=— 
 
 But it has been shown already (Art. 136), that when the un- 
 known quantity takes, this form, it is not satisfied by any finite 
 value, or, it is infinitely great. This evidently corresponds to the 
 circumstances of the problem; for, if the couriers travel at the 
 same rate, the one can never overtake the other. This is some- 
 times otherwise expressed, by saying they only meet at an infinite 
 distance from the point of starting. 
 
118 RAY'S ALGEBRA, PART SECOND. 
 
 4th. Let US suppose a=:0. 
 
 rru , 
 
 Then 3C= , and x — a= -. 
 
 m — n m — n 
 
 When the unknown quantity takes this form, it has been 
 shown already (Art. 135) that its value is 0. This corresponds 
 to the circumstances of the problem; for, if the couriers are no 
 distance apart, they will have to travel no (0) distance to bo 
 together. 
 
 5th. Let us suppose m=n, and (2=0. 
 
 In this case, x=^, and x — a=8. But when the unknown 
 quantity takes this form, it has been shown (Art. 137) that it 
 may have any finite value whatever. This, also, evidently corres- 
 ponds to the circumstances of the problem; for, if the couriers 
 are no distance apart, and travel at the same rate, they will be 
 always together; that is, at any distance whatever from the point 
 of starting. 
 
 6th. Let us suppose 7J=0. 
 
 In this case a=^=o; that is, the first courier travels from 
 
 m 
 
 A to B, overtaking the second at B. 
 7th. Lastly, let us suppose ra= — . 
 
 In this case x= =2a; that is, the first travels twice the dis>- 
 
 m 
 
 tance from A to B, before overtaking the second. The results 
 
 in the last two cases evidently correspond to the circumstances 
 
 of the problem. 
 
 IV. CASES OF INDETERMINATION IN EQUATIONS OF THE FIRST 
 DEGREE, AND IMPOSSIBLE PKOBLEMS. 
 
 Aet. 167. An equation is termed independent, when the rela- 
 tion of the quantities which it contains, cannot be obtained 
 directly from others with which it is compared. Thus, the 
 equations 
 
 a;+3y=19, 
 and 2a:-l-5j/=33, 
 •re independent of each other, since the one carjiot be obtained 
 from the other in a direct manner. 
 
 The equations a;-f-3y=19, 
 2x4-6 j^=3 8, 
 are not independent of each other, the second being derived 
 directly from the first, by multiplying both sides bv 2. 
 
EQUATIONS OF THE FIRST DEGREE. 119 
 
 Art. 16§. An equation is said to be indelerminate, when it 
 can be verified by different values of the same unknown quantity. 
 Thus, if we have the equation 
 
 X — j'=3, by transposition we find 
 a;=3-l-y. 
 If we make 3/=!, then a!=4; if we make jr=2, then a:=5 
 and so on; from which it is evident, that an unlimited number of 
 values may be given to x and y, that will verify the equation. 
 
 If we have two equations containing three unknown quantities, 
 we may eliminate one of them; this will leave one equation con- 
 taining two unknown quantities, which, as in the preceding ex- 
 ample, will be indeterminate. Thus, if we have the following 
 equations, 
 
 1+33/— 5«=20, 
 X — y-["32=16. 
 If we eliminate x, we have, after reducing, 
 y— 22=1; 
 whence y=\-\-2z. 
 If we mal^e z=X, then y=i, and x=QQ-\-bz — 3y=16. If we 
 make 2=2, then J/=5, and x=15. 
 
 In the same manner, an unlimited number of values of the 
 three unknown quantities may be found, that will verify both 
 equations. Other examples might be given, but these are suffi- 
 cient to establish the following general principle. 
 
 When the number of uriknovm quantities exceeds the nuwier of in- 
 dependent equations, the problem is indeterminate. 
 
 A question is sometimes indeterminate that involves only one 
 unknown quantity; the equation deduced from the condition* 
 being of that class denominated identical. (Art. 145.) The fol- 
 lowing is an example: 
 
 What number is that, whose 5 increased by the g is equal to 
 the 1 1 diminished by the fg? 
 Let x= the number. 
 
 Then ?4-?=il^_?^; 
 4^6 20 15 
 
 clearing of fractions, 15a;-|-10a:=33a! — 8x; 
 or, 25K=:25a;; 
 which will be verified by any value whatever of ' 
 
 A more simple example is the following: — W-nat number is 
 that whose half, third, and fourth, taken together, is equal to the 
 number itself increased by its one-twelfth. 
 
120 RAY'S ALGEBRA, PART SECOND. 
 
 Art. 169. The reverse of the preceding case requires to be 
 considered; that is, when the number of equations is greater than 
 the number of unknown quantities. Thus, we may have 
 2ar-j-3y=;i3, (1) 
 3x— 2y=2, (2) 
 
 5x+4i/=40. (3) 
 Each of these equations being independent of the other two, 
 one of them is unnecessary, since the values of x and y, which 
 are 4 and 5, may be found from either two of them. 
 
 When a problem contains more conditions than are necessary 
 for determining the values of the unknown quantities, those that 
 are unnecessary are termed redundant. 
 
 The number of equations may exceed the number of unknown 
 quantities, so that the values of the unknown quantities shall be 
 Incompatible with each other. Thus, if we have 
 a;+!/=12, (1) 
 
 2x+y==ll, (2) 
 
 3x+2y=20. (3) 
 The values of x and y, found from equations (1) and (2), are 
 x=5, y=T; from equations (1) and (3), a;=6, a^d y=Q; and 
 from equations (2) and (3), x=i, and j/=9. From this, it is 
 manifest that only two of these equations can be true at the same 
 time. 
 
 A question is sometimes impossible that involves only one un- 
 known quantity. The following is an example: 
 
 What number is that whose j'j diminished by 5 is equal to the 
 difference between its | and | increased by 7. 
 
 Let as=the number, then — —5=—— ^+7; 
 12 4 6^ 
 
 clearing of fractions, 7a: — 60=9a; — 2a;-|-84, 
 
 reducing, 0=144, 
 
 which shows that the question is absurd. 
 
 Remark. — Problems from which contradictory equations are deduced, 
 are termed irrational or impossible. The pupil should be able to detect the 
 character of such questions when they occur, in order that his efforts 
 may not be wasted in attempting to perform an impossibility. 
 
 EXAMPLES TO ILLUSTRATE THE PRECEDINO 
 PRINCIPLES. 
 
 1 . What number is that, which being divided successively by 
 m and n, and the first quotient subtracted from the second, the 
 remainder shall be gl . mnq 
 
EQUATIONS OF THE FIRST DEGREE. 123 
 
 What supposition will give a negative solutiou? An infinite solution 1 
 An indeterminate solntlon? Illustrate by nu:uber9. 
 
 2. Two boats, A and B, set out at the same time, one from C to 
 L, and the other from L to C; the boat A runs m miles, and the 
 boat B, n miles per hour. Where will they meet, supposing it to 
 oc a miles from C to L! 
 
 Ans. miles from C, or — ^ miles from L. 
 
 m+re m-\-n 
 
 Under what circumstances will the boats meet half way between C 
 and L? Under what circumstances will they meet at C7 At L? Un- 
 der what circumstances will they meet above C? Below L? Under 
 what circumstances will they never meet? Under what circumstances 
 will they sail together? Illustrate each of these questions by using 
 numbers. 
 
 3. Whnt number is that which, being multiplied by 8, the pro- 
 duct increased by 16, and the sum divided by 4, will give a quo- 
 ■.ient equal to twice the number diminished by 7? 
 
 Resulting eq. 11=0. 
 What does this result show? (Art. 169.) 
 
 4. There are three persons, A, B, and C, whose ages are so re- 
 lated, that B is 6 years younger than A and 4 years older than 
 C; and 3 of A's age increased by -] of C's, is equal to -^ii of B's, 
 increased by one year. Required their ages. 
 
 Resulting equation. 0=0. 
 If A's age was 15 years, B's was 9, and C's 5; if A's was 16, 
 B's was 10, and C's 6. 
 
 5. Given 2a: — 2/=2, 
 
 5a:— 3j/=3, 
 
 3a;+2j,=17, ^^- ^3, y=4. 
 
 4a:+32/=24; 
 to find X and y, and show how many equations are redundant. 
 (Art. 169.) 
 
 6. Given a;-j-2jr=ll, 
 
 a:-f3y=19; 
 to show that the equations are incompatible. (Art. 169.) 
 
 V. AN EQUATION OF THE FIRST DEGREE HAS BUT 
 ONE ROOT. 
 
 Aet. 1¥©. In any equation of the first degree involving only 
 one unknown quantity (a;), if a represents the sum of the positive, 
 
 and — c the sum of the negative coefficients of x; b the siiiii of 
 
122 RAVS ALGEBRA, PAKT SECOND. 
 
 the positive, and — d the sum of the negative known quantities, il 
 will evidently reduce to the following form: 
 ax — cx=b — d, 
 or (a — c)x^b — d. 
 Let a — c=jn, and b — d:=n, we then have 
 mx^n, 
 
 whence x= — . 
 m 
 
 Now since n divided by m can give but one quotient, we infer 
 
 that, an equation of ilie first degree has but one root; that is, in an 
 
 equation of the first degree, involving but one unknown quantity, 
 
 there is but one value that will verify the equation. 
 
 VI. EXAMPLES INVOLVING THE SECOND POWER OF 
 THE UNKNOWN QUANTITY. 
 
 Art. IVl. It sometimes happens in the solution of an equa- 
 tion of the first degree, that the second or some higher power of 
 the unknown quantity occurs, but in such a manner that it \a 
 easily removed, after which the equation may be solved in tha 
 usual manner. 
 
 The following equations and problems belong to this calss. 
 
 1. (44-a;)(j;— 5)=(a:— 2)=. 
 
 Performing the operations indicated, we have 
 a:2— a— 20=x2— 4a:+4 . 
 
 Omitting a:' on each side and transposing, we have 
 3a;=24, or ac=S. 
 
 2x-\-l ' 3a; 
 
 =x-\-\ . Ans. x=l . 
 
 o 4a: 20— 4a: 15 , o o 
 
 o X XX '' 
 
 3x^-2x+l _ (!7:^2)(2x-e) , 
 
 5 35 -t-T%- Ans.x-l^^ 
 
 5. 3+2» 5+2x_^ _ 4a:^-2 
 
 l+2a; 74-2a; 7+16a;+4a;»- ' ^' 
 
 6. i^^(3a:— 19)=2a:+19. Am. x=8. 
 6x — 43 
 
 7 'i,j_'^^+9 Q ,1022—18 . 
 4a;+3 2a;-|-3 
 
 i ax A b 
 
 =ac+—.^ Am. x=-. 
 
 bx 
 
 c 
 
 C3l^ rfa?» . ad—ce 
 
 = Ans. x=——Z. 
 
 a-\-bx e-\-fx' cf—M 
 
FORMATION OF POWERS. 123 
 
 10. ia-\-x)(b+x)—a(b+c)=!^+x^. Ans. x=-. 
 
 b b 
 
 11. It is required to find a number which being divided into 
 two, and into three equal parts, four times the product of the 
 two equal parts, shall be equal to the continued product of the 
 three equal parts. Ans. 27. 
 
 12. A rectangular floor is of a certain size. If it were 5 feet 
 broader and 4 feet longer, it would contain 116 feet more; but 
 f it were 4 feet broader and 5 feet longer, it would contain 113 
 feet more. Required its length and breadth. 
 
 Ans. Length, 12 feet; breadth, 9 feet. 
 
 CHAPTER VI. 
 
 FORMATION OF P OWERS— EXTRACTION 
 OF ROOTS— RADICALS— INEQUALITIES. 
 
 I. INVOLUTION OR FORMATION OF POWERS. 
 
 Aet. 1?2. The power of a number is the product obtained by 
 multiplying it a certain number of times by itself. 
 Any number is the Jirsl power of itself. 
 
 When the number is taken twice as a factor, the product is 
 called the second power or square of the numbei. 
 
 When the number is taken three times as a factor, the product 
 is called the third power or cube of the number. 
 
 In like manner, the fourth, fifth, &.C., powers of a number, are 
 the products arising from taking the number /our t\mea,five times, 
 &c., as a factor, the power being always denoted by the nurabei 
 of times the number is taken as a factor. 
 
 The number which denotes the power is called the index or ex 
 fonent of the power, and is written to the right of the numbei 
 and a little above it. 
 Thus, 3=3'= 3, is the 1 st power of 3. 
 
 3X3=3== 9," 
 3X3X3=3'= 27," 
 3X3X3X3=3"= 81," 
 3X3X3X3X3=3'=243,'- 
 4X ^x sy s=f 31"^ 8' " 
 
 S'^ft'^s'^S ^5' — SIS' 
 
 2nd 
 
 C( 
 
 ■' 3. 
 
 3rd 
 
 (1 
 
 « 3. 
 
 4th 
 
 It 
 
 " 3. 
 
 5th 
 
 ft 
 
 « 3. 
 
 4th 
 
 <( 
 
 " I- 
 
124 RAY'S ALGEBRA, PART SECOND. 
 
 From the preceding we Bee, that the nth power of a quantity it 
 the product of n factors each equal to tlie quantity. Hence we have 
 the following 
 
 Rule for raising a quantity to ant kequiked power. — Mvl- 
 tiply the given quantity by itself, until it is taken as a factor as 
 many times as there are units in the exponent of the required 
 power. 
 
 Remark. — This rule is perfectly general, and applies either to monO' 
 niials or polynomials, whether integral or fractional. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Find the square of 5axh'. Ans. 25a^x*z'. 
 
 2. Find ihe square of —Zb^cd. Ans. 9JV=(J'. 
 
 3. Find the cube of 2x'z. Ans. BxH'. 
 
 4. Find the cube of —Sa^c'. Ans. —^Ta^c'. 
 
 5. Find the fourth power of — 2xz^. Ans. IGx'z'. 
 
 6. Find the fifth power of — 3o'i'. Ans. —2A2a'<>b'K 
 
 7. Find the seventh power of — m'n. Ans. — m"n^. 
 S. Find the square and the cube of "^a^al" r^yp-^. 
 
 (1) Aws. 5'*ga6a:2'» + ''2/=y-2. (2) i-|5oV"-fyf-', 
 
 9. Find the square of a — x. Ans. a' — 2ax-\-x'. 
 
 10. Find the square of mx — 7w'. Ans. m^x' — 2mnx'-\-n''x*. 
 
 11. Find the cube of 2x—z. Ans. Sx'—lZx'z+exz'—zK 
 
 12. Find the cube of 3a;+2y. 
 
 Ans. 27x'-\-54x^y+26xy'-{Sy'. 
 
 13. Find the fourth power of m — n. 
 
 Ans. m'* — im^n-{-6mV — imn'-\-n*. 
 
 14. Find the square of o-f-J — c. 
 
 Ans. o'+2a6+6'— 2ac— 2ic+c». 
 
 15. Find the cube of a — J+c. 
 
 Ans. a'— 3o'6+3a62— i'+Sa^o— 6aJc4-362c+3ac'— Sic'+e'. 
 
 16. Find the square of a-\-b — c-f-(i. 
 
 Ans. a^+2ab+b^—2ac—2bc+c^+2ad+2bd—2cd+d^. 
 
 17. Find the square of f? Ans. "2^ 
 
 18. Find the square of ?=? Ans. "'—2'"+^' 
 
 a-\-x a'+2aa;+x2' 
 
 19. Find the cube of ^ Ans. ^ 
 
 20. Find the cube of ?fL' Ans ^ 
 
 3c»' 27r»' 
 
EXTRACTION OF THE SQUARE ROOT. 125 
 
 21. Find the cube of ^!=!^ Ans. rn^-'^mH+Zmn^-n^ 
 
 -2n' m? — ^m-mAA.'i.mm? — 8n^ 
 
 22. Find the square of \a—lb. Am. ^^a?—lab-\-\l^. 
 
 23. Find the cube of {a— lb. Ans. ia^—i^b^—la^-[-'jab'. 
 
 24. Find the square of i>—-—l. Ans. x'—2x+}-^'-i—l 
 
 X x' X 
 
 25. Find the cube of x— 1 Ans. a'— i 3 / x—l. ) 
 
 X x' \ , X f 
 
 26. Find the cube of x" — 1. Ans. a;'f — 3a;'«-|-3a?' — 1 
 
 27. Find the cube of e» — e'". Ans. e'' — e"'' — 3(6^ — e'') 
 
 28. lfx+l=:p, show that x'+l =p'—3p. 
 
 X x^ 
 
 29. If two numbers differ by unity, prove that the difference 
 of their squares is equal to the sum of the numbers. 
 
 30. Show that the sum of the cubes of any three consecutive 
 integral numbers is divisible by the sum of those numbers. 
 
 Note. — For a more ijere-al method of raising a binomial to any re- 
 quired power, see the Biaomial Theorem, Art. 310, page 264. 
 
 II. EXTEACTIOn Cr THE SaUARE ROOT. 
 EXTRACTION OF THE SIJUAKK.- BOOT OF NUMBERS. 
 
 Art. l^S. The root of a nuriiber, i.' a factor which multiplied 
 by itself a certain number of time? will produce the given 
 number. 
 
 The second root or square root of a number is that number which 
 multiplied by itself, that is, taken twi^f. as •> factor, will produce 
 the given number. 
 
 The process of finding the second root of a number, is called 
 the extraction of the square root. 
 
 Art. 174. To show the relation thatexii-ta between the num- 
 ber of figures in any given number, and th« nrwber of figures in 
 its square root. 
 
 The first ten numbers are 
 1, 2, 3, 4, 5, 6, 7, 8, 9, 10; 
 and their squares are 
 
 1, 4, 9, 16, 25, 36, 49, 64, S.T, XOO. 
 
 The numbers in the first line are also the Bqvjpre roots of the 
 numbers in the second line. 
 
 We see, from this, that the square root of a nunb?T between 1 
 and 4, is a number between 1 and 2; the square roo* o£ a num- 
 ber hetween 4 and 9, is a number between 2 and .S; the square 
 
126 RAY'S ALGEBRA, PART SECOND. 
 
 root of a number between 16 and 25, is a number between 4 
 and 5, and eo on. 
 
 Since the square root of 1 is 1, and the square root of any 
 number less than 100 is either one figure, or one figure and a frac- 
 tion, it is evident that when the number of places of fig^ares in a 
 number, is nol more than two, the number of places of figures in 
 the square root will be one. 
 
 Again, take the numbers 
 10, 20, 30, 40, 50, 60, 70, 80, 90, 100; 
 their squares are 
 100,400,900,1600,2500, 3600, 4900, 6400, 8100, 10000. 
 
 From this we see, that the square root of 100 is 10; and of 
 any number greater than 100 and less than 10000, the square 
 root will be less than 100, that is, it will consist of two places of 
 figures; hence, when, the number of places of figures is more than 
 TWO, and not more than four, the number of places of figures in the 
 square root will be two. 
 
 In the same manner it may be shown, that when the number 
 of places of figures in a given number is more than /our, and not 
 more than sir, the number of places in the square root will be 
 three, and so on. 
 
 Or, as the same principle may be expressed otherwise, thus: — 
 When the number of places in the given number is either OTje or 
 tiDO, there will be one figure in the root; when the number of 
 places is either three or four, there will be two figures in the root; 
 when the number of places is either five or six, there will be three 
 figures in the root, and so on. 
 
 Art. 175. Investigation of a rule for extracting the square 
 root. 
 
 Every number may be regarded as being composed of tens 
 and units. Thus, 76 consists of 7 tens and 6 units; and 576 
 consists of 57 tens and 6 units. Therefore, if we represent the 
 tens by t, and the units by u, any number will be represented bv 
 t-\-u, and its square by the square of t-\-u, or (t-\-uy. 
 (,t-lruy=t^-\-2tUr^u'=t^+(2t+u)u. 
 
 Hence, the square of any number is composed of two quantities, 
 one of which is the square of the tens, and the other twice the tens phu 
 Ihe units multiplied by the units. 
 
 Thus, the square of 25, which is equal to 2 tens and 5 unitB, ii 
 2 tens squared =(20)2=400 
 (■4 tens 4 5 units) multiplied by 5=(40+5)5=225 
 
 625 
 
EXTRACTION OF THE SQUARE ROOT. 121 
 
 1. Let it now be required to extract the squai-e root of 625. 
 Since the number consists of three places 625|25 
 
 uf figures, its root will consist of two places, " 400i 
 
 according to the principle established in Art. 20 X2=40 225 
 174; we, therefore, separate it into two pe- 5 225 
 
 riods, as in the margin. 45 
 
 Since the square of 2 tens is 400, and of 3 tens is 900, it is 
 txldent that the greatest square contained in 600 is the square 
 ol 2 tens (20); the square of 2 tens (20) is 400; and subtract- 
 ing this from 625, the remainder is 225. 
 
 Now, according to the preceding theorem, the remainder 225, 
 consists of twice the tens plus the units, multiplied by the units; 
 that is, by the formula, it is (2t-\-u)u, of which t is already found 
 to be 2, and it remains to find u. Now the product of the tens 
 by the units cannot give a product less than tens; therefore, the 
 unit's figure (5) forms no part of the double product of the tens 
 by the units. Hence, if we divide the remaining figures (22) by 
 the double of the tens (4), the quotient will be the unit's figure, 
 or a figure greater than it. 
 
 We, therefore, double the tens, which makes 4 (2i), and divide 
 this into 22, which gives 5 (u) for a quotient; this is the upic's 
 figure of the root. This unit's figure (5) is to be added to the 
 double of the tens (40), and the sum multiplied by the unit's 
 figure. The double of the tens plus the units is 40-|-5^45 
 (2(-|-K); multiplying this by 5 (m), the product is 225, which is 
 the double of the tens plus the units, multiplied by the units. As 
 there is nothing left after subtracting this from the first remainder, 
 we conclude that 25 is the exact square root of 625. 
 
 In squaring the tens, and also in doubling them, . . 
 
 it is customary to omit the ciphers, though they d^Oj^O 
 
 are understood. Also, the unit's figure is added ^^0 
 
 to the double of the tens, by merely writing 451225 
 it in the unit's place. The actual operation is 1^'^" 
 
 usually performed as in the margin. 
 
 2. Let it be required to extract the square root of 59049. 
 Since this number consists of five places of figures, its square 
 root will consist of three places. (Art. 174.) We, therefore, 
 lep'irate it into three periods. Kon/iQiojq 
 
 In performing this operation, we find the square _ 
 
 root of the number 590, on the same principle _L:;^1; 
 
 as in the precedins example. ,, 
 
 '^ - ^ 483 1449 
 
 1449 
 
128 RAY'S ALGEBRA, PART SECOND 
 
 We next consider 24 aa bo many tens, and proceed to find 
 the unit's figure (3) in the same manner as in the preceding 
 example. 
 
 From these illustrations, we derive the following 
 
 •IVTE FOR THE EXTK ACTION OF THE SQUAEE KOOT OF NUMSEES. — 
 
 I St. Separate the given number into periods of two places each, he- 
 ginning at the unit's place. (The left period will often contain 
 but one figure.) 
 
 '■iad. Find the greatest square in tlie left period, and place its root on 
 the right, after the manner of a quotient in division. Subtract 
 tlie square of the root from the left period, and to tlie remainder 
 bring down the next period for a dividend. 
 
 3rd. Double t!ie root already found, and place it on the left for a divi- 
 sor. Find how many times the divisor is contained in the divi- 
 dend, exclusive of the right hand figure, and place the figure in the 
 root and also on the right of the divisor. 
 
 4 th. Multiply the divisor thus increased by the last figure of tlie root, 
 subtract the product from the dividend, ami to the remainder bring 
 down tlie next period for a new dividend. 
 
 ■)\h. Double the whole root already found for a Tiew divisor, and con- 
 tinue tlie operation as before, until all the periods are brought down. 
 
 Note. — If, in any case, the dividend will not contain the divisor, the 
 right hand figure of the former being omitted, pf:ine a cipher in the 
 root and also at the right of the divisor, and bring down the next period. 
 
 Akt. 176. In division of numbers, when the remainder is 
 greater than the divisor, the last quotient figure may be increased 
 by at least 1 ; but in extracting the square root of numbers, the 
 remainder may, sometimes, be greater than the divisor, while the 
 last figure of the root cannot be increased. To know when any 
 figure may be increased, the pupil must be acquainted with the 
 relation that exists between the squares of two consecutive 
 numbers. 
 
 Let a and a-\-l, be two consecutive numbers. 
 
 Thus, {a-\-iy=a'-^2a-{-l, is the square of the greater, 
 
 and (g)'— fflS " " " " " less. 
 
 Their difiference is 2a-f-l . 
 
 From which we see, that the difference of the squares of Iko con- 
 ixutive numbers, is equal to twice the less number, increased b} 
 unity. 
 
 Consequently, when any remainder is less than twice th-e part 
 of the root already found, plus unity, the last figure cannot be 
 increased 
 
EXTRACTION OF THE SQUARE ROOT. 129 
 
 EXAMPLES, 
 
 In extracting the square root of whole numbers. 
 
 1. 2601. Arts. 51. 
 
 2. 7220. Ans. 85. 
 
 3. 9801. A.-w. 99. 
 
 4. 47089. Ans. 217. 
 
 5. 138384. Ans. 372. 
 
 6. 390625. Ans. 625. 
 
 7. 553030. Ans. 744 
 
 8. 5764801. 4ns. 2401 
 
 9. 43046721. Ans. 65C1 
 
 10. 49042009. Ans. 7003 
 
 11. 1061326084. Ans. 32578. 
 
 12. 943042681. Ans. 30709. 
 
 EXTRACTION OF THE SQUARE ROOT OF FRACTIONS. 
 
 Art. X77. Since §Xf=5\, therefore, the square root of ^^ 
 is I ; that is V2''5=^=i. Hence, 
 
 When both terms of a fraction are perfect squares, its square root 
 will he found by extracting the square root of both terms. 
 
 Before attempting to extract the square root of a fraction, it 
 should be reduced to its lowest terms, unless both numerator and 
 denominator are perfect squares. The reason for this will be 
 seen by the following example. 
 
 Find the square root of |o. 
 
 Here |-5=:1^. Now, neither 20 nor 45 are perfect squares; 
 but, by canceling the common factor 5, the fraction becomes |, of 
 which the square root is -f . 
 
 When both terms are perfect squares, and contain a common 
 factor, the reduction may be made either before, or after the 
 square root is extracted. 
 
 Thus,V|f=^==; or, M=|, and Ji=l 
 
 EXAMPLES, 
 
 In extracting the square root of fractions. 
 
 5- tVt)V5- -Ans. II . 
 
 6. jUhnv Ans. j%%\. 
 
 1. ||. Ans. f . 
 
 2. Jj\. Ans. t\. 
 
 3. fig. Ans. l 
 Art. lyS. A number whose square root can be ascertained 
 
 exactly, is termed a. perfect square. Thus, 4, 9, 16, &c., are per- 
 fect squares. Such numbers are comparatively few. 
 
 A number whose square root cannot be ascertained exactly, is 
 termed an imperfect square. Thus, 2, 3, 5, 6, &,c., are imperfect 
 squares. 
 
 Since the difference of two consecutive square numbers, a' and 
 it'-j-2a-|-l , is 2a+l ; therefore, there are always 2a imperfect 
 
130 RAY'S ALGEBRA, PART SECOND. 
 
 squares between them. Thus, between the square of 5 (25) and 
 the square of 6 (36), there are 10 (2a=:2x5) imperfect squares, 
 
 A root which cannot be expressed exactly, is called a radical, 
 or surd, or irrational root. The root obtained is also called an 
 approximate value, or approximate root. Thus, J2 is an irrational 
 root; it is 1.414-|-. 
 
 The sign + is sometimes placed after an approximate root, to 
 denote that it is less than the true root; and the sign — , that it ii 
 greater than the true root. 
 
 Art. IVO. To prove that the square /oot of an imperfect 
 square cannot be a fraction. 
 
 Remark. — It might be supposed, that when the square root of a 
 whole number cannot be expressed by a whole number, that it might be 
 found exactly equal to some fraction. That it cannot, will now be 
 shown. 
 
 Let c be an imperfect square, such as 2, and, if possible, let its 
 
 square root be equal to a fraction _, which is supposed to be in 
 
 b 
 
 its lowest terms. 
 
 Then Jc= -; and c=—, by squaring both sides. 
 b b^ 
 
 Now, by supposition, a and J have no common factor, therefore 
 
 their squares, a' and b", can have no common factor, since to 
 
 square a number, we merely repeat its factors. Consequently, 
 
 _ must be in its lowest terms, and cannot be equal to a whole 
 
 b' ^ 
 
 number. Therefore, the equation c=_, is not true; and hence 
 the supposition on which it is founded is false, that is, the suppo- 
 sition that .Jc= - is not true; therefore, the square root of an 
 b 
 
 imperfect square cannot be a fraction. 
 
 APPROXIMATE SQUARE ROOTS. 
 
 Art. ISO. To explain the method of finding the approximate 
 square root of an imperfect square, let it be required to find the 
 square root of 5 to within \. 
 
 If we reduce 5 to a fraction whose denominator is 9 (the square 
 of 3, the denominator of the fraction J), we have 5=-g. 
 
 Now the square root of 4.5 is greater than 6 and less than 7 , 
 therefore the square root of --^ is greater than |, and less thanfi 
 deuce f , or 2, is the square root of .5 to within g. 
 
EXTRACTION OF THE SQUARE ROOT. 131 
 
 To generalize this explanation, let it be required to extract the 
 
 square root of a to within a fraction _. 
 
 n 
 
 We may write a (Art. 127) under the form — , and if we de- 
 
 note the entire part of the square root of an" by r, the number 
 
 an" 
 n' 
 
 an' will be comprised between r" and (r-\-iy; therefore 
 will be comprised between — and ^ ^^ ■' ; hence the square 
 
 Ti' n' 
 
 root of — will be comprised between - and ^1-. 
 n" n n 
 
 But the difference between _ and _!_- is - , therefore _ rep- 
 n n n n 
 
 resents the square root of a to within -. From this we derive 
 
 n 
 the following 
 
 Rule for extkactiho the squabe koot of a whole numbek 
 TO WITHIN A GiVEH FKACTiON. — Multiply the given number by 
 the square of the denominator of the fraction which determines the 
 degree of approximation; extract the square root of this product to 
 the nearest unit, and divide the result hy the denominator of the 
 fraction. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Find the square root of 3 to within 3. Ans. If. 
 
 2. Find the square root of 10 to within j. Ans. 3. 
 
 3. Find the square root of 19 to within g. Ans. 4|. 
 
 4. Find the square root of 30 to within j';;. Ans. 5.4. 
 
 5. Find the square root of 75 to within yj^. Ans. 8.66. 
 Since the square of 10 is 100, the square of 100, 10000, and 
 
 60 on, the number of ciphers in the square of the denominator of 
 a decimal fraction, is equal to twice the number in the denomina- 
 tor itself. Therefore, when the fraction which determines the degree 
 of approximation is a decimal, it is merely jiecessary to add two 
 ciphers for each decimal place required; and, after extracting the 
 square root, to point off from the right one place of decimals for eacji 
 'too ciphers added. 
 
 (J. Find the square root of 3 to live places of decimals. 
 
 Ans. 1.73205. 
 
132 RAY'S ALGEBRA, PART SECOND. 
 
 7. Find the square root of 7 to five places of decimals. 
 
 Ans. 2.64575. 
 3. Find the square root of 50. Am. 7.071067+. 
 
 9. Find the square root of 500. Ans. 22. 360670+. 
 
 Aet. 181. To find the approximate square root of a fraction. 
 
 1. Let it be required to find the square root of 4 to within 7. 
 
 4 — 4 V''' — 28 
 f — 7'^'? — 4"S- 
 
 Now, since the square root of 28 is greater than 5 and lesi 
 than 6, the square root of f | is greater than f and less than ^i 
 therefore f is the square root of 4 to within less than ^. 
 
 From this it is evident, that if we multiply the numerator of a 
 fraction h/ its denominator, then extract the square root of tlie pro- 
 duct to the nearest unit, and divide the result by the denominator, the 
 quotient itill be the square root of the fraction to within one of its 
 equal parts. 
 
 2. Find the square root of /y to within y\. Ans. y*j. 
 
 3. Find the square root of \\ to within yg. Ans. y|. 
 
 4. Find the square root of ig to within j'p. Ans. y'jj. 
 
 It is obvious that any decimal may be written in the form of a 
 common fraction, and having its denominator a perfect square, by 
 adding ciphers to both terms. Thus .3=y'5=y^(/'5; .156=yififi)'o> 
 and so on. Therefore, the square root of a decimal may be 
 found, as in the method of finding the approximate square root of 
 a wliole number (Art. 180), by anneooing ciphers to the given deci- 
 mal, until the number of decimal places shaU be equal to dovMe the 
 number required in the root. Then, after extracting the root, point- 
 ing off from the right the required nurriber of decimal places. 
 
 5. Find the square root of .4 to six places. Ans. .632455+. 
 
 6. Find the square root of .35 to six places. Ans. .591607+. 
 
 The square root of a whole number and a decimal may be 
 found in the same manner. Thus, the square root of 1.2 is the 
 same as the square root of 1.20^y§8) which, extracted to five 
 places, is 1.09544+. 
 
 7. Find the square root of 7.532 to five places. 
 
 Ans. 2.74444+. 
 
 When the denominator of a fraction is a perfect square, its 
 square root may be found by extracting the square root of the 
 numerator to as mar.y places of decimals as are required, and 
 dividing the result by the square root of the denominator. 
 
i!,ATRACTION OF THE SQUARE ROOT. 133 
 
 Or, by reducing the fraction to a decimal, and then extracting 
 its square root. When the denominator of the fraction is not a 
 perfect square, the latter method should be used. 
 
 8. Find the square root of /g to five places. 
 ^5=2.23606+, V16=4, ^-^==^^^3""°+— .55901+. 
 Or, /g=.3125, and V-3125=.559014-. 
 
 9, Find the square root of ?. Am. .774596-j-. 
 
 10. Find the square root of 1^. Ans. 1.11803+. 
 
 11. Find the square root of 3f . Ans. 1.903943+. 
 
 12. Find the square root of llf . Ans. 3.349958+. 
 
 13. Find the square root of y*iv. Ans. 0.645497+. 
 
 14. Find the square root of 17f . Ans. 4.168333+. 
 
 EXTRACTION OF THE SttUAEE ROOT OF ALGEBRAIC 
 aU ANTITIES. 
 
 EXTRACTION OP THE SQUARE ROOT OF MONOMIALS. 
 
 Art. 182. From Art. 172, it is evident that to square a 
 monomial, we must square its coefficient, and multiply the expo- 
 nent of each letter by 2. Thus, 
 
 (3 mra2)2=3 mra2 X 3m7i2=9m2m^ 
 
 Therefore J9ni'n^=Smn'. Hence, we have the following 
 
 RtTLE FOR EXTRACTING THE SQUARE ROOT OF A MONOMIAL. — Ex- 
 tract the square root of the coefficient and divide the exponent oj 
 each Utter hy 2. 
 
 Since +flX+o^+o', — aX — a=+a'; 
 
 therefore, ^a'=-\-a, or — a. 
 
 Hence the square root of any positive quantity is either ^Ziis, 
 or minus. This is generally expressed by writing the double 
 sign before the square root. Thus, ^4La^=-±S'a; which is read 
 flus or minus 2a. 
 
 If a monomial is negative, the extraction of the square root is 
 impossible, since the square of any quantity, either positive or 
 
 negative, is necessarily positive. Thus ^ — 4, ^ — a^b^, ^ — b, 
 are algebraic symbols, which indicate impossible operations. 
 Such expressions are termed imaginary quantities. An example 
 of their occurrence always arises, in proceeding to find the value 
 of the unknown quantity in an equation of the second degree, 
 when some absurdity, or impossibility exists in the equation, or 
 in the problem from which it was derived. 
 
134 RAY'S ALGEBRA, PART SECOND. 
 
 EXAMPLES FOR PRACTIC] 
 
 1. 16xY. Ans. ±4ij/'. 
 
 2. 25mW. Ans. ±5m7!. 
 
 3. 36j:V. Ans. ±QxH^. 
 
 4. Sla'iV. ^7«. rfc9ai;l 
 
 5. m^x^y^z'. Ansdtmx^fz*. 
 
 6. 1024a2i«2"'. Atw. 32ai'i' 
 
 Hence, iojZrad iAe square root of a monomial fraction, extract Ihf 
 ijuare root of holh terms. 
 
 7. Find the square root of -— -. Ans. db— r.. 
 
 8. Find the square root of -^—„ Arts. ±-^,. 
 
 25o'6^ aab' 
 
 EXTRACTION OF THE SaUAKE ROOT OF POLYNOMIALS. 
 
 Art. 183. In order to deduce a rule for extracting the square 
 root of polynomials, let us first find the relation that exists be- 
 tween the several terms of any quantity and its square. 
 
 (a+J)'=a2+2o6+i2=a='+(2a+i)i. 
 
 (a-i-6 + cf = 02 + 2aft + i24-2ac-|-2ic+c2=a2+(2a+i)J+(2a 
 +2i-l-c)c. 
 
 (a _(_ i_|_c + d)2 = «= 4- 2a5+6=+2ae+2 Jc+c=+2ai+2fid+2cd 
 +d2=o24-(2a+&)H-C2a+26+c)<;+(2o+2i+2c+<i)d. 
 
 Or, by calling the successive terms of a polynomial, r, r', r'', 
 r'", and so on, we shall have {r-\-r' -\-r" -\-r"'y=r^-\-(^r-\-r')r' 
 4-(2r+2r'+r")r"+(2r+2r'+2r"+r"')r"', where the law ia 
 manifest. 
 
 In this formula, r, r', r", r'", may represent any algebraic quan- 
 tities whatever, either integral or fractional, positive or negative. 
 
 This formula gives the following law: 
 
 The square of any polynomial is equal to the square of the first 
 term — plus turice the first term, plus the second, multiplied by the 
 second — plus twice the first and second terms, plus the third, multi- 
 plied by the third — plus tioice the first, second, and third terms, plus 
 the fourth, multiplied by the fourth, and so on. Hence, by reversing 
 the operation, we have the follow ing 
 
 Rule fok extracting the square root of a poltnomiai,. 
 
 1st. Arrange the polynomial vrilh reference to a certain letter; then 
 find the first term of the rQot by extracting the square root of thi 
 first term of the polynomial; place the result on the right, and sub- 
 tract its square from the given quantity. 
 
EXTRACTION OF THE SQUARE ROOT. 135 
 
 2nd. Divide the first term of the remainder by double the part of the 
 root already found, and annex the result to both the root and the 
 divisor. Multiply the divisor thus increased by the second term 
 of the root, and subtract the product from the remainder. 
 
 3rd. Double the terms of the root already found for a partial divisor, 
 tlien divide the first term of the r^emainder by the first term c,f Iht 
 divisor, and annex the result to both the root and the partial divisor. 
 Multiply the divisor thus increased by the third term of the root, 
 and subtract the product from the last remainder. Then proceed in 
 a similar manner to find the other terjns. 
 
 Remark. — If in the course of the operations in any example^ we find 
 a remainder of which the first term is not exactly divisible by double 
 the first term of the root, we may conclude that the polynomial is not a 
 perfect square. 
 
 1. Find the square root of 4:X^y^-\-12x'y-\-9x- — 2Qxy' — 20xy' 
 +2by\ 
 
 Arranging the polynomials with reference to y, we have 
 
 ROOT. 
 
 25y''—20xy^+4xY—30xy^+12x^y-\-9x^\5y^—2xy—3x 
 'i^ 
 
 lQy''—2xy —2Qxy^-\-ixY 
 -2i)xy^-\-ixY 
 
 IQy-i—Axy—Zx — 300^2+1 2ar2^+9a;2 
 — 30aj/^+12a:'y+9j' 
 
 The square root of the first term is by'', which we write as the 
 first term of the root. We next subtract the square of by'' from 
 the given polynomial, and then divide the first term of the re- 
 mainder, — 2Qxy^, by lOj^^, which gives — 2ocy, the second term 
 of the root. We then place — 2ay in the root and also in the 
 divisor, and multiply the divisor, thus increased, by — ~xy, and 
 subtract the product from the first remainder. We then double 
 by'' — 23nj, the terms of the root already found, for a partial divi- 
 sor, and divide the first term, lOy' of the divisor, into — 30xi/', the 
 first term of the remainder, which gives — 3a; for the third term 
 of the root. Completing the division, multiplying by — 3a;, end 
 subtracting, we find there is nothing left. 
 
 Note. — The first remainder consists of all the terms after the 
 first subtraction, and the second, of all the terms after the second 
 subtraction. It is useless to bring down more terms than have 
 corresponding terms in the quantity to be subtracted. 
 
 If the preceding example be arranged according to the powers 
 
136 RAY'S ALGEBRA, PART SECOND. 
 
 nf X, the root found will be 3x-f-2iry — 5y'. This is correct, also 
 ss may be shown (renerally, thus, 
 
 ,y(polynomial)'=±(polynoniial). 
 Thus, V(a'+2aa;+a;2)=^(o+a;)'=±(<r+a;)=a+a; or — a— j: 
 
 EXAMPLES FOR PRACTICE. 
 
 2. a:'+6<M:4-9o=. Ans. a;+3o. 
 
 3. 16a;2— 40xi^+25y'. An^. 4a!— 5y. 
 
 4. 4x'z'— 12iy2+9y^ Aras. 2a;z— 3jf. 
 
 5. 49a<'»-» — 42a«'"-'-[-9o«'»+'. Atw. Ta^*""-'— Sa^^+i. 
 
 6. l+2x4-7a:'+6a;5+9a:'. Atw. l+ar-fSa;!. 
 
 7. 9a<— 12a'6+34a2i2— 20aJ'4-25J4. Atm. Sa^— 2aJ+5J3. 
 
 8. x«+4a;'4-10a:<+20x'+25a;»+24a;+16. 
 
 Atw. a:'+2a:'4-3j:+4. 
 
 9. 9i»— 6a3/+30aa+6xt4-y'— lOj^jf— 23//+25z2+102<+/2, 
 
 Ans. 3a!— y+5z-|-'. 
 
 10. x<— 2a!»+?^'_f+ 1 . Ans. a!'— a:+-i 
 
 2 2 TB Tf 
 
 4 3 ^9 2 3 
 
 12. x-'-{-Ux-\--—bii^lab-iJ'l. Ans. x-{-- — ^ 
 
 jg 1051^'_6^_l^'4.9+49^. A^. 7;.=-? +3. 
 
 14. ?!_2+i' Ans.f-* 
 
 15. ?!4.5i4.^Li^+3. An.. ?+*+!. 
 
 16. Reduce the following expression to its simplest form, and 
 extract the square root. 
 
 (a— J)^— 2Ca2+i2)(a— i)=+2(a4+6*). Ans. a'+b'. 
 
 17. Find the square root of 1 — ^ai' to five terms. 
 
 Ans. l-^_?L-^_5^- &c. 
 2 8 16 128 
 
 18. Find the first five terms of the square roots of ai^+a', and 
 of ,'-2aa!. ^^ x+?l_^ . ^_^. + &c. 
 
 2a: 8x'^16a!5 128x'^ 
 
 J a^ a* 5a^ o 
 
 and X — a — - " ^ &c. 
 
 2a: 2a;2 8a:' 
 
 Akt. 181. The following remarks will be found useful 
 
 1st. /Vo binomial can be a perfect square; for the square of e 
 
 Kionomia) is a monomial, and the square of a binomial is a tri- 
 
EXTRACTION OF THE CUBE ROOT. 13T 
 
 nomial. Tims a'-\-b^ is not a perfect square, but if we add to it, 
 or subtract from it, 2ab, it becomes the square of a-{-h or of a — b. 
 2nd. In order that a trinomial may be a perfect square, the two 
 extreme terms must be perfect squares, and the middle term the 
 double product of the square roots of the extreme terms. Hence, 
 to find the square root of a trinomial when it is a perfect square 
 extract the square roots of the extreme terms, and unite them by the 
 tign plus or minus, according as the second term is plus or minus. 
 Thus, o^"* — 4o'"+"-|-4a2" is a perfect square, since ^a''"=a'", 
 ^4<z'"=2a", and -(-a^X— 2a''X2= — la^t". But 4a;2+8a^+9!/' 
 is not a perfect square, since ,JAx^=2x, ^9y^=3y, and 2xX3y 
 X2=12ay, which is not equal to the middle term Sxy. 
 
 III. EXTRACTION OF THE CUBE ROOT. 
 EXTRACTION OF THE CUBE ROOT OP NUMBERS. 
 
 Art. 185. The cube or third power of a number, is the pro- 
 duct arising from taking it three times as a factor. (Art. 172.) 
 The cube root, or third root is one of three equal factors into which 
 it may be resolved; hence, to extract the cube root of a number, is 
 to find a number which, taken three times as a factor, will produce 
 the given number. 
 
 Art. 186. To show the relation that exists between the num- 
 ber of figures in any given number, and the number of figures in 
 its cube root. 
 The first ten numbers and their cubes are: 
 roots, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10; 
 cubes, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000. 
 The numbers in the second line are the cubes of those in the 
 first; and, reciprocally, the numbers in the first line are the cube 
 roots of those in the second. We see from this that the cube of 
 a number consisting of one place of figures, does not exceed 
 three places. 
 
 Again, comparing the numbers 10 and 100 with their cubeSj 
 we have, 
 
 numbers, 10, 100; 
 
 cubes, ioOO, iOOOOOO. 
 
 Since the cube of 10 is 1000, and the cube of 99, which is 
 less than 100, is less than iOOOOOO; therefore, the cube of a 
 number concisting of two places of figures, has more than three 
 places and not more than site places of fiarures. 
 12 
 
138 RAY'S ALGEBRA, PART SECOND. 
 
 Again, since the cube of 100 is lOOOOOO, and the cube of 
 1000 is iOOOOOOOOO; therefore, the cube of a number consist- 
 ing of three places of figures has more than six places, and not 
 more than niTie places of figures. Therefore, if we begin at llie 
 unit's place of a number, and separate it into periods of three 
 places each, the number of periods will show the number of places 
 of figures in the root. The left period will often contain only 
 ane or two figures. 
 
 Abt. 187. To investigate a rule for the extraction of the cube 
 root. 
 
 The first step in this investigation is to show the relation that 
 exists between any number composed of units and tens, and its 
 cube. 
 Let t^ the tens and u= the units of a given number. 
 Then t-\-u= the number, 
 and (_t-\-uy= the cube of the number. 
 
 Hence, the cube of any number consisting of tens and units, j'j 
 equal to the cvhe of the tens, — plus three times the square of tlie lens, 
 plus three times the product of the tens and units, plus the square of 
 the units, all three multiplied by the units. 
 
 With this principle, let us proceed to extract the cube root of 
 13824. 
 
 . tu 
 13824|24 
 8 
 3<2=1200 
 3tu= 240 
 M»= 16 
 
 1456 
 
 5824 
 5824 
 
 We commence, by separating the number into periods, by 
 placing points over the figures in unit's and thousand's places; 
 and as there are two periods, there will be two figures in the root. 
 We find the greatest cube in 13 (thousand), which is 8 (thousand); 
 the cube root of this is 2 (t); and its cube, 8 (thousand), corres 
 •ponds to t' in the formula. 
 
 We then subtract this from the given number, and find a re 
 mainder 5824, which corresponds to (2t^-\-2tu-^u^)u in the fot- 
 mula. The first term, 3t', of this formula, is sometimes termed 
 the trial divisor, as it is used to find the unit's figure u. 
 
 If the remaining terms were only St'u, we could readily find a 
 by dividing by 3i', but if we divide by St^, we may obtain a figure 
 
EXTRACTION OF THE CUBE ROOT. 139 
 
 too large, on account of omitting the terms 3iu-{-u', of which « 
 is as yet unknown. But if we first obtain a figure too large, at a 
 second trial we must take one that is less. 
 
 Since the square of tens is hundreds, therefore, in using three 
 limes the square of the ten's figure as a trial divisor, we must 
 omit the figures (24) in the unit's and ten's places ol the 
 dividend. 
 
 In this case we find 12 is contained in 58 four times. This 
 gives 4 (u) for the required unit's figure, and we now proceed to 
 complete the divisor by first adding to 3/', three times the pro- 
 duct of the tens by the units (Ztu), and writing the product in 
 ten's place, since the product of tens by units gives a pro- 
 duct of tens. We next write the square of the unit's figure 
 («'), and then taking the sum, find the complete divisor 1456 
 which corresponds to Zt^-\-Ztu-\-u'. Multiply this by 4 (u) 
 the product is 5824, which subtracted from the first remain- 
 der leaves zero (0), and shows that 24 is the exact cube root 
 required. 
 
 In cubing the tens, it is customary to omit the ciphers; but in 
 taking three times the square of the tens, also in taking three 
 times the product of the tens by the units, it is best to write 
 ciphers in the vacant orders. 
 
 2. Let it be required to find the cube root of 44361864. 
 
 , htu 
 443618641354 
 
 27 
 
 3A'==2700 
 
 3ht= 450 
 
 t'= 25 
 
 3175 
 
 17361 
 
 15875 
 
 3(J+<)==367500 
 
 3(,h-\-t)u— 4200 
 
 u^= 16 
 
 371716 
 
 1486864 
 
 1486864 
 
 After separating the number into periods, we find the cube root 
 (35) of 44361 on the same principles as in the preceding example. 
 Then considering 35 (lOi-j-') as so many tens, we find the unit'« 
 figure (4), as in the preceding example. 
 
 In dividing by the trial divisor 27, to find the second figure (5), 
 we first obtain 6, but as this is found by trial to h» too large, we 
 take a less number. 
 
140 RAY'S ALGEBRA, PART SECOND. 
 
 From the preceding we derive the following 
 
 Rule for the extraction of the cube root of numbers.— 
 
 1st. Separate tlie given number into periods of three places each, hegin- 
 ning at the unit's place. (The left period will often contain but 
 one or two figures.) 
 
 2nd. Find the greatest cube in the left period, and place its root on 
 the right, as in division. Subtract the cuJbe of the root from the 
 left period, and to the remainder bring down the next period for a 
 dividend. 
 
 3rd. Square the root already found, and muUiply it by 3 for a trial 
 divisor. Find how many times this divisor is contained in the 
 dividend, omitting the unit's and ten's figures, and write the resuU 
 in the root. Add together, the trial divisor with two ciphers an- 
 nexed; three times the product of the last ^figure of the root by the 
 rest, with one cipher annexed; and the square of the last figure; the 
 sum will be the complete divisor. 
 
 4th. Multiply the complete divisor by the last figure of the root, and 
 subtract the product from the dividend, and to the remainder bring 
 down the next period for a new dividend, and so proceed until all the 
 periods are brought down. 
 Extract the cube root of the following numbers. 
 
 3. 
 
 12167. 
 
 Ans. 23. 
 
 9. 
 
 127263527. Ans. 503. 
 
 4. 
 
 39304. 
 
 Ans. 34. 
 
 10. 
 
 403583419. Ans. 739. 
 
 5. 
 
 493039. 
 
 Ans. 79. 
 
 11. 
 
 1883652875. 
 
 6. 
 
 2097152. 
 
 Ans. 128. 
 
 
 Ans. 1235, 
 
 7. 
 
 14348907. 
 
 Ans. 243. 
 
 12. 
 
 158252632929. 
 
 8. 
 
 43614208. 
 
 Ans. 352. 
 
 
 Ans. 5409. 
 
 Since a fraction is cubed by cubing its numerator and denomi- 
 nator, therefore, the cube root of a fraction may be found by ex- 
 tracting the cube root of both terms, the fraction being in its 
 lowest terms before commencing the operation, for reasons similar 
 to those given in Art. 177. 
 
 13. Find the cube root of j'W'j. Ans. %. 
 
 14. Find the cube root of aVj^. Ans. ^. 
 
 Art. 188. A number whose cube root can be exactly ascer- 
 tained is a. perfect cube. Thus, 8, 27, 64, &c., are perfect cubes. 
 These numbers, like perfect squares, are comparatively few. 
 
 A number whose cube root cannot be exactly ascertained is 
 termed an imperfect cube. Thus, 2, 3, 4, &c., are imperfect cubes. 
 
 It may be shown, by a course of reasoning precisely similar to 
 that employed in Art. 179, that the cube root of an imperfect cubt 
 tannot be a fraction.. 
 
EXTRACTION OF THE CUBE ROOT. 141 
 
 APPROXIMATE CUBE ROOTS. 
 
 Art. 1§9. To illustrate the method of finding the approxi- 
 mate cube root of an imperfect cube, let it be required to find the 
 cube root of 6 to within |. 
 
 Reducing 6 to a fraction whose denominator is 64 (the cube of 
 4 the denominator of the fraction I), we have 6=-'//-. 
 
 Now the cube root of 384 is greater than 7 and less than 8; 
 therefore the cube root of -^//- is greater than -J and less than f," 
 hence f is the cube root of 6 to within less than \. 
 
 To generalize this method, let it be required to extract the 
 
 cube root of a number a, to within a fraction _. 
 
 n 
 
 Let r be the root of the greatest cube contained in an'; then 
 ^ is comprised between !L and ^'"~'~ ■' ; hence its cube root is 
 
 comprised between - and ~ ■; and since the difference of these 
 
 n n 
 
 \ r 1 
 
 fractions is _, therefore _ is the cube root of a to within -. 
 n n n 
 
 From this we derive the following 
 
 Rule foe extracting the cube root of a whole itomber to 
 WITHIN A given fraction. — Multiply the given number by the 
 cube of the denominator of the fraction which determines the degree 
 of approximation; extract the cvhe root of this product to the nearest 
 unit, and divide the result by the denominator of tlte fraction. 
 
 2. Find the cube root of 5 to within -g. Ans. 1|. 
 
 3. Find the cube root of 10 to within g. Ans. 2g. 
 
 Since the cube of 10 is 1000, the cube of 100, 1000000, and 
 so on, the number of ciphers in the cube of the denominator of 
 a decimal fraction, is equal to three times the number in the de- 
 nominator itself. Therefore, when the fraction which determines 
 the degree of approximation is a decimal, it is merely necessary to 
 add three ciphers for each decimal place required; and- after extracting 
 the root, to point off from the right one place of decimals for each 
 three cipliers added. 
 
 4. Find the cube root of 2 to five places. Ans. 1.2.5992. 
 
 5. Find the cube root of 9 to five places, Ans. 2.08008. 
 
 6. Find the cube root of 37 to six places. Ans. 3.332222. 
 
142 RAY'S ALGEBRA, PART SECOND. 
 
 By adding ciphers to both terms, any decimal may be written 
 in the form of a fraction, having its denominator a perfect cube; 
 thus, .2=f\P5%, .25=f\f(P0, and so on. Therefore the cube root 
 may be found, as in the preceding examples, by annexing ciphers 
 to the given decimal, until the number of decimal places shall be equal 
 to three times the number of decimal places required in the root. 
 Then, after extracting the root, painting off from the right the required 
 number of decimal places. 
 
 7. Find the cube root of .4 to four places. Ans. .7368. 
 
 8. Find the cube root of .25 to five places. Ans. .62996. 
 The cube root of a whole number and a decimal may be found 
 
 in the same manner. Thus, the cube root of 6.4, is the same as 
 the cube root of f J88, which is 1.85663-|-. 
 
 9. Find the cube root of 12..5 to five places. Ans. 2.32079. 
 10. Find the cube root of 34.3 to six places. Ans. 3.249112. 
 
 To find the cube root of a fraction or a mixed number, reduce 
 the fraction to a decimal, and then proceed as in the preceding 
 examples. 
 
 11. Find the cube root of |. Ans. .82207+. 
 
 12. Find the cube root of 5|8f. Ans. 1.816+. 
 
 13. Divide the cube root of ^^^^iW^ ^^ *^ square root of the 
 square root of 8.3521. Ans. .25. 
 
 14. Add together the cube roots of .059319 and 4.173281; 
 and multiply the sum by the square root of 105yg. Ans. 20.5. 
 
 EXTRACTION OF THE CUBE ROOT OF ALGEBRAIC ttUANTITlES. 
 EXTEACTIOH OF THE CUBE BOOT OF MONOMIALS. 
 
 Art. 190. If we cube a monomial, for example, 2fla;', we 
 have 
 
 (2ax2)3_2aa;2x2(Kc=X2(u:'=23aVX5=8a3a:». 
 
 That is, to cube a monomial, we must cube the coefiicient, and 
 multiply the exponent of each letter by 3. Hence, by a converse 
 operation we have the following 
 
 Rule foa extracting the cube root of a monomial. — Extract 
 the cube root of the coefficient, and divide the exponent of each 
 letter by 3. 
 
 Find the cube root of each of the fc^owing monomials. 
 
 1. 
 
 Si^z'. 
 
 Ans. 2xz^. 
 
 4. 
 
 — 64o'm«. 
 
 Ans. —iam\ 
 
 2. 
 
 27a:y* 
 
 Ans. 3xy. 
 
 5. 
 
 ^3m-f-3cj,6_ 
 
 Ans. a" 'a' 
 
 a. 
 
 — "««. 
 
 Ans. —2a'. 
 
 6. 
 
 —x^nz^. 
 
 Ans. —xp^z^. 
 
EXTRACTION OF THE CUBK ROOT. 149 
 
 Since ( ? ) =°x-X-=-; therefore, = /^=1 
 
 Hence, to find the cube root of a monomial fraction, extract tht 
 tube root of both terms. 
 
 7. Find the cube root of . Ans. — ? 
 
 27a» 3x'- 
 
 8. Find the cube root of — ^l^y_ Xn». — -i^' 
 
 126mW 5mn'' 
 
 EXTRACTION OF THE CUBE KOOT OF POLTNOMIALS. 
 
 Art. 191. To investigate a rule for extracting the cube root 
 of polynomials. 
 
 Let us first examine the relation that exists between a poly- 
 nomial and its cube. 
 
 (a+6)3=o3+3a'i+3a5»+i'=a'+(3a'+3ai+J2)6. 
 
 (^a + b + cy=l(,a + b) + cy = (_a+by+ |3(<^f 6)=+3(a+i)c 
 
 +c^c. 
 
 (a+b+c+dy= \ (,a+b+e)+d 1 3=(a+b+cy+ 
 
 j 3 (a+i+c) 2+3 (,a-{■b-\-c)d+d^d. 
 
 Hence, the cube of a polynomial is formed according to the 
 following law : 
 
 The cube of a polynomial is equal to the cube of the first term — 
 pZus three times tlie square of the first term, plus three times the pro- 
 duct of the first term by the second, plus the square of the second, all 
 ihree.multiplied by the second — plus three times the square of the first 
 two terms, plus three times the product of the first two terms by the 
 third, plus the square of the third, all three multiplied by the third 
 and so on. 
 
 From this law, by reversing the process, we derive the fol- 
 lowing 
 
 Rule for the extraction of the cube root of a poltno- 
 MIAL. — 1st. Arrange the polynomial with reference to u, certain 
 letter. Extract the cube root of the first term, this will give the first 
 term of the root, and subtract its cube from the given polynomial. 
 
 2nd. Take three times the square of the first term of tlie root, and call 
 it a trial divisor for finding each of the remaining terms of the 
 root. Find how often the trial divisor is contained in the first term 
 of the remainder, this will give the second term of the root. Then 
 form a complete divisor by adding together three times the square of 
 the first term of the root, plus three times the product of the first term by 
 the second, plus tki square cf the second. Multiply these by Ihi 
 
144 RAY'S ALGEBRA, PART SECOND. 
 
 second term of the root and subtract the prodvct from the firsi 
 remainder. 
 
 3rd. Again find how often the trial divisor is contained in the first 
 term of the remainder, this will give the third term of the root. 
 Then form u complete divisor as before, hy adding together three 
 times the square of the first and second terms, plus three times the 
 product of the first and second terms hy the third, plus the square 
 of the third. Multiply these hy the third term of the root and sub- 
 tract the product from the last remainder. 
 
 4 th. Continue this process till all the terms of the root are found. 
 
 Note. — The remainder in each case, is all the terms left after each 
 sabtraction. 
 
 1. Find the cube root of x«— 6a;'+]2a!<-|-3o2a;'— 8a;'— 12aV 
 -|-12o=x2-f So"!'— eoix+o'. 
 
 a.6_-6a;'+12i*+3oV— 8a'— 12aV-^12oV+ao<a;'— 6o<a;+o« 
 
 a;'' |a;°— 2j:+a' 
 
 3 , <_6a;'+4a;2)— 6i*+12x<— 8a;' 
 
 — 6j'4-12^— 8^:' 
 
 3x<— 1 2a;'+l 2a;2+3 a=a;'— 6a=x+o<)+3 o V— 1 2a=a;'+l ^a'x^ 
 
 I +3a''a;2— 6a''a;+o«. 
 
 To bring the work within the page, the last _L3o2^4 i2a'x'+12a'x' 
 
 remainder and subtrahend are each written in i q d 9 c d i « 
 
 ,. -'r-Ca'X' Dffi'X+B''. 
 
 two hnes. 1 ! ' 
 
 We first extract the cube root of x", which gives x' for the 
 first term of the required root. Then 3 times the square of this, 
 3(x')'=3x*, constitutes the trial divisor for finding the remaining 
 terms. To find the second term of the root we divide 3x^ into 
 — 6xS the first term of the remainder, which gives — 2x, the sec- 
 ond term of the root. We then form the complete divisor by 
 adding together 3(x=)2+3(xX— 2x)+(— 2x)2=3xi— 6x'-l-4x'. 
 Multiplying this by the second term, — 2x, and subtracting the pro- 
 duct from the first remainder, the first term of the second remain- 
 der is -t-3a'x', which, divided by the trial divisor, gives -\-a^, for 
 the third term of the root. We next find the complete divisor by 
 adding together 3(x2— 2x)2 + 3(x'— 2x)o2-(-(o')2=3x4— 12x' 
 -l-12x2-|-3o'x' — 6a'x-\-a*. Multiplying this by a' and subtract- 
 ing, there is no remainder ; hence the root obtained is exact. 
 
 Find the cube root 
 
 2. Of o'-f24o'J+192ai2+512J'. Ans. a+8i. 
 
 3. Of 80'— 84a'x+294ax'— 343x'. Ans. 2o— 7x. 
 
 4. Of o«— 6o'+15a<— 20a'+15<i2— 6a-f 1. " 
 
 Anji. a' — So-j-I. 
 
EXTRACTION OF THE CUBE ROOT. 145 
 
 5. Of i«— 9x*+39a;«— 99a:»+156x»— 144a;+64. 
 
 Ans. x^ — 3a;+4. 
 
 6. Of (a+l)«"i'— 6coP(a+l)<"a;= + 12cV''(a+l)="a:— 8c'aS' 
 
 Ans. (o-fl)'"a;— 2coP. 
 
 7. Find the first three terms of the cube root of 1 — a;. 
 
 X 
 
 '3' 9" 
 
 IV. EXTRACTION OF THE FOURTH ROOT, SIXTH 
 ROOT, n'° root, &C. 
 
 Art. 192. The fourth root of a number is one of four equal 
 factors, into which the numljer may be resolved ; and in general, 
 the n"" root of a number is one of the n equal factors into which 
 Ihe number may be resolved. 
 
 When I'le degree of the root to be extracted is a multiple of 
 two or more numbers, as 4, 6, &e., the root can he obtained by ex- 
 tracting the roots of more simple degrees. 
 
 To explain this we remark, that 
 
 (a'y=a'Xa'Xa^Xa'=a^+'+'+'=a''<*=a" 
 and in general 
 
 (o'")"=a'"Xa'"Xa'"Xa'". • • • =a'"X"=a"'». 
 Hence, the u'* potver of the m" power of a nimiber, is equal to 
 the mn'* power of the number. 
 
 Reciprocally, the mn" root of a number, is equal to the n" root 
 of the m" root of that number ; that is 
 
 To prove this, let \l'ija=a' ; 
 
 raising both members to the n"' power, we have 
 
 and by raising both members of the last equation to the m* 
 power, a=a''^ ; 
 
 extracting the mn"' root of both members, 
 
 But, by supposition \J'i]a^a' ; 
 
 thsrefore, '^Ja=')J'iJa. 
 
 It may be proved similarly, that "iya="y'^a. 
 
 From this it follows that ija=\j^a; and ^a=\V>o' 
 
 \}l/a; in like manner 5/"=\ V V**' ^""^ ' 
 
146 RAY'S ALGEBRA, PART SECOND. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Find the 4th root of 65536. Ans. 16. 
 
 2. Find the 4th root of 1500625. Ans. 35. 
 
 3. Find the 4th root of 13107.9601. Ans. 10-7. 
 
 4. Find the 6th root of 2985984. Ans. 12. 
 
 5. Find the 6th root of 11390625. Ans. 15. 
 
 6. Find the 8th root of 214358881. Ans. 11 . 
 7 Find the 4th root of 81a*xK Ans. Sax'. 
 
 8. Find the 4th root of 1^|^. Ans. if' 
 
 625o«c'' 5a'c 
 
 9. Find the 4th root of a*-\-4a?bx-\-6a'b'x^+iab'x>+b*x*. 
 
 Ans. a-\-bx, 
 
 10. Find the 4th root of ^J+lfL'+JI-Vl^+e. Ans. ^-f? 
 
 y* y' X* X' y X 
 
 11. Find the 4th root of a"— 4a;8+10a^— 16a:2+19— ?f+l^ 
 
 x' X* 
 
 -i-i-L 1 
 
 a;»~a;8- Ans. a'— 1+- 
 
 12. Find the 6th root of a«+I— 6 f a*+\ ) +15 (a'+-] 
 
 a' V a* ' \ a? I 
 
 —20. 1 
 
 Ans: a — _ 
 a 
 
 Art. 193. It has been shown already (Arts. 182, 183,) that 
 the square root of a monomial, or a polynomial, may be preceded 
 either by the sign +, or — ; we shall now explain the law ir 
 regard to the roots generally. 
 
 If we take the successive powers of -\-a, we have 
 
 +o, +0', -\-a?, -|-oS 
 
 the successive powers of — a, are 
 
 -a, +a', —a?, +aS . . . +0'", — a="+'. 
 
 From this we see that every eoen power is positive, and that an 
 aid, power has the same sign as the root. 
 
 In general, let n be any whole number, then every power of 
 an even degree, as 2», may be considered as the n"* power of 
 the square, that is, o^"=:(a2)n_ 
 
 Hence, cmry 'power of an even degree is essentially positive, 
 HihetJier the quantity itself be positive or negative. 
 
 Thus, (±3a)''=+81o«; (±262)«=+64i". 
 
 Again, as every power of an odd degree (271-4-1) is tlie product 
 of a power of an even degree, 2n, by the first power, it follows, 
 
EXTRACTION OF THE FOURTH ROOT. 147 
 
 that every power of an uneven degree of a monomitd has the same 
 sign as the monomial itself. 
 
 Thus, (_+2ay=+8a\ (— 2a)'=— 80'. 
 Hence, it is evident, 
 
 1 St. That every odd root of a monomial must have the same ttgn 
 as the monomial itself. 
 
 Thus, V+8a'=+2a, XISa'=—2a, «/— 32o'»=— 2o». 
 2nd. That an even root of a positive monomial may be either posi- 
 tive or negative. 
 
 Thus, */81aW=±3a62, {/6ia^^=-±2a^. 
 
 3rd. That every even root of a negative monomial is impossible , 
 since no quantity raised to a power of an even degree can give 
 a negative result. Thus, ij — a', {J — b, \J — c, are symbols of 
 operations which cannot be performed. They are imaginary 
 expressions like ^ — a, J — 6. (Art. 182.) 
 
 TO EXTRACT THE n"^ HOOT OF A MONOMIAL. 
 
 Akt. 194. In raising any monomial to the n" power accord- 
 ing to the rule, Art. 172, it is obvious that the process consists 
 in raising the numeral coefficient to the w'* power, and multiply- 
 ing the exponent of each letter by n, thus, (2a'J<)'=2a2i*X2a'i* 
 X2a2M=23a2".'fx3=8a«i'2. 
 
 Hence, conversely, to find the m"* root of a monomial. 
 Extract the n'* root of the coefficient, and divide the exponent oj 
 each Utter by n. 
 
 Remark. — In the following examples, the pupil is expected to fine 
 the root of the numeral coefficient by inspection, as we have given na 
 rules for extracting the 5th, 7th, &c., roots of numbers. Indeed, in the 
 present state of science such rules are useless, for when the operations 
 are required they are readily performed by means of Logarithms. 
 
 1. Find the 5th root of — 32a*a;"i. Ans. —2ax^. 
 
 2. Find the 6th root of 729Mc". Ans. d=36c». 
 
 3. Find the 7th root of 128x'y". Ans. 23^" 
 
 4. Find the 8th root of 6561a8Ji«. Ans. ±3ai'. 
 
 5. Find the 9th root of —512x^z". Ans. ~2xz\ 
 
 6. Find the 10th root of 1024i'»2". Ans. ±2bz'. 
 
 7. Find the m" root of ab^e^. Ans. ici{";i/a). 
 
 8. FiXtract '^ a^'b^c^". Ans. a^hc'. 
 
148 RAY'S ALGEBRA, PART SECOND. 
 
 V. RADICAL QUANTITIES. 
 
 Note. — These quantities are generally called surds by English writers j 
 while the French more properly term them radicals, from the Latin 
 word radix, a root, because they express the roots of quantities. The 
 Germans also distinguish them by a synonymous term, umrzel groasm, 
 {root quantities)^ 
 
 Art. 195. A rational quantity is one either not affected by the 
 radical sign, or of which the root indicated can be exactly ascer 
 tained ; thus, 2, a, ^4, and 1/8 are rational quantities. 
 
 A radical quantity is one of .which the root indicated cannot be 
 exactly expressed in numbers ; thus, ^5 is a radical ; its value is 
 2.23606797 nearly. 
 
 Radicals are frequently called irrational quantities, or svrds. 
 
 Art. 196. From Art. 193 it is evident that when a monomial 
 is a perfect power of the m'* degree, its numeral coefficient is a 
 perfect power of that degree, and the exponent of each letter is 
 divisible by n. Thus 4a' is a perfect square, and 8a' is a per- 
 fect cube ; but 6o' is not a perfect square, because 6 is not a per- 
 fect square, and 3 is not divisible hy 2 ; also, 8a^ is not a perfect 
 cube, for, although 8 is a perfect cube, the exponent 4 is not 
 divisible by 3. 
 
 In extracting any root, when the exact division of the expo- 
 nent cannot be performed, it may be indicated by writing the 
 divisor u'nder it in the form of a fraction. Thus, Ja' may be 
 
 5 — 4 
 
 written o^, and f/a^ may be written a» ; and in general the »" 
 root of the m'* root of any quantity, is expressed either by ^o", 
 
 m 
 
 or an. 
 
 Since a is the same as a' (Art. 19), the square root of a may 
 
 be expressed thus, a' ; the cube root thus, a' ; and the «"* root 
 
 thus, a". Hence, the following expressions are to be considered 
 equivalent : 
 
 J a and a', 
 %Ja and a', 
 "^a and a". 
 Also, %la? and o', 
 
 m 
 
 Vji^ and o"' 
 From this we see, that the numerator of the fractional exponent 
 ^•niotes the power of the quantity, and the denominator the root ofOtat 
 power to ie extracted. 
 
RADICALS. 149 
 
 Art. 197. Theorem. Any quantity affected with a fractional 
 ixponent, may he transferred from one term of a fraction to the other, 
 if, at the same time, the sign of its exponent be changed. This pro- 
 position has already been established (Art. 81) when the ex- 
 ponent is integral ; we will now prove it when the exponent ie 
 fractional. 
 
 1 
 
 Let it be required to extract the cube root of — ,, and of its 
 
 equivalent a~'. 
 To extract the cube root of a fraction, we extract the cube root 
 
 3 A — 1 
 
 of each term (Art. 190), hence, W-j r. But, to extract the 
 
 cube root of a~' we must divide the exponent — 2 bv 3 (Art. 
 1 94), hence, 
 
 S/«-»=a-t 
 1 -f 
 
 Similarly, — =a~'". 
 Extracting the re'* root of each side, 
 
 ° " which establishes the theorem. 
 
 Art. 198. The quantity which stands before the radical sign 
 
 is called the coefficient of the radical. Thus, in the expressions 
 
 a^l, and 2l/c, the quantities a and 2 are called coefficients. 
 
 Radicals are said to be of the same degree when they have the 
 2 t — — 
 
 same index ; thus, o' and 5^, or ^a' and 5/5', are of the same 
 
 degree. 
 
 Similar radicals are those which have the same index, and the 
 
 same quantity under the radical sign ; thus, a^b and c^b are 
 
 similar radicals ; so, also, are 3lja' and S^o'. 
 
 REDUCTION OF RADICALS 
 
 Case I. — To reduce radicals to their most simple form. 
 
 Art. 199. Reduction of radicals consists in changing the foim 
 of the quantities, without altering their value. Reduction of 
 radicals of the second degree is founded on the following prin- 
 ciole : 
 
150 RAY'S ALGEBRA, PART SECOND. 
 
 The square root of the product of two or more factors is equal ir 
 the product of the square roots of those factors : 
 
 That is, ^ai=VaXV*^; which is thus proved; 
 (^'aby=ab. 
 And (VoXV6)'=(VaX V'*)X(V"^X V*)=o/a X Ja X 
 V^X ^b=ab. 
 Hence, ^ab and ,JaJb, are equal to each other j since the 
 square of each is equal to db. 
 
 By this principle, ^36=^4X9=2X3; ^144=^9X16 
 =3X4. 
 
 Any radical of the second degree can be reduced to a more 
 simple form when it can be separated into factors, one of which 
 is a perfect square. 
 
 Thus, V8=V4X2=V4XV2=2V2. 
 
 ^ m.V= ^ m*n''y,mn:=^ m*n''y, ,Jmm=rr?nJmn. 
 
 ^28^== V4aVxTa=V4^=X J^a=^acjla. 
 Hence, we have the following 
 
 Rule for the reduction or a radical or the second degree 
 TO ITS SIMPLEST FORM. Ist. Separate the quantity to he reduced, 
 into two parts, one of which shall contain all the factors that are 
 perfect squares, and the other the remaining factors. 
 
 8nd. Extract the square root of the part thai is a perfect square, arid 
 prefix it as a coefficient to the other part placed under tlie radical 
 sign. 
 
 To determine if any quantity contains a numeral factor that is 
 a perfect square, ascertain if it is divisible by either of the per- 
 fect squares 4, 9, 16, 25, 36, 49,64, 81,100, 121, 144, &c. If 
 not thus divisible, it contains no factor that is a perfect square, 
 and the numeral factor cannot be reduced. 
 
 Reduce to their simplest forms, the radicals in each of the fol- 
 lowing 
 
 EXAMPLES FOR PRACTICE. 
 
 1. >/12, ^18, ^45, V32, V50a', jn2a'bK 
 
 Ans. 2^3, 3^2, 3^5, 4^2, 5a,J2i, eabjTh. 
 
 2. ^245 ^448, ^810, JdOlbV, ^l8Q5a^'. 
 
 Ars 7^5, 8J1, 9^10, 13ic^3i, 19a'i/5 
 
RADICALS. 151 
 
 In a similar manner polynomials may sometimes be simplified. 
 Thus, 
 
 J(3a»— 6ffl=c+3fl;c=)=V3ffl(a^— 3ac+c')=(o— c)^3a. 
 
 3. V(o3— a^i), J ax^—Qcuc-\-9a, J (x^—y^Xx-^). 
 
 Ans. aj(_a—b), (ai— 3)^a, (a;+^/)^/(a^— y). 
 
 To reduce a fractional radical to its most simple form by the 
 game principle : Render the denominator of the fraction a per- 
 fect square by multiplying or dividing both terms by the same 
 quantity. Then separate the fraction into two factors, one of 
 which is a perfect square. Extract the square root of this fac- 
 tor, and write it as a coefficient to the other factor placed under 
 the rafdical sign. 
 
 _ .— » 
 
 4. Reduce ^i, and /?, to their simplest forms. 
 
 »• ^h Vf. ^/^f. 6^/^. 30^"^, 18^^. 
 
 Ans. ^J2, 1J6, 1^2, ^3, 3^30, |^10. 
 
 Q [^ J3a /ffl5^ / 3xj/' \ ^ 
 
 Art. 200. To reduce radicals of any degree to the most sim- 
 ple form. 
 
 The principle of Art. 199 may be generalized thus : 
 
 The n'" root of the product of two or more factors, is equal to tht 
 product of the n'* roots of those factors. 
 
 That is, Tjab^yaX'iJi; which is thus proved : 
 {"Jab)''=ab ; 
 and ('VaXV*)"=(V"a)"X(V*)"=aZ.. 
 
 Since the n" powers of these expressions are equal, the quanta 
 ties themselves are equal, that is, 
 
 ':j'ab=l/aXlJb. 
 
152 RAY'S ALGEBRA, PART SECOND. 
 
 1. Reduce ^54 to its most simple form.. 
 
 s/54= V27 X 2= 5/27 X V5=3 V^- 
 
 Similarly, »/|=V|x|xi = Vl|=VlV><^=l ^18- 
 
 Hence, for the eimplification of monomial radicals, we have the 
 following 
 
 General rule. — Separate the quantity into two factors, one of 
 which is a perfect power of the given degree ; extract its root and 
 prefix the result as a coefficient to the other factor placed under the 
 radical sign. 
 
 Reduce the radicals in each of the following examples to the 
 most simple form : 
 
 2. V40, X/80aV, %/81c*, \/128a'c>, »/162m<?i'. 
 
 Ans. 23/5^ 2ah%/i0a', Sc^Yc, ia'c^2?, ZmnlfQiimK 
 
 3. V320, s/2808, IJaV, V32, V144, V128. 
 
 Ans. 4S/5, 6»/l3, ah^'^, 2^2, 24/9, 2»/4. 
 
 4. VI. VI. VI. VI, Vh V75- 
 
 Ar^. ]/Jl, iV6, JV36, -JViS. ^V^ iV45. 
 
 5. «/162, ♦/'768, V1250, V3888. 
 
 4ns. 3V2, 4V3, 5V2, 6V3. 
 
 6. V64flS V32a»i', V48m«wS i/|- 
 
 Ans. 2alJIa, 2abiJ2^, 2m^nif3n, \ijE4. 
 
 7. V64, fJ^2Qa', «/A. Vf. V|. 
 
 4n«. 2^2, 3aV3a. iV32, -' V486, ^V'^- 
 
 Art. 201. Since the mn'" root of a is equal to the m" root of 
 the n" root, or the n'" root of the m" root (Art. 192), therefore the 
 tnn"' root of any algebraic expression may be simplified when it 
 is a complete power of the m"' or n"' degree. 
 
 Thus, V9a==\/V9a^V3a- 
 
 Also, !^a»— 2aA+6'=VVa'— 2ai-l-*'=Va^- 
 
 In general, '■^J^= yj :Jar='^'a. 
 
RADICALS. 158 
 
 Reduce each of the following radicals to its most simple form 
 
 1. V36a'c^ iJ81m'n\ X/25a^b\ «/4ffl=. 
 
 Ans. iJQac, Zn^m, a^bh, %/2a. 
 
 2. ^16aVi %/2Ta\ %l\2b¥, %f6iaK 
 
 Ans. s/ioc^, jJZa, J 5b, 2ja. 
 
 Case 11. To reduce a rational quantity to the form of a rad- 
 ical. 
 
 Akt. 202. — If we square a, and then extract the square root 
 of the square, the result is evidently a ; that is, a=:Ja?=^d^' 
 
 In like manner the cube root of the cube, of a, is a ; that is, 
 
 Generally, a='J/a'"=:a'»' 
 
 Hence, to reduce a rational quantity to the form of a radical of 
 any degree, we have the following 
 Rule. — Raise the quantity to a power correspcnding to the given 
 
 root, and write it under the radical sign. 
 
 E X AM PL E S. 
 
 1. Reduce 6 to the form of the square root. Ans. ^36. 
 
 2. Reduce 2 to the form of the cube root. Ans. %/d. 
 
 3. Reduce 3 oa; to the form of the square root. Ans. JQa^x^. 
 
 4. Reduce — 3a to the form of the cube root. 
 
 Ans. V— 27a', or (— 27o')5' 
 
 5. Reduce m — n to the form of the square root. 
 
 Ans. /Jm?- — 2TOra+n2. 
 
 By the same principle the coefficient of a radical may be passed 
 under the radical sign. 
 
 Thus, 2 ^3= V4X ^3=VJ2. 
 
 So, also, a^b^Ja^y.Jh=^a'b. 
 
 Generally a'i[b='^'^y,'i[h='%[cH). 
 
 6. Express 3^6 entirely under the radical sign. Ans. ^54. 
 7 Express 5^7, and a^ijb, entirely under the radical sign. 
 
 Ans. jnS), and ^a'^b 
 
154 RATS ALGEBRA, PART SECOND. 
 
 8. Pass the coefficient of the quantity 2^5, under the radical 
 sign. Ans. ^40. 
 
 Case III. — To reduce radicals having different indices to equiva- 
 lent radicals having a common index. 
 
 Akt. 203. Since 'i/a='"^'a" (Art. 192), or aS=o"" (An. 
 1 18) ; therefore, we may multiply the index of a radical hy any 
 number, provided we elevate the quantity under the sign to a power oj 
 the same degree denoted by the radical. This is really only multi- 
 plying both terms of the fractional exponent by the same num- 
 ber, which does not change its value. (Art. 118.) 
 
 Let it be required to reduce %/2a, and 1/36, or (2a)» and (3J)* 
 to quantities of equal value, having the same index. 
 
 Reducing the fractional exponents to the same denominator, we 
 
 have l=j\, anil=j\; hence, (2a)i =(2o)T*Jr, or 'V(2ay, and 
 
 (3J)i=(3i)^'f, or 'S/(3^- 
 Hence, we have the following 
 
 Rule. — Rediice the fractional exponents to a common denominator ; 
 them, the numerator of each fraction will represent the power to which 
 the corresponding quantity is to be raised, and the common, denom- 
 inator the index of the root to be extracted. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Reduce ^S and ^2, or 3* and 2», to a common index. 
 
 Ans. «/27 and ^4, or 27* and 4*. 
 
 2. Reduce ^5 and ^4 to a common index. 
 
 Ans. 6/25, and ^64 
 
 3. Reduce o' and b' to a common index. Ans. J a* and ^Jb. 
 
 4. Reduce J^o, ^56, and ydc, to a common index. 
 
 Ans. "^a^, 'y625b\ and ",J2Y6?. 
 
 5. Reduce ,Ja', i/a^, and ija' to a common index. 
 
 Ans. 'V^, 'Va*, and 'Vo". 
 
 6 . Reduce 3', 2*, and 5^ to a common index. 
 
 Ans. 3", 2^^, and 5^', or 'V656i, '5/512, and 'VI5625 
 
RADICALS. 155 
 
 ADDITION AND SUBTRACTION OF EADICALS. 
 
 Aei 204. Let it be required to find the sum of 3 J/ a, and 
 
 It is evident that 3 times and 5 times any quantity whatever, 
 •nust make 8 times that quantity ; therefore, 
 
 3i/o+5V^=8Va. 
 
 But, if it were required to find the sum of ^Ja and 5^0, since 
 (he square root of a and the cube root of a axe different quantities, 
 we cannot add them together and call them by the same name. 
 Therefore, we can only indicate their addition ; thus. 
 
 From this we see that to add similar radicals we must find the 
 sum of their coefficients, and place it before the common radical, 
 and that to add radicals which are not similar, they must be con- 
 nected by their proper signs. 
 
 Radicals that are not similar may often be rendered similar ; 
 thus, ^12 and ^27 are equal to 2^3 and 3^3, and their sum 
 is bji. 
 
 It is evident that the subtraction of radicals may be performed 
 in the same manner as addition, except that the signs of the sub- 
 trahend must be changed. (Art. 44.) 
 
 From the preceding we derive the following 
 
 RtJLE FOR THE ADDiTioH OF EADICALS. — 1st. Reduce (he radicals 
 to their simplest forms. 
 
 2nd. If the radicals are similar, find (he sum of their coefficients and 
 prefix it to the common radical ; hut if they are not similar, connect 
 them by their proper signs. 
 
 Rule for the subtraction of radicals, — Change the sign of 
 the subtrahend and proceed as in addition of radicals. 
 
 EXAMPLES FOR PRACTICE. 
 
 1 . Find the sum of ^448 and ^112. Ans. l%Jn. 
 
 2. Find the sum of ?/24 and V8T. Ans. 5s/3. 
 
 3. Find the sum of' V48 and »/l62. Ans. 6^6. 
 
156 RAY'S ALGEBRA, PART SECOND. 
 
 4. Find the sum of ^ISa^b' and JdOo'V. 
 
 Am. (3a'6+5ai)J2ai. 
 
 5. Subtract ^TSO from ^405. Ans. Sj'E. 
 
 6. Subtract ^40 from ^135- -A^w. %/b. 
 Perform the operations indicated in each of the following ex- 
 amples : 
 
 7. ^243+^27+^/48. Ans. 16 J3. 
 
 8. ^24+^54—^96. Ans. ^6. 
 
 9. ^"128— 2V50+V72— V18. Am. J2. 
 10. 2J8—1^18-\-5jT2—J5b. *Ans.8^2. 
 
 11. V48o6=+6,/75a+^3a(a— 9ft)'. Ans. a^'da. 
 
 12. 2^1+1^60+ JlE+Jl. Ans. Jf ^^5. 
 
 13. »/128— ^686— V16+4V250. Ans. ld^2. 
 
 14. 2s/|'+8»/'^. Ajw. 3»/2. 
 
 15. 7^54+3 s/16+?/2—5»/i28, Ans. 8^2. 
 
 16. 6«/4^+2V2a+V8o'. Ar«. 9^2^ 
 
 17. 3VI+7VII-V54. Ans. J^^-Q 
 
 18. 2^3— 1 VT2+4^"27— 2^"^. Ans. \^^'3 
 
 19. »/40— >»/320+>/T35. Ans. 3^5. 
 
 20. V16+S/81—V— 512+^/192— 7«/9- Ans. 10. 
 
 21. 8(|.)*+ixl2*— 1X27*— 2(^3^)1, Atw. ^^3. 
 
 22. 6(8a6i)i+4a(o'ft0*— (125a«i*)i Ans. a'J*. 
 
 23 . ^ /^+i J (a'ft— 4a'ft'+4a6»). Ajw. -?- Va*. 
 \ c' 2c 2c 
 
 MULTIPLICATION AND DIVISION OF RADICALS. 
 
 Art. 205. The rule for the multiplication of radicals is founded 
 on the principle (Art. 200) that tlie product of the n'* root of two 
 or more quantities,is equal to the n'* root of their product ; that is, 
 
RADICALS. 167 
 
 Hence, (Art. 53), a:;/by,c1Jd=:aXcX'iJbX'iJd=ac:;Jbd. 
 
 The rule for division is founded on the principle that the qnch 
 lient of tJie n" roots of two quantities is equal to the n'* root of thei* 
 quotient ; that is, 
 
 5!L?=" /? ; which is thus proved. 
 
 ya 
 
 to the »' 
 
 If we raise „n to the »" power, we have 
 
 and if we raise " /_ to the «'* power, we have 
 
 (^/DH 
 
 Since the same powers of the two quantities are equal, we in- 
 fer that the quantities themselves are equal ; that is. 
 
 
 Vi 
 
 Hence, acJbd^ajT^"^^'^Jl=cJl. 
 
 Therefore, we have the following 
 
 Rules for the m0ltipi.ication and division of eadicals. — 
 I. If the radicals have different indices, reduce them to the same 
 index, 
 
 II. To Mui-TIFLT. — Multiply the coefficients together for the coeffi- 
 cient of the product, and also the parts under the radical for the 
 radical part of the product. 
 
 III. To Divide. — Divide the coefficient of the dividend by the coeffi- 
 cient of the divisor for the coefficient of the quotient, and the radical 
 part of the dividend by the radical part of the divisor for the radi- 
 cal part of the quotient. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Multiply 2^06 by Za^'c^c. 
 
 2jabXZa^'abc=2xZaJaby.abc=SaJa^c=6a^bJc 
 
158 RAY'S ALGEBRA, PART SECOND. 
 
 2. Divide 4a^oJ by 2;^ac. 
 
 2V^ 2Nfac \/c VC c" 
 
 3. Multiply 2^3 by 372. 
 
 2»/3=2»/3^; 3^2=3^2' 
 
 2«/32x3V2'=2X3V3'X2'=6V72. 
 4. Divide 672" by 3 »/2. 
 
 6V2=6V2''=6s/8 
 35/2=3 8/2'=3V4. 
 
 
 3V4 " 
 
 
 5. 
 
 Multiply 3^12 by 5^18. 
 
 Ans. 90^ 
 
 6. 
 
 Multiply 4^12" by 3^4. 
 
 Ans. 24 yc 
 
 7. 
 
 Multiply 15/18 by 7^15. 
 
 Atis. 7S/I0 
 
 8. 
 
 Multiply together 573, 7^"|, and ^2- 
 
 Atos. 140. 
 
 9. 
 
 Multiply V3 by \J2. 
 
 Ans. V108. 
 
 10. 
 
 Multiply3 9/iby 4Va- 
 
 Aw. 12' V'^*. 
 
 11. 
 
 Multiply together ^2, ^3, and «/5- 
 
 Ans. 'V648000. 
 
 12. 
 
 Multiply V2X J/3 by Vlx V]- 
 
 Ans. V2. 
 
 13. 
 
 Multiply together «Vi^ ^^^^ and '"Jl^. 
 
 Ajm. 'V^- 
 
 14. 
 
 Divide ^40 by J2. 
 
 Ans. 2^5. 
 
 15. 
 
 Divide 6^54 by 3 V2. 
 
 Ares. 6V3. 
 
 16. 
 
 Divide 105/108 by 5»/4. 
 
 Am. 6. 
 
 17. 
 
 Divide70V^by7»/18. 
 
 A»w. 5»/4. 
 
 18. 
 
 Divide %pT2 by ^2. 
 
 Atjs. V3. 
 
 19. 
 
 Divide 4V9 by 2^3. 
 
 Ans. 2V3. 
 
 20. 
 
 Divide 20«/200 by 4^2. 
 
 Ans. 55/5. 
 
 91. 
 
 Divide 5/72 by »/3. 
 
 Ans. J^. 
 
RADICALS. lf>9 
 
 22. Divide ^4 by i ^6. Arts, f »/18 
 
 23. Divide Xjl by XJl An. ^l 
 
 24. Divide 1^1 by V2+3^/^. An*, -^ig. 
 
 When one or more of several polynomial factors contains tadi. 
 cals, they may be multiplied together by observing the rule for 
 the exponents in the case of monomial factors. 
 
 25. Multiply 3+^5 by 2— ^¥. Ans. 1— .^5. 
 
 26. Multiply J2+'^ by ^2—1. Ans. 1. 
 
 27. Multiply 1172—4715 by JG+^5. Ans. 2^3— jTO. 
 
 28. Raise J2+JS to the fourth power. Ans. 49+2076. 
 
 29. Multiply 3^4+6 72 by 572. Ans. 30^/2+3 7^2. 
 
 30. Multiply !/l2+719 by ?/l2— 7T9. Ans. 5 
 
 31. Multiply a!^— X72+I by a:2+a;72+l. Ans. x^+1. 
 
 32. (,x^+l)(,x'—xJS+l-)(x'+xJS^l). Ans. x^+1. 
 
 33. (271+375— 772x772— 5720— 272). 
 
 jItw. 427IO— 174. 
 
 Art. 206. To reduce a fraction whose denominator contains 
 radicals, to an equivalent fraction having a rational denominator. 
 
 When the denominator of the fraction is a monomial, as _ _, if 
 we multiply both terms by 7*> the denominator will be rational. 
 Thus, "_- "XV^ -^V* 
 
 7* 7*x7* * 
 
 Again, if the denominator is ^a, if we multiply both terms by 
 l/a', the denominator will become 5/aX ^a^=a. 
 
 In like manner, if the denominator is '7a", it will become ra- 
 tional by multiplying it by '!i/a'"~", since 
 
 '!;/o"X'7a'"~"=Va"Xa"'^"="7 "'"=«• 
 
160 
 
 RAY'S ALGEBRA, PART SKCOND. 
 
 Therefore, when the denominalor of the fraction is a monomial 
 multiply hoth terms hy such a factor as will renda- the exponent of tht 
 given radical equal to un ity. 
 
 Since the sum of two quantities, multiplied by their difference, 
 is equal to the difference of their squares (Art. 80) ; if the frac- 
 
 tion is of the form 
 
 , and we multiply both terms by i — ^e, 
 
 the denominator will be rational. 
 
 Thus, " — "(^—•J'^) _ ab—ajc 
 
 For the same reason, if the denominator is b — ,Jc, the multi- 
 plier will be b-\-Jc. If the denominator is Jb-[-Jc, the multi- 
 plier will be Jb — Jc ; and if the denominator is Jb—^fc, the 
 multiplier will be Jb-\-Jc. 
 
 If the denominator is of the form ^ a-\- ^b-\- ^ci it may be 
 rendered rational by two successive multiplications. Thus, (Ja 
 + VH- ^/c)( Va+ V*— V^)=( ^a+^'^—c=a+b—c+2 ^'ab. 
 
 Let 2d=a-^b—c, then 2d-\-2,Jttb will become rational if it be 
 multiplied by d — ^aJ ; 
 
 .-. the whole multiplier is (_J^-\.Jb—ji) ( '^±tll—^'^b ) . 
 
 Reduce the following fractions to equivalent ones having ra- 
 tional denominators. 
 
 1. J_ 
 
 2. ^ 
 
 Ans. !^=ijl. 
 
 ^6 6 S^'' 
 
 5. 8-^Vg 
 3—2^2 
 
 6. V"^Vg 
 V3— a/2' 
 
 7. 
 
 V3+1 
 2—J3' 
 
 2_ 
 
 V3" 
 6 
 
 Am. 1^9. 
 Ans. |V16- 
 
 Aru. 4+^2. 
 
 An. 54-2^5 
 
 Ans. 5-|-& ^"3 
 
RADICALS, 
 
 161 
 
 8.1=^ 
 3+V5 
 
 9. 3Vg-2^/2 
 2^5— Vl*^' 
 
 IC, 
 
 11. j+4V"3"_ 
 V6+V2-J5' 
 
 1 
 
 12. 
 
 1 
 
 13. ^/(^+'^)+^/(''^o) 
 
 .^ {x+a) — J (x — a) 
 
 Ans. 2—^6 
 Ans. 9+§VlO- 
 
 A^ ^/36+3V2+2V3 
 12 
 
 Atw. V6+V2+V5. 
 
 Ans. 2x. 
 
 Ans. ?±5^^?!l^ 
 
 4 Va:°+1+Va;'-1 ^ ^x'+l-Jar'-l 
 
 Ans. 2a;'. 
 
 Remark. — The utility of most of the preceding transformations con- 
 sists in shortening the calculations necessary to find the numerical value 
 of a fractional radical. Thus, if it be required to find the value of 
 
 2 *2 
 
 -^, we may divide 2 by the square root of 5. But __ is equal to 
 
 ^/5 ^5 
 
 ?V5, where it is merely necessary to extract the square root of 5 and 
 take two-fifths of the result. A comparison of the two methods of oper- 
 ation will show that the latter is much shorter than the former. 
 
 Reduce each of the following fractions to its simplest form, and 
 find the numerical value of the result. 
 
 15. ^ ,and ^ 
 V5 V2 
 
 Ans. 
 
 .894427+, and .707106+, 
 
 16. 1 
 
 2+V3' 
 
 
 Ans. .267949+. 
 
 17. l+>/^ 
 2— J2 
 
 
 Ans. 4.12132+, 
 
 ,Q ^/20+J^2 
 
 V A 
 
 Ans. 15.745966+ 
 
162 RAY'S ALGEBRA, PART SECOND. 
 
 POWERS OF RADICALS. 
 
 Art. 207. Let it be required to raise ^a to the n'* power. 
 
 By the rule for the multiplication of radicals (Art. 205), we 
 
 have (':yo)"=^aX 'i/aX'i/a to re factors, 
 
 ='V"XaX<' to 7! factors =^'!ja". 
 
 Hence, to raise a radical quantity to any power, we have the 
 following 
 RnLE. — Raise the quantity under the radical to the given power, and 
 
 affect the result wilh the primitive radical sign. 
 
 Thus, ( </4o')2= i/(iay=ijl6a^=flaija'=^aja. 
 
 If the quantity have a coefficient, it must also be raised to the 
 given power. Thus, 
 
 (2V3^)s=2'X t/(3o)'=8 V27?. 
 If the index of the radical is a multiple of the exponent of the 
 power, the operation may be simplified. Thus, 
 
 ( V2a)2=Cv/v2a)' (Art. 192); 
 
 and since the operation of squaring removes the first radical, we 
 have 
 
 ( V2o)2= ( V V2^ ) '= V2i. 
 
 In general, (r^/T)'^ ( "/'^a ) »='!y7. 
 
 Hence, if the index of the radical it divisible ly the exponent of 
 the power, we may perform this division, and leave the quantity under 
 the radical sign unchanged. 
 
 E XAMP LE 8. 
 
 1. Raise s/2o to the 4th power. Ans. 2a»/2a. 
 
 2. Raise 3l/2ab' to the 4th power. Ans. 162aJ2»/2a5^ 
 
 3. Raise 2Jxy' to the 5th power. Ans. 32xyjxy. 
 
 4. Raise i/ac' to the 2nd power. Ans. cja. 
 
 5. Raise ^ac^ to the 4th power. Ans. aV. 
 
 6. Raise 3 =/2a to the 5th power. Ans. i86ai/4d'. 
 
 7. Raise ^3c2 to the 3rd power. Ans. cJZ 
 
 8. Raise Jx—y to the 3rd power. Ant. {x—y)^^^. 
 
KADICALS. 163 
 
 ROOTS OF RADICALS. 
 
 . Art. 208. Since •?/Va='"V" (Art. 192), therefore, to ex. 
 
 tract the roots of radicals, we have the following 
 
 Rule. — Multiply the index of (ke radical hy the index of the root to 
 he extracted, and leave the quantity under the radical sign un- 
 changed. 
 
 Thus, the square root of Sj2a is .I%j2a=^ij2a. 
 
 If the radical has a coeiEcient, its root must he extracted by the 
 rule (Art. 194). Thus, 
 
 ^9a^l/¥c=^Qd'y. J 5/3c=3o«/3c. 
 
 If the quantity under the radical is a perfect power of the same 
 degree as the root to be extracted, the process may be simplified. 
 Thus, 
 
 iJt/S^'iB equal (Art. 192) to \JJ/E^'=i/2i. 
 
 E X AMFL £S. 
 
 1. Extract the cube root of ^a'J. Atis. 'ija'b 
 
 2. Extract the 4th root of IQa'y^ Am. 2o' '\/2c. 
 
 3. Extract the cube root of \/2Ta'. Ans. 4/3a. 
 
 4. Extract the square root of ^49o'. Ans. %J1a. 
 
 5. Extract the cube root of 64«/8ffl». Ans. 4*/2a». 
 
 ■6. Extract the cube root of (ni+ra)^m-f-Ji. Ans. ^m-\-n. 
 
 IMAGINARY, OR IMPOSSIBLE aUANTITIES. 
 
 Art. 209. An imaginary quantity (Arts. 182, 193,) is an even 
 root of a negative quantity. 
 
 Thus, ^ — a, and \l—^b*, are imaginary quantities. 
 
 The rules for the multiplication and division of radicals (Art 
 205) require some modification when imaginary quantities are to 
 be multiplied or divided. 
 
 Thus, by the rule (Art. 205), V— "X <J^^= >J —aX—a= 
 Ja'=:-±ft. But, since the square root of any quantity multiplied 
 
164 RAY'S ALGEBRA, PART SECONC. 
 
 by the square root itself, must give the original quantity, there- 
 fore, ^ — aXV — a= — a. 
 
 Akt. 210. Every imaginary quantity may be resolved into two 
 factors, one a real quantity, and the other the imaginary expression, 
 ,J — 1 ; or an expression containing it. 
 
 This is evident if we consider that every negative quantity may 
 be regarded as the product of two factors, one of which is — 1 
 Thus, — a=oX— 1. — 6^=t'X— 1, and so on. 
 
 Hence, ^ — a=JaX — l—^/aXj—^- 
 
 V_a== Va'X— 1=V''"X V— 1=±<»V— 1- 
 
 </—a=iJaX—l=i/aX 4/— l=±V''-Nf V— 1- 
 
 Since the square root of any quantity, multiplied by the square 
 root itself, must give the original quantity ; 
 
 therefore, (V— i)'=V— IX V^=— 1- 
 
 also, u—iy=(^^—iyx ^—i=—i V^=— >/— i- 
 
 Attention to this principle will render all the algebraic opera- 
 tions, with imaginary quantities, easily performed. 
 
 Thus, ,/^ X ^—b=^aX J—i X JiX V— 1 = V^* X 
 
 OPERATION. 
 
 If it be required to find the product of "-{-^'J — 1 
 
 o-l-ft^/ — 1 by a — 6^ — 1, the operation is a — b^ — 1 
 
 performed as in the margin. o'+aJ^ — 1 
 
 —abJ—i-^-b' 
 
 Since (,a + bj—i){a—bj'—i) = a' + b'; therefore, a'-|- J' 
 
 =(a-\-b ^^-i)(a — b^ — 1) ; hence, any binomial whose terms 
 are positive, may be resolved into two factors, one of which is the 
 sum and the other the difierence of a real and an imaginary quan< 
 tity. Thus, 
 
FRACTIONAL EXPONENTS. 166 
 
 EXAMPLE S. 
 
 1. Find the sum and difference of a-,ij — 1. and a — b,J — 1. 
 
 Ans. 2a, and 2b J— -t, 
 
 2. Multiply V— «^ by V— *^- ^'^- — «*• 
 
 3. Find the 3rd and 4th powers of a J — I. 
 
 Ans. — a's] — 1; and a*. 
 
 4. Multiply 2j^ by 3V^2. Ans. —6^6. 
 
 5. Find the cube of — i+i V— 3> ^'^^ — 3— aV"-^- ^'^- 1- 
 
 6. Divide 6^^ by 2 V^- A.ns. 1^3. 
 
 7. Simplify the fraction .T^~ . Ans. J— I. 
 
 8. Find the continued product of x-\-a, x-\-a^ — 1, x — a, and 
 
 X — a J — 1 . Ans. x* — a*. 
 
 9. Of what number are 24-|-7^ — 1, and 24—7^^, the im- 
 
 aginary factors ? Ans. 625. 
 
 VI. THEORY OF FRACTIONAL EXPONENTS. 
 
 Art. 211. The rules for the exponents in multiplication and 
 division (Arts. 56 and 70), have been demonstrated, under the 
 supposition that the exponents were integral. These rules, as 
 well as those which relate to the formation of powers (Art. 172), 
 and the extraction of roots (Art. 194), are equally applicable 
 when the exponents are fractional. 
 
 Fractional exponents have their origin (Art. 196) in the ex- 
 traction of roots, when the exponent of the power is not divisible 
 by the index of the root. Thus, in extracting the n"' root of a"* 
 the operation requires that the exponent m should be divided by 
 the index n. When m is divisible by n the exact root of a" is 
 obtained, but when m is not divisible by n, the operation is indi- 
 eated by indicating the division of the exponents, Thus, 
 
 Ij'ce^a'^' 
 
 As has been shown already (Art. 196), every radical quantity 
 may be represented by the same quantity with a fractional ex- 
 ponent, the numerator of the exponent denoting the power of the 
 
166 RAY'S ALGEBRA, PART SECOND. 
 
 given quantity, and the denominator the index of the required 
 root. 
 
 Thus, s/^=a5, »/^=a5, :/-l=^/F^i=a~?- 
 Ma"" 
 
 As a' is called u to the p power, when ^ is a positive whole 
 
 number ; so, by analogy, a', a' and a~ ", are called respectively. 
 
 I to the I power, a to the -| power, and a to the minus _ power. 
 
 But it would, perhaps, be more accurate to say, a exponent |, a 
 
 exponent -^ , a exponent — - ; and reserve the term porcer to de- 
 n 
 
 note the product arising from multiplying a quantity by itself one 
 
 or more times (Art. 19). 
 
 MULTIPLICATION AND DIVISION OF QUANTITIES WITH FRAC- 
 TIONAL EXPONENTS. 
 
 Akt. 312. It has been shown (Art. 56) that the exponent of 
 any letter in the product is equal to the sum of its eicponents in tkt 
 two factors. It will now be shown that the same rule applies 
 when the exponents are fractional. 
 
 a 4 
 
 1 . Let it be required to multiply a^ by o> • 
 
 a^=r5/«'— V^; a^=V«'='V«'^. (Art. 205.> 
 
 But this result is the same as that obtained by adding the ex- 
 ponents together. Thus, 
 
 2. Let it be required to multiply a* by a*. 
 
 \ a» \ a' 
 
 rn p 
 
 And in general, the product of o " by aj is, 
 
 _m 2. —-J./ ry—mq 
 
 '/ "X<M=o "■'■«=o m ■ 
 
 Hence, 1» multiply quantities affected with fracliondl exponents, 
 apply the rule given (Art. 56) in the case of entire exponents^ 
 
FRACTIONAL EXPONENTS. 167 
 
 Abt. 213. In the preceding article (212) it has been shown 
 that when the exponents axe fractional, the exponent of any letter 
 in the product is equal to the sum of its exponents in the two fac- 
 tors ; and since division is the reverse of multiplication, there- 
 fore the exponent of any letter in the quotient must be equal to 
 the excess of its exponent in the dividend over that in the divisoi 
 
 m ji m_p mg — np 
 
 That is, an.i-af:=fl!" f=o~»r~' 
 Perform the operations indicated in each of the following 
 
 E SAMPLES. 
 
 1. a'Xo'i and o~*Xa'' Ajw. o', and o«' 
 
 2. a^e-'Xa^c^' Ans. a^c'^ 
 
 M?)'x(^)*x(^)'- ^.(1)' 
 
 4. (a^+a^ji+5^)(o*— j4). Ans. a—b. 
 
 5. (x*y-\-y^)(x* — y'^). Am. x^y — y^ 
 
 1 I I 1 nv^n 
 
 e. (a+i)sx(o+6)»X(a— i)iSX(a— i)»- Ans. (o'—js)^^- 
 T. x^-^x*, and xmnf-^ny^. Ans. x'^, and x "<•> -if 
 
 8. (0?— 6^)^(ai— i^). Ans. a^J^ah^-{-hi. 
 
 9. ((f— i2)j.(ffli+aV-l-oi6+it). Ans. a^—b^. 
 
 POWERS AND ROOTS OF QUANTITIES WITH FRACTIONAL 
 EXPONENTS. 
 
 Art. 214. Since the m" power of a quantity is the product 
 of m factors, each equal to the quantity (Art. 172) ; therefore, 
 
 to raise an to the m** power, we must find the product 
 
 i 1. L 
 a" Xa» Xo" ... . to jn factors. 
 
 Here it is evident the exponent i must be taken m times , 
 
 -1 - 
 
 hence, (o")"'=o"' 
 
 Therefore, to raise a quantity affected with a fractional expo- 
 
 I— m 
 
168 RAY'S ALGEBRA, PART SECOND. 
 
 nent to any power, mvlUply the exponent of the qmntily by the 
 exponent of the power. 
 
 Thus, (^ahiy=ah^=a'b^. 
 
 Aet. 215. We have just seen in the preceding article, that in 
 finding the m"* power of any quantity, we must multiply the ex- 
 ponent of the quantity by the exponent of the power. Hence, 
 conversely, in extracting the m" root, we must divide the ex- 
 ponent of the quantity by the exponent of the root ; that is, 
 
 Since (o^)'"=(to ■ '"=a» ; therefore, 
 
 '"/o"=o'~-r-'»=a" "•=a«" 
 
 From the pri:;eding it is obvious that the rules in Arts. 172 
 and 1 94 apply, without any change, to quantities having frac- 
 tional exponents. 
 
 EXAM FLES. 
 
 1. Raise o*6» to the 4th power. Ans. a'b^- 
 
 2. Raise — 2x'y^z* to the 3rd, 4th, and 6th powers. 
 
 Ans. —Bx^yz* ; IQx^y^z ; 64x^yH^. 
 
 3. Find the square of a — (ax — a^)^. Ans. ax — 2a(ax — o')*. 
 
 4. Find the square of ( 1±^ ) V ( i=^ ) ^• 
 
 Ans. 1+(1— m=)*. 
 
 5. Find the cube of o'K~'+o~'a;. 
 
 Ans. aar'4-3o^ar'+3o-*a;+o-'a:' 
 
 6. Find the square roots of 3(5)5^; and Jlfi^^L^. 
 
 9(3436=)* 
 
 Ans. (135)*; !Li-. 
 3J* 
 
 7. Find the cube roote of (27o»a;)* and (27o'a)i 
 
 Ans. Sio^x* or (Sai*)* ; and (Soar*)*. 
 
KyUATlONS CONTAINING RADICALS. 169 
 
 8. I'ind the square root of 5x' — 4a;(5cx)'-^c. 
 
 Ans. 6'x' — 2c' 
 
 9. Find the square root of 1+JJa — ~|~ '' a'-\-a^. 
 
 Ans. 1 — _+a. 
 4 
 
 10. Find the cube root of loS— 3a'J^+6a6— 66^. 
 
 Ans. |a— 26*. 
 
 Rehase. — In solving examples 8, 9, and 10, the pupil is expected to 
 combine the rules, Arts. 183 and 191, with those for fractional 
 exponents. 
 
 VII. EaUATIONS CONTAINING RADICALS. 
 
 Note to Teachers. — This part of the subject of Equations of the 
 First Degree could not be treated till after Radicals, as the operations 
 oeccssarily Involve the formation of powers and the multiplication of 
 radicals. 
 
 Akt. 216. In the solution of questions containing radicals, 
 the method to be pursued will often depend on the judgment of 
 the pupil, as many of the questions can be solved in diflferent 
 ways, and the shortest processes can only be learned from frac- 
 tice. 
 
 1st. When the equation to be solved contains only one radical 
 expression, transpose it to one side of the equation and the ra- 
 tional terms to the other ; then involve both sides to a power cor- 
 responding to the radical sign. 
 
 Ex. Given, l/(.a'-{-x) — o=c, to find x. 
 Transposing, ^(^a'-\-x)=c-\-a ; 
 Cubing, a'-|-a;=c'-|-3ac'4-3a'c+a' ; 
 Whence, x=c'4-3<K'+3a'c. 
 
 2nd. When a radical expression occurs under the radical sign, 
 fe operation of involution must be repeated. 
 
 Ex, Given Nix — ^1 — a;=l — Jx, to find x. 
 
 Squaring, x — Jl — i=l — 3^x-\-x; 
 1.5 
 
170 RAY'S ALGEBRA. PART SECOND. 
 
 Canceliog z on each side and squaring again, 
 
 l-ni=l— 4Vx4-4a. 
 Canceling 1 on each side, transposing, squaring, and reducing 
 We find, x=^B, 
 
 3rd. When there are two or more radical expressions, it ia 
 generally preferable to make one of them stand alone, before per- 
 forming the process of involution. 
 
 Ex. Given, ,Jx-{-9 — ,^x=l, to find x. 
 
 Transposing — ^x, we have ^x-^9=:l-\-,Jx. 
 
 Squaring each side, a+9=l-|-2^a:-j-a; ; 
 
 Canceling x on each side, transposing and dividing by 2, 
 
 Jx=A: ; hence, a:=16. 
 
 In some cases, however, it is preferable, when an equation con- 
 tains two radical expressions, to retain them both on the same 
 side. Thus, the equation 
 
 V(S)+V(S)='' 
 
 x-\-a 
 
 tyill be cleared of radicals at once, by squaring each side, tiM 
 
 ab 
 value of X being .,, . • 
 
 EXAMPLES FOR PRACTICE. 
 
 4. V(^+5)+3=8— VZ Ans.x=\. 
 
 5. -v/ l+V(3+^6i)=2. Ans. x=&. 
 
 6. V^+^«=V^+fl. Ans.x=J^d}^. I 
 
 T. ^2x—3a+^2x=3^a. 
 
 8. J113+^/['^+^/(3+V■^)]?=4• 
 
 
 4 ■ 
 
 Ans. . 
 
 x=2a. 
 
 Ans. 
 
 1=1. 
 
 Ans. 
 
 .*=!■ 
 
 
 a 
 
 9. J2+x+^x=rj^^. 
 10 J^x^-^^Jx. Ans ^ __^„ 
 
INEQUALITIES. 171 
 
 11. Jx+lS—Jx—U=1i. Ans.x=S6 
 
 — — — d' 
 
 12. ajx+bjx — cJx=i. Ans. x=- 
 
 ^ ^ {a+b—cf 
 
 X — ax Jx 1 j 
 
 "• V^ * 1— a / 
 
 14. a;+a=Va2+iV(*'+*')- ^'^- as='''"~^''' 
 
 4a 
 
 jt^-:-4 _ 3^a; 
 
 ^^- Jx+JT' 3 +^^^''• A«s.as=16a. 
 
 3ar— 1 J3x— 1 
 
 18. V4a+^=2J6+i— Jx. Ans. js=(i!i:?2! 
 
 
 21. J^x+Z—Jjx—2=j2^x. Ans. x=9. 
 
 
 23. J(l+ffl)=+(l_a)a;+J(l— a)2+(l+a)i=2a. 
 
 ' Atw. x=3. 
 
 Vllt. INEQUALITIES. 
 
 Art. 817. In the discussion of problems it often becomes 
 necessary to compare quantities that are unequal, and to operate 
 upon them so as to determine the values of the unknown quanti- 
 ties, or to establish certain relations between them. 
 
 In most cases the methods of operating on equations apply to 
 inequalities ; still there are some exceptions, which render if 
 expedient to present the principles and rules of operation in one 
 fiew. 
 
1T2 RAY'S ALGEBRA, PART SECOND. 
 
 Art. sis. Def. — In the theory of inequalities, it is conven 
 ient to consider negative quantities less than zero. Also, in com- 
 paring two negative quantities, that is considered the least which 
 contains the greatest number of units; thus, 0]> — 1, and 
 — 3>— 5. 
 
 Two inequalities are said to subsist in the same sense when the 
 ^eater quantity stands on the right in both, or on the left in 
 both ; thus, 
 
 5>3, and 2<5, 
 7>4, 3<8. 
 
 are said to subsist in the same sense. 
 
 Two inequalities are said to subsist in a contrary sense, when 
 - the greater stands on the riffht in one and on the left in the other ; 
 thus, 5>1 and 4<^8 are inequalities which subsist in a contrary 
 sense. 
 
 Aet. 219. Peop. I. — TTi£ same quantity, or equal quantities, 
 may be added to or svhtraded from hath members of an inequality, 
 and the resulting inequality will continue in the same sense. 
 
 Thus, 7>5, and by adding 4 to each member, 
 
 11^9 ; or by subtracting 4 from each member, 
 3>1. « 
 
 Also, — 5<^ — 3, and by adding 4 to each member, 
 
 — 1<C+1 ; °^ ^y subtracting 4 from each member, 
 — 9<— 7. 
 Similarly, if o>J, then 
 
 a-\-c'^b-\-c, or a— c>J — c. 
 
 It follows from this proposition, that any quantity may be trans- 
 posed from one side of an inequality to the other, if at the same timt 
 its sign be changed. Thus, if 
 
 a2+62>2o6+c=, 
 
 as-|-i!_2at>2dj— 2ai+c2, 
 
 or o2— 2ai+Z.2>c'. 
 
 Aet. 220. Peop. II. — If two inequalities exist in the same sense, 
 tJie corresponding members may be added together, and the resulting 
 inequality wiU exist in the same sense. Thus, 
 
 7>6, and 5>4, and 
 74..5>6+4, or 12>10. 
 
INEQUALITIES. 173 
 
 But when two inequalities exist in thie same sense, if we sub- 
 trad the orresponding members, the resulting inequality will 
 sometimes exist in the same sense, at other times in a contrary 
 sense. 
 
 First, 7>3 By subtracting, we find the resulting inequality 
 4^1 exists in the same sense. 
 3>2 
 
 Second, 10]>9 In this case, after subtracting, we find the 
 8'>3 resulting inequality exists in a contrary 
 2<^6 sense. 
 
 In general, if 0^6 and c'^d, then, according to the particular 
 values of a, b, c, and d, we may have a — c^b — d, a — c<[J — d, or 
 
 a — c=b — d. 
 
 Art. 821. Prop. III. — If the two members of an inequality be 
 multiplied or divided by a positive number, the resulting inequality 
 will exist in the same sense. Thus, 
 
 8>4 and 8x3>4x3, or 24>12. 
 Also, 8^3>4-^2, or 4>2. 
 
 This principle enables us to clear an inequality of fractions by 
 multiplying both sides by the least common multiple of the 
 denominalors. 
 
 But, if the two members of an inequality be multiplied or 
 divided by a negative number, the resulting inequality will exist 
 in a contrary sense. Thus, — 3<^ — 1, but — 3X — 2>^lX 
 —2, or 6>2. 
 
 From this principle we derive 
 
 Art. 222. Prop. IV. — Tlie signs of all the terms of both mem- 
 bers of an inequality may be changed, if at the same time we establish 
 the resulting inequality in a contrary sense, because this is the same as 
 multiplying both members by — 1 . 
 
 Art. 223. Prop. V. — Both members of a positive ineqiuility 
 may be raised to the same power, or have the same root extracted, and 
 the resulting inequality will exist in the same sense. Thus, 
 
 2<3 and 2»<32, 2'<33 ; or 4<9, 8<27 ; and so on. 
 
 Also, 25>16, and jMyjlQ, or 5>4 ; and so on. 
 But if the signs of both members of ar inequality are not pos- 
 itive, after raising both members to the same power, or extracting 
 the same root, the resulting inequality will sometimes exist in the 
 same sense, and at others in a contrary sense. 
 
1'74 RAYS ALGKHkA, part SECOND. 
 
 Thus, 3>— 2, and 32>(— 2)2, or 9>4. 
 But, — 3<— 2, and (— 3)=>(— 2)», or 9>4. 
 
 EXAMPLES INVOLVING THE PRINCIPLES OF 
 
 INEftUALITIES. 
 
 1. Five times a certain whole number added to 4 is greatei 
 than twice the number added to 19 ; and 5 times the numbei 
 Jiminished by 4 is less than 4 times the number increased by 4. 
 Required the number. . 
 
 Let x= the number. 
 
 Then, 5i-j-4>2a;+19, (1) 
 
 and 51— 4<4a;+4. (2) 
 
 5j: — 2ar>19 — 4, from eq. (1) by transposing, 
 3a;>15, by reducing, 
 x^5, by dividing both members by 3. 
 5a: — 4x<^4-|-4, from eq. (2) by transposing, 
 a;<;8, by reducing. 
 
 Hence, the number is greater than 5 and less than 8, conse- 
 quently either 6 or 7 will fulfill the conditions. 
 
 2. If 41— 7<2x+3, and 3a:+l>13— a;, find a;. 
 
 Ans. x=4: 
 
 3. Find the limit of x in the equation 7a: — 3>32. 
 
 Ans. a;]>5. 
 
 4. Find the limit of x in the equation 54--|a<8+|a;. 
 
 Ans. a<.36. 
 
 5. Show that ''+^+^ > the least, and < the greatest of the 
 
 b+d+f 
 
 fractions, -, 1, -, each letter representing a positive quantity. 
 b d f 
 
 Let G be a quantity greater, and g a quantity less than any of 
 the fractions, " %, %. Then, 
 
 h' d f 
 
 %<G, '5<G, 1<G, 
 b a J 
 
 p9, p9, )>9. 
 
EQUATIONS. 175 
 
 .•. a<JG, c<dG, e</Gr. 
 
 «>*^> <^'>i9> C>fg- 
 
 ... o+c+e<(J-H!+/)G, 
 and o+'^+eX^-l-'H"/).?' 
 .-. ?+^<Gand>i,. 
 
 6. It is required to prove that the sum of the squares of any two 
 unequal magnitudes is always greater than twice their product. 
 
 Since the square of every quantity, whether positive or nega- 
 tive, is positive, it follows that 
 
 (o— i)», or a'— 2aJ-|-i2>0 ; 
 Adding, +2aJ to each side (Art. 219), 
 a2— 2ffl&-f6'-l-2aJ>0+2a4, 
 or a'+6'>2oi, which was required to te proved 
 
 Most of the inequalities usually met with, are made to depend 
 ultimately upon this principle. 
 
 r. Which is greater, ^1+^14 or ^3+3^21 
 
 Atis. the former. 
 
 8. Given l(iK+2)+ia<|(a>-4)+3 and >J(a;+l)+-J, to 
 find X. Ans. x=b. 
 
 9. The double of a certain number increased by 7 is not 
 greater than 19, and its triple diminished by 6 is not less than 
 13. Required the number. Ans.Q. 
 
 10. Show that re'+l is greater than n^-\-n, unless n=l. 
 
 11. Show that every fraction + the fraction inverted ia greater 
 
 than 2 ; that is, that -+->2. 
 a 
 
 12. If x>y, show that x — y^i^x — JvY- 
 
 13. Show that ^-|-l.>l-f-l, unless a=b. 
 
 b^ a' a b 
 
 14. Show that a='+6'+c'>aJ4-ao+6c, unless a=b=c. 
 
 1 5. Show that the ratio of a'+6' to a'+J' is greater than the 
 ratio of o-|-J to o'-|-6'. 
 
 16 If a!==ffi2+i=, and y'=c'-^^, which is greater, xy, or ae 
 -\-hd "! Ans. xy. 
 
 17. Show that aJ<:>(a+i — c)(o +c— J)(i -|-c — «), unless a 
 =b=c. 
 
176 RAY'S ALGEURA, PART SECOND. 
 
 CHAPTER VII. 
 
 EQUATIONS OF THE SECOND DEGEEE 
 
 Article 224. An Equation of the Second Degree (see Art 
 i4 3) is one in which the greatest exponent of the unknown 
 quantity 's 2. Thus, 
 
 x^=:a, and 
 
 ax'-^hx=c, are equations of the second degree. 
 
 An equation containing two or more unknown quantities, in 
 which the greatest exponent, or the greatest sum of the expo- 
 nents of the unknown quantities in one term is 2, is also an 
 equation of the second degree. 
 
 Thus, iy=a, x^-\-x!/=h, xy — a; — y=c, are equations of the 
 second degree. 
 
 Equations of the Second Degree are frequently termed Quad 
 ratic Equations. 
 
 Art. 225. Equations of the second degree, containing onl) 
 one unknown quantity, are divided into two classes ; viz. : incom- 
 plete, and complete. 
 
 An incorhplete equation of the second degree contains only the 
 second power of the unknown quantity and known terms. Thus, 
 
 a;J-f2=47— 4i', and 
 ar'-l-J=:ca;' — d, 
 are incomplete equations of the second degree. 
 
 A complete equation of the second degree contains the first a» 
 well as the second power of the unknown quantity, and knowp 
 terms. Thus, 
 
 5a;'-f 7a;=34, and 
 ax^ — ix'-\-cx — dx=e — -f, 
 are complete equations of the second degree. 
 
 Remark. — Incomplete equations are sometimes termed Pure Quadrat- 
 ics; and complete equations, Affected, or Adjected Quadratics. 
 
 Art. 226. The general form of an incomplete equation of the 
 second degree is ax^=b. 
 
 The general form of a complete equation of the second degree 
 is ax^-\-hx=^c. 
 
EQUATIONS OF THE SECOND DEGREE. 1 /T 
 
 Every equation of the second degree containing only one 
 unknown quantity may be reduced to one of these forms. For, 
 in the case of an incomplete equation, all the terms containing 
 i' may be collected together, and then, if the coefficient of x' 
 contains more than one term, it may be assumed equal to a single 
 quantity, as u, and the sum of the known quantities to another 
 quantity, 6; and the equation then becomes, 
 0x^=1, or ax^ — 6=0. 
 
 A complete equation may be reduced in like manner ; for, all 
 the terms containing x^ may be reduced to one term, as ax'' ; 
 and those containing x to one, as ix ; and the known terms to 
 one, as c ; the equation then is, 
 
 aa;'-[-iar=c, or ax^-\-hx — c=0. 
 
 Hence, we infer, that every equation of the second degren contain- 
 ing only one unknoten quantity, may be reduced to an incomplete equa- 
 tion containing two terms, or to a complete equation containing three 
 terms. 
 
 Frequent illustrations of these principles will occur hereafter. 
 
 INCOMPLETE EaUATIONS OF THE SECOND DEGREE. 
 
 Aet. aaV. 1. Let it be required to find the value of x in thfl 
 equation, 
 
 .Ij;2_3^_5^a;2=12|— X'. 
 
 Clearing of fractions, 4x'— 36+5x2=153— 12x' ; 
 Transposing and reducing, 21x'=189 ; 
 Dividing, x'^9 ; 
 
 Extracting the square root of both members ; 
 
 x=:±3, that is, x=+3, or x= — 3. 
 Verification. i(+3)'— 3+-Sj(+3)==12|— (+3)=. 
 or, 3— 3+3|=12|— 9 ; 
 3|=3|. 
 
 Since the square of — 3 is the same as the square of -j-3, the 
 value x= — 3, will give the same result as x=-|-3. 
 2. Given ax^-]-b^=d-\-cx^, to find the value of x. 
 Transposing, ox' — cx'=ci! — J ; 
 Factoring, (a — c)x'=d — b ; 
 _, d~b. 
 
 Dividing, 
 
 -^i 
 
 ' d—b 
 a — e' 
 
^'78 RAT'S ALGEBRA, PARI SECOND. 
 
 From the preceding examples, we derive the following 
 
 Rule foe the solution of an incomplete equation of the 
 SECOND DEGREE. — Reduce the equation to the form ax'=b. 
 Divide by the coefficient of 'X?, and extract the square root of both 
 members. 
 
 Akt. 228. If we solve the equation a3?=A, we have, 
 
 a 
 
 and as=±,^/_ ; that is, 
 r=+^Und«=-^? 
 
 If we substitute each of these values of x in the equation aa? 
 =6, we find, 
 
 ay.(+^-y=h, or ay!L=h; 
 
 and ox ( — .^^ ) '=i, or aX-=i. 
 
 Since each of these roots or valvics of x, verifies the equation, 
 and since the square root of - can 07dy be+ /-, or — /l 
 
 therefore we infer, 
 
 1st. That every iTicomplete equation of the second degree has two 
 roots, and only two. 
 2nd. That these roots are equal in value, but have cc/nirary signs. 
 
 Note. — Let the pupil recollect that the term root, in refereuce to an 
 equation, is equivalent to the value of the unknown quantity. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Ill'— t4=:5a;2+10. An*. i=±3. 
 
 2. ^(i'— 12)=la~'— 1. Ans. x=±ld. 
 
 3. (a;+2)==4x+5. Ans. x=±l. 
 
 4. |«'— (2«'— 3)=l^f!±?. Ans. ar=±V§i. 
 
 7_65x 
 
 'x 7 
 
 5, 8x+i=^. Ans. «=±2f 
 
EQUATIONS Jt THE SECOND DEGREE. 179 
 
 Ans. a:=db.3. 
 
 Ajis. a;=±5i. 
 
 Ans. x^zh2. 
 
 Ans. x=±9. 
 x'—tx x''+lx «'— 73' 
 
 10. _^+^=o. Ans. x=±ljb^c^—2abc. 
 
 o-\-x — a; c 
 
 g 
 
 S J_ ^ — ?fi 
 
 
 1— 2x^l+2a 
 
 7. 
 
 (5^+i)'=756-|+5a:. 
 
 a 
 
 7x'+8 x'+4 _a' 
 
 
 21 8a;2— 11 3" 
 
 Q 
 
 2,4.7 a;_7 _ 7 
 
 11. a:V6+a'=l-f*'- Ans.x=±^. 
 
 The methods of clearing an equation of radicals have been 
 already explained in Art. 216. 
 
 2fl' 
 
 12. x+Ja^+x^ — "" . Ans. as=±- J3. 
 
 .-/^ 
 
 Ans. d=^2ab—b'. 
 
 X X b' 
 
 2 2 
 
 15. , , -, — 5=6. Ans. x=:± i_ii • 
 
 a-\-Ja^ — a-" -^0-t-l 
 
 QUESTIONS PRODUCING INCOMPLETE EaUATIONS OF THE 
 SECOND DEGREE. 
 
 Art. 329. In the solution of problems producing equations 
 of the second degree, the equation is found on/ the same principle 
 as in questions producing equations of the first degree. See Art. 
 154. 
 
 1. What two numbers have the ratio of 2 to 5, and the sum 
 of vphose squares is 261 T 
 
 Let 2x and 5x:= the numbers. 
 
 Then, 4a;24-25a'=29x2=261; 
 
 Whence, a;'=9 and x=3. 
 
 Hence, 2x=Q and 5as=15, the required numbers. 
 
 2. The square of a certain number diminished by 17, is equal 
 to 130 diminished by twice the square of the number. Required 
 the number. Ans. 7. 
 
180 RAT'S ALGEBRA, PART SECOND. 
 
 3. There is a certain number, which being subtracted from 10 
 and the remainder multiplied by the number itself, gives the same 
 product as 10 times the remainder left after subtracting 6g from 
 the number. Required the number. Ans. 8. 
 
 4. What number is that, the third part of whose square beinj 
 subtracted from 30, leaves the same remainder as one-fourth ol 
 the square increased by 9 1 Ans. 6. 
 
 5. There are two numbers whose difference is |thB of the 
 greater, and the difference of dieir squares is 128 ; find them. 
 
 Ans. 18 and 14. 
 
 6. Divide the number 21 i.nto two such parts, that the squart 
 of the less shall be to that of the greater as 4 to 25. 
 
 Let X and 21 — a=z the parts. 
 Then, x' : (21— xy : : 4 : 25 ; 
 or, ( Arith., Art. 209,) 25x'=i(21—xy ; 
 Extracting the square root of both sides, 
 5x=2(21— a;) ; 
 Whence, as=6, and 21 — a:=15. 
 
 7. Divide the number 14 into two such parts, that the quotient 
 of the greater divided by the less, shall be to the quotient of the 
 less divided by the greater, as 16 to 9. Ans. 6 and 8. 
 
 8. What number is that which being added to 20 and subtracted 
 from 20, the product of the sum and difference shall be 3] 9 ? 
 
 Ans. 9. 
 
 9. What two numbers are they, whose product is 126, and 
 the quotient of the greater divided by the less, 3' 1 
 
 Ans. 6 and 21. 
 
 10. The product of two numbers is p, and their quotient g. 
 
 Required the numbers. . i~ j Ip 
 
 Ans. ^pq and /i- 
 
 11. The sum of the squares of two numbers is 370, and the 
 difference of their squares 208. Required the numbers. 
 
 Atis. 9 and 17. 
 
 12. The sum of the squares of two numbers is c, and the differ- 
 ence of their squares, d. Required the numbers. 
 
 Ans. l^2{c-\-d), and ^J2(c^). 
 
 13. A certain sum of money is lent at 5 per cent, per annum 
 If we multiply the nvmber of dollars in the principal by tha 
 
EQUATIONS OF THE SECOND DEGREE. 181 
 
 number of dollars in the interest for 3 months, the product is 720 
 What is the sum lent 1 Ans. $240. 
 
 14. It is required to find three numbers, such that the product 
 of the first and second =a, the product of the first and third =b, 
 and the sum of the squares of the second and third =c. 
 
 -■^/("-^T"^/(.^.).■«-^(4p)■ 
 
 in. The spaces through which a body falls in different periods 
 of time, being to each other as the squares of those times, in how 
 many seconds will a body fall through 400 feet, the space it falls 
 through in one second being 16.1 feet? 
 Let x= the required number of seconds, then 
 
 16.1 :400 : :1= :a;« ; whence, a:=4.97+ sec. 
 In what time will a body fall through a hight of 1000 feet 1 
 
 Ans. 7.88+ sec. 
 
 16. What two numbers are as 3 to 5, and the sum of whose 
 cubes is 1216 1 
 
 Let 3a; and 5iB= the numbers ; 
 Then 27a;'+125a;'= 152^3=1216, 
 whence, a;3=8, 
 
 and a;=»/¥=2. 
 Hence, the numbers are 6 and 10. 
 
 RiMARK. — This is properly a pure equation of the third degree ; but 
 questions producing such equations are generally arranged with tliose of 
 tlie second degree. 
 
 17. A money safe contains a certain number of drawers. In 
 each drawer there are as many divisions as there are drawers, and 
 in each division there are four times as many dollars as there are 
 drawers. The whole sum in the safe is $5324 ; what is the 
 number of drawers'! Ans. 11. 
 
 18. Two travelers, A and B, set out to meet each other ; A 
 leaving the town C at the same time that B left D. They trav- 
 eled the direct road from C to D, and on meeting it appeared 
 that A had traveled 18 miles more than B ; and that A could 
 have gone B's journey in 15| days, but B would have been 28 
 days in performing A's journey. What is the distance between 
 C and D 1 Ans. 126 miles. 
 
 19. Two men, A and B, engaged to work for a certain number 
 of days at different rates. At the end of the time, 4, who had 
 
182 RAY'S ALGEBRA, PART SECOND. 
 
 played 4 of those days, received 75 shillings ; but B, who had 
 played 7 of those days, received only 48 shillings. Now had R 
 played only 4 days, and A played? days, they would have received 
 the same sum. For how many days were they engaged "! 
 
 Ans. 19 days. 
 
 20. A vintner draws a certain quantity of wine out of a full 
 vessel that holds 256 gallons ; and then filling the vessel with 
 water, draws off the same number of gallons as before, and so 
 on for four draughts, when there were only 81 gallons of pure 
 wine left. How much wine did he draw each time 1 
 
 Ans. 64, 48, 36, and 27 gaUons. 
 
 COMPLETE EQUATIONS OF THE SECOND DEGREE. 
 
 AsT. 230. 1. Let it be required to find the value of x in the 
 equation, 
 
 a;2— «a;+9=4. 
 
 It is evident, from Art. 184, that the first member of this equa- 
 tion is a perfect square. By extracting the square root of both 
 members, we find 
 
 X— 3=±2; 
 Whence, as=3±2=3+2=5, or 3—2=1. 
 
 Verification. (5)=— 6(5)+9=4, that is, 25—30+9=4. 
 (1)S— 6(l)+&=4, that is, 1—6+9=4. 
 
 Hence, x has ivx> valves, +5, and +1, either of which verifies 
 the equation. 
 
 2. Let it be required to find the value of x in the equation, 
 
 a;2_6a=27. 
 
 If the left member of this equation were a perfect square, we 
 might find the value of x by extracting the square root, as in the 
 preceding example. To ascertain what is necessary to render 
 the first member a perfect square, let us compare it with tlie 
 square of x — a, which is, 
 
 x' — 2ax-\-a'. 
 We have, a' — 6a; =27. 
 
 From this we see that 2a corresponds to 6 ; hence, a corres- 
 ponds to 3, and a' to 9. Hence, by adding 9, which is the square 
 of half the coefficient of x, to each member, the equation becomes 
 
 a:'— 6a!+9=36. 
 Extracting the square root, x — 3 =±6. 
 
EQUATIONS OF THE SECOND DEGREE. 183 
 
 Whence, a;=3±6=4-9, or — 3, either of which values of x 
 will verify the equation. 
 
 Art. 231. We will now proceed to explain the method of 
 completing the square. 
 
 Since every complete equation of the second degree (Art. 226) 
 may be reduced to the form, ( 
 
 ax^-{-bx=c ; if we divide both sides 
 
 oy a, we have x^-\--x==t 
 
 a a 
 
 h c 
 
 For the sake of simplicity, let -=2/i, and -=?. The equa- 
 
 tion then becomes 
 
 x'-\-2px=q. (1) 
 
 b c 
 
 If - is negative, and - positive, the equation becomes 
 a a 
 
 x^ — 2^r=j. (2) 
 
 h c 
 
 If _ is positive, and _ negative, the equation becomes 
 a a 
 
 x'+2px=—g. (3) 
 
 4 c 
 
 Lastly, if _ and _ are both negative, the equation becomes 
 a a 
 
 x' — 2px= — q. (4) 
 
 Hence, every complete equation of the second degree, may he reductA 
 to the form x'-f-2px=q, in which 2p and q may he either positive w 
 negative, integral or fractional quantities. 
 
 We will now proceed to explain the principle by which tl ^ 
 first member of this equation may always be made a perfect 
 square. 
 
 Since the square of a binomial is equal to the square of the 
 first term, plus twice the product of the first term by the second, 
 plus the square of the second ; if we consider x^-\-2px as the 
 first two terms of the square of a binomial, we find x^ is the square 
 of the first term ; hence, the first term must be x ; we next 
 observe that 2jt!x is the do^ible of the product of the first term by 
 the second ; therefore, if we divide 2px by x, the quotient 2p is 
 double the second term. Hence p, which is half the coefficient of 
 X, ip the second term of the binomial ; therefore, its square, p'^, 
 added to x-+2px, will render it a perfect square. But, to preserva 
 
184 RAY'S ALGEBRA, I ART SECOND. 
 
 khe equality, we must add the same quantity to both sides. This 
 fives, 
 
 Extracting the square root, x-\-p=±^q-\-p^ ; 
 
 Transposing, x= — y±V94"P°" 
 
 It is obvious that in each of the remaining three forms, the 
 square may be completed on the same principle ; that is, by tak- 
 ing half the coefiScient of the first power of x, squaring it, and 
 adding it to each member. 
 
 Solving equations (2), (3), and (4), and collecting together the 
 four different forms, and the values of x in each, we have the fol- 
 lowing table. 
 
 (1) x'+2px=q. a=-i7±V9+7'- 
 
 (2) x' — 2px=q. ':=+P±<Jq+p''. 
 
 (3) x^-\-2px= — q. x=—pdz,J — q+P' 
 
 (4) 3p — 2px=: — q. x=-\-p±^ — ?+?'• 
 
 Although the method of finding the values of x is the same in 
 each of these forms, it is convenient to distinguish between them. 
 See Art. 235. 
 
 From the preceding we derive the following 
 
 Rule for the solution of a complete equation of the sec- 
 ond DEGREE. — 1st. Reduce the equation, by clearing of fracticms 
 and transposition, to the form ax^+bx=c. 
 
 2nd. Divide each side of the equation by the coefficient ofx^, and add 
 to each member the square of half the coefficient of the first power 
 of X. 
 
 3rd. Extract the square root of both sides, and transpose the known 
 term to the second member. 
 
 Remakes. — 1st. When the coefficient of x* is negative, as in the 
 equation — x^'-\-mx=n, the pupil may not perceive that it is embraced 
 In the four general forms. This difficulty is obviated by multiplying 
 both sides of the equation by — 1. 
 
 9nd. Since the sign of the square root of xt, or of (x-j-js)', is ±, II 
 might seem that when x^:=m'', we should have I j, — -l-m, that is, -\-x= 
 -\-m(l), -|-x= — m(2), — x=4-n»(3), and — x= — m(4). But it is evi- 
 dent that equations (1) and (4) are the same equation, as also (3) and 
 (3). Hence, -[-a:^±m, embraces all the values of x. For the same 
 reason it is necessary to take only the plus sign of the square root of 
 
EQUATIONS OF THK SECOND DEGREI- 185 
 
 1. Given nx—'!ix^=22~Zx, to find k. 
 
 Transposing, — 2a;2+20a;=32 ; 
 Reducing, x' — 10i= — 16 ; 
 
 Coinpleting the square by adding (y)'=25 to both Bidea of 
 the equation, 
 
 aJ_10a:4-25=— 1 6+25=9 ; 
 Extracting the root, x — 5=±3; 
 Whence, x=5±Z=a, or 2. 
 
 Verification. 17(8)— 2(8)2=32— 3(8), or +8=+8. 
 17(2)— 2(2)2=32— 3(2), or +26=4-26. 
 
 2. Given 3a;'— 2a:=65, to find x. 
 
 Dividing by 3, i'— |a:=^6 ; 
 
 Completing the square, a:'- |a;+(J)'=6jS+(|)2=lM. 
 
 Extracting the root, x — 1^±*3*. 
 Whence, x=l±^-^=5, or — i 
 
 Both of which values verify the equation. 
 
 3. Given 4d'—2x'+2ax=18ab—18b\ to find a;. 
 
 Transposing, — 2x2+2aa::=— 4a'+18o6— 18J' ; 
 
 Dividing by —2, x^—ax=2a'—9ab+9b^ ; 
 
 Completing the square, x'—ax+—=——9a>-{-9b' ; 
 
 Whence, x=-±: ( ?f— 35 ) ; 
 
 ;c=?+(^-3j) =20-36; 
 a;=°-(?^-3i)=-a+3i, 
 
 4. Given ar+;y(5i+i0)=8, to find x. 
 
 By transposition, ,^(5a;+10)=8 — x; 
 
 By squaring, 5a:+10=64 — 16x+a;' ; 
 
 or, x' — 21at= — 54 ; 
 10 
 
186 RAY'S A.LGEBRA, PART SECOND. 
 
 Completing the square, x'—21x+(\iy=i±l.—5i=?l^ 
 
 Extracting the root, a: — 2^'=±'j* ; 
 Whence, 1=2^1 ±1^6 =3^6=1 8, or ^=3. 
 
 These two values of a: are the roots of the quadratic equation, 
 c' — 21a;= — 54 ; but they will not both verify the proposed equa- 
 tion x-)-^(5a;+10)=8, from which the former was derived, for 
 Ihe following reasons. Since the square root of a quantity may 
 have either the sign + or — prefixed to it, the proposed equation 
 might have been x^hj (px-\-i0)=8 ; because by the operations 
 which have been employed, the same resulting equation, a;' — ^21a; 
 = — 54, would be obtained, whether the sign of the radical pa t 
 be + or — . 
 
 Hence, in the equation x+iJ(px-\-10)=8, the value of a: is 'f . 
 but in the equation x — ^(5a;-i-10)=S, the value of x is 18. 
 
 EXAMPLES FOR PRACTICE. 
 
 5. a:=-|-4a;=60. Ans. x=6, or —10. 
 
 6. a;2 — 4aK=60. Ans. a;=10, or —6. 
 
 7. a;'4-16as=— 60. Ans. a;=— 6, or —10. 
 
 8. a!^— 16a;=— 60. Ans. x—&, or 10. 
 
 9. a;'— 6a;=6a;+28. Ans. a=14, or —2. 
 
 10. ?l4-350— 12aj=:0. Ans. a;=70, or 50. 
 10 
 
 11. l^'+8»— 50^=429|. Ans. x=2Q, or —30. 
 
 12. 2a!=4+-. Ans. ax=3, or —1 
 
 X 
 
 13. 3*=+10as=57. Ans. x=Z, or — 6| 
 
 14. (a;— l)(a:— 2)=1 Ans. a;=i(3±^5). 
 
 15. 4a;2— 3a;— 5=80. Ans. x=5, or — 4|. 
 
 16. ia;2— 11+2=9. Ans. as=4, or — 3 >. 
 
 17 a:=l+il2. Ans. a=ll, or —10. 
 
 X 
 
 18. 3(a— 2)'=8(a>-2)+3. Ans. x—5, or If. 
 
 19. ^»+22_4^?^z5 Ans. x=1, or '|. 
 
 3 a; 2 • '28 
 
EQUATIONS OF THE SECOND DEGREE. 
 
 181 
 
 20. 
 21. 
 
 22. 
 
 23. 
 
 24. 
 
 25. 
 26. 
 27. 
 28. 
 29. 
 30. 
 31. 
 
 32. 
 
 33. 
 
 34. 
 
 35. 
 36. 
 87. 
 
 ^+3^=1+8. 
 17irM-19a!=1848. 
 
 3 10^^ 
 
 ff— 5- 
 
 3a>— ia;»=10. 
 
 4 
 
 2^— 10_H:3_2 
 8— X x—2 
 2x , 2a; — 5^gj 
 
 a^ a;— 3 ^' 
 
 1 
 
 x—1 x+Z 
 1 , 1 
 
 = 1 . 
 
 15 
 
 9 
 
 x'—3x x'+ix 8x 
 48 ^ 165 _q 
 1+3 x+10 
 »-j-4_7— x_4a-f7_j 
 ^~ a;— 3 9 
 
 4a;+7 j_ 5 — a!^ 4a; 
 ~15''^9+x~~9- 
 
 a;+-_-— . 
 
 a; a; lo 
 
 X — - 1 — _ 
 
 X X 
 
 a X a 
 
 2a;(a — x) a 
 
 3a— 2a! ~4' 
 
 :=2aa; — ca;'. 
 
 c 
 
 «2— (a+6)a:+oi=0. 
 
 (a— 6)a:2— (a+6)j;4-2fc=0 . 
 
 Atw. a;=3, or — 2|_ 
 Ans. as=9l6, or — 11. 
 
 Atw. z=^j or — J. 
 
 Atw. a:=6±2V— 1- 
 Ans. x=7, or |. 
 
 A?M. a:^6, or Sy'^. 
 
 A?M. as=ll, or —13. 
 
 Atu. x=5|, or 5. 
 
 Ams. a;=21, or 5. 
 Ans. x=3, or — 8.7. 
 
 Am. x=^Z, or -g^S 
 
 Ares. as=l±>/(l— o2). 
 
 A»!x. a:=|a, or |a, 
 Ans. x= 
 
 .c 
 
 An«. x=a, or A, 
 2J 
 
 Ajis. as=l , or 
 
 a — V 
 
188 RAY'S ALGEBRA, PART SECOND. 
 
 88. mqx' — mnx-\-p^ — np=0. ' Aris. x=-, or — ". 
 
 q m 
 
 39. 1+hx-^=c Am. ^°g±>/(°'c'-4'^)- 
 a^ • 2 
 
 40. _?!__(a*— 4^ar= 
 
 1 
 
 Ans. x^a, or — b, 
 
 41. adB — ac3?=bcx — hd. Arts. as=-, or — - 
 
 c a" 
 
 19 
 
 42. V(«+5)= ,/tio v ^"*- *=*• •"■ -21- 
 
 V(«+12) 
 
 43. V^+V(a-*)=V*- Ans. a=f±5!(M=*!- 
 
 j4^+2 4— Ji . . 
 
 44 -=! i=_ ^. Atw. a!=4. 
 
 4+^1 Jx 
 
 45. ^a:' — 2^x=x. Ans. x=i. 
 
 46. ^x+a—J^h=J2x. Ans. x=—'^±:^J2a'+2b\ 
 
 •a 
 
 47. {x—c)jVb—(,(i—b)Jcx=0. Ans. x=—, or *? 
 
 6 a' 
 
 12a ^^4a ^^Za 
 
 48. 'Ja+''+'Ja—x=^j^- Am. 5' 5- 
 
 Aet. 232. Hindoo method of solving quadratics. — When 
 Bn equation is brought to the form ax'-\-lx=:c, it may be reduced 
 to an equation of the first degree, without dividing by the coeffi- 
 cient of a:' ; thus avoiding fractions. 
 
 If we multiply every term of the equation ax''-\-bx=c, by four 
 times the coefficient of the Jirst term, and add to both sides the 
 square of the coefficient of the second term, we shall have, 
 ia^x^+iabx+b'—ltac+b^. 
 
 Now the first member of this equation is a perfect square, and 
 by extracting the square root of both sides, we have 
 
 2ax-\-b=d:z,Jiac-\-b^, which is an equation 
 cf theirs* degree. This gives the following, called the 
 
EQUATIONS OF THE SECOND DEGREE. 189 
 
 Hindoo eule "or the solution of equations of the seconb 
 DEGKEE. — Reduce the equation to the form ax'-(-bx=c. Multiply 
 loth sides by four times iJie coefficient of x'. Add the square of 
 the coefficient of x to each side, and then extract the square root. 
 This will give an equation of the first degree, from which tlie valua 
 of X is easily found, 
 
 1. Given 2x'' — 5a;=3, to find a;. 
 
 Multiplying both sides by 8, which is four times the coefficient 
 »f x^, we have 16i' — 40r=24. 
 
 Adding to each side 25, which is the square of the coefficient 
 of X, we have 
 
 16a;'— 40a:+25=49 ; 
 Extracting the root, 4a; — 5=^7 ; 
 Whence, x=3, or — i. 
 
 Find the value of the unknown quantity in each of the follow- 
 ing examples by the Hindoo rule. 
 
 2. 3a:2-f 5a;=2. Ans. a;=-i, or —2. 
 
 3. a:'+as=30. Ans. as=5, or —6. 
 
 4. a:'— a;=72. Ans. x=Q, or —8. 
 
 40 27 
 
 5. — -+ — =13. Ans. x=9, or i|-, 
 X — X ' ' 
 
 PROBLEMS PRODUCING COMPLETE EaUATIONS OF THE SECOND 
 DEGREE. 
 
 Art. 233. 1. A person bought a certain number of sheep for 
 40 dollars, and if he had bought 2 more for the same sum they 
 would have cost a dollar apiece less ; required the number of 
 sheep, and the price of each. 
 
 Let X be the number of sheep, then . — is the price of one, 
 
 X 
 
 and __ is the price of one on the second supposition. 
 
 40 40 , , ,, 
 
 . . = — — 1, by the question. 
 
 a;+2 X 
 
 Solving this equation, we find x= — l=t:9=8, or — 10, th» 
 number of sheep ; and _==4^o=5 dollars, the price of each. 
 
 Also, 1?=J!L=_4. 
 X —10 
 
190 RAY'S ALGEBRA, PART SECOND. 
 
 Now either of these values of x satisfies the equation, but thg 
 negative value, — 10, does not fulfill the conditions of the ques- 
 tion in an arithmetical sense. But, since the subtraction of a 
 negative quantity produces the same result as the addition of a 
 positive quantity of the same numerical value, the question may 
 be so modified that the value, — 10, will be a correct answer to 
 it, the 10 being reckoned positive. The question thus modified 
 is : A person sells a certain number of sheep for 40 dollars. If 
 he had sold 2 feiver for the same sum he would have received a 
 dollar apiece more for them ; required the number sold. 
 
 Remark. — In the preceding, and in many other cases, especially in 
 the solution of philosophical questions, we derive answers which do not 
 correspond with tlie conditions. The reason is that the algebraical 
 expression is more general than the common language ; and the equa- 
 tion, which is a proper representative of the conditions of the given 
 question, also expresses other conditions ; and hence, when it is solved, 
 answers should be obtained, fulfilling all the conditions expressed by the 
 equation. 
 
 2. Find a number such, that if 17 times the number be dimin- 
 ished by its square, the remainder shall be 70. 
 
 Let ar= the number. 
 Then 17a;— a;'=70. 
 or, a;2— 17j:=— 70. 
 Whence, x=%, or 10. 
 
 In this case both values of x satisfy the question in Its arith- 
 metical sense. Thus, 
 
 17X7— 7==119— 49=70. 
 or, 17X10— 105=170-100=70. 
 
 3. Of. a number of bees, afler eight-ninths, and the square root 
 of half of them, had flown away, there were two remaining; 
 what was the number at first ? 
 
 To avoid radicals, let 2x' represent the number of bees at first ; 
 then, 
 
 ]^+x+2=2x^ ; 
 
 Whence, x=6, or — 15 ; but the latter value, being fractional, 
 is excluded by the nature of the question ; the number of been 
 is 2x6='=72. 
 
 4. Divide a into two parts, whose product shall be 5'. 
 
 Let x= one part, then a — x= the other ; 
 .-. a:(a — x), or ax — a:'=6'. 
 
EQUATIONS OF THE SECOND DEGREE. 191 
 
 Whence, x=yia±,J a' — iV) ; that is, 
 
 as=^(ffld=V'»' — **')• ^nd ar—x=yi(e^Ja^ — W), are the 
 parts required, and the two parts are the same, whether the upper 
 or lower sign of the radical quantity be used. Thus, if the num- 
 ber a is 20, and 68, the parts are 16 and 4, or 4 and 16. 
 
 The forms of these results enable us to determine the limits 
 under which the problem is possible ; for it, is evident that if 46' 
 be greater than a', ^a? — 46' becomes imaginary, and thus the 
 two parts are unassignable, according to the principles of arith- 
 metic ; that is, no such parts can be found. It is also easily seen 
 
 that the extreme possible case will be, when ^a'— 4i'=0, in 
 which case x=^hi, and a — fC=io ; also, ft'=io^. 
 
 Reuakk. — In the following examples, that value of the unknown 
 quantity only is given, which satisfies the conditions of the question in 
 an arithmetical sense. 
 
 5. What two numbers are those whose sum is 20 and product 
 36 1 Ans. 2 and 18. 
 
 6. Divide the number 16 into two such parts that their product 
 shall be to the sum of their squares, in the ratio of 2 to 5. 
 
 Ans. 5 and 10. 
 
 7. ririd a number such, that if you subtract it from 10, and 
 multiply the remainder by the number itself, the product shall be 
 21. Ans. 7 or 3. 
 
 8. It is required to divide the number 24 into two such parts 
 that their product shall be equal to 35 times their difference. 
 
 Ans. 10 and 14. 
 
 9. Divide the number 346 into two such parts that the sum of 
 their square roots shall be 26. Ans. IP and 15=. 
 
 ition. — Let x= the square root of one of the parts, and 
 26 — X, the square root of the other part. 
 
 10. What number added to its square root gives 132 ? 
 
 Ans. 121. 
 
 11. What number exceeds Its square root by 48| I 
 
 Ans. 561. 
 
 4 
 
 12. What two numbers are those, whose sum is 41, and the 
 sum of whose squares is 901 1 Ans. 15 and 26. 
 
 13. What two numbers are those, whose difference is 8, and 
 the sum of whose squares is 544 I Ans. 13 and 20. 
 
192 RAY'S ALGEBRA, PART SECOND. 
 
 14. A merchant sold a piece of cloth for 24 dollars, and gained 
 as much per cent, as the cloth cost him. Required the first cost. 
 
 Ans. 20 dollars. 
 
 15. Two persons, A and B, had a distance of 39 miles to travel, 
 and they started at the same time ; but A, by traveling J of a 
 mile an hour more than B, arrived one hour before him ; find 
 their rates of traveling. Ans. A 3^, B 3 miles per hour. 
 
 16. A and B distribute 1200 dollars each among a number 
 of persons ; A gives to 40 persons more than B, and B gives 5 
 dollars apiece to each person more than A ; find the number of 
 persons. Ans. 120 and 80. 
 
 17. From two towns, distant from each other 320 miles, two 
 persons, A and B, set out at the same instant to meet each other 
 A traveled 8 miles a day more than B, and the number of days 
 in which they met was equal to half the. number of miles B went 
 in a day ; how many miles did each travel per day 1 
 
 Ans. A 24, and B 16 miles. 
 
 18. A set out from C towards D, and traveled 7 miles a day. 
 After he had gone 32 miles, B set out from D towards C, and 
 went every day y^ of the whole journey ; and after he had trav- 
 eled as many days as he went miles in one day, he met A. Re- 
 quired the distance of the places C and D. 
 
 Ans. 76, or 152 miles. 
 
 19. A grazier bought a certain number of oxen for $240 
 and after losing 3, sold the remainder for $3 a head more than 
 they cost him, thus gaining $59 by his bargain. What number 
 did he buy % Ans. 16, 
 
 20. Divide the number 100 into two such parts that their pro- 
 duct may be equal to the difiTerence of their squares. 
 
 Ans. 38.197, and 61.803 nearly. 
 
 21. Two persons, A and B, jointly invested $500 in business, 
 each contributing a certain sum ; A let his money remain 5 
 months, and B only 2, and each received back $450, capital and 
 profit. How much did each advance'! 
 
 Ans. A $200, B $300. 
 
 22. It is required to divide each of the numbers 11 and 17 
 into two parts, so that the product of the first parts of each may 
 be 45, and of the second 43. Atis. 5, 6, and 9, 8. 
 
 23. Divide each of the numbers 21 and 30 into two parts, s» 
 that the first part of 21 may be three times as great as the first 
 
EQUATIONS OF THE SECO>fD DEGREE. 199 
 
 part of 30 ; and that the sum of the squares of the remaining 
 parts may be 585. Ans. 18, 3, and 6, 24. 
 
 24. Divide each of the numbers 19 and 29 into two parts, so 
 that the difierence of the squares of the first parts of each may 
 be 72, and the diiFerence of the squares of the remaining parts 
 180. Ans. 7, 12, and 11,18. 
 
 BISCUSSION OF THE GENERAL EdUATION OF THE SECOND 
 DEGREE. 
 
 Akt. 254. The discussion of the general equation of the sec- 
 ond degree, consists in investigating the general properties of the 
 equation, and in interpreting the results, which are derived from 
 making particular suppositions on the diiferent quantities which 
 it contains. 
 
 The general form, to which every complete equation of the sec- 
 ond degree, containing one unknown quantity, may be reduced 
 (Art. 226), is 
 
 x'-\-2px=:q, 
 in which 2p and q may be either both positive or both negative, 
 or one positive and the other negative. 
 Completing the square, we have 
 
 x^-\-2px-\-p':=q-\-p^. 
 Now, x'-{-2px-\-p^=(x-\-py. For the sake of simplicity, put 
 q-\-p'=zm', that is, ^q-\-p'=m, then 
 (x+pY^m' ; 
 Transposing, (x-\-py^-m^=0. 
 
 But, since the left member of this equation is the difference of 
 two squares, it may be resolved into two factors (Art. 93); this 
 gives (x-{-p-\-m)(x-\-p — m)=0. 
 
 Now this equation can be satisfied in two ways, and in only two ; 
 that is, by making either of the factors equal to 0. If we make 
 tlie second factor equal to zero, we have 
 
 x-\-p — m=0 ; 
 
 Or, by transposing, x= — p-\-m= — pJ^,JqJ^, 
 
 If we make the first factor equal to zero, we have 
 
 j;_L.o+TO=0 r 
 
 n 
 
194 RAY'S ALGEBRA, PART SECOND. 
 
 Or, by transposing, a:= — -p — m= — j> — >/?H~P'- 
 Hence, we have 
 
 Peopeett 1 ST. Every equation of the second degree has two root) 
 (or valves of the unknoum qaantity), and only two. 
 Prom the equation (x-\-p-{-m)(x-{-p-^m)=Q, we derive 
 
 Phoperty 2nd. Every complete equation of the second degree, 
 reduced to the form x'-|-2px=q, may be decomposed into two bino- 
 mial factors, of which the first term in each is x, and the second, the 
 two roots with the signs changed. 
 
 Thus, the two roots of the equation, x' — ^7x+10=0, are x=2, 
 andx=:5; hence, i' — Tx+lO^Qc — 2)(x — 5). 
 
 It is now evident that the direct method of resolving a quad- 
 ratic trinomial into its factors, is, to place it equal to zero, and then 
 find the roots of the resulting equaticn. 
 
 In this manner let the learner solve the examples in Art. 94, 
 page 50. 
 
 By reversing the operation, we can readily form an equation 
 whose roots shall have any given values. 
 
 Thus, let it be required to form an equation whose roots shall 
 be — 3 and 4. 
 
 We must have 
 
 a;=— 3, or a;+3=0, 
 
 And 
 
 x=4t, or X — 4=0. 
 
 Hence, 
 
 (x+3)(a>-4)=0; 
 
 Or, 
 
 x'-^— 12=0; 
 
 Or, 
 
 a:' — x=:12, which is an equation whose 
 
 roots are 
 
 —3 and -+4. 
 
 1. Find an equation whose roots are 4 and 5. 
 
 Ans. x'— 9as=— 20. 
 
 2. Find an equation whose roots are — I and -j-l. 
 
 Ans. a;2-|-^i=>. 
 
 S. Find an equation, without fractional coefficients, whose roots 
 are | and -|. Ans. ISa;'— 22a:=— 8. 
 
 4. Find an equation whose roots are m-\-n, and m — n. 
 
 Ans. x' — Zmx=n'' — m'. 
 
 Resuming the equation x^-\-Zpx=^q, and denoting the two rooti 
 by i! and a/', we have 
 
EQUATIONS OF THE SECOND DEGREE. 19S 
 
 a'=— p+V7+i'' 
 
 Adding; x'+x"= — 2p. But, — 2p is the 
 
 coefficient of x, taken with a contrary sign. Hence, we have 
 
 Peopektt 3kd, TAe sum of the two roots of an equation of the 
 second degree, reduced to the form x'-f-2px=q, is equal to the coeffi- 
 cient of (he first power of x, taken with a contrary sign. 
 
 If we take the product of the roots, we have 
 
 *——P+ •/?+;''. 
 
 x"=— p— ^54-p» 
 
 i>'— PVj+i"' 
 
 -\-pJq-^^—(.q-\-f) 
 
 a:V'=;)'. . . . -(?-f:P=)=-9. 
 
 But — q is the known term of the equation, taken with a coiv 
 trary sign. Hence, we have 
 
 Peoperty 4th. The product of the two roots of an equation of 
 the second degree, reduced to the form x'-4-2px=q, is equal to th6 
 knoum term taken with a contrary sign. 
 
 Note. — In the preceding demonstrations, we have regarded 2? and q 
 as both positive ; but the same conclusions will l)o obtained by taking 
 them both negative, or one positive and the other negative. 
 
 Art. 235. We shall now proceed to determine the essential 
 sign of the roots in each of the four different forms, and to com- 
 pare the two roots in each form, in regard to their numerical mag- 
 nitude. 
 
 To do this, it is necessary first to compare p with »Jq-\-p', ani 
 also with ^ — 9+P'- 
 
 If we examine iJq-{-p', we see that its value must be a quan- 
 tity greater than p, since the square root of p^ alone, is p. 
 
 But the value of ,/ — ?+?'> must be less than p, since it is the 
 square root of a quantity less than p'. 
 
 With these principles, a careful consideration of the roots, oi 
 values of x in each of the four different forms will render the fol- 
 lowing conclusions evident : 
 
196 RAi b ALGEBRA, tART bUCOKU. 
 
 1st form x'-\-2px:=q. 
 
 x'=—p+^q+p' ; 
 
 The first root is essentially positive, and the second essentially 
 oegative ; and the first root is numerically less than the second. 
 
 2nd form, x' — 2px:=q. 
 
 x"=p—Jq-{-p>. 
 
 The first root is essentially positive, and the second essentially 
 negative ; and the first root is numerically greater than the 
 second. 
 
 3rd form, x^-\-2px= — q. 
 
 '>^=~P+'J—q+P^ 
 
 r"=— ;p— ^— 9-4-p'. 
 
 Both roots are essentially negative, and the first root is numeri- 
 cally less than the second. 
 
 4tli form, a;' — 2px= — q. 
 
 x"=p—J—q+p'. 
 
 Both roots are essentially positive, and the first root is numeri- 
 cally greater than the second. 
 It is obvious that in each of the forms, the exact numerical 
 
 value of the roots can be found, only when iJq-\-p^, or J—q-\-p^ 
 I's a perfect square. 
 
 Note. — Questions 5, 6, 7, 8, page 186, are specially adapted to illus- 
 trate the four difiereut forms. See, also, Ray's Algebra, Fart 1st, 
 Art. 217. 
 
 Aet. 286. We shall now proceed to show when the roots 
 become imaginary, and why. 
 
 In the third and fourth forms, the radical part is ,J — j-f-p'- 
 Now when q is greater than p", this is essentially negative, and 
 we are required to extract the square root of a negative quantity, 
 which is impossible (Art. 193). Hence, when q is greater than 
 p', that is, when the known term is negative, and greater than the 
 square of half the coefficient of the first power of x, the roots an 
 imaginary. 
 
EQUATIONS OP" THE SECOND DEGREE. 197 
 
 To show why the roots, are imaginary, we must inquire, into 
 what two parts a number must be divided, that the product of the 
 parts shall be the greatest possible. 
 
 Let 2p represent any number, and let the parts, into which it 
 is supposed to be divided, be p-\-x, and p — 2. The product of 
 these parts is 
 
 (p-\-z){p—z)—p^—z^. 
 
 Now this product is evidently the greatest, when i' is the least ; 
 that is, when z''=Q, or z=0. But when z is 0, the parts are p 
 and p. Hence, 
 
 When a number is divided into two equal parts, their product is 
 greater than that of any other two parts into which the number can be 
 divided. Or, as the same principle may be otherwise expressed, 
 
 The product of any two unequal numbers, is less than the square of 
 half their sum. 
 
 Now it has been shown (Art. 234, Properties 3rd and 4th), 
 that 2p, the coefficient of the first power of x, is equal to the sum 
 of the two roots, and that q is equal to their product. But, when 
 q is greater than p^, we have the product of two numbers, greater 
 than the square of half their sum, which, by the preceding princi- 
 ple, is impossible. If, then, any problem furnishes an equation of 
 the form i^±2pa^= — q, in which the known term is negative and 
 greater than the square of half the coefficient of the first power 
 of the unknown quantity, we infer, that the conditions of the 
 problem are iTWompatible with each other. The following is an 
 example. 
 
 Let it be required to divide the number 8 into two parts, whose 
 product shall be 18. 
 
 Let X and 8 — x represent the parts. 
 Then, a;(8— a;)=18; or i'— 8a;=— 18; 
 
 Whence, a:=4+V-^. o>" 4— V^- 
 These expressions for the values of x, show that the problem is 
 impossible. This we a,lso know from the preceding; theorem 
 since the number 8 cannot be divided into any two parts whose 
 product will be greater than 16. Thus, the algebraic solution 
 renders it manifest that the problem is impossible. 
 
 Aet. 337. Examination of particular cases. 
 
 1st. If, in the third and fourth forms, where q is negative, we 
 suppose 5=p', the radical, ^ — q+p't becomes 0, and x= — p, in 
 the third form, or -\-p in the fourth form. It is then said, the two 
 roots are equal. 
 
198 RAY'S ALGEBRA, PART SECOND. 
 
 In fact, if we Bubstitute p' for q, the equation in the third form 
 becomes x^-^2px-\-p'=0. 
 
 Hence, {x-\-pf, or, {x-\-p){x-\-p)=0. 
 
 In this case, the first member is the product of (wo equal factors 
 Hence, the roots of the equation are equal, since either of the 
 two factors, being placed equal to zero, gives the same value for 
 X. A similar conclusion is obtained by substituting p^ for q in 
 the fourth form. 
 
 2nd. If, in the general equation, x'-\-2px=^, we suppose 9=0, 
 the two values of x reduce to, 
 
 *= — p-\-p=0, and x= — p — p= — 2p. 
 In fact, the equation is then of the form 
 
 x'-\-2px=0, or x(_x+2p)=z0, 
 which can be satisfied only by making 
 
 x=0, or x-{-2p=0; 
 Whence, x=0, or 1= — 2p. 
 
 3rd. If, in the general equation, x^-\-2px=q, we suppose 2^=0, 
 we have x'^=q, 
 
 Whence, x^zt.>Jq- 
 
 In this case, Ihe two values of x are equal and have contrary signs, 
 real, if q is positive, as in the first and second forms, and imagin- 
 ary, if q is negative, as in the third and fourth forms. 
 
 Under this supposition the equation contains only two terms, 
 and belongs to the class treated of Art. 228. 
 
 4th. If we suppose 2p=Q, and 5=0, the equations reduce to 
 jt'^0, and give the two values of x, in all the forms, each equal 
 toO. 
 
 Akt. 23S. There remains a singular case to be examined, 
 vhich is sometimes met with in the solution of problems produc 
 'ig equations of the second degree. 
 
 To discuss it, take the equation 
 
 (m;'+Jx=c. 
 (So.ving this equation, the values of x are 
 
 2a 2a 
 
EQUATIONS OF THE SECOND DEGREE. 199 
 
 If, now, we suppose 0=0, these values become 
 
 — i+i_0 —b—b — 26_ 
 
 x=—Q =0' x=—Q— =-Q-=a} . 
 
 That Is, one value of x is indeterminate and the other infiniii 
 (Arts. 136, 137). 
 But if we suppose a=0 in the given equation, we have 
 
 bx=c, and x=- 
 b- 
 
 But X can have only two values, (Art. 234) ; heijce, there is «( 
 least an apparent contradiction ; how can it be reconciled 1 
 
 Let us first examine the value of x=- 
 
 0' 
 
 If wo multiply both terms of the second member of the equa 
 lion x=1UJZy- — it — by — b — iJb'-{-iac, we have 
 
 J2_(i2_|-4ac) — 4ac 
 
 x= 
 
 2a(_6— ^ft2+4ac) 2a{—b—Jb'+iacy 
 or, by dividing both terms by 2a 
 
 —2c 
 
 —b—Jb'+iac ' 
 
 Whence, ar=-, by making a=0. 
 J 
 
 Hence we see, that the value of x=-, is really _, and arises 
 
 ^ J 
 
 from having made a factor zero, that was common to the numera- 
 tor and denominator. 
 
 We shall now examine the value of 3;= — _=ao 
 
 
 
 By supposing a=0, the equation ax^-{-bx=e, reduces to bx=c, 
 an equation of the^rs< degree. 
 
 It is, therefore, impossible that it can have more than OTie root 
 (Art. 170.) Hence, the supposition that it has two, gives one 
 value infinite, which is equivalent to saying, the equation has but 
 one finite root. 
 
 If we had at the same time 
 
 a=0, 6=0, c=0, the equation would be 
 altogether indeterminate. This is the only case of indetermina- 
 tion presented by the equation of the second degree. 
 
200 
 
 RAV'S ALGEBRA, PART SECONI/. 
 
 Art. 239. We shall now apply the preceding principles in the 
 discussion of a problem, which presents most of the circum- 
 stances commonly met with, in problems producing equations of 
 the second degree. 
 
 PROBLEM OF THE LIGHTS. 
 
 It is required to find, in a line BC, which joins two lights, B ant" 
 C, of different intensities, a point which is illuminated equally 
 by each. 
 
 F' S P Z P' 
 
 It is a principle in optics that, the intensity of the same light a. 
 different distances, is inversely as the square of the distance. 
 
 Let a be the distance BC between the two lights. 
 
 Let b be the intensity of the light B at the distance of one foot 
 
 from B. 
 Let c be the intensity of the light C at the distance of one foot 
 
 from C. 
 Let P be the point required. 
 Let BP =x, then CP =a—x. 
 
 By the optical principle above stated, since the intensity of the 
 light B at the distance of 1 foot, is b, its intensity at the distance 
 
 of 2,3, 4 feet, must be -,,....; hence, the 
 
 intensity of B, at the distance of x feet, must be _ In like man- 
 ner, the intensity of the light C, at the distance of a — x feet, 
 
 must be , But, by the conditions of the problem, these 
 
 (<z — xy 
 
 two intensities are equal ; hence, we have for the equation of the 
 
 problem. 
 
 6 
 
 c 
 
 . , which easily reduces to (" "0 — c . 
 
 i' (a — xy 3? 
 
 a—x -\-Jc — Jc 
 Whence, x "^^.or -=-• 
 
 This gives the following results : 
 
 1 » "V^ , ajc' 
 
 let. x= — = --, whence a — xz= 
 
 ajb —aJc 
 
 — = —I whence a — as=— = — zi. 
 
 Jb — Jc V* — >/c 
 
EQUATIONS OF THE SECOND DEGREE '■iOl 
 
 We shall now proceed to discuss these values. 
 I. Let 6>c. 
 
 The first value of x, 7F, r, is positive and less than a, foi 
 
 -^ ■y= is a proper fraction ; hence, this value gives for the 
 
 point illuminated equally, a point P situated between B and C. 
 We perceive, also, that the point P is nearer to C than B ; for 
 since i>c, we have jT+ y/'^ Jb-\- ^ c, or 2^by^b-\-^c^ 
 
 Jj ^ ajb a 
 
 and .-. ,y, ,->5. and, consequently, Jbj^Jc^2' This is 
 
 manifestly correct, for the required point must be nearer the light 
 
 aye 
 of less intensity. The corresponding value of a—x, -r. f^, 
 
 is positive, and evidently less than -, 
 
 ajb 
 The second value of x, JF^ j-, is positive, and greater than a 
 
 for Jb-yjb-jc; .: J^ZJ>'' -" J^IJi^- 
 
 This value gives a point P', situated on the prolongation of 
 BC, and to the right of the two lights. In fact, we suppose that 
 the two lights emit rays in all directions ; there will, therefore, 
 be a point P', to the right of C, and nearer the light (jf less inten- 
 sity, which is illuminated equally by the two lights. 
 
 It is easy to perceive, why the two values thus obtained, are 
 expressed by the same equation. If, instead of assuming BP for 
 the unknown quantity x, we take BP', then CP'=a! — a, and the 
 equation of the problem is 
 
 a;2 (ar — a)' 
 but (x — ay=(a — xy. Hence, the new equation is the same as 
 that already found, and, consequently, ought to give BP', as well 
 asBP. 
 
 — a^ c 
 
 The second value of a — x, .r- r, is negative, as it ought 
 »Jb—~isJc 
 
 to be, because x'^a, but changing the signs of the equation 
 
202 KAY'S ALGEBRA, PART SECOND. 
 
 ^— , we find X — o=— = —• and this value of x — a. 
 
 represents the distance CP'. 
 
 II. Let *<<:• 
 
 I I I 
 
 P" B P 
 
 The first value of x, .r, f is positive, and less than -, lot 
 
 _ _ _ _ _ ^h 
 
 Jb+^c>^b+Jb, or >2Jb; .: ^Jf:^<i' ""* 
 
 a^b a 
 -^9 
 
 aye 
 The corresponding value of a — x, .j-, .-, is positive, and 
 
 greater than f . These values of x, and of a — x, show that the 
 
 point P is situated between B and C, but nearer to B than C. 
 This is evidently a true result, since, under the present supposi- 
 tion, the intensity of the light B is less than that of the light C. 
 
 a^b — ajb 
 
 Tke second value of X, .^-_ .-, or .-^ .y, is essentially neg- 
 ative. To interpret this result, we must recollect that if distance 
 to the right of a certain point is reckoned positive, then distance 
 to the left is negative (Art. 47); hence, if we consider P" on the 
 left of B, as the point illuminated equally, we ought to represent 
 the distance BP" by — x, and then the distance CP" would be 
 represented by BP"4-BC=: — x-\-a=a — x. Under this supposi- 
 tion, the equation of the problem would be 
 
 * - = -, that is, *= . <= 
 
 ( — xy (o — xy x' (a — xy 
 
 and the solution of this equation, which is the same as that ob- 
 tained for P, ought to give the point P". 
 
 — a^c ajc 
 The corresponding value of a — x, is .r- .-= .-_ .-y. It is 
 
 - - - V"^ ^, 
 
 positive and greater than a, for Jc^Jc — Jb . . ,— /T-^ 
 _ •J''' — v* 
 
 and ,- ,-T'^''' This represents the distances CP", and is 
 Jr—Jb 
 
EQUATIONS OF THE SECOND DEGREE. 203 
 
 merely the sum of the distances CB and BP". These results are 
 manifestly correct, and correspond to the circumstances of the 
 problem. 
 
 III. Let b=c. 
 
 The first values of x, and of a — x, reduce to - , which shows 
 
 2 
 
 (hat the point illuminated equally, is at the middle of the line 
 
 3C, a result manifestly true, upon the supposition that the inten- 
 
 tities of the two lights are equal. 
 
 The other two values are reduced to ?^=ao . (Art. 136). 
 
 This result is manifestly true, for the intensities of the two liorhta 
 being supposed equal, there is no point at any Jinite distance, 
 except the point P, which is equally illuminated by both. 
 
 IV. Let h=c, and a=0. 
 
 The first system of values of x and a — x, become 0. This is 
 evidently correct, for when the distance BC becomes 0, the dis- 
 tances BP and CP also become 0. 
 
 The second system of values of x and a — x, become 9 ; this io 
 
 the symbol of indetermination (Art. 137). 
 
 This result is also correct, for if the two lights are equal, and 
 placed at the same point, even/ point on either side of them will 
 be illuminated equally by each. 
 
 v. Let 0=0, 6 not being =c. 
 
 In this case, all the values of x and of a — x reduce to 0, which 
 shows tha't there is only one point equally illuminated by each; 
 viz : the point in which the two lights are placed. 
 
 The preceding discussion, affords an example of the precision 
 with which algebra answers to all the circumstances included ii 
 the enunciation of a problem. 
 
 Art. 239". Examples for the discussion and illustration of 
 principles. 
 
 1 . Required a number such, that twice its square, increased by 
 S times the number itself, shall be 90. Ans. 5, or — 9. 
 
 How may the question be changed, that the negative answer, taken 
 positively, shall be correct in an arithmetical sense ? 
 
 2. The difference of two numbers is 4, and their product 21. 
 Required the numbers. Ans. -\-S, -f-7, or — 3 and — 7. 
 
204 RAY'S ALGEBRA, PARI SECOND. 
 
 3. A man bought fi watch, which he afterward sold for 16 dol- 
 lars. By the sale his loss per cent, on the first cost of the watch: 
 was the same as the number of dollars which he paid for it. 
 What did he pay for the watch 1 
 
 Ans. 20 dollars, or 80 dollars. 
 
 4. Required a number such, that the square of the number 
 increased by 6 times the number, and this sum increased by 7 
 the result shall be 2. Alls. x= — 1, or — 5. 
 
 What do the values of x show ? How may the question be changed 
 to be possible in an arithmetical sense 7 
 
 5. Divide the number 10 into two such parts, that the product 
 shall be 24. Ans. 4 and 6, or 6 and 4. 
 
 Is there more than one solution ? Why 7 
 
 6. Divide the number 10 into two such parts that the product 
 shall be 26. Ans. 5-\-J^\, and b—^'—i. 
 
 What do these results show 1 
 
 7. Divide the number a into two such parts, that their squares 
 shall be to each other as 1 to n. 
 
 ajn a 
 Ans. x= ::^, and a — x= =. 
 
 aJn a 
 
 Or ar= — —, and a — x= =^. 
 
 1 — ^n 1 — Jn 
 
 What are the parts when a=12 and n=4? When fl=10 and n=11 
 
 8. The mass of the earth, according to astronomers, is 80 times 
 tliat of the moon, and their mean distance asunder 240000 miles. 
 Now the attraction of gravitation being directly as the quantity 
 of matter, and inversely as the square of the distance from the 
 center of attraction, it is required to find at what point on the 
 line passing through the centers of these bodies the forces of 
 attraction are equal ? 
 
 Ans. 21586.5.5+ miles from the earth, 
 and 24134.5— " " " moon. 
 Or, 270210.4+ " " " earth, 
 and 30210.4+ " beyond the moon from th« 
 earth. 
 
 This question involves the same principles as the Problem of 
 the lights, and may be discussed in a similar manner. The 
 required results, however, may be obtained directly from the 
 values of x, page 200, calling o=240000, 4=80, and =1 
 
QUATIONS OF THE SECOND DEGREE. 205 
 
 TRINOMIAL EOUATIONS. 
 
 Akt. S40. A trinomial equation is one consisting of three 
 lernis. The general form is ar'"+Aa;"=c, in v/hich all the quan- 
 tities are supposed to be known, except x. 
 Every trinomial equation of the form 
 
 a;'"+2pa"=5, that is, every equation ot 
 three terms containing only two powers of the unknown quantityi 
 and in which one of the exponents is dovble the other, can be 
 solved in the same manner as a complete equation of the second 
 degree. Thus, 
 
 let x'^y then x^''=y^, and the equation becomes 
 y^-\-^fy=q ; 
 Whence (Art. 231), y==-^>±Jq+f; 
 Substituting a?" for y, and extracting the re" root of both sides, 
 
 As an example, let it be required to find the value of x in the 
 «^uation x"^ — 2px^=q. 
 
 Let x^=y, the equation then becomes 
 y''—2py=q. 
 
 Whence, y=-h?±V9+i''=*'°- 
 
 x=±^p±^q+p'. 
 
 Art. 241. Binomial Surds. — Expressions of the form A=h 
 ^B, or VAi^B, like the value of x just found, are called 
 Binomial surds ; they are sometimes found in the solution of 
 Trinomial equations of the fourth degree, and as it is Bometimes 
 possible to reduce them to a more simple form by extracting the 
 square root, it is necessary to consider them here. 
 
 We shall first show that it is sometimes possible to extract the 
 square root of AdziJB, or ,JA±aJB- 
 
 (2±^/3■)'=7=t4 J3 ; . . ^7±4V3=2±V3. 
 
 (V2±V"3 )'=5±2 V6 ; •■• ^5±2V6=^/2±^/3'. 
 
 We shall now proceed to show that it is always possible to 
 extract the square root of A±V^' or JA.dcz,JB, if A^ — B is a 
 perfect square. To do this it is necessary to prove the following 
 theorems. 
 
206 RAY'S ALGEBRA, PART SECOND. 
 
 Theorem I. — The square root of no (juantity can he partly 
 ratUmal and partly a radical of the second degret. 
 
 For, if possible, let ^x=a-\-^b ; . . squaring both sides, 
 
 it=a'+2a;^6-|-i ; .-. ^A= — , that is, an irra- 
 
 2a 
 
 tional quantity is eufual to a rational quantity, which is impossible; 
 hence, the supposition is impossible and the theorem is true. 
 
 Theorem II. — In any equation consisting of rational quantities 
 and radicals of the second degree, the ratitmal quantities on each side 
 are equal, and also the irrational quantities. 
 
 If x-\-^y=a-{-^b, then x=a, and ^y=:Jb. 
 For if X does not ^a, let x=:a-\-m ; 
 
 .•. a-\-m+Jy=a->!-Jb; .-. m-\-^y=^'b; 
 
 that is, the square root of a quantity is partly rational and partly 
 irrational, wfiich has been shown by Th. I, to be impossible ; 
 hence, x=a, and Jy=Jb. 
 
 We shall now proceed to find a formula for extracting the square 
 root of A+ Jb. 
 
 Assume jA.-\-J'&=J^-\-^y, 
 
 A-\-^S=x-\-y-{-2 ^xy, by squaring. 
 
 By Th. II, x+y=A(l) ; and 2 ^^=^¥(2) ; 
 Squaring equations (1) and (2), we have 
 a:=+2a;y+3,==A', 
 4xy =B ; 
 Subtracting, x'—2xy-{-y'=A^—B; or (x-</)'=A'—B 
 
 Let A' — B be a perfect square =C', then C=^A'— B. 
 •■• (a;— y)'=C2, or x-^y=C ; 
 But a;-l-y=A ; 
 
 Whence, x= ^+^ ; and v=^^ 
 
 2 ' 2 ■ . 
 
 and ^c^±^^; and ^^=±^^-0 
 
 'A— C 
 
 2 • 
 
EQUATIONS OF THE oECOND DEGREE. 207 
 
 A— C 
 
 Similarly, ^x-Jy==^A-^B=±^^^^: 
 
 These formulas are easily verified by squaring each side, and 
 substituting A»— B for C. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Extract the square root of Sl-j-lO^yS. 
 
 Here A=31, JB—lO^e'; .-. A2—B=C2=961— 600=361, 
 and C=19. 
 .-. A4-C=50, A— C=12; .-. a;=25, y=6; 
 
 • ■ V'^+Vy=^A+VB=V25+^/6"=5+V6. 
 
 2. Reduce l?ip-\-2m' — 2m^np-{-m^, to its simplest form. 
 
 Here A=rep+2m', and B=4m2(7ip4-'«')- 
 
 A' — B;= n^p', and C=np, (formula L). 
 .•. A+C=2np+2m', A— C=2m2 .-. x^np+m^, y=mK 
 
 Formula (L) gives Jx — ^y=±i^np-\-m^ — m). 
 
 Proof. ±{Jnp-\-m} — my=np-\-2m' — 2miJnp-\-m'. 
 
 3. Find the square root of 11+6;^^. Ans. B+J2. 
 
 4. Find the square root of 7 — 4;^3. Ans. 2 — ^3. 
 
 5. Find the square root of 3=1=2^2^. Ans. V2±l. 
 
 6. Find the square root of 13+2 J30. Ans. ^ 10+^3". 
 
 7. Find the square root of 17+2760. Ans. 2^3-{-^5. 
 
 8. Find the square root of x — ZJx^-1. Ans. Jx — 1— ;1. 
 
 9. Find the square root of 2a J— i. (A=0). 
 
 Ans. ^7(1+7'^. 
 10. Find the square root of x-\-y-\-z-\-2 ^ xz-\-yz. 
 
 Ans. Jx-\ry+J'i. 
 
208 HAY'S ALGEBRA, PART SECOND. 
 
 11. Reduce to its simplest form lbc-\-2hJhc — 6' + 
 
 lbo—2bJbc^\ Ans. dz^Jbc—b'. 
 
 Art. 242. We shall now resume the subject of Trinomial 
 Equations. The general form of Trinomial equations ia 
 j;2n_|_2pa;n^g ; but there are several varieties of this form, of which 
 
 the following are the principal: viz: x-\-^x=q; x*-^px'=q, 
 3f-\px'i=q, x'"-'rpx^=q, x*"+px'''=q, (x'-^x+qy+Ux'-{-px 
 +q)=r, and (x'+px+q)^+h(x^+px+qy=k. 
 
 It is easily seen that some of these varieties, if developed, 
 would produce very complicated expressions. Yet they may all 
 be solved by the general method given above, that is, by plac- 
 ing the lowest power of the compound term equal to an un- 
 known quantity, and then substilviing the latter in the given 
 equation. In some cases it is necessary to substitute more than 
 once, as in the following, which is one of the most complicated 
 
 forms : 
 
 (3?-\-px+qy+h{x^-{-px-\-q)<'=k. 
 
 Let x^-\-px-\-q=y, then {x'^-\-px-\-qy''=y^'', and the equa- 
 tion becomes y'"-|-i^=fc. 
 Let y=z, the equation then becomes z'-{-iz=k, from which i 
 
 i, found =r^±V*!S^;whencey==n/^i^^\ 
 
 Calling this last expression h, for simplicity, and we have 
 x'-\-px-\-q=h. 
 
 Whence, x=—P-±\h — g+?- ; 
 2 N 4 
 
 EX AUFL E S. 
 
 1 . Given, a;' — 6a:'=:16, to find the value of x. 
 Assume, iK'=y, then x^=y^, and 
 
 y5— 63^ =16; 
 Whence, y =8, or — 2. 
 
 Thereftre, a;'=8, or — 2. 
 and X =2, or — %I2. 
 
EQUATIONS OF THE SECOND DEGREE. 209 
 
 It will be shown hereafter, (Art. 396), that every equation haa 
 as many roots as there are units in the expqpent of the highest 
 power of the unknown quantity. We do not, therefore, by this 
 method, in all cases, obtain all the values of the unknown quan- 
 tity. Thus, in the preceding example, there are four values of x 
 Dot determined. 
 
 2. Given 5x—4tJx=SS, to find the value of a;. 
 Assume, ,Jx=y, then x=i/^, and 
 
 5^'— 4y=;33 ; 
 Whence, y=3, or — y ; 
 
 Consequently, a;=9, or jj*. 
 
 3. Given, Vx+12+«/^+12=6, to find the value of ar. 
 
 Assume, i/x-^12=:y; then ;^x-|-12=y', and 
 Whence, y=2, or — 3; 
 
 .-. Va+12=2, or —3; 
 Whence, a!-|-12=16, or 81; 
 
 and a:=4, or 69. 
 
 Or, without introducing a new letter y, we may consider the 
 whole expression under the radical as the unknown quantity, and 
 proceed to complete the square thus, 
 
 V^+r2+v^+i2-H=6+i=V ; 
 
 Extracting the root, Va:+12+2=±5 ; 
 
 Vi+T2=— Jiti=+2, or —3. 
 1+12=16, or 81. 
 Whence, x=4:, or 69. 
 
 The principle of both operations is the same, but the use ot 
 the new letter renders the process more easily understood by 
 beginners. 
 
 4. Given 3a:'+73a;'+l=55, to find the value of x. 
 Adding 1 to each member, the equation becomes 
 
 3x'4-l+J3a;'+l=56. 
 The equation may now be solved like the preceding. 
 
 The values of x are +4, — 4, +^21 and — ^21. 
 18 
 
210 RAY'S ALGEBRA, PART SECOND. 
 
 Find the values of x ii 
 
 1 each of the following examples. 
 
 5. a:*— 25a;='=— 144. 
 
 Ans. a 
 
 Avs. K=±3, or ±4 
 
 6. 5a;<+7x2=6732. 
 
 :=±6, or ±ToV— 3T40. 
 
 1 9a;»— lla:'=488. 
 
 
 Ans.x=2, or iiJ—19S. 
 
 8 e»-<:?=15500. 
 
 
 Ans. x=25, or (—124)*. 
 
 9 a;^+a:^=1056. 
 
 
 Ans. 1=64, or (—33)*. 
 
 10. a;+5=Va;+5+6. 
 
 >-Jax-\^. 
 
 Ans. x=4, or —1. 
 
 11. 2^*=— 3x+ll=i= 
 
 
 
 Ans 
 
 .x=2, 1, or ^±W— 31- 
 
 12. a;=—7i4-^x»— 72+18=24. 
 
 Arw. 1=9, —2, or iOzkjTTS). 
 
 13. (x'— 9)==34-ll(a:'— 2). Ans. a:=d=5, or ±2. 
 
 14. (x+- ) +1=42— §. 
 \ X / X 
 
 Ans. x=A, 2, or ^(— 7±^"l7). 
 
 15. a* f 1+i- ) — (3a!'+i)=70. 
 
 \ 3a; / 
 
 Ans. as=3, — 3^, or J(— ld=V^=:55l). 
 
 16. arVli-«j=-^. Ans. as=±V(l±i72). 
 
 Aet. 243. In the preceding examples the form of the trino- 
 mial equation, is either given or easily ascertained ; but it some- 
 times happens that questions are given, in which the compmtnd 
 term is not presented to view, but which may be reduced to the 
 form of a trinomial equation by the following method : 
 
 If the greatest exponent of the unknown quantity is not even, 
 it must be made even, by multiplying both members of the equa- 
 tion by the unknown quantity. Then extract the square root to 
 two or three terms, and if we find a remainder (omitting known 
 terms if necessary), which is any multiple or any part of the root 
 already found, the given equation may be reduced to a trinomial, 
 of which the compmind term will be the root already found. 
 
 Example. Given, a;'— 4(m2— 2a'a:+12a'=l^'', to find z. 
 
 X 
 
 Multiplying both sides by x, and transposing, we have 
 x*—4iLc>—2aV+12a'x—16a*x=0. 
 
EQQATIONS OP THE SECOND DEGREE. 211 
 
 Proceeding to extract the square root, we have the following 
 
 OPEBATIOIf. 
 
 «<— 4<u;'— 2a V+1 2a'x—l 6a<|j;'— 2aa;. 
 
 'Zx'—2ax\ — iax^—2a.^x' 
 — ia3fi-\-iaV 
 
 Remainder — 6o'a;'+12a'x — 16a* ; 
 
 or, —6a\x^—2ax)—16a*. 
 
 Hence, the given equation may be written thus : 
 (a;2— 2aa;)»— 6 a\x'—2(a)—16a'>=0. 
 Assuming x^ — 2ax=y, we find y=Sa', or — 2a' ; then from 
 the equation x' — 2ax=8a', or — 2o^, we find 
 
 x=4a, — 2a, or a-±:aj — 1. 
 
 2. jf— 2*1— 2a:2+3as=108. 
 
 Ans. a;=4, —3, or ^(ld=>/— 35). 
 
 3. a;<— 2a;'+a;=30. Ans. x=Z, —2, or i{l±J—iQ 
 
 4. a'— 6a;2+lla!— 6=0. Ans. x=l, 2, or 3, 
 
 5. ««— 6a!»+5a;'+12a:=60. 
 
 ilns. 0=5, —2, or 4(3±V— 1-5)- 
 
 6. !<— 8a;'+10a;'+24a;=— 5. Ans. a=:5, —1, or 2=bV5- 
 
 7. 4a:<+|=4jc'+33. Aras. ar=2, — |, or i(l±^i:43). 
 
 2 
 30 
 
 -i±Hi-=L+ii. 
 
 ■ li laP^ Zx 2x' 
 
 Ans. x=i:, 3, or 5(7±^"69). 
 
 SIMULTANEOUS EaUATIONS OF THE SECOND DEGREE CON- 
 TAINING TWO OR MORE UNKNOWN aUANTITIES. 
 
 Art. 244. Equations of the second degree, containing two or 
 more unlcnown quantities, may be divided into two classes, 
 1st. Pure Equations. 
 2nd. Adfected Equations. 
 
 The first class embraces those equations that may be solved 
 without completing the square ; the second, tliose in the solution 
 of which it is necessary to complete the square. The same 
 equations, however, may sometimes be solved by both methods. 
 
212 RAY'S ALGEBRA, PART SECOND. 
 
 Akt. 245. Puke Equations. — Pure equations may in general 
 be reduced to the solution of one of the following forms, or pairs 
 of equations. 
 
 -ci-)^+^?|. (2.)^^s. (3-)^;j?;sj. 
 
 We shall explain the general method of solution in each of 
 these cases. 
 
 1. To solve i-)-y=a (1), and xy=h (2), we must find a; — y. 
 Squaring Eq. (1), a;'-|-2ay-|-y°=''' ! 
 
 Multiplying Eq. (2) by 4, 4ay =4& ; 
 
 Subtracting, x' — ^2i^+y'=s' — 46, 
 
 or, (a; — y)'=a' — 46 ; 
 
 Whence, x — y=±^o' — 46 ; 
 
 But, x-\-t/=a ; 
 
 Adding and dividing by 2, a;=ja±2«/''' — ^^• 
 
 Subtracting and dividing by 2, y=^a':^\^a^ — 46. 
 
 The pair of equations (2) is solved in the same manner, except 
 that in finding aj-f-j/ we must add 4 times the second equation 
 -to the square of the first. 
 
 The pair of equations (3) is solved merely by adding and sub- 
 tracting, then dividing by 2 and extracting the square root. 
 
 EXAMPLES IN PURE EQUATIONS. 
 
 1. Given, x''-\-tf=d (1), and x-\-y=:a (2), to find x and y. 
 Squaring Eq. (2), ar'+Say+y'so^ ; 
 
 But, a' -t-y'=d (1). 
 
 Subtracting, 2xy=a.'' — d, (3). 
 
 Take (3) from (1) ; a;'— 2ay-|-y'=2d— o', 
 
 .-. x-^y=:db^2d—a'. 
 
 Whence, x=had=^J2d—a', y=^a=p^ J2i— o^. 
 
 2. Given, x^+3ci/+y^=91(l), and a;+^'^4.y=13(2), to find 
 X and y. 
 
 Divide Eq. (1) by (2), x—J^-\-y= 7. (3). 
 
 But, x+jl^-\^y=\Z. (2). 
 
 By Bobtracting, 2Jay=S. (4). 
 
EQUATIO^S OF THE SECOND DEGREE 218 
 
 By adding, a:+y=10. (5) 
 
 Squaring, (5), x^+2xy+y^=100 : 
 Squaring, (4), txy = 36; 
 
 a:' — 2xy-\-y^=6'l:, .•. x — y=±8. 
 But, x-\-y=:10, whence, x=9 or 1, andj/=l or 9. 
 Equations of higher degrees than the second, that can be 
 solved by simple methods, are usually classed with pure equations 
 of the second degree. 
 
 3. Given, a:*-|-y5=6, and a;*-|-^=126, to find a and y. 
 
 In all cases of fractional exponents, it renders the operations 
 more simple to learners, to make such substitutions as will render 
 
 I 1 
 
 the exponents inteffral. In this example, let a;*=P, and y^=Q; 
 
 then a;*=P', and j(»=Q,'. The givei) equations then become, 
 P+Q,=6 (1), 
 
 P»-|-Q,3=126 (2). 
 
 Dividing Eq. (2) by (1), P=— PQ4-a'=21; 
 Squaring Eq. (1), P=+2Pa+Q,'=36; 
 
 Subtracting, 3PQ,=15, .-. PQ,=5. 
 
 Having P-|-Q=6, and PQ=5, by the method explained ia 
 form (1 ), we readily find P=5 or 1 , and Q=l or 5 . 
 Whence, a;=625 or 1, and y=l or 3125. 
 
 4. Given, (av-y)(a;'— j/=)=160 (1), 
 
 (x-\-yXx'+f)=580 (2), to find x and y. 
 
 x3 — x^y — xyf-]-y^=160 (1), by multiplying. 
 x^+x'y-\-xy'+f=580 (2), " 
 
 2x'y+2xy'=i20 (3), by subtracting. 
 Add (3) to (2), a!'+3a;'ji+3OT/2_[^'=1000. 
 Extract cube root, x-\-y=10. 
 
 From (3),, OT/(a:+3/)=210; .-. xy=21. 
 
 From x-\-y=lO, and a^=21, we readily find x='7 or 3, and 
 y=3, or 7. 
 
 Let the following examples be solved by the preceding or simi- 
 lar methods. 
 6. X — y=2, Ani. i=15, or — 13; 
 
 a,2_|.y2=394. 3^=13, or —1.5. 
 
 6. a;2-|-^'=13. Am. a;=i:±3; 
 
 a^ = 6. v=±2. 
 
*14 RAY'S ALGEBRA, 
 
 PART SECOND. 
 
 7. 2x+y=7. 
 
 Ans. 1=2 or |; 
 
 4x»+y'=25. 
 
 3/=3, or 4. 
 
 8. x'—y'=ie, 
 
 Atw. as=5; 
 
 x—y= 2. 
 
 2/=3. 
 
 9 x-\^= 11, 
 
 Atw. a: — 7, or 4; 
 
 a!"+3('=:407. 
 
 y=4, or 7. 
 
 10. 7(x'4-y')=9(^-ry"). 
 
 Ans. a;=4; 
 
 a^y— y2as=16. 
 
 i/=2. 
 
 ]l. a;»+2y=84. 
 
 .4nj!. a=±7; 
 
 aJ_^y2:r-24. 
 
 y=±5. 
 
 12. a:'+y'=152, 
 
 An*. as=5, or 3; 
 
 ar'— jey-|-y==19. 
 
 y=3, or 5. 
 
 13. i'+i/'+ay=208, 
 
 Ans. a;=12, or 4; 
 
 x-Hy = 16. 
 
 y— 4, or 12. 
 
 14. a;'— y'=7ay, 
 
 Ans. x=i, or — 2; 
 
 I— y== 2. 
 
 y=2, or —4. 
 
 15. a:«4-a;y+y<=91, 
 
 Ans. as=±3, or ±1; 
 
 a;»+xj/-|-y'=13. 
 
 y=±l, or ±3. 
 
 16. x—y=Jx-\-^y, 
 
 Ans. as=16, or 9; 
 
 J—yi=3n. 
 
 y= 9, or 16. 
 
 17. a;i+yi=5. 
 
 Ans. a:=16, or 81; 
 
 x^j*=13. 
 
 y=27, or 8. 
 
 18. x^+y^= 5 
 
 Ans. x= 8, or 27; 
 
 a: 4i/ =35. 
 
 y=27, or 8. 
 
 19. ar*+y*= 4, 
 
 Ans. as=9, or 1 ; 
 
 a;?+y^==28 
 
 y=l, or 9. 
 
 20. x'+f=Z51, 
 
 Ans. x=^, or 2; 
 
 xy= 14. 
 
 y— 2, or 7. 
 
 21. a+y=4, 
 
 Ans. x=2, or 1; 
 
 ar<-|-y<=82. 
 
 y=l, or 3. 
 
 22. *(y-|-z;=o, A«s 
 
 • =^\/ 2(J+o-fl) 
 
 y(a;+z)=6. 
 
 
 «(»+y)=e- 
 
 V 2(<t4-6-c^ 
 
EQUATIONS OF THE SECOND DEGREE. 215 
 
 Aet. 246. A»FECTED Equations. — The most general form 
 of an equation of the second degree, containing two unknown 
 quantities, is, 
 
 ax'-!f-bxy-\-cx-\-dy'-^ey-\-f=0 . 
 
 By arranging the terms according to the powers of x, and divid- 
 ing by the coefficient of the first term, two equations of the sec- 
 ond degree containing two unknown quantities, may be reduced 
 to the following forms : 
 
 a;s+(ffl 3,4-6 )a;+cy»+dy+e=0 (1), 
 x^+(,a'i/+V)x+cy+d'!/-\-e'=0 (2). 
 
 To find the values of either of the unknown quantities, we 
 must eliminate the other. We shall now show that this operation 
 produces an equation of the fourth degree. 
 
 By subtracting the second equation from the first, and mak- 
 wg a — a'=a", b — h'=A", &c., we have 
 
 (o'V+6")a:+c>»+d"y+e"=0. 
 Whence, ^_c"y°+'^"y+e" _ 
 
 Substituting this.value of x in the first equation, we get 
 
 (a y-\-h y a 'y+b 
 
 and multiplying by (a"y-\-b"y, 
 
 {cY+d"y+e'y-(.ay+bXa"y+V'){c'y+d"y+e")+icy'+dy+e) 
 (a"y+i")'=0, 
 an equation of the fourth degree. 
 
 Hence, in general, the solution of two equations of the seumd 
 degree, containing two unknown quantities, depends upon the solution 
 of an equation of the fourth degree, containing one unknown 
 quantity. 
 
 Aet. 247. There are, however, two cases in which two equa- 
 tions of the second degree, containing two unknown quantities, 
 may always be solved as equations of the second degree ; viz : 
 
 Case I. — When one of the equations, containing two unknown 
 quantities, rises only to \he first degree, and the other to the sec- 
 ond degree, the values of the unknown quantities may be found 
 by the solution of an equation of the second degree. 
 
 Given, ax-\-hy:=c (1), 
 
 dx^-\-exy-\-fy^-\-gx-\-hy==k (2), io find x and y. 
 
■<J16 RAY'S ALGEBRA, PART SECOND. 
 
 Here equation (1) is evidently the general equation of the first 
 degree between x and y, and equation (2) the general equation of 
 the second degree between x and y. 
 
 From eq. (1), we have x= ^ ^ ; and substituting this value of 
 a 
 
 K in equation (2), and multiplying every term by o', we have 
 
 d(c^—2bci/+bY)+aey(c—b!/)+a^fy'+aff(c—bi/)+a%y=a% 
 
 an equation of the second degree, from which the value of y may 
 
 be found by the rule (Art. 231). Having the value of y, that of 
 
 X may be found from the equation x= ^, 
 
 a 
 
 Case H. — When both equations of the second degree are 
 komoyeneous, that is, have the sum of the indices of the unknown 
 quantities the same in every term which contains unknown quan- 
 tities, they may be solved by an equation of the second degree. 
 Given, ax'-\-bxy-{-cy^=d (1), 
 
 a'x'-{-b'xy-\-c'y'=d' (2), to find x and y. 
 Let y=:tx, where « is a third unknown quantity, termed an 
 auxiliary quantity. Substituting this value of a; in the two equa- 
 tions, we have 
 
 ax''-\-btx^-\-cPx^=x\a-\-bt-\-cf')=d (3), 
 
 a'x^-{-b'tx'-\-c'tV—x\a'-\rb't-\-c'P)=d' (4). 
 
 Prom eq. (3) we find x^^:^^^^. 
 
 d' 
 and from eq. (4) '''=^r^:^^:^r 
 
 • d _ d' 
 
 a+bt+cP a'+b't-\-cV 
 or, d(a'+b't+c'P)=dXa-\-bt+ct^, 
 
 in equation of the second degree, from which the value ot t may 
 Ds found, (Art. 231), and thence x from the equation 
 
 a;'= ; — ; — ■„. and thence y from the equa- 
 
 a-{-bt-\-cP 
 
 lion y=tx. 
 
 Akt. 24S. When two equations of the second degree are sy7t> 
 metrical with respect to the two unknown quantities, that is, when 
 the two unknown quantities are similarly involved, they may fre- 
 quent ty be solved by substituting for the unknown quantities, the 
 sum s 4d difTerenee of two others. 
 
EQUATIONS OF THE SECOND DEGREE. 217 
 
 Example. Given a; -|--y ='» (1)> 
 
 a;5-)-j/5=J (2), Xo find x and ^ 
 
 Let x=s-{-z, and y=s — z, then s=_. 
 
 a;s=s'+5s'2+10s'254-10s'z3_|_5s24_|_25^ 
 
 j,5=s5— 5s<z+lOs'02— lOs^e'+Ssz"— 2'; 
 '"a;54.ys=2s5+20s32=+l Osz-'=6. ' 
 
 By substituting the value of s=-, and reducing, wo find 
 
 , , a' , 166 — o' 
 
 z^-i-- z'= 
 
 ^2 80o ■ 
 
 From this equation, by completing the square, we find 
 
 Art. 249. An artifice that is often used virith advantage, con- 
 sists in adding such a number to both members of an equation, as 
 will render the side containing the unknown quantities, a trino- 
 mial equation that can be resolved by completing the square, 
 (Art. 240). The following is an example : 
 
 2.Given,^+f+?+2^=^ (1), 
 y^ x' y a; 4 
 
 and x'-|-y2=20 (2), to find x and y. 
 
 Since, ( --^ ) =^+2+?!! ; add 2 to each side of 
 
 \ y X I y' X' 
 
 «q (1), and then ^ to complete the square. 
 
 Whence, ^+?!=±3— ^=f 
 
 or- 
 
 7 
 
 y X 
 
 Let ?+^=|; then, ^ifX', or £?=§ ; 
 
 y ^ ^y "y 
 
 whence, xy=8, and 2xy=16 
 
 If) 
 
218 RAVS ALGEBRA, PART SECOND. 
 
 From the equation a;'-l-y':=20, and 2ay=16, we readily find 
 »=±4, and y=±2. 
 
 By taking --^= — ', two other values of x and y may be 
 y X 2 
 
 found. 
 
 Abt. 230. Another artifice consists in considering the sum, 
 
 difference, product, or quotient of the two unknown quantities, as 
 
 a single unknown quantity, and first finding its value. Thus in 
 
 example 9 following, the value of xy should be found from the 
 
 first equation ; and in example 10, that of -. 
 
 y 
 
 Other unknown auxiliaries may also sometimes be employed 
 with advantage, but their use, as well as that of various other 
 expedients that may be en^loyed, can only be learned by experi- 
 ence, while much will always depend on the judgment and tacl 
 of the operator. 
 
 EXAMPLES FOR PRACTICE. 
 
 Note. — In some of the examples all the values of the unknowi 
 quantities are not given ; those omitted are generally imaginary. 
 
 3. x^+y^+x+y=SSO, Ans. a=15, or —16; 
 
 
 a;2_2/5_|-a;— 2/=150. 
 
 y= 9, or —10. 
 
 4. 
 
 X -|-4j/=14, 
 y2-|-4a=2y+ll 
 
 
 Ans. i=:2, or — 46; 
 y=3, or 15. 
 
 5. 
 
 2y —2x =14, 
 
 
 Ans. x= 2, or IJ ; 
 
 
 3a;2_|_2(j,— 11)2= 
 
 =14. 
 
 y=10, or 84. 
 
 6. 
 
 x-^=2, 
 
 
 Ans. as=5, or | ; 
 
 
 _— 2=1t5- 
 
 y X 
 
 
 y=3, or — li. 
 
 7. 
 
 2ix^-\- xy=\B, 
 4y=!+3ay=54. 
 
 Art. 247, 
 Case II. 
 
 Ans. ir=±2, or ±2^3; 
 y=±3, or :f3V3. 
 
 8. 
 
 x'-\-xy =10, 
 
 
 Ans.x=±2, or ±5^2; 
 
 
 ay+2y'=24. 
 
 
 j,=±3, or ±4V2. 
 
 9. 
 
 4.xy=QG—xhi\ 
 a:-fy=6. 
 
 
 Ans. x=2, 4, or Z±J2l ; 
 y=4,2, or 3=fV21- 
 
 10 
 
 x^jAx 85 
 
 y' y 9' 
 
 »— y=2. 
 
 
 Ans. x=5, or |J ; 
 V=3, or -8 
 
EQUATIONS OF THE SECOND DEGREE. 219 
 
 u. 
 
 xY=ldO—8xy, 
 x+Zy=ll. 
 
 Ans. x=5, or 6; 
 y=2, or |. 
 
 12. 
 
 x-\^+^x+y=i2, 
 
 x2-|-y3=41. 
 
 Ans. x=5, or 4; 
 y=4, or 5. 
 
 13. 
 
 x-{-i/+x^+y^=:18, Am. 
 
 . as=3, 2, or — 3±V3^; 
 
 
 xy=6. 
 
 y=2, 3, or —S::p^3. 
 
 14. 
 
 x'+Zx+y=nS—2xy, 
 
 Ans. x=i, or 16; 
 y=5, or — ^7. 
 
 15. 
 
 11 
 x+y-"' 
 
 2 
 
 ^715* a? — , 
 
 a±J2b—a' 
 2 
 
 
 ^ a=FV24— a= 
 
 16. 
 
 xy+xy'=l2, 
 
 X +xy'=lS. 
 
 Atm. x=2, or 16; 
 y=2, or 1. 
 
 17. 
 
 x^+y'—x—y=78, 
 
 ^+y+'y =39. 
 
 Anj. a!=9, or 3 ; 
 y=S, or 9. 
 
 18. 
 
 (,x+yy—2y=28+3x, 
 2xy-{-Sx=S5. 
 
 Ans. x=5, or 3|; 
 y=2, or 31. 
 
 19. 
 
 (.f,)*+(^)'-. 
 
 Ans. x= 6, or — Ig! 
 
 
 ^— (^+y)=54. 
 
 j/=12, or — 9. 
 
 20. 
 
 x^+i(,x^+Zy+5)i=55-Zy, 
 
 Ans. x=S, or — s^s. 
 
 
 6x—7y=16. 
 
 y-2, or -Vs". 
 
 21. 
 
 y ^J==+y IT 
 
 ,(x+y)^ y ijx+y 
 x=f+2. 
 
 Ans. 1=6, or 3; 
 y=2, or 1. 
 
 aOESTIONS PRODUCING SIMULTANEOUS EQUATIONS OF THB 
 
 SECOND DEGREE, INVOLVING TWO OR MORE UNKNOWN 
 
 QUANTITIES. 
 
 Art. 251. 1. There are two numbers, whose sum multiplied 
 by the less, is equal to 4 times the greater, but whose sum multi- 
 plied by the greater is equal to 9 times the less. What are the 
 numbers'! Am. 3.6, and 2.4. 
 
 2. There is a number consisting of two digits, which being 
 multiplied by the digit in tens place, the product is 46 ; but if 
 
220 RAVS ALGEBRA, PART SECOND. 
 
 the s'lm of the digits be multiplied by the same digit, the pro- 
 duct is only 10. . Required the number. Arts. 23. 
 
 3. What two numbers are those whose difference multiplied by 
 the difference of their squares, will produce 32, and whose sum 
 multiplied by the sum of their squares, is 272! Aas. 5 and 3. 
 
 4- The product of two numbers is 10, and the sum of their 
 ubes 133. Required the numbers. Ans. 2 and 5. 
 
 5. If the sum of two numbers be multiplied by the greater 
 ind that product be divided by the less, the quotient will be 24 ; 
 but if their sum be multiplied by the less, and that product be 
 divided by the greater, the quotient will be 6. Required the 
 numbers. Ans. 4 and 8. 
 
 Note. — Tha preceding problems may be solved by pure equations. 
 
 6. The difference of two numbers is 15, and half their product 
 is equal to the cube of the less number ; find them. 
 
 Ans. 18 and 3. 
 
 7. The product of two numbers is 24, and their sum multiplied 
 by their difference is 20; find them. Ans. 4 and 6. 
 
 8. What two numbers are those whose sum multiplied by the 
 greater is 120, and whose difference multiplied by the less is 161 
 
 Ans. 2 and 10. 
 
 9. What two numbers are those whose sum added to the sum 
 of their squares is 42, and whose product is 15? 
 
 Ans. 3 and 5. 
 
 10. Find two numbers such, that their product added to their 
 sum shall be 47, and their sum taken from the sum of their 
 equares shall leave 62. Ans. 5 and 7. 
 
 11. Find two numbers such, that their sum, their product, and 
 the difference of their squares shall be all equal to each other. 
 
 Ans. i+hj5, and s+i^^- 
 
 12. Find two numbers whose product is equal to the difference 
 of their squares, and the sum of their squares equal to the differ- 
 ence of their cubes. Ans. hj^t and K^+VS)- 
 
 13. A grocer sold 80 pounds of mace and 100 pounds of 
 floves for 65 dollars ; but he sold 60 pounds more .jf cloves for 
 20 dollars than he did of mace for 10 dollars. Required the 
 price of a pound of each. Ans. Mace 50 cts, cloves 25 cts. 
 
 14. A and B gained by trading 100 dollars. Half of A's stock 
 was less than B's by 100 dollars, and A's gain was three twen- 
 
EQUATIONS OF THE SECOND DEGREE. 221 
 
 tieth'B of B's stock. Supposing the gains in proportion to the 
 stock) required the stock and gain of each. 
 
 Arts. A's stock $600, B's $400; 
 A's gain $60, B's $40. 
 
 15. The product of two numbers added to their sum is 23 j 
 and 5 times their sum taken from the sum of their squares leaves 
 8. Required the numbers. Aras. 2 and 7. 
 
 16. There are three numbers, the difference of whose differ 
 ences is 5; their sum is 44, and continued product 1950; finii 
 the numbers. Ans. 25, 13, 6. 
 
 17. Divide the number 26 into three such parts that their 
 squares shall have equal differences, and that the sum of those 
 squares shall be 300. Atis. 14, 10, 2. 
 
 18. The number of men in both fronts of two columns of 
 troops, A and B, where each consisted of as many ranks as it had 
 men in front, was 84; but when the columns changed ground, 
 and A was drawn up with the front that B had, and B with the 
 front that A had, then the number of ranks in both columns was 
 91. Required the number of men in each column. 
 
 Ans. 2304, and 1296. 
 
 Art. 253t Formula — General Solutions. — A gejieral so- 
 lution to a problem producing an equation of the second degree, 
 like one of the first degree, gives rise to a, formula (Art. 162), 
 which expressed in ordinary language, furnishes a rtde. We shall 
 illustrate the subject by a few examples. 
 
 1. Two men, A and B, bought 300 (o) acres of land for 600 
 (b) dollars, of which A paid 300 (c) dollars, and B 300 (b—c) 
 dollars. For certain reasons they agreed to divide the land so 
 that B should pay 75 cents (d dollars) per acre more than A. 
 How much land did each man get, and what did he pay per acre 1 
 
 Gekekal solution. Let x= cost of A's land per acre, 
 
 then a;-|-d= cost of B's land per acre ; 
 
 c b — c 
 
 also, -= acres A got, and . , = acres B got. 
 
 c b — c 
 •'• i+i+rf^"' ^y *® problem. 
 Clearing of fractions and reducing, 
 
 6 — ad cd 
 
 x' — — „ X — — • 
 a a 
 
 b—ad (6— ad)'_(i— a<i)=+4acii 
 
 ^'~~r''"+ 4ffl' 4^» ' 
 
2'<ia RAYS ALGEBRA, PART SECOND. 
 
 b—ad J(b—ad)^+iacd 
 
 or, a:=i^\^(b—ady+icad-\-b—adl. 
 
 This formula gives the following rule for finding the amount 
 p '.icl per acre, by him who paid least per acre : 
 
 P ULE. — Find the cost of the whole number of acres at the differerux 
 between the prices per acre of the different pieces of land; svilract 
 this from iJie amount paid for the whole land ; square this remain- 
 der and add to it the cost of the whole number of acres at the dif- 
 ference between the prices per acre, multiplied hy four times the sum 
 of money paid by him who paid least per acre ; extract the square 
 root of this sum, add to the square root thus found, the remainder 
 that was squared, and divide the sum by twice the whole number of 
 acres; the quotient will be the amount paid per acre by him who 
 paid least per acre. Having this, every otlier requirement in the 
 question is easily found. 
 
 For the particular case the results are, 
 
 A paid $2,443 per acre, and got 122.8 acres nearly. 
 
 B paid $1,693 per acre, and got 177.2 acres nearly. 
 
 2. Investigate a formula for finding two numbers, x and y, at 
 which the sum of the squares is s, and difference of the squares d. 
 
 Ans. x=li^2(^s-{-d) ; y=i^'2(s — d). 
 
 3. Investigate a formula for finding two numbers, a; and y, of 
 which the diflerence is d, and the product p. 
 
 Ans. x=l(,d-\-Jd^A'py, 
 
 4. Investigate a formula for finding a number, x, of which the 
 sum of the number and its square root is s. 
 
 Ans.x=s+^—^s+l. 
 
 5. The same when the diflerence of the number x, and iw 
 square root is d. Ans. x=d-\-^-\- «/<i-|-4- 
 
 6. Given x-\-y=s, and xy=p, to find the value of x'4-v' 
 t'+y', and x^-\-y*, in terms of s and p. 
 
 Ans. x'-\-y^=s^ — 2p ; 
 x'-\-y'=s' — Sps ; 
 x^-\-y*=s* — ips^+2p'. 
 
EQUATIONS OF THE SECOND DEGREE. 223 
 
 Let the student express each of the preceding formulae in the 
 form of a Rule, and exemplify its use, by forming examples with 
 particular numbers, and then solving them. 
 
 Note. — In the great variety of equations that occur, which may ba 
 solved as equations of the second degree, it is not to be supposed tliat 
 Rules can be given for every operation necessary for their solution. 
 The artifices by which algebraic calculations are abridged, are numerous, 
 a id their successful application can be learned only by practice. In the 
 Collowiug article, which is intended only for advanced students, we shall 
 exhibit some of these artifices. 
 
 Akt. 853. Special soLtmoNs and examples. — If an equa- 
 tion can be placed under the form 
 
 (a;+a)X=0, 
 
 in which X represents an expression involving x, the unknown 
 quantity ; since the equation will be satisfied by making either 
 factor =0, we have x-\-a=0, and X=0. Therefore, x= — a, 
 is one solution of the equation, and the other values of x will be 
 found by solving the equation X=0. Hence, whenever an equa- 
 tion is simplified by division, or the omission of a factor, if the 
 divisor or factor contains the unknown quantity, one solution, at 
 least, of the equation will be found by putting that divisor or 
 factor equal to 0. Thus, the equation x' — x^ — 4x+4=0, may 
 be placed under the form (x — 2){x''-\-x — 2)=0. Hence, x — 2=0, 
 or x^-\-2, and from the other factor we find x=:-\-l, or — 2. 
 
 The difficulty to be overcome in applying this artifice, consists 
 in finding the factors of the given equation, or in transforming it 
 BO that it can be readily separated into factors. 
 
 2 
 Ex. 1. Given, a^-l=2-|-^• to find x. 
 Jx 
 
 _ _ 2 2 
 
 Since, x—l=(Jx+lX^x—l) and 2^ — ==-^(Vx+l), 
 
 aJx Jx 
 
 .: (VS+l)(v'"i-l)=4^(V'^4-l) ; 
 ijx 
 
 .-. Jx+1=0, and a:=(— 1)'=1. 
 
 - 2 _ 
 
 Also, ^x — 1 = ^=, by dividing by Jx-{-l. 
 ijx 
 
 Whence, ^x=2, or — 1; and a;=4, or 1. 
 2. x^—Zx=2. (Add 2a! to eaih side.) Ans. x— — 1, or 2, 
 
224 RAT'S ALGEBRA, PART SECOND. 
 
 3. «»— ^=14. I Transpose * and —_ \ 
 
 Ans. x=—~s, or i(l±VTO). 
 
 4. Zxi—x^^l. Ans. x—l, or -1(— 1±V-^- 
 
 5. t>—Sx^+x+2=0. Ans. x=2, or aCli^/S)- 
 6 a!»=6x+9. Ans. x=S, or ^(— 3±V-^)- 
 
 7. *4-7a;i=22. Ans. a=8, or 29^=7^—10. 
 i4-7a:i— 22=(x— 8)+7(x*^— 2). 1^—2 is a divisor. 
 
 8. ai^+V**— 39«=81- 
 
 Ans. x=±S, or b(— 13±J— 155- 
 
 An artifice that is frequently employed, consists in adding to 
 each side of the equation, such a number or quantity as will ven- 
 der both sides perfect squares. , 
 
 9. Given, i=L?±^^, to find x. 
 
 X — 5 
 
 Clearing of fractions, x' — 5a;=124-8^a;. 
 
 Add jt4-4 to each side, and extract the square root. 
 
 a:— 2=±(4+J^). 
 
 From vi^hich we easijy find x=9, 4, or J( — 3±V — ^^)' 
 
 10. x-Z=^±^, Ans. a;=i(7±Vl3), i(— li^^^j 
 
 a: -' - . 
 
 11. *5^+^— 49=9+-. Add 1 to each side. 
 
 4 a:' a; x^ 
 
 Ans. x=2, — f , or 4(— 3drV^3) 
 
 12. x*+^—9ix=16. Ans. a=±2, —8, or ->. 
 
 — J to each side. 
 
 13. x^ ( 1+1 ] '— (3a:»+a;)=70. 
 
 9 
 
 DiTide by x* and add — to each side. 
 4a;'' 
 
 Ans. x=3, — 3j, or b(— 1±V— 251). 
 
EQUATIONS OF THE SECOND DEGREE. 225 
 
 14. — , 81— a'^ a'— 6.5 jiyiti 1 by 2, and add |l+f4 
 10 each side. Ans. 9, — 4, or — 9. 
 
 1,5. 27x2— ^-il+i7=^_^_J__|_5. 
 
 Multiply both sides by 3, transpose and _ and add i 
 
 x^ x^ 
 
 to each side to complete the square. 
 
 Ans. x=2, —V. or i(— 2±V— 266). 
 
 We shall now present a few solutions giving examples of other 
 artifices. 
 
 16. i±^=a ; to find x. 
 
 a+xy 
 
 1 +x*=a(l +xy=aO.+4:X+6x^+ix^+x*), 
 (1— a)(l+x'')=4a(x+x')+6aa:=, 
 
 dividing by a=, (1— o) ( a;=+l ] =4a / x+l | -i-6a, 
 
 x-' 1 — ffl \ X / 1 — a 
 
 (x+l)'-J^(x+l)=-^+2=2_±4_. 
 \ X / 1 — a \ X I 1 — a 1 — a 
 
 Complete the square, and find the value of x-\--, which is 
 
 X 
 
 2 g±V2(l+a) ^ call this 2p, and we then find x=p±^f—\ . 
 1 — a 
 
 17. x«+y=3/''', and ^+!'=x«, to find x and y. 
 
 i+V 
 
 From 1st equation 2/=r4a ' 
 « 2nd " 3/=a£Fii; 
 
 .•. ar4o =x»+j, and — L£= ; 
 
 4a x-\-y 
 
 (z+y)^=4a', or x-\-y=2a ; 
 
 but x°=j/''', since x4-y=2a, 
 
 x=y'', and y'-|-y=2a. 
 
 Whence, 3/.-i(— l±>/8a+l). and x=l(4a+l=F^/8a-1-i). 
 
226 RAY'S ALGEBBA, PART SECOND. 
 
 When two unknown quantities are found in an equation, in the 
 form of x-\-y and xy, it is generally expedient to put their sum 
 x-\-y=s, and their product xy=^p. 
 
 18. Given {x+y){xy+l)='iaxy (1), 
 
 (,x^-\-y^)(xY+'i-')=^^OBxY (2), to find x and y. 
 Let x-\-y^=s, and xy=p, then 
 
 s(p+l)=18p (1). 
 
 and (s=— 2;))(p'+l)=208j9S (2). 
 
 From the square of (1) take (2), and after dividing by 2p, we 
 have s^+p^+l=5ep. (3), 
 
 but 2s{p+l)=36p for (2), 
 and 2p= 2p, 
 
 Adding, {s+p+iy=96p, 
 
 but jH-i=^ ; 
 
 .-. s=4 J6^— 1^, or s»— 4sV6p=— 18o, from 
 
 which, s=SJGp, or ^6/>. 
 But, p+l=s, =3^6^, or J6p, 
 Whence, p=26±^615, or 2±^3', 
 and s=±3VJ6(26±V675)J, or ±SJ]6(2±/3)]. 
 Having x-i-y, and ot/ the values of x and y are easily found 
 (Art. 246) ; two of the values are x=7rh4^3', y—2:^JS. 
 
 A similar substitution may be used in solving the following 
 example : 
 
 19. 2(:x+yy+l=(x'+y^Xxy+x^+y') (1), 
 
 x+y=2 (2). 
 
 Ans. x=2, y=l. 
 
 20. l+x'=:a(l+xy. Ans. ,.^^ l+2g±V12g-3 
 
 ^ 2Cl-a) ■ 
 
 21. ^-llx-2a-^l. 
 
 X^ X\ X 
 
 Ans. a;=iJl±Vl— 8a±\/2±2(l— 8o)*+8a|. 
 
 22. x-\-y-\-xy{3e '-y)+xY=Q5, Ans. as=6, or 1. 
 
 a^-(-(a;+f -)-a^(a;+y)=97. 3/=!, or 6. 
 
RATIO AND PROPORTION. 327 
 
 . 2c-\-ad c 
 Ans. x= — J — , or 
 
 {a-\-b)d ^a—b)d 
 
 24. (x'+l)(x24-l)(a;+l,c=30x^ Ans. x=liZdzj'5y 
 
 25. a;'+y'=35, Ans. x=S, 2, or liaVSa"; 
 a;=+^'=13. y=2, 3, or l=FiV22. 
 
 26. ^=a. A7«. a:= / ^ 2abc(,ac+hc-ab) ) . 
 
 i+y \ ((ab-\-ac — hc)(_ab-{-bc — ac)S 
 
 ^^=b, y= I \ SaieCaft+fe— ae) { . 
 
 fy±—c j_ / i 2aic(a64-ac — Jc) ) 
 
 y+z V ((flc+ic — ai)(afi-|-ic — acp 
 
 27. (a«+l)y=(y'+l>', 
 (y«+l)i=9(a:'+l)2/5. 
 
 Ans. a;=i;^V3+3+^y3-l J; 
 y=^lV3"..^V3+3±^3^9-] J. 
 
 CHAPTER VIII. 
 RATIO, PROPORTION, AND PROGRESSIONS. 
 
 Aet. 254. Two quantities of the same kind, may be compared 
 in two ways : 
 
 1st. By considering how much the one exceeds the other. 
 2nd. By considering how many times the one is contained in the 
 other. 
 
 The first method is termed comparison by Difference ; the sec- 
 ond, comparison by Quotient. The first is sometimes called Arith- 
 metical ratio, the second. Geometrical ratio. 
 
 If we compare 2 and 6, we find that 2 is four less than tj, oi 
 that 2 is contained in 6 three times. 
 
228 RAY'S ALGEBRA, PART SECOND. 
 
 Also, the arithmetical ratio of a to i is b — a, the gcometrica! 
 
 ratio of a to J is -. The term Ratio, unless it is otherwise 
 a' 
 
 stated, always signifies geometrical ratio. 
 
 Aet. 235. Ratio is the quotient which arises from dividing one 
 quantity by another of the same kind. Thus, the ratio of 2 to 6 
 is 3, and the ratio of a to ma is m. 
 
 Art. 256. When two numbers, as 2 and 6, are compared, the 
 JirsS is called the antecedent, and the second the consequent. An 
 antecedent and consequent, when spoken of as one, are called a 
 zouplet. When spoken of as two, they are called the terms of 
 the ratio. Thus, 2 and 6 togellier form a couplet, of which 2 is 
 \he first term, and 6 the second term. 
 
 Art. 257. Ratio is expressed in two ways : 
 1st. In the form of a fraction, of wliich the antecedent is the 
 denominator, and the consequent the numerator. Thus, the ratio 
 
 of 2 to 6 is expressed by 6 ; the ratio of a to i, by - 
 ^ a' 
 
 2nd. By placing two points vertically between the terms of 
 the ratio. Thus, the ratio of 2 to 6, is written 2:6; the ratio 
 of a to 2>, a : h, &c. 
 
 Art. 258. The ratio of two quantities, may be either a whole 
 number, a common fraction, or an interminate decimal. 
 
 Thus, the ratio of 2 to 6 is |, or 3. 
 The ratio of 10 to 4 is x^s, or f . 
 
 The ratio of 2 to ^5 is ^, or ^•^^^+, or 1.118+. 
 
 We see, from this, that the ratio of two quantities cannot al- 
 ways be expressed exactly, except by symbols ; but, by taking a 
 sufficient number of decimal places, it may be found to any 
 •equired degree of exactness. 
 
 Art. 259. Since the ratio of two numbers is expressed by a 
 fraction, of which the antecedent is the denominator, and the con- 
 sequent the numerator, it follows, that whatever is true with 
 regard to a fraction, is true with regard to the terms of a ratio. 
 Hence, 
 
 1st. To multiply the consequent, or divide the antecedent of a ratie 
 by any number, multiplies the ratio by that number. (Arith., Pari 
 Srd Arts 142, 14.5.) 
 
RATIO AND PROPORTION. 229 
 
 2nd. To divide the consequent, or to multiply tlie antecedent of a 
 ratio lyy any number, divides the ratio by that number. (Arith., Part 
 3rd., Arts 143, 144.) 
 
 3rd. To multiply, or divide, both the antecedent and consequent of 
 a ratio by any number, does not alter the ratio. (Art. 118.) 
 
 Art. S60. When the terms of a ratio are equal to each other, 
 tlie ratio is said to be a ratio of equality. When the second term 
 ii greater than the first, the ratio is said to be a ratio of greater 
 inequality, and when it is less, the ratio is said to be a ratio of 
 less inequality. 
 
 Thus, the ratio of 2 to 2 is a ratio of equality. 
 
 The ratio of 2 to 3 is a ratio of greater inequality. 
 
 The ratio of 3 to 2 is a ratio of less inequality. 
 
 Hence, a ratio of equality may be expressed by 1 ; a ratio of 
 greater inequality, by a number greater than 1 ; and a ratio of 
 less inequality, 'by a number less than 1. 
 
 Akt. 261. When the corresponding terms of two or more 
 ratios are multiplied together, the ratios are said to be com- 
 pounded, and the result is termed a compound ratio. Thus, the 
 
 ratio of a to b, compounded with the ratio of c to d is _X-=— 
 
 a c ac 
 
 A ratio compounded of two equal ratios is called a duplicate ratio. 
 
 Thus, the duplicate ratio of - is _X-=— . 
 a a a a" 
 
 A ratio compounded of three equal ratios is called a triplicate 
 
 ratio. Thus, the triplicate ratio of - is _X-X-=- 
 
 ■ a a a a a^' 
 
 The ratio of the square roots of two quantities is called a sub- 
 duplicate ratio. Thus, the subduplicate ratio of 4 to 9 is | ; and 
 
 JF 
 the subduplicate ratio of o to A is -^. 
 
 ^a 
 
 The ratio of the cube roots of two quantities is called a subtrip- 
 
 licate ratio. Thus, the subtriplicate ratio of 8 to 27 is | ; and 
 
 %/b 
 the subtriplicate ratio of a to i is -^. 
 
 Va 
 
 Akt. 262. Ratios may be compared with each other by reduc- 
 ing the fractions which represent them to a common deuomiiiator. 
 Thus, to ascertain whether the ratio of 2 to 7 is less or wreater 
 than the ratio of 3 to 10, we have the two fractions j, and JJ> . 
 
230 RAY'S ALGEBRA, PART SECOND. 
 
 which being reduced to a common denominator are %' and ^^ , 
 and since the first is greater than the second, we conclude that 
 the ratio of 2 to 7 is greater tlian the ratio of 3 to 10. 
 
 PROPORTION. 
 
 Art. 263. Proportion is an equality of ratios. That ia, 
 \Dhen txoo ratios are equal, their terms are said to be proportional. 
 Thus, if the ratio of a to b, is equal to the ratio of c to d, that is, 
 
 if -=-, then a, b, c, d, form a proportion, and we say that a is to 
 a c 
 
 6 as c is to d. 
 
 Proportion is written in two ways : 
 
 Ist. By placing a double colon between the ratios ; thus, 
 
 a -.b : :c id. 
 2nd. By placing the sign oi equality between tlie ratios ; thiia, 
 
 a : b=c : d. 
 The first method is the one commonly used. 
 
 From the preceding definition, it follows, that when four quan- 
 tities are in proportion, the second, divided by the first, must give 
 the same quotient as the fourth divided by the third. This is the 
 primary lest of the proportionality of four quantities. Thus, if 
 3,5,6,10,arethefourtermsof a proportion, so that 3 : .5 : : 6 : 10, 
 we must have |='g'. 
 
 If these ftactions are equal to each other, the proportion is 
 trne ; if they are not equal to each other, it is false. Thus, let it 
 be required to determine whether 3 : 8 : : 2 : 5. The ratios are 
 I and §, or 'g* and '/ ; hence, the proportion is false. 
 
 Remark. — In common language the words ratio and proportion ar* 
 sometimes confounded witli each other. Thus, two quantities are saia 
 to be in the proportion of 2 to 3, instead of in the ratio of 2 to 3. A 
 ratio subsists between two quantities, a proportion only between four. 
 It requires two equal ratios to form a proportion. 
 
 Art. 364. Each of the four quantities in a proportion is called 
 a term. The first and last terms are called the extremes; and the 
 second and third terms, the means. 
 
 The terms of a proportion may be either monomials or polyno- 
 mials. 
 
 Art. 265. Of four quantities in proportion, the first and third 
 ore called the antecedents, and the second and fourth, the come- 
 
RATIO AND PROPORTION. 231 
 
 juents (Art. 257) ; and the last is said to be a fourth proportional 
 to the other three taken in their order. 
 
 Art. 266. Three quantities are in proportion when the first 
 has the same ratio to the second, that the second lias to the third. 
 In this c!>.se the middle term is called a. mean proportional between 
 the other two. Thus, if we have the proportion 
 a : b : : b : c, 
 
 then b is called a Tnean proportional between a and c ; and c la 
 called a thirct proportional to a and b. 
 
 When several quantities have the same ratio between each two 
 that are consecutive, they are said to form a continued proportion. 
 
 Art. se?. Pbofosition I. — In every proportion, the product of 
 the means is equal to the product of the extremes. 
 
 Let o : 6 : : c : d. 
 
 Since this is a true proportion, the ratio of the first term to the 
 second, is equal to the ratio of the third term to the fourth (Art. 
 263). Therefore, we must have 
 
 . i d 
 
 Multiplying both sides of this equality by ac, to clear it of frac- 
 ahc adc 
 
 tions, we have 
 
 a c 
 bc=ad. 
 
 Illustration by numbers. 2 : 6 :5 : 15 ; and 6x5=2x15. 
 
 Prom the equation hc=ad, we find d^L, 0="—, i=_, and 
 
 a b c 
 
 be 
 a=— . Hence, if any three terms of a proportion are given, the 
 d 
 
 fourth may be found. 
 
 Ex. 1 . The first three terms of a proportion are x-\-j/, x' — y^^ 
 and X — y ; what is the fourth "i Ans. x' — 2xy-\-y^. 
 
 2. The first, third, and fourth terms of a proportion are 
 (m — ny, m' — n', and m-\-n ; required the second. 
 
 Ans. m — n. 
 
 Remark. — This proposition furnishes a more convenient test ot the 
 proportionality of four quantities, than the method given in Art. 263. 
 Thus, to ascertain vifhether 2 : 3 : ; 5 : 8, it is merely necessary to com- 
 pare the prcduct of the means and extremes ; and since 3X5 is no( 
 e^Ufxl Id 2X.^. wo infer that 2, 3, 5, and 8, are not in n'-onortinn. 
 
23i8 RAY'S ALGEBRA, PART SECOND. 
 
 Art. 268. Peoposition II. Conversely, IJ the product of two 
 quantities is equal to the product of two others, two of them may he 
 made the means, a?id the other two the extremes of a proportion. 
 
 Let bc=ad. 
 Dividing each of these equals by ac, we have 
 
 be ad 
 
 ac ac 
 
 or ^_=1 
 a c 
 
 That IB, (Art. 263), a -.b : -.c -.d. 
 
 By dividing each of the equals by db, we may prove that 
 
 a '.c : :b :d. 
 
 lUust. 3X12=4X9, and 3 : 4 : >9 : 12 ; also, 3 : 9 : : 4 : 12. 
 
 Art. 269. Proposition III. If three quantities are in propor- 
 tion, tlie product of the extremes is equal to the square of the mean. 
 
 If a : b : : b : c, 
 
 then (Art. 267), ac=JJ=i». 
 
 It follows from Art. 268, that the converse of this proposition 
 is also true. Thus, if ac=V, 
 
 then, a :&::&: c. 
 
 That is, if the product of the first and third of three quantities is 
 equal to tJie square of the second, tlie first is to the second, as tlie second 
 ~ to the third. 
 
 Note. — It is recommended to the teacher to require the pupils to 
 illustrate all the propositions by numbers. (See Ray's Algebra, Part 
 1st., Proportion.) 
 
 Art. 2'S'O. Proposition IV. — If four quantities are in propor- 
 tion, they will he in proportion by Alternation ; that is, the first 
 will be to the third, as the second to the fourth. 
 
 Let a lb : ic :d. 
 
 Then, (Art. 263), * —^ 
 
 Multiply both sides by c, 
 
 tlivide both sides by b, 
 
 That is (Art. 263), c 
 
 NoTF. — This proposition is true, only when the four quantities are of 
 tlie same kind. 
 
 a 
 
 c 
 
 be. 
 
 =d; 
 
 a 
 
 
 c . 
 a 
 
 _d 
 
 c : : 
 
 :b:d. 
 
RATIO AND PROPORTION. 233 
 
 Aet. ft71. Proposition V. — If four quantities are in propor- 
 tion, they will be in proportion by Inversion ; that is, the second wiU 
 be to the first, as the fourth to the third. 
 
 Let a -.b : -.c \d. 
 
 Then (Art. 263), -=^ ; 
 
 a c 
 
 dividing 1 by each aide, r^^- • 
 
 d 
 
 or, _=_ 
 
 b d 
 
 That is, (Art. 263), b -.a : -.d -.c. 
 
 Art. 272. Proposition VI. — If two sets of proportions have 
 an antecedent and consequent in the one, equal to an antecedent and 
 consequent in the other, the remaining terms will be proportional. 
 
 Let a:b::c:d (1), 
 
 and o : ft : : e :/ (2) ; 
 
 then will c:d::e:f. 
 
 From the 1st proportion -=-, from the 2nd, -=-. 
 a c a e' 
 
 A f 
 Hence, _=i ; which gives c:d::e:f. 
 c e 
 
 Art. 273. Proposition VII. — If four quantities are in pro- 
 portion, tltey will be in proportion by Composition ; that is, the sum 
 of the first and second will be to the second, as the sum of the third and 
 fourth is to the fourth. 
 
 Let a -.b : : c : d. 
 
 Then will a+6 : 6 : : c+d : d. 
 
 From the 1st proportion, bc=:ad, (Art. 267) ; 
 bd=bd ; 
 Adding the two equations together, bc-{-bd=ad-{-bd ; 
 factoring, i(c+d)=d(a-(-6) ; 
 
 dividing each side by e+d i=^(?+?) ; 
 
 c-\-d 
 
 by a+b J-=zJ_ 
 a-{-b c-\-d' 
 
 This gives, o+i : 6 : : c+d : d. 
 
 NoTK. — In a similar manner let the student prove that the sum of the 
 first and second of two quantities is to theirs*, as the sum of the third 
 snd fourth is to the third. 
 20 
 
234 RAYS ALGEBRA, PART SECOND. 
 
 Art. 274. Proposition VIII. — If four quantilies are in pro- 
 portion, they wiU lie in proportion by Division ; that is, the difference 
 of the first and second wiU be to the second, as the difference of the 
 third and fourth is to the fourth. 
 
 Let a:b::c:d (1), 
 
 then wil. a — b : 6 : : c — d : d. 
 
 From the Ist proportion, bc=ad (Art. 267). 
 bd=bd ; 
 aubtracting, be — bd=ad — bd ; 
 
 factoring, b(c — d)=d{a — i). 
 
 Dividing each side by c — d, b=— — -2 ; 
 
 c — d 
 
 by a~i J_= J_ 
 
 a — b — d 
 
 This gives a — b : J : : c — d : d. 
 
 Note. — In a similar manner, let the student prove that the difference 
 of the first and second is to the ^rst, as the difference of the third and 
 fourth is to the third. 
 
 Art. 275. Pkoposition IX. — If four quantities are in propor- 
 tion, the sum of the first and second vdU be to their difference, as the 
 sum of the third and fourth is to their difference. 
 
 Let a:b::c:d (1), 
 
 then will a-l-J : a — b : : c-\-d : c — d. 
 
 From the 1st, by Composition, (Art. 273), 
 
 a-\-b : 6 : : c-\-d : d ; • 
 
 By Alternation. a+5 : c-\-d •.•.h:d; 
 
 ... . c-\-d d 
 
 this gives, — h:=r- 
 
 a-\-b b 
 
 From the 1st, by Division, 
 
 a — b : 6 : : c — d ; d ; 
 
 by Alternation, a — 6 : c — d : : 6 : d ; 
 
 , . . c — d d , c4-d c—d 
 
 this gives, ^=-; hence, _!_= — 
 
 a — b b a-\-b a — o 
 
 That is, o+i : c+d : : a — b : c — d, 
 
 or by Alternation, a+6 : a — 6 : : c-\-d : o — d. 
 
ftATIO AND PROPORTION. 235 
 
 Art. 2YC. Pkoposition X. — If four quantities are in propor- 
 tion, like powers or roots ofiJiose quantities will also be in proportion 
 
 Let a ; J : : c : d, 
 
 then will o" : 6" : : c" : d". 
 
 Prom the 1st, -=-. Raising each of these equals 
 
 a c 
 
 to the n"* power. 
 
 lp_d'^ 
 
 That is. 
 
 o» : 6» : : c" : d", 
 
 where n may be either 
 
 a whole number or a fraction. 
 
 AST. 27'!'. Peoposition XL — If two sets of quantities are in 
 
 proportion, the products 
 
 1 of the corresponding terms wiU also be in 
 
 proportion. 
 
 
 Let 
 
 a:b::c:d, (1), 
 
 and 
 
 m:n : :r :s (2), 
 
 then will 
 
 am : bn : : cr : ds. 
 
 For from the Ist, 
 
 b d 
 , 
 
 a c 
 
 and from the 2nd, 
 
 -=1. Multiplying these equals 
 m r 
 
 together, - X 
 a 
 
 m c r am cr 
 
 this gives am -.hn : -.cr : ds. 
 
 Aet. 278. Pkoposition XIL — In any numier of proportions 
 having the same ratio, any antecedent is to its consequent, as the sum 
 of aU the antecedents is to the sum of all the consequenif- 
 
 Let a:b : -.c-.d : -.m in, &c. 
 
 Then a : 6 : : a-\-c-\-m : b-\-d-\-n. 
 
 Since a:b: :c -.d, we have bc=ad (Art. 267). 
 Since o : 5 : : m : n, we have bm^=an 
 
 ab=ab. The sum of these equal- 
 ities gives ab-\-bc-\-bm=ab'\-ad-^an. 
 Factoring, b{a-\-c-\-m)=a(b-{-d-\-n). 
 
 Dividing by a+c+m, i=^^+^)_ 
 a-\-c-{-m 
 
 Dividing both sides by a, ^_= H^n _ 
 a a+c-j-ro 
 
 This gives o ; i : : a-\-c-\-m : b-\-d-\-n. 
 
'36 RAY'S ALGEBRA, PART SECOND. 
 
 Remark. — In most of the preceding demonstrations, the contlnsljn 
 has been derived directly from the equality of ratios. In Heveral cases, 
 however, it may be derived more easily from Art. 268, Proposition II j 
 but the method here given is considered the most satisfactory, as it keeps 
 before the student, theprinc^le on vrhich proportion depends. 
 
 EXERCISES IN RATIO AND PROPORTION. 
 
 1. Which is the greater ratio, that of 3 to 4, or 3= to 4' 'i 
 
 Ans. last. 
 
 2. Compound the duplicate ratio of 2 to 3; the triplicate ratio 
 9f 3 to 4; and the subduplicate ratio of 64 to 36. 
 
 Ans. 1 to 4. 
 
 3. What quantity must he added to each of the terms of the 
 
 -atio m : n, that it may become equal to ^ : g ■! Ans. S — c 
 
 p—q 
 
 4. If the ratio of a to 5 is 2f , what is tlie ratio of 2o to b, 
 und of 3o to 4J ? , Ans. 1 J, and 3|. 
 
 5. If the ratio of a to 76 is 5;J, what is the ratio of a to i, and 
 '.f 5o to 46 ? Ans. |, and |. 
 
 6. If the ratio of a to 6 is if, what is the ratio of o-j-i to b, 
 and of b — a to a? Ans. |, and |. 
 
 7. If the ratio of m to » is 4) what is the ratio of m — n to 6m, 
 and also to 5n 1 Ans. 14, and 6|. 
 
 8. If the ratio of m to 2m+3n is 2-|, what is the ra'lio of m to 
 n J Ans. 5 to 1 . 
 
 9. If the latio of m to n is 3^, what is the ratio of 12m to 
 m-J-n, and of 12n to n — 2m. Ans. |, and jg. 
 
 10. If the ratio of 5y — 8x to Ix — 5y is 6, what is the ratio of 
 ttoy'i Ans. T toll. 
 
 11. What is the proportion deducible from the equation 
 06^0' — x'. Ans. a : a-\-x : : a — x : b. 
 
 12. What is the proportion deducible from the equation 
 «'-|-y'=2aa; ? Ans. x:y::y: 2a — a. 
 
 13. Four given numbers are represented by a,b,c,d; what 
 quantity added to each will make them proportionals 1 
 
 Ans.-tz^, 
 a — b — c-\-d 
 
 14. If four numbers are proportionals, show that there is no 
 number which, being added to each, will leave the resulting four 
 numbers proportionals. 
 
RATIO AND PROPORTION. 23T 
 
 15. Find x in terms of y from the proportions x:y::a?:b', 
 
 and a:b: : %jc-\-x : %ld-\-y, Ans. a;=^. 
 
 d' 
 
 16. Prove that equal multiples of two quantities are to each 
 other as the quantities themselves. 
 
 17. Prove that like parts of two quantities are to each other aa 
 tlie quantities themselves. 
 
 18. Prove that in any proportion, if there be taken equal mul- 
 tiples of the antecedents, and equal multiples of the consequents, 
 the resulting quantities and the antecedents will be proportional. 
 
 19. If o : fi : : c : d, prove that ma : mb : : tic : nd, and also that 
 ma : Tib ■.: mc : Tid, m and n being any multiples. 
 
 20. If a : 4 : : c : d, prove that - :_::_: ^ ; and also that 
 
 mm n n 
 
 a _b _ _ c _d 
 m n m n 
 
 21. Prove that the quotients of the corresponding terms of 
 two proportions are proportional. 
 
 22. Prove that if two sets of proportions have their antece- 
 dents proportional, their consequents will also be proportional. 
 
 23. Prove that if the antecedent and consequent of a ratio be 
 increased or diminished by like parts of each, the resulting quan- 
 tities and the antecedent and consequent will be proportional. 
 
 24. If {a+by : (,a—by : : b+c : b—c, show that 
 
 a : J : : ^2a — c : ^c. 
 
 Akt. ^79. The preceding exercises are designed merely to 
 make the student acquainted with the principles of ratio and pro- 
 portion. The following are intended as exercises in the applica- 
 tion of the principles of proportion to the solution of problems. 
 
 1 . Resolve the number 24 into two factors, so that the sum of 
 Iheir cubes may be to the difference of their cubes, as 35 to 19. 
 
 Let X and y denote the required factors ; then a^=24, and 
 
 x'+i/ : x'—y' : : 35 : 19; 
 -• (Art. 275), 2x' : 2^' : : 54 : 16 ; 
 
 or, x':y'::21:8; 
 
 or, (Art 276), x ly : -.S -.2. 
 
 From which y=|a; ; then substituting the value of y in the 
 equation xy=2i, we find i=rt:6; hence, v=±4. 
 
238 RAY'S ALGEBRA, PART SECOND. 
 
 2. Given, V^+l+V^-1 ^2. to find x. 
 %lx+\—Xlx—l 
 
 Resolving this equation into a proportion, we have 
 %jl^\—%[^l : Xl'^-i^-U'^—i- : : 1 :2; 
 
 . . (Art 275), 2yx-\-\ : 2»/x— 1 
 
 L : :3 
 
 :li 
 
 
 or, \Jx+l : ^x—1 
 
 ::3: 
 
 1; 
 
 
 or, (Art. 276), a;-|-l : i— 1 : : 
 (Art. 275), 2a; : 2 : : 28 
 
 :27 : 
 :26; 
 
 1; 
 
 
 whence, 52x=56, or 
 
 .r— 1 
 
 t'^. 
 
 
 3. x-\-y :x—-y : :3 :1, 
 aJ— y3=56. 
 
 
 
 Ans. x=4, 
 
 y=2. 
 
 4. X — y :a; : :5 :6, 
 
 iy2=384. 
 
 
 
 Ans. x=2i, 
 
 y=4. 
 
 5. x-\-y:x: :7 :5, 
 
 Ij,+yJ=126. 
 
 
 
 Ans. a:=±15, 
 !/==h6. 
 
 6. (x+j,)':(a^^)'::64:l, 
 xy=63. 
 
 
 
 Ans. x—±9, 
 
 y=±7. 
 
 7. '^n/"'-^ -ft. 
 a-hJa^—x' 
 
 Am. «=+2«V*_ 
 J+1 
 
 8. ^a+x-^ar-x _l ^^_ ^ 
 
 2a& 
 
 ^a-\-x-\-iJa — a: 
 
 V J'+l' 
 
 9. °+^ =1 Am. x=±a {±1±J^^^ ] . 
 a-\-x-\-^-2ax+x^ ^ ^ ±^2b—b^ ' 
 
 10. It is required to find two numbers whose product is 320, and 
 the difference of whose cubes is to the cube of their difference, 
 as 61 is to 1. Am. 20 and 16. 
 
 Art. 280. Hakmonical Pkopoktion. — Three or four quanti- 
 ties are said to be in harmonical proportion, when the first has the 
 same ratio to the last, that the difference betweeen the first and 
 second has to the difference between the last, and the last except 
 one. Thus, a, b, c, are in harmonical proportion when a : c ; : 
 a — b : b — c ; and a, b, c, d, are in harmonical proportion when 
 a :d : : a — 6 ■ " — d. 
 
 1. Let ''. be required to find a third harmonical proportional a 
 to two piven numbers a and b. 
 
RATIO AND PROPORTION. 239 
 
 We have, a:x:: a — 6 : b — x ; 
 
 .-. (Art. 267), a(i— x)=a;(a— i) ; 
 
 whence, 
 
 ab 
 
 '2a — 6' 
 
 2. Find a third harinonical proportional to 3 and 5; Ans. 15. 
 
 3. Find a fourth harmonical proportional x, to three given num- 
 bers, o, b, and c. Ans. x=. "^ 
 
 2a— J' 
 
 VARIATION. 
 
 Aet. 281. Variation, or as it is sometimea termed, Generai 
 Proportion, is merely an abridged form of common Proportion. 
 
 Variable quantities are such as admit of various values in the 
 same computation. Constant, or invariable quantities have only 
 one fixed value. 
 
 One quantity is said to vary directly as another, when the two 
 quantities depend upon each other in such a manner, that if one 
 be changed the other is changed in the same proportion. 
 
 Thus, if A and B are two variable quantities, mutually de- 
 pendent on each other, in such a way, that if A be changed to 
 any other value a, B must be changed to another value b, such 
 that A : o : : B : J, then A is said to vary directly as B. 
 
 This relation is expressed thus, A ccB, the symbol cc being 
 used instead of varies, or varies as. 
 
 From this it will be seen that variation is merely an abridg 
 ment of Proportion, and that four quantities are understood 
 although only two are expressed. 
 
 Note. — When it is simply stated that one quantity varies as anothei 
 tt is always meant that the one varies directly as the other. 
 
 Art. 282, There are four diiferent kinds of Variation, which 
 are distinguished as follows : 
 
 (1). AaB. Here A is said to vary directly as B. 
 
 Ex. If a man works for a certain sum per day, the amount of 
 his wages varies as the number of days in which he works. 
 
 (2). Aoc __ Here A is said to vary inversely as B. 
 
 Ex. If a man has to perform a journey of a certain number of 
 miles, the time in which he performs it will vary inversely as the 
 rate of traveling. Thus, if he doubles his speed, he will perfoTn» 
 the journey in hdf the time. 
 
240 RAY'S ALGLBRa, PART 6EC0ND. 
 
 (3). AorBC. Here A is said to vary as B and C jointly. 
 Ex. The wages to be received by a workman will vary jointly 
 as the number of days he works, and the wages 'per day. 
 ■p 
 (4). Acx_ Here A is said to vary directly as B, and inversely 
 
 as C. 
 
 Ex. The base of a triangle varies as the area directly, and the 
 altitude imxrsely. 
 
 Let the pupil give other examples of each kind of Variation. 
 
 In the following articles, A, B, C, represent corresponding 
 values of any variable quantities, and a, b, c, any other corres- 
 ponding values of the same quantities. 
 
 Art. 2 §3. If one quantity vary as a second, and that second as a 
 third, the first varies as the third. 
 
 Let AocB, and BocC, then shall AocC. For A:a::B:J, 
 and B : 6 : : C : c, therefore, (Art. 272), A : a : : C : c ; that is 
 AccC. 
 
 In a. similar manner it may be proved that if AocB, and 
 
 Bk^ , that Aocl.. 
 C C 
 
 Art. 284. If each of two quantities vary as a third, their sum, 
 yr their difference, or the square root of their product, wUl vary at 
 Che third. 
 
 Let AaC, and BaC, then A±BocC ; also, ^ABaC. 
 
 By the supposition, A:a::C:c::B:6; 
 .. A:a::B:6; 
 alternately, (Art 270), A : B : : o : J ; 
 by Composition or Division, A±B : B : : a±6 = b ; 
 alternately, A±B : a±5 : : B : i : : C : c ; 
 
 that is, A±BaC. 
 
 Again, A : a : : C : c ; 
 
 and B : J : : C : c ; 
 
 .-. (Art. 277), AB : a6 : : C : c= ; 
 
 and, (Art. 276), JA3 : ^A: : C : c ; 
 
 that is, ^"ABocC. 
 
 Art. 285. If one quantity vary as another, it wiU also vary a* 
 any mttK ipfe, or any part of the other. 
 
RATIO AND PROPORTION. 241 
 
 Let AorB, and m be any constant quantity, then since 
 : n 
 B 
 
 A : o : : B : i, A : a : mB : m6, or A : a : : ?- : -, (Art. 260, 3rd) ; 
 
 m m 
 
 that iSv AccmB, or ex 
 
 Am. 286. If one quantity vary as another, any power or root of 
 tht former will vary as the same power or root of the latter. 
 
 Let A cB, tlien A : a : : B : 6, and (Art. 276), A" : o» : : B" : 6"; 
 that is, A"xB", where n niay be integral or fractional. 
 
 Akt. aSV. If one quantity vary as another, and each of them it 
 multiplied or divided by any quantity, variable or invariable, the 
 products, or quotients, will vary as each other. 
 
 Let A vnry as B, and let T be any other quantity. Then, by 
 
 the supposition, A : a : : B : i ; 
 
 . . AT ■■ at ■■< BT :bt; that is, ATaBT. 
 
 ,, Aa Bill,.. AB 
 
 Also, _:_::_:_; that is, —oc — 
 
 T t T t T T" 
 
 Art. 28§. If one quantity vary as two others Jointly, either of 
 (lie latter varies as the first directly, and the other inversely. 
 
 V PT V 
 
 Let VocFT, then (Art. 287), loc_, or Fozl, and simi- 
 larly, T«Y 
 
 Art. 289. If A vary as B, A is equal to B multiplied hy some 
 constant quantity. 
 
 Since, by supposition, A : a : : B : 6, therefore, A=?B ; but a 
 
 b 
 and b are supposed to be constant, being certain corresponding 
 
 values of A and B. Hence, if we denote _ by m, we have 
 
 — b 
 
 A=mB. 
 
 It is evident that if we know any corresponding values of A 
 ind B, that the constant quantity m may be found. 
 
 Art. 990. In general the simplest method of treating varia- 
 tions, is to convert them into equations. 
 
 Ex. 1. Given, that yoc the sum of two quantities one of which 
 varies as x, and the other as a', to find the corresponding 
 equation. 
 
 21 
 
242 RAY'S ALGEBRA, PART SECOND. 
 
 Because one part oci, let this =mx, 
 and the other " oci=, " " =na;' ; 
 
 y=mx-\-nx', where m and n are two un. 
 known invariable quantities which can only be found when we 
 know two pairs of corresponding values of x and y. 
 
 2. li y=r-\-s, where roca; and soc-, and if, when a;=:l, 
 
 5=6 and when as=2, y^9, what is the equation between x 
 and y'i 
 
 Let r=mx, and s=_ .•. y=mx-\-- 
 
 X X 
 
 But if x=l, y=6, .•. 6=)n+n ; 
 and if x=2, y=9, .: 9=2m+^, 
 
 Hence, »»=4; n=2, and y=ix-{--, 
 
 X 
 EXAMPLES FOR PRACTICE. 
 
 3. If yoci ; and when as=2, y=4o ; find the equation be- 
 tween X and y. Ans. y=2ax. 
 
 4. If ycc - ; and when ir=l, ^=^8; find the equation between 
 
 X 
 
 X and y. . Ans. y=-. 
 
 X 
 
 6= 
 
 5. If v'oco'-^<e' ; and when x=Ja' — 4^, y=— ; find the 
 
 a 
 
 equation between x and y. Ans. y=^-^a'' — x'. 
 
 a 
 
 6. If y is equal to the sum of two quantities, one of which 
 varies as x, and the other varies inversely as x^ ; and when x=\, 
 y=6, and when x=2, y=5; find the equation between x and y. 
 
 Ans. j/=2x+_. 
 
 7. Given that y is equal to the sum of three quantities, of 
 which the 1st is invariable, the second varies as x and the third 
 varies as a;^. Also when x=l, 2, 3, j/s=6, 11, 18, respectively ; 
 find y in terms of x. Ans. y=2-\-Qx4-x'. 
 
ARITHMETICAL PROGRESSION. 243 
 
 8. Given that socfi, when / is constant ; and soc/, when t 
 IB constant ; also, 2s=/, when t=l. Find the equation between 
 /, i, and t. Ans. s=^ft^. 
 
 9. If y=r-{-s where rocar, and sccjx; and if when as=4, 
 j/=5, and when a;=9, y=10; find y in terms of a:. 
 
 Ans. y=\Qe-\-Jx). 
 
 10. Given that xoc — , and yoc— ; also, when x=a, z=c , 
 find the equation between x and z. Ans. a2'""=:c'""a;. 
 
 ARITHMETICAL PROGRESSION. 
 
 Art. 291. Quantities are said to be in Arithmetical Progres- 
 iion,when they increase or decrease by a Common Difference. 
 
 Thus, 1, 3, 5, 7, 9, &c., ii, a-]-d, a-\-2d, &c., a, a — d, a — 2d, 
 &c., are quantities respectively in Arithmetical Progression. 
 
 The series is said to be increasing or decreasing, according as d 
 is positive or negative. 
 
 Art. 292. To investigate a rule for finding any term of an 
 arithmetical progression, take the following series, in which the first 
 line denotes the number of each term, the second an increasing 
 arithmetical series, and the third a decreasing arithmetical series 
 
 12 3 4 5 
 
 a, a-\-d, a-\-2d, a-\-2d, a+4d, &c., 
 
 a, a — d, a — 2d, a — 3(Z, a — ^d, &c. 
 
 It is manifest that the coefficient of d in any term is less by 
 unity than the number of that term in the series ; therefore, the 
 b" term =a-\-{n — \)d. 
 If we designate the n" term by I, we have 
 
 Z=o+(?i — l)d, when the series is increasing, 
 and t=o — (ji — \)d, when the series is decreasing. 
 
 Hence, we have the following 
 
 Rule, for finding ant term of an aeithmeiical series.— 
 Multiply the common difference by the number of terms less oni 
 and add the product to the first term when the serigs is increasiny 
 but subtract it from the first term when the series is decreasing. 
 
 The equation Z=«+(7S — l)d, contains four variable quantities, 
 any one of which may be found when the other three are known 
 
244 RAY'S ALGEBRA, PART SECOND. 
 
 Art. 293. Having given the first term a, the common differ, 
 ence d, and the number of terms n, to find S, the sum of the 
 series. 
 
 If we take any arithmetical series, as the following, and write 
 the same series under it in an inverted order, we have 
 S= 1+3 + 5+ 7+ 9+11, 
 S=ll+9 + 7+ 5+ 3+ 1. 
 Adding, 28=12+12+12+12+12+12. 
 28=12 X the number of terms. 
 28=12x6=72. 
 Whence, 8=5 of 72=36, the sum of the series. 
 
 To render this method general, let 1= the last term, and write 
 the series both in a direct and inverted order. 
 
 Then, S=a+(a+<i)+(a+2d)+(a+3rf)- • • +h 
 and S=Z + (/— d)+(Z— 2d)+(i— Sfi). ■ • +«• 
 By adding the corresponding terms, we have 
 
 2S=Q+a)+Q+a)+Q+a)+(.l+a). . . +Q+a), 
 2S=(H-<') taken as many times as there are terms (n) in 
 
 the series. 
 Hence, 2S=(Z+a)re; 
 
 S=G+a,l=(bt^)n. 
 
 This formula gives the following 
 
 Rule, fok finding the sum of ah abithmetical series. — 
 Multiply half the sum of the two extremes, by the number of 
 terms. 
 
 Prom the preceding it appears, that the sum of the extremes u 
 equal to the sum of any other two terms equally distant from the 
 
 Kxtremes. 
 
 Aet. 294. The equations l=a-\-{n — l)d, 
 and S=(a+Z)|, 
 
 furnish the means of solving this general problem : Knowing any 
 three of the five quantities, a, d, I, n, S, which enter into an arith- 
 metical series, to determine the other two. 
 
 The following table contains the results of the solution of all 
 the different cases. These formulae should be verified by the 
 student. 
 
ARITHMETICAL PROGRESSION- 
 
 24S 
 
 No. 
 
 GiTen. 
 
 Hequired. 
 
 Formulse. 
 
 1. 
 
 2. 
 3. 
 
 4. 
 
 a, d, n 
 a, d, S 
 
 a, n, S 
 d, n, S 
 
 1 
 
 l=a+{n-l)d, 
 
 h=-id±^{2dS+(,a-kdyi. 
 1= — — a, 
 
 ,_S,(«^l)d 
 ra"^ 2 • 
 
 5. 
 6. 
 
 7. 
 8. 
 
 a, d, n 
 a, d, I 
 
 a, n, I 
 d,n,l 
 
 s 
 
 S=in\2a+(_vr-l)d\, 
 ^2 + 2d • 
 
 S=ln\21—(n—l)d\. 
 
 9. 
 10. 
 11. 
 12. 
 
 a, n, 1 
 a, n, S 
 a, 1, S 
 n, I, S 
 
 d 
 
 d= -, 
 
 n— 1 
 ^_2(S— an) 
 
 n(™— 1) ' 
 
 2S—l—a 
 
 ^_2(nZ-S) 
 
 n(ra— 1)' 
 
 13. 
 14. 
 15. 
 16. 
 
 a, d, I 
 a, d, S 
 a, I, S 
 
 d, I, S 
 
 n 
 
 «=^%1. 
 
 „ ±J(2a-dy+8dS-2a+d\ 
 
 " 2d 
 
 _21+d±:^(2l+i)'—8dS 
 2d 
 
 17. 
 18. 
 19. 
 20. 
 
 d, n, I 
 d,n,S 
 d,l, S 
 n,l, S 
 
 a 
 
 a=l—(,nr-l)d, 
 
 S (n-l)d 
 
 a — — — — , 
 
 n 2 
 
 a=i^±^Q+^dy~2d8, 
 n 
 
 EXAMPLES FOE PRACTICE. 
 
 1. Find the 15" term of the series 3, 7, 11, &o. Ans. 59. 
 
 2. Find the 20» term of the series 5, 1, — 3, &c. 
 
 Am. —71. 
 
'■^46 . RAT'S ALGEBRA, PART SECOND. 
 
 3. Find the 8'* term of the series |, j'ji 5, &c. Ans. y\. 
 
 4. Find the 30" term of the series —27, —20, —13, &c. 
 
 Ans. 176. 
 
 5. Find the re"* term of 1+3+5+7. . . . A7is.2n~l. 
 
 andof 2+2-J+2J+. . . . Ans. i(n+5) 
 and of 13+12f+12i+. . . . Ans. i(40— ra). 
 Find the sum of 
 
 6. 1+2+3+4 &o., to 50 terms. (See Formula 5). 
 
 Ans. 1275. 
 
 7. 7+2,9+y+, &c., to 16 terms. Ans. 142. 
 
 8. 12+8+4+, &c., to 20 terms. Ans. —520, 
 
 , „ n 
 
 9. 2+23+2f+, &c., ton ternis. Ans. g(re+ll)- 
 
 10. 13+12f+12|+,&c., ton terms. Ans. g(79— n). 
 
 11. ^— |— y — , &c., to re tenns. Ans. —(IS—rin). 
 
 12. ^+^.^Z??+, &c., to re terms. 
 a+J a+i 
 
 Ans. -^ \na-^I!±^l 
 a+b ( 2 5 
 
 13. ^+^=^+'5=5+, &o., to n terms. Ans. ^. 
 
 re 71 re 2 
 
 14. If a falling body descends 16^3 feet the first second, 
 three times this distance the next, five times the next, and so on, 
 how far will it fall the 30 th second, and how far altogether in 
 half a minute 1 Ares. 948H, and 14475 ft. 
 
 15. Two hundred stones being placed on the ground in a 
 straight line, at the distance of 2 feet from each other ; how far 
 will a person travel who shall bring them separately to a basket, 
 which is placed 20 yards from the first stone, if he starts from the 
 ipot where the basket stands ■! Ans. 19 miles, 4 fur., 640 ft. 
 
 16. Insert 3 arithmetical means between 2 and 14. 
 
 To solve this problem generally, let it be required to insert m 
 irithmetical means between a and b. 
 
 Since the required terms, and those which are given, form an 
 arithmetical series, if we insert m terms between a and b, we 
 
ARITHMETICAL PROGRESSION. 247 
 
 shall have a series consisting of (m+2) terms. Then to find d, 
 the common difference, substitute m+2 instead of n in Formula 
 
 9, page 245, and we have d=tll=z i—a_^b—a rj,^^^^^ 
 
 n — 1 m-\-2 — 1 m-\-\ ' 
 fore, the common difference {d) will be equal to the difference of the 
 extremes (b — a) divided by the number of terms plus one. 
 
 In the particular example we find (2=3 ; hence, the terms are 
 5, 8, and 11. 
 
 17. Insert 4AR. means between 3 and 18. 
 
 Ans. 6, 9, 12, 15. 
 
 18. Insert 9AR. means between 1 and — 1. 
 
 Ans. |, I, &c., to — |. 
 
 19. How many terms of the series 19, 17, 15, &c., amount 
 to 911 Atis. 13, or 7. 
 
 Explain this result. 
 
 20. How many terms of the series .034, .0344, .0348, &c., 
 amount to 2.748? Ans. 60. 
 
 21. The sum of the first two terms of an arithmetical pro- 
 gression is 4, and the fifth term is 9; find the series. 
 
 Ans. 1,3, 5,7, 9, &c. 
 
 22. The first two terms of an arithmetical progression being 
 together =18, and the next three terms =12, how many terms ' 
 must be taken to make 28? Ans. 4, or 7. 
 
 23. The ra" term of an arithmetical series is g(37t — 1), find 
 the first term, the common difference, and the sum of n terms. 
 
 Ans. J, |, and ^(2n^l). 
 
 24. In the series 1, 3, 5, dz.c., the sum of 2r terms : the sum 
 of r terms : : x : 1 ; determine the value of a;. j4?is. 4 . 
 
 25. Find the ratio of the latter half of 2re terms of any AR. 
 series, to the sum of 3ra terms of the same series. Ans. 3. 
 
 26. The sum of n arithmetical means between 1 and 19 : 
 turn of the first n — 2 of them : : 5 : 3; required n. Ans. 8. 
 
 27. A traveler sets out for a certain place, and travels 1 mile 
 the first day, 2 the second, and so on. In 5 days afterward 
 another sets out, and travels 12 miles a day. How long and how 
 for must he travel to overtake the first? 
 
 Ans. 3 days or 10 days, and travel 36 miles, or 120 miles. 
 Explain these results. 
 
248 RAY'S ALGEBRA, PART SECOND. 
 
 GEOMETRICAL PROGRESSION. 
 
 Art. 295. A Geometrical Progression is a series of terms 
 each of which is derived from the preceding, by multiplying it by 
 a constant quantity termed the ratio. 
 
 Thus, 1, 2, 4, 8, 16, &c., is an increasing 
 
 geometrical progression, whose common ratio is 2. 
 
 54, 18, 6, 2, &c., is a decreasing geometrical 
 progression, whose common ratio is ] . 
 
 In general, a, ar, ar', ar', &c., is a geometrical progression, 
 whose common ratio is r, and which is an increasing series when 
 r is greater than 1 ; but a decreasing series when r is less than 1 . 
 
 It is evident that in any given geometrical series, the common 
 ratio will he found hy dividing any term by the term next preceding. 
 
 Re.mare. — An Arithmetical Progression may be defined to be a series 
 in which the difference between any two consecutive terms is the same ; 
 and a Geometrical Progression a series in which the ratio of any two 
 consecutive terms is tlie same. Hence, a. geometrical progression is a 
 continued proportion. (Art. 266.) 
 
 Art. 296. To find the last term of a geometrical progression. 
 
 Let a denote the first term, r the common ratio, I the ra" term, 
 and fcj the sum of n terms ; then the respective terms of the series 
 will be 
 
 1, 2, 3, 4, 5 . . . . n — 3, n — 2, n — 1, n, 
 a, ar, ar', ar', ar^ . . . . ar"'*, ar"~^, ar"'', ar"~'. 
 
 That is, the exponent of r, in the second term is 1, in the tliird 
 term 2, in the fourth term 3, and so on ; hence, the n"' term of 
 the series will be l=af"~^ ; that is, any term of a geometrical 
 series is equal to the product of the fixst term, by the ratio raised to a 
 power, whose exponent is one less than the number of terms. 
 
 Ex. Let it be required to find the 6'* term of the geometrica! 
 progression whose first, term is 7, and common ratio 2. 
 
 25=2X2X2X2X2=32; and 7x32=224, the 6"- term. 
 
 Art. 297. To find the sum of all the terms of a geometrical 
 progression. 
 
 If we multiply any geometrical series by the ratio, the product 
 will be a new series, of which every term, except the last, will 
 have a corresponding term in the first series. 
 
GEOMETRICAL PROGRESSION. 249 
 
 ThuB, take the series, 1, 3, 9, 27, 81, and represent its sum 
 by S ; then 
 
 8=1+3+9+27+81. 
 Multi'iilying each term by the ratio 3, we have 
 3S= 3+9+27+81+248. 
 
 The terms of the two Buries are identical, except thejJrsi term 
 of the first series, and the last term of the second series. If 
 then, we subtract the first equation from the second, all the otlier 
 terms will disappear, and we shall have 
 
 3S— S=243— 1; 
 whence, S=121. 
 
 To generalize this method, let a, or, ar^, ar^, &c., be any 
 geometrical series, and S its sum ; then, 
 
 S=:a-{-ar-\-ar^-\-ar' . . . . -^ar"~'-\-ar"-K 
 Multiplying this equation by r, we have 
 
 rS= ar-\-ar^-^ar' +ar"~'+ar". 
 
 Subtracting, rS — S=ar'' — a ; ' 
 
 or, S(?-— l)=a(r"— 1) ; 
 
 whence, ^_ o(r"-l) 
 
 r— 1 ■ 
 
 Since, Z=:ar"~', we have rb^ar" ; 
 
 therefore, B=''r=^=.±:3 . 
 
 r — 1 r — 1 
 
 This formula gives the following 
 
 Rule, foe findins the sum of a geometkicai, sebies. — 
 Multiply the last term by the ratio, from the product subtract the 
 first term,a7id divide the remainder by the ratio less one. 
 
 Ex. Find the sum of 6 terms of the progression 3, 12, 48. 
 &c. 
 The last term =3 X4'=3 X 1024=3072. 
 
 g_Zr-«_3_072><4-3_4095 A». 
 r— L 4—1 
 
 Akt. 298. If the ratio r is less than 1, the progression is de- 
 creasing, and the last term I, or ar"^^, is less than a. In order 
 
 that both terms of the fraction . , or may be positiv >. 
 
 r — 1 r — 1 
 
250 RAY'S ALGEBRA, PART SECOND. 
 
 the signs of the terms may be changed, (Art. 1 24), and we have 
 
 S= ° , or — ° "'"" Therefore, the sum of the series, when 
 
 1 — r 1 — r ' 
 
 the progression is decreasing, is found by the same rule, as when 
 it is increasing, except that the product of the last term by the 
 ratio, is to be subtracted from the first term, and the ratio sub- 
 tracted from unity. 
 
 Akt. 299. The formula S= , by separating the numera- 
 
 1 — r 
 
 tor into two parts, may be placed under the form 
 
 g___a_ gr" 
 
 1— r 1— r- 
 NoTO when r is less than 1 , it must be a proper fraction, which 
 may be represented by " ; then r''= I - I =^. 
 
 Since p is less than q, the higher the power to which the frac- 
 tion is raised, the less will be the numerator compared with the 
 denominator ; that is, the less will be the value of the fraction ; 
 
 therefore, when n becomes very large, the value of <-, or r", will 
 
 9" 
 be very small ; and when n becomes infinitely great, the value of 
 
 ?-,orr", will be infinitely small; that is, 0. But, when the 
 9" 
 numerator of a fraction is zero (Art. 135) its value is 0. This 
 
 reduces the value of S to 
 
 1 — r 
 
 Hence, when the number of terms of a decreasing geometrical 
 series is infinite, the last term is zero, and the sum is equal to the first 
 term divided by one minus the ratio. 
 
 Ex. Find the sum of the infinite series l+3+5+i+. &c. 
 
 Here a=l, r=:i, and S= = =2. Ans. 
 
 That the sum of an infinite number of terms of a geometrical 
 progression may be finite, will easily appear from the following 
 illustration : 
 
 Take a straight line, AK, and bisect it in B ; bisect BK in C ; 
 CK in D, and so on continually ; then will 
 
 B CD 
 A 1 ' ' K 
 
 AK=AB+BC-fCD4-, &c., in infinitum, =AB-|-4aB+JAB 
 &c., in infinitwm, =2AB, which agrees with the example. 
 
GEOMETRICAL PROGRESSION. 
 
 251 
 
 Art. 300. The two equations, l=ar"-\ and 8=°''" ", fur- 
 
 r — 1 
 nish this general problem : Knowing any three of the five quantities 
 a, r, n, I, and S, of a geometrical progression, to determine the other 
 two. Tlie following table contains all the values of each unknown 
 quantity", or the equations fr-^m which it may be derived. 
 
 No. 
 
 Uivcn. 
 
 1. 
 
 a, r, n 
 
 2. 
 
 a, r, S 
 
 3. 
 
 li, n, S 
 
 4. 
 
 >-, n, S 
 
 5. 
 
 a, r, n 
 
 6. 
 
 a, r, I 
 
 7. 
 
 a, n, I 
 
 8. 
 
 r, n, I 
 
 9. 
 
 r, n, I 
 
 10. 
 
 r, n, S 
 
 11. 
 12. 
 
 r, I, S 
 n, I, S 
 
 13. 
 
 a, n, I 
 
 14. 
 
 a, n, S 
 
 15. 
 
 a, I, S 
 
 16. 
 
 n, I, S 
 
 17, 
 
 a, r, I 
 
 18 
 
 a, r, S 
 
 19 
 
 a, I, S 
 
 20. 
 
 r, 1. S 
 
 Kiiquired. 
 
 
 Z(S— 0"-'— a(S— o)»-'=0, 
 ,_ (r— l)Sr"-' 
 
 r"— 1 
 
 r — 1 
 
 c<_rl — a 
 
 O— r-, 
 
 r — 1 
 
 "~'VF— "-iy"g» 
 
 Ir^—l 
 
 S= 
 
 r"—\ 
 a=ri— (r— 1)S, 
 a(S— a)''-'^;(S— 0"-'=0. 
 
 „ S , S— a n 
 
 " — r-f- ^=0. 
 
 a a 
 _S— a 
 
 "— _^r"-i +_L =0. 
 S—l ^S—l 
 
 ^^ log. I— log, a I ^ ^ 
 toy. r 
 Joy. [a-\-{r — 1)8] — log, a 
 
 log. r 
 log. I — log. a 
 
 +1- 
 
 log. (S—a)—log. (S— 
 Jog.l—log. [Zr— (r— 1)S] _^_, 
 log. r 
 
252 
 
 RAY'S ALGEBRA, PART SECOND. 
 
 Remark. — To determine the value of the unknown quantity In Nos. 
 3, 12, 14, and 16, may require the solution of an equation higher than 
 the second degree. The values of n in the last four Nos. are ob- 
 tained from the solution of an exponential equation (see Art. li&Sj. 
 Although the method of solving these equations has not been given, it 
 was deemed proper to complete the table for the convenience of refer- 
 ence. The pupil should be required to verify all the values except those 
 here referred to. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Find the 8"* term of the Beries 5, 10, 20, &c. Aiis. 640. 
 
 2. Find the 7" term of the series 54, 27, 13^, &c. 
 
 Am. ||. 
 
 3. Find the 6" term of the series 3|, 2i, 1^, &c. Arts, f 
 
 4. Find the 7» term of the series —21 14, — 9|, &c. 
 
 3"-2 
 
 5. Find the »" term of the series i, 3, |, &c. -An*- ;^^. 
 
 Find the sum 
 
 6. Of 1+3+9+, &c., to 9 terms. Ana. 9841. 
 
 7. Of 1+4+16+, &c., to 8 terms. Avs. 21845. 
 
 8. Of 8+20+50+, &c., to 7 terms. Am. 3249g. 
 
 9. Of 5+20+80+, &c., to 8 terms. Arts. 109225. 
 
 10. Of 1+3+9+, &c., to » terms. Ans. 3(8"— 1). 
 
 11. Of 1— 2+4— 8+, &c., to n terms. Ans. i(l=F2»). 
 
 12. Of X — y-\-^— — -A-i &c., to n terms. 
 
 Am.j^\\-{-iyK 
 
 a;+y f V xlS' 
 
 13. The first term is 4, the last term 12500, and the numbn 
 of terms 6. Required the ratio and the sum of all the terms. 
 
 Am. Ratio =5; Sum =15624. 
 
 4. Find the geometrical progression, when the sum of the finrt 
 and second terms is 9, and the sum of the first and third 15. 
 Am. 3+6+12+, &c., 13A— 4i+li— , &,c. 
 
 Find the sum of an infinite number of terms of each of the 
 following series : 
 
 15. Of I+3+B+. «&c. Am. J. 
 
GEOMETRICAL PROGRESSION. 253 
 
 16. Of 9+6+4+, &c. Ans. 27 
 
 17. Of 6+2+§+, &o. Ans. 9. 
 
 18. Of I— l+B— , &c. Ans. |. 
 
 19. Of 100+40+16+, &c. Ans. 166f . 
 
 20. Of a+J+'-!+^+, dz-c Ans. -^ 
 
 a a' a — b' 
 
 21. Of l+2a+2o2+2aS+, &c. Aras. 1±? 
 
 1 — a 
 
 22. The sum of an infinite \geometric series is 3, and the sum 
 of its first two terms is 2| ; find the series. 
 
 Ans.2+i+»g+ . . or 4-^1— • • 
 
 23. Find a geometric mean between 4 and 16. Ans. 8. 
 Let a:=4, c=16, and m the required mean ; then a '.m::m -.c; 
 
 whence m=Jac. 
 
 24. The first terra of a geometric series is 3, the last term 96, 
 and the number of terms 6 ; find the ratio, and the intermediate 
 terms. 
 
 Byformulal3,page251, wefind r="-'/L, which in this case 
 
 becomes r=«/32=2; hence, the intermediate terms are 6,12, 
 24, 48. 
 
 If it be required to insert m geometrical means between two 
 numbers a and b, we have n, the whole number of terms, =m+2 ; 
 
 hence, n — l=m+l, and r~^-\-\IL_ 
 
 25. Insert two geoijietric means between jf, and 2. 
 
 Ans. |, |. 
 
 26. Insert seven geometric means between 2 and 13122. 
 
 Ans. 6, 18, 54, 162, 486, 1458, 4374. 
 
 Art. SOI. To find the value of Circulating Decimals, that is, 
 decimals in which one or more figures are continually repeated. 
 
 Circulating decimals are quantities in geometrical progression, 
 where the common ratio is yj,, y J-jj, y^'j^r, &c., according as one, 
 two, or three figures recur ; thus the circulating decimal 
 
 .253131 ... is equal tp 2= + ( ^+^+11+, &c. ) ; 
 and the part within the bracket is a geometrical series, of which 
 
254 RAT'S ALGEBRA, PART SECOND. 
 
 1 31 
 
 the common ratio is — =:_!, ; we have, therefore, a= 1 
 
 102 ■'"' 10* 
 
 '^Tk; i^=^,-^l%%=^Uo' ''"d the sum of the whole 
 
 series =i%%+^U^=UU=4Ui- 
 
 This operation may be performed more simply, as follows : 
 
 Let S=.253131. . . . Multiply by 100, in order to remove 
 the decimal point to the commencement of the Jirst period of 
 decimals, we have 
 
 1008=25.3131. . . . 
 
 Again, multiplying by 100 to remove the decimal point to the 
 commencement of the secmd period of decimals, we have 
 100008=2631.3131. . . . 
 Subtracting the preceding pquation from the last, we get 
 99008=2506; .-. S=|i8g. 
 
 1. Find the value of .636363. . . . Ajw. j\. 
 
 2. Find the value of .54123123. , . . Ans.-ll°U 
 
 HARMONIC AL PROGRESSION. 
 
 Akt. 302. Three or more quantities are said to be in Har- 
 monica! Progression, when their reciprocals are in arithmetics! 
 progression. 
 
 Thus, 1, >, J,4, &c.; and I, |, -|, |, &.C. 
 
 are in harmonica! progression, because their reciprocals 
 
 1, 3, 5, 7, &c. ; and 4, 3^, 3, 2 J, &c. 
 are in arithmetical progression. 
 
 Art. 303. Proposition. — If three quantities are in harmonical 
 progression, the first term is to the third, as the difference of the firsi 
 and second, is to the difference of the second and third. 
 
 For if a, b, c, are in harmonical progression, -, _, _, are in 
 
 a b c 
 
 srithmetical progression, 
 
 .■. - — -=- — -_ Hence, multiplying by aift 
 b a c b 
 
 ac — bc=ab — ac; or c{a — b)=:a(b — c). 
 
HARMONICAL PROGRESSION. 255 
 
 Dividing both sides by a — b, and by a, we have 
 
 c b — c . 
 
 o a — b 
 this gives a : c : : a — h : b — c. 
 
 Therefore, u Harmonical Progression is a series of quantities in 
 harmonica! proportion (Art. 280) ; or such that if any three con- 
 secutive terms he taken, the first is to the third, as the difference 
 of the first and second is to the difference of the second and 
 third. 
 
 From this proposition it follows, that all problems with respect 
 to numbers in harmonical progression, may be solved by inverting 
 them, and considering the reciprocals as quantities in arithmeti- 
 cal progression. This renders it unnecessary to give any special 
 rules for the solution of problems in harmonical progression. 
 
 EXAMPLES FOR PRACTICE. 
 
 1 . Given the first two terms of a harmonical progression, u 
 and b, to find the n"' term. 
 
 Let I be the n"' term, then (Art. 302), _ and - are the first 
 
 a b 
 
 two terms of an arithmetical progression, and it is required to 
 
 find -, the m'* term. 
 I 
 
 <i=l— -=?=?, and ?L=?i+(ra— l)d, (Art. 292) ; 
 b a ab la 
 
 la ab ab 
 
 whence, l=- 
 
 (n— l)a— (7!--2)6' 
 
 By means of this formula, when any two successive terms of a 
 harmonical progression are given, any other term.may be found. 
 
 2. Insert m harmonic means between a and b. 
 Here, if d be the common dilference of the reciprocals of the 
 terms, we have 
 
 L=l+(_nr-l)d, and d=_f=L = ""^ : 
 i a {n — l)ai (m-\-V)ao 
 
 whence the arithmetical progression is found ; and by inverting 
 its terms, the harmonicals are also found. 
 
256 RAY'S ALGEBRA, PART SECOND. 
 
 3. Insert two harmonic means between 3 and 12. 
 
 Arts. 4 and 6. 
 
 4. Insert two harmonic means between 2 and j. 
 
 Ans. 2 and 7. 
 
 5. The first term of a harmonic series is 5, and the 6" j^j ; 
 find the intermediate terms. Ans. j, g, g, -j-'j. 
 
 6. a,b, c, are in arithmetical progression, and 5, c, d, are in 
 harmonical progression ; prove that a -.b : -.cid. 
 
 PROBLEMS Df ARITHMETICAL AND GEOMETRICAL PRO- 
 GRESSION. 
 
 Art. 304. The sum of five numbers in arithmetical progres- 
 gion is 35, and the sum of their squares 335; find the numbers. 
 
 4ns. 1,4,7, 10, 13. 
 Let X — 2y, x — y, x, a;+V> ^+2^1 be the numbers. 
 
 2. There are four numbers in arithmetic progression, and the 
 sum of the squares of the extremes is 68, and of the means 52; 
 find them. Am. 2,4,6,8. 
 
 Let X — 3y, X — y, x-\-y, x-\-Zy, be the numbers. 
 
 Suggestion. — When the number of terms in an arithmetic progression 
 Is odd, the common difference should be called y, and the middle term x; 
 bat when the number of terms is even, the common difference must be 
 2y, land the two middle terms x — y, and x-\-y. 
 
 3. The sum of 3 numbers in arithmetical progression is 30, 
 and the sum of their squares 308; find them. Ans. 8, 10, 12. 
 
 4. There are 4 numbers in arithmetical progression, their suin 
 is 26, and their product 880; find them. Ans. 2, 5, 8, 11. 
 
 5. There are 3 numbers in geometrical progression, whose sum 
 is 31; and the sum of the 1st and 2nd : sum of 1st and 3rd : : 
 3 : 13; find them. Ans. 1, 5, 2!^, 
 
 6. The sum of the squares of three numbers in arithmetic pr» 
 gression is 83 ; and the square of the mean is greater by 4 than 
 the product of the extremes. Required the numbers. 
 
 Ans. SiS,"" 
 
 7. Find 4 numbers in arithmetical progression, sue^ that tha 
 product of the extremes =27; of the means =35. 
 
 Ans. 3, 5, 7, 9. 
 
 8. There are 3 numbers in arithmetical progression, whose 
 
ARITHMETIC AND GEOMETRIC PROGRESSION. 257 
 
 sum is 18; but if yon multiply the first term by 2, the second by 
 3, and the third by 6, the products will be in geometrical pro- 
 gression ; find them. Ans. 3, 6, 9. 
 
 9. The sum of the 4"" powers of three successive natural num- 
 bers is 962; find them. Ans. 3, 4, 5. 
 
 10. The product of four successive natural numbers is 840; 
 find them. Ans. 4, 5,6,7. 
 
 11. The product of four numbers in arithmetical progression 
 [g 280, and the sum of their squares 166; find them. 
 
 Ans. 1,4, 7, 10. 
 
 12. The sum of 9 numbers in arithmetical progression is 45, 
 and the sum of their squares 285; find them. 
 
 Ans. 1, 2, 3, &c., to 9. 
 
 13. The sum of 7 numbers in arithmetical progression is 35, 
 and the sum of their cubes 1295; find them. 
 
 Ans. 2, 3, &c., to 8. 
 
 14. Prove that when the arithmetical mean of two numbers 
 is to the geometric mean : : 5 : 4; that one of them is 4 times 
 the other. 
 
 15. The sum of 3 numbers in geometrical progression is 7; 
 and the sum of their reciprocals is j ; find them. Ans. 1, 2, 4. 
 
 16. There are 4 numbers in geometrical progression, the sum 
 of the first and third is 10, and the sum of the second and fourth 
 is 30; find them. Ans. 1, 3, 9, 27. 
 
 17. There are 4 numbers in geometrical progression, the sum 
 of the extremes is 35, the sum of the means is 30; find them. 
 
 Ans. 8, 12, 18,27. 
 
 18. There are 4 numbers in arithmetical progression, which 
 being increased by 2, 4, 8, and 15 respectively, the sums are in 
 geometrical progression ; find them. Ans. 6, 8, 10, 12. 
 
 19. There are 3 numbers in geometrical progression whose 
 continued product is 64, and the sum of their cubes 584; find 
 them. I Ans. 2, 4, 8. 
 
 SnGGESTioN. — In solving difficult problems in geometrical progression, 
 instead of denoting the terms by x, xy, xif, &o., it is sometimes prefer- 
 able to express them by other forms. Thus, three terms may be ex- 
 
 by X, yjx7j, y, or, x'^, xy, y^ ; four terms by f-, x,y, ^ ; fivo 
 
 terms by f., x% xy, y^, ?L ; six terms by ^, ?^ , x, y, L, ?_. In all 
 y X y^ y X x^' 
 
 these cases the product of the first and third of any three consecutive 
 terms, is equal to the square of the middle term. 
 22 
 
858 RAT'S ALGEBRA, PART SECOND. 
 
 CHAPTER IX. 
 
 PERMUTATIONS, COMBINATIONS, AND 
 BINOMIAL THEOREM. 
 
 Aat 305. The different orders in which quantities can be 
 trranged, are called their Permutations. Quantities may be ar- 
 ranged in sets of one and one, two and two, three and three, and 
 80 on. Thus, if we have three quantities, a, h, c, we may arrange 
 [hem in sets of one, of two, or of three, thus : 
 
 Of one a, , b, c. 
 
 Of two ah, ac ; ba, be ; ca, cb. 
 
 Of three dbc, acb ; bac, bca ; cab, cba. 
 
 Remare. — Some writers, confine the term permutations to the class 
 where the quantities are taken all together, and give the title of ar- 
 rangements, or variations, to those groups of one and one, two and two, 
 three and three, &c., in which the number of quantities in each group Is 
 /ess than the whole number of quantities. 
 
 Akt. 306. To find the number of permutations that can be 
 formed out of n letters, taken singly, taken two together, three 
 together. . . . and r together. 
 
 Let u,b,c,d,. . . . k, be the n letters ; and let P, denote 
 the whole number of permutations where the letters are taket 
 singly; Pj the whole number of permutations taken 2 to- 
 gether .... and P, the whole number of permutations taken 
 r together. 
 
 The number of permutations of n letters taken singly, or one 
 and one, is evidently equal to the number of letters, that is n ; 
 therefore, 
 
 P,=n. 
 
 The number of permutations of n letters, taken two together, 
 is n(n — 1). For since there are n quantities 
 
 a, h, c, d, , . . . k, 
 if we remove a, there will remain (n — 1) quantities, 
 b, c, d, , . . . /:. 
 Writing a before each of these (n — 1) quantities, we shaP 
 have 
 
 ab, ac, ad . . . .ah 
 
 That is, (» — 1) permutationB in which a stands jCra^ 
 
PERMUTATIONS AND COMBINATIONS. 259 
 
 *In the same manner there are (n — 1) permutations in which b 
 stands first, and so of each of the remaining letters c, d . . . k. 
 And since there are n letters, there are n(ji — 1) permutations 
 taken two together ; that is, 
 
 P,=7!(?l-1). 
 
 Hence, the number of permutations of n letters taken two together, 
 is equal to the number of letters, multiplied by the number less one. 
 
 For example, if 7i=4, the number of permutations of the four 
 letters, a, i, c, d, taken two together, is 4X(4 — 1)=4X3=12. 
 Thus, ab, ac, ad, |{ ba, be, bd, \\ ca, cb, cd, \\ da, db, dc. 
 
 The number of permutations of n letters, taken three together, 
 is n(n — l)(n — 2). For if we take (n — 1) letters 
 
 b, c, d, . . . . k, the number of permuta- 
 tions taken tux) together, by the last paragraph, is 
 (71— l)(n— 2). 
 
 Let a be placed before each of these permutations ; then there 
 are (n — l^{n — 2) permutations of n letters, taken three together, 
 in which a stands first. Proceeding in the same manner with b, 
 thare are (n — l)(n — 2) permutations in which b stands first; and 
 60 for each of the n letters. Hence, the whole number of per- 
 mutations of n letters, taken three together, is n(n — l)(n — 2); 
 that is, 
 
 Pj=»(7l— 1)(»— 2). 
 
 Hence, the number of permutations of n letters taken three together, 
 is equal to the number of letters, multiplied by the number less one, 
 multiplied by the number less two. 
 
 For example, if ra=4, the number of permutations of the four 
 letters, a, b, c, d, taken three together, is 4(4 — 1)(4 — 2)=4x3 X2 
 =24. Thus, 
 
 dbc, dbd, acb, acd, adb, adc, bac, bad, bca, bed, bda, bdc, cab, cad, 
 cba, cbd, cda, cdb, dab, doc, dba, dbc, dca, deb. 
 
 By following the same method, we can prove that the number 
 of permutations of n letters taken four together, is 
 
 P^=n(7i— 1)(»— 2)(7i— 3). 
 
 By examining each of the preceding results, we see that tne 
 negative number in the last factor is less by unity, than the cum- 
 ber of letters in each permutation. Thus, 
 
 Pi=B= n^i—i. 
 
 P,=»(ra— 1>= n(»— -2— 1). 
 
280 RAY'S ALGEBRA, PART SECOND. 
 
 P3=n(7l— 1)(TO— 2)= «(?!— l)(7l— 3— 1). 
 
 Pt=n(n—l )(ra— 2)(n— 3)= n(n—].)(Ti^2)(n—i—i). 
 
 Hence, from analogy, we conclude, that the number of permu- 
 tations of n things taken r together, is 
 
 P,=n(ra— l)(n— 2) (ip-r—i). 
 
 Aet. 306a, Corollary. If all the letters be taken together 
 Iben r becomes equal to n, and the last factor becomes 1 ; that is, 
 
 P„=7l(7l— l)(?i— 2) (jir-ii^), 
 
 or P„=n(n— l)(w— 2) 1. 
 
 Or, inverting the order of the factors, 
 P.=1X2X3 (n— 1>. 
 
 Hence, the number of permutations of n letters taken n together, 
 is equal to ike product of the natural numbers from 1 up to n. 
 
 Ex. The permutations of three letters, a, b, c, taken three 
 together, is 1 X2X3=6. 
 
 Akt. 307. If the same letter occur p times, the number of 
 permutations in n letters, taken all together, is 
 
 1X2X3 (w— l)ro 
 
 1X2X3 . . . p 
 
 Suppose these p letters to be all different. Then for any par- 
 ticular position of the other letters, these p quantities, taken p 
 together, will form (1x2x3 . . . p) permutations from their 
 interchange with each other ; and when these letters are alike, 
 these permutations are all reduced to one. And as this is true for 
 every position of the other letters, there will be altogether 
 (1X2x3 . . . ^) times fewer permutations when they arealike 
 than when they are all different. 
 
 Thus, in the letters A, I, D, there are 1x2x3=6 permuta- 
 tions taken all together, but if I becomes D, then three of these 
 permutations become identical with the remaining three, and the 
 whole number of permutations of the letters ADD taken all 
 
 together, is 1X2X3_3 
 
 1X2 
 
 Art. S07a, Corollary. In like manner, if the same letter oc- 
 cur p times, another letter q times, a third letter r times, and bo 
 on, the number of permutations taken all together, is 
 
 1X2X3 (iir-l )n 
 
 (1X2 . . p)(lX2 . . 9X1X2 . r)X.&c- 
 
PERMUTATIONS AND COMBINATIONS. 281 
 
 For by the last article, if p letters be alike, there will be 
 (1X2X3 ■ ■ ■ p) fewer permutations than when they are all differ- 
 ent ; also if q other letters be alike, but different from the first, 
 there will be (1X^X3 . . q) times fewer permutations, and so 
 on; hence, there will be altogether (1X2X3 . . .jp)(lX2 
 X3 . . . q), &c., times fewer permutations than when the letters 
 are all different, and consequently the general expression will be 
 as announced. 
 
 Art. 308. Combinations. — The Combinations of quantities 
 are the different collections that can be formed out of them, with- 
 out reference to the order in which they are placed. Thus, ai, ac, 
 be, are the combinations of the letters o, b, c, taken two together ; 
 ab and ha, though different permutations, forming the same com- 
 bination. 
 
 Pkofosition. — To find the number of combinations that can be 
 formed out of n letters, taken singly, taken two together, three together, 
 and r together. 
 
 Let C, denote the number of combinations of n things taken 
 singly ; C^ the number of combinations taken two together, 
 and C, the number of combinations taken r together. 
 
 The number of combinations of n letters taken singly is evi- 
 dently n ; that is, 
 
 C,=?!. 
 
 The number of permutations of n letters, taken two together, is 
 n{n — 1) ; but each combination, as ai, admits of (1X2) permuta- 
 tions, ab, ba ; therefore there are (1x2) times as many permuta- 
 tions as combinations. Hence, 
 
 1X2 • 
 
 - Again, in n letters taken three together, the number of permu- 
 tations is n{n — l)(ra — 2) ; but each combination of three letters, 
 as abc, admits of 1x2x3 permutations; therefore, there are 
 1X2X3 times as many permutations as combinations. Hence, 
 
 p _ TOrro— l)(n— 2) 
 ° 1X2X3 • 
 
 And in the same manner it appears that in n letters, the num- 
 6er of combinations, each of which contains r of them, is 
 
 ^_ n(re— l)(m— 2) . . . [w— (r— 1)] 
 1X2X3 r 
 
282 RAY'S ALGEBRA, PART SECOND 
 
 Ex. The number of combinations of 5 letters, taken three 
 
 together, is £21i21?=10. 
 
 1X2X3 
 
 Aht. 309. The number of combinations of n thinffs iahen r to- 
 gether, is the same as the number of combinations of n things taken 
 n — r together. 
 
 The truth of this proposition is evident from the following con- 
 eideration : if out of n things r be taken, (n — r) things will al- 
 ways be left ; and for every different parcel containing r things, 
 there will be a different one left containing (n — r) ; therefore, the 
 number of parcels containing r things, must be equal to the num- 
 ber containing (n — r). 
 
 For example, in the letters abcde, for each combination of three 
 letters, there is a different one of two letters. Thus, 
 
 ahc, abd, abe, acd, ace, ade, bed, bee, bde, cde. 
 de, ce, cd, be, bd, be, ae, ad, ac, ab. 
 Hence, in finding the number of combinations taken r together, 
 when r'^ln, the shorter method is to find the number taken 
 (n — r) together. 
 
 EXAMPLES FOR PRACTICE. 
 
 1 . How many permutations of two letters each, can be formed 
 out of the letters a, b, c, a!, e 1 How many of three t How many 
 of four ? Ans. (1) 20. (2) 60. (3) 120. 
 
 2. How many combinations of two letters each, can be formed 
 out of the letters a, h, c,d,e'i How many of three 1 How many 
 of four 1 How many of five 1 
 
 Ans. (1)10. (2)10. (3)5. (4)1. 
 
 3 . In how many ways, taken all together, may the letters in 
 the word NOT be written ?. In the word HOME. 
 
 Ans. 6, and 24. 
 
 4. How often can 6 persons change their places at dinner so 
 as not to sit twice in the same order ■! Ans. 720. 
 
 5. In how many different ways, taken all together, can the 
 seven prismatic colors be arranged ? Ans. 5040. 
 
 6. In how many different ways can six letters be arranged 
 when taken singly, two by two, three by three, and so on, til 
 they are all taken I Ans. 1956. 
 
 Suggestion. — Take the sum of the different peimutatioos. 
 
PERMUTATIONS >ND COMBINATIONS. 263 
 
 7. How many different products can be formed with any two 
 of the figures 3, 4, 5, 61 Am. 6. 
 
 8. How many different products can be formed with any three 
 of the figures 1, 3, 5, 7, 91 Ans. 10. 
 
 9. The number of permutations of n things taken four to- 
 gether = six times the number taken three together ; And n. 
 
 Ans. 71=9. 
 
 10. The number of permutations of 15 things taken r together 
 = ten times the number taken (r — 1) together ; find r. 
 
 Ans. r=6. 
 
 11. How many different sums of money can be formed with a 
 cent, a three cent piece, a half dime, and a dime"? 
 
 SnooESTioN. — Take the sum of the different combinations of foui 
 things taken singly, two together, three together, and four together. 
 
 Ans. 15. 
 
 12. With the addition of a twenty-five cent piece, and a half 
 dollar, to the coins in the last example, how many different sums 
 of money may be formed ? Ans. 63. . 
 
 13 . At an election, where every voter may vote for any num- 
 ber of candidates not greater than the number to be elected, there 
 are 4 candidates and only 3 persons to be chosen ; in how many 
 ways may a man vote ? Ans. 14. 
 
 14. Of the combinations of 5 letters, a, b, c, d, e, taken three 
 together, in how many will a occur 1 
 
 Suggestion. — First find the combinations of /our letters taken tux 
 together 7 
 
 Ans. 6. 
 
 15. On how many nights may a different guard be posted of 4 
 men out of 161 and on how many of these will any particulai 
 man be on guard 1 Ans. 1820, and 455. 
 
 16. The number of combinations of n quantities /our together, 
 is to the number two together, as 15 to 2; find n. Ans. ra=12. 
 
 17. How many changes may be rung with 5 bells out of 8, and 
 how many with the whole peal 1 Ans. 6720, and 40320. 
 
 18. Find the number of permutations taken all together, that 
 can be made out of the letters of the word Algebra. (See Art 
 307.). Ans. 2520. 
 
 19. In how many ways can we write the term o'6*c' 1 
 SnaaiEBTiON. — There are 3 a'a 4 b'a, and 3 c's. (See Art. 307a.) 
 
 Atu. 1260. 
 
264 RAY'S ALGEBRA, PART SECOND. 
 
 20. In how many terms in the preceding example, will a' stand 
 first? 
 
 Suggestion. — The number will be equal to the permutations taken all 
 together, of the letters in b*c^. 
 
 Ans. 15. 
 
 21. In the permutations formed out of a, b, c, d, e,f, g, talcen 
 all together, how many begin with db ? How many with abc 1 
 [low many with abed 1 Ans. (1)120. (.2)24. (3)6. 
 
 22. Out of 17 consonants and 5 vowels, how many words can 
 be formed, having two consonants and one vowel in each 1 
 
 Ans. 4080. 
 
 23. Find the number of combinations that can be formed out 
 of the letters of the word " Nbialion," taken 3 together. 
 
 Ans. 22. 
 
 BINOMIAL THEOREM, 
 WHEN THE EXPONENT IS A POSITIVE INTEGER. 
 
 Akt. 310. We have already explained (Art. 172) the method 
 of finding any power of a binomial, by repeated multiphcation ; 
 and we shaH now proceed from the theory of Combinations (Art. 
 308), to derive a general rule, which is called the Binomial Theo- 
 rem, and sometimes Sir Isaac Newton's Theorem, from the name 
 of the inventor. 
 
 In its most general form the Binomial Theorem teaches the 
 method of developing into a series any binomial whose index is 
 either integral or fractional, positive or negative ; that is, quanti- 
 ties of the form 
 
 (a+x)», (o+x)-", (o+x)^, (a+x)-^, 
 where a or x may be either plus or minus. 
 
 The following Investigation applies only to the case where tne 
 exponent is positive and integral, the other cases will be considered 
 hereafter. (See Art. 319.) 
 By actual multiplication it appears that 
 
 (x-t-a)(x+&)=a;'+a|a:-t-o6. 
 
 In like manner {x-\-a)(x-\-b)(x-\-c) 
 
 ^x^-\-a 
 +6 
 
 x'-^-ab 
 -\-ac 
 -\-bc 
 
 x-\-abc. 
 
BINOMIAL THEOREM. 
 
 265 
 
 Also, (_x-\-a)(,3C-\-b)(x-{-c)(x-\-d) 
 
 
 x^-{- ah 
 -(- ac 
 -{-ad 
 + bc 
 + bd 
 + cd 
 
 x^-\-abc 
 -{-abd 
 -\-acd 
 -{-bed 
 
 x-{-abcd. 
 
 An examination of either of these products, shows that it is com- 
 posed of a series of descending powers of a;, and of certain coeffi- 
 cients, formed according to the following law : 
 
 let. The exponent of the highest power of x is the same as the 
 number of binomial factors, and the other exponents of x decreasB 
 by 1 in each succeeding term. 
 
 2nd. The coefficieni of the first term is 1 ; of the second, the 
 sum of til:' quantities a, b, c, &c.; of the third, the sum of the pro- 
 ducts of every two of the quantities a, b, c, &,c. ; of the fourth, the 
 sum of the products of every three, and so on ; and of the last, 
 the product of all the n quantities a, b, c, &c. 
 
 Suppose, then, this law to hold for the product of n binomial 
 
 factors x-{-a, x-\-b, x-\-c, x-{-k ; so that (x-(-a)(a;+J) 
 
 (s+c) (a;+i)=x»+Aic"-'+Ba^-2+Cx»-3+ . . . +K, 
 
 where A=a-\-b-{-c-{- .... -\-k. 
 
 B^=ab-\-ac-\-ad-{- 
 
 C=aic-\-abd-\- 
 
 &c. = &c 
 
 K=abcd k. 
 
 If we multiply both sides of this equation by a new factor x-f-J, 
 we have 
 
 {x-{-d)(x+b)(x-\-c) (x+Q(x+0 
 
 — a?'+i+A|ar"-|-B|x"-'+Cla;»- 
 + l\ +AZ| BZl 
 
 Here 
 
 +kl. 
 
 A+Z =ffl-fj+c4- .... +i-f-Z; 
 B-l-At=a6+oc+ad . . . -i-al-{-bl 
 &c. == &c 
 
 Kl =abcd . . . . kl. 
 
 [t is evident the same law still holds ; that is, 
 
 Ist. The exponent of the highest power of x is the same as the 
 ^ui/mber of binomial factors ; and the other exponents of x decrease 
 by 1 in each succeeding term. 
 23 
 
266 RAY'S ALGEBRA, PART SECOND. 
 
 2nd. The coefficient of the first term is 1 . 
 
 A-\-l, the coefficient of the second term, is the sum of the sec 
 ond terms, a,b,c, . . . . k, and I of the binomial factors. 
 
 B-j-AZ, the coefficient of the third term, is the sum of the pro. 
 ducts of the second terms of the binomial factors taken two to- 
 gether ; for by hypothesis, B is the sum of the products of n 
 binomial factors, taken two together, and Al is the product of the 
 second terms of the preceding n binomials by the second term ! 
 of the new binomial ; therefore, B+AZ is the product of the 
 second terms of all the binomial factors taken two together. 
 
 Kl, the last term, is the product of aU the second terms of the 
 n+l binomial factors. 
 
 Hence, if the law holds when n binomial factors are multiplied 
 together, it will hold when n-\-l factors are multiplied together ; 
 but it has been shown by actual multiplication to hold up to 4 
 factors ; therefore it is true for 4-|-l, that is 5; and if for 5, then 
 also for 5-|-l , that is 6 ; and so on generally, for any numbei 
 whatever. 
 
 Now let b, c, d, &c., each =a, 
 
 then A=a+a-)-a+<H~> ^^-t to n terms =na. 
 
 B=a'-\-a'-\-&c., =a' taken as many times as i - , .. , 
 
 is equal to the No. of combinations of n things^ = — * ^ — , 
 taken two together, which is (Art. 308), 3 
 
 C=ffl^+a'4-. &-C., =a' taken as many ] 
 
 ames as is equal to the No. of combinations 1 n(ji — 1 )(« — 2)a' 
 
 of the things taken three together, which is [ 1 ■ 2 • 3 
 
 (Art. 308), J 
 
 &c. = &c. 
 K=aaa .... to n factors =a". 
 
 Also, (x-^a)(x-\-b)(x-\-c ) (ir+Z) becomes 
 
 (x-\-a)(x-\-a)(x-{-a) (a:+a)=(a;-l-a)". 
 
 ^ ^ 1-2. ^l-2-3 
 
 + +1". 
 
 By changing a; to a and a to oc, we have 
 
 (a-|-a;)"=a"+Ka'-'a;+^^^:=:Da"-V+ 
 1-2 
 
 ;,(;^l)(^-2) 
 
 1-2-3 ^^' 
 
BINOMIAL THEOREM. 267 
 
 Let a=l , tlien since every power of 1 is 1 , 
 
 Cor. 1 . — It is obvious that the sum of the exponents of a and x 
 in each term =n. 
 
 Cor. 2. — If either term of the binomial is negative, every odd 
 power of that term will be negative (Art. 193) ; therefore the 
 igns of the terms in which the odd powers are found will be 
 negative. 
 
 ... (i-^)»=i-^+'^^!^)x'-<!^=D^?=±)x»+,&c. 
 
 ■' ^1-2 1.3.3 ^ 
 
 > 
 
 Cor. 3. — The general term of the series is 
 
 n(.7v-l)(.n^2) (a_r+2) ^„__^ ,^, 
 
 1 • 2 ■ 3 (r— 1) 
 
 For the 1st term is a", 
 
 2nd " " 110."-% 
 
 3rd " « "'(^^'^) „n-2^i • 
 
 1-2 
 
 4th " " n(„— l)(n-2) ^„.,^ 
 
 1.2.3 
 &c., &c. 
 
 Here it is evident the coefBcient of any term is formed of the 
 
 product of the factors -, , ", &c., in number, one less 
 
 than the number which denotes the place of the term ; therefore, 
 the coefficient of the r'* term will be 
 
 n(,n—l)(n—2) [w— (r— 2)] 
 
 1.2-3 (r— 1) 
 
 Also, the exponent of x is the same as tlie denominator of the 
 last factor of the coefficient ; and the exponent of a is equal to n 
 minus the exponent of x, (Cor. 1) ; therefore, the whole r" term 
 is, 
 
 «(re— l)(>t— 2) (ro— r+2) ^„_,^ ,^, 
 
 1 • 2 • 3 (r— 1) 
 
 This is called tTie general term, because by making r=2, 3, 4 
 &c., all the others can be deduced from it. 
 
 Ex. Reouired the 5'* term of (o — xY 
 
i^^ RAY'S ALGEBRA, PART SECOND. 
 
 Here r=5, and n=7; 
 
 .-. term required =!L!Al5_L^(o)sC— j;)'=35a'ar<. 
 ^ 1 • 2 ■ 3 • 4 ^ 
 
 Cor. 4. — If n be a positive integer, and r=n-\-2, then 
 (n- — 7-(-2) becomes 0, and the (n-|-2) term vanishes; therefore, 
 the series consists of (re+1) terms altogether ; that is, in raising 
 a binomial to any given power, the number of terms is one greater 
 than the exponent of the power to which the iinomial is to he raised. 
 
 Cor. 5. — When the index of the binomial is a positive integer, 
 the coefficients of the terms taken in an inverse order from the 
 end of the series, are equal to the coefficients of the correspond- 
 ing terms taken in a direct order from the beginning. 
 
 If we compare the expansion of (a-f-i)" , and (i-f-a)", we have 
 
 .-^ ^ ^1-2 ^ 1 -2 • 3 ^ 
 
 Since the binomials are the same, the series resulting from 
 their expansion must be the same, except that the order of the 
 terms will be inverted. It is clearly seen that the coefficients 
 of the corresponding terms are equal. 
 
 Hence, in expanding a binomial, whose index is a positive inte- 
 ger, the latter half of the expansion may be taken from the first 
 half. 
 
 Ex. Expand (a — 6)'. 
 
 Here the number of terms (n+1) is equal to 6; therefore, it 
 will only be necessary to calculate the coefficients of the first 
 three, thus : 
 
 (a— i)'=a'— 5a<6+5-lio'J2— 10a'i'-|-5aZ.<— J». 
 
 Cor. 6. — The sum of the coefficients of any expanded bino- 
 mial whose index is n, and where both terms are positive, is 
 always equal to 2". 
 
 For if a:=a=l, then (a:+a)"=(l+l)«=2'' 
 ^r+T^2-+ 1.2-3 + 1-2-3-4 '^' 
 
BINOMIAL THEOREM. 26sl 
 
 Thus, the coefficients of 
 
 a-\-x =1+1=2=2', 
 (a+a:)='=l+24-l=4=2', 
 (a+xy=l +3+3 +1 =8=2», 
 (a+a;)''=l+4+6+4+l=16=2<, 
 &c., &c. 
 
 Abt. Sll. In the application of the Binomial Theorem, it ii 
 convenient to observe, that if the coefficient of any term be multi- 
 plied by the exponent of the first Utter of the binomial in that term, 
 aitd the product be divided by the number of the term, the quotient 
 win be the coefficient of the next term. Thus, in raising a — x to the 
 7'* power, the terms without the coefficients, are 
 
 a', a°x, oV, a*a^, a'x*, a'x^, ax', a;' ; 
 and the coefficients are 
 
 , 1X7 7X6 21X5 35X4 35x3 21X2 7x1 
 ' -[ 2-' -3-' -^' -5-' -&-' -!-■ 
 
 And since the signs of the terms are alternately plus and 
 minus, (Art. 310, Cor. 2), we have 
 
 (a-^r)'=o'—7a«a;+21(i'a:2—35a''a;'+35aSa;<— 210=1' 
 +7aar«— a'. 
 
 Aet. 312. If the terms of the given binomial are affected 
 with coefficients, or exponents, they must be raised to the re- 
 quired powers, by the rule for the involution of monomials (Art. 
 172). Thus, 
 
 (2a'— 3J»)''=(2a')'— ^(2a')'(3i»)+i^(2a»)>(3J»)», 
 
 — ^ ' ^ ' ^ (2a=)(3y)'+^ • 3 • 2 • 1 ^g J , 
 1 • 2 • 3 ' -^ ^1 • 2 ■ 3 • 4 ' ' 
 
 =16o»— 4x8o8x3i'+6X4a''X9i«— 4x2a'X276»+81J" 
 ,=16a«— 96a«J»+216a<i«— 216aW+81J". 
 
 Aet. sis. By means of the Binomial Theorem we can raise 
 any polynomial to any power. Thus, let it be required to raise 
 — 6+c to the third power. 
 
 Let a^-^=m, then (a— 6+c)>=(m+c)'=m«+3m=c+3j7M;»+<^ 
 Substituting for m its equal a — b, we find 
 
 (a— J+o)»=fa— i)3+3(ff— t)'c+3(a— 5)c»-l-c'. 
 
270 RAY'S ALGEBRA., PART SECOND. 
 
 Developing the powers of a — J, and performing the operations 
 indicated, we finally obtain 
 
 — 3ic'-f-c». 
 
 KXAMFLES FOR PRACTICE. 
 
 1. Expand (a+J)«, (a— i)', (2x—Syy, and (5— Ax)*. 
 
 (1) Ans. o»-l-8a'64-28a66»-|-56a'i3_^70o''6'+56o'J'+28a%« 
 
 (2) Ans. o'— 7o«J4-21a'i2_35ai6J4-35a'M— 21o'i'+7o4« 
 
 —6'. 
 
 (3) Ans. Z1i3f—2i0x*y+T20x'y''—1080xy-i-810xy* 
 
 — 243y'. 
 
 (4) Ans. 625— 2000a;+2400j;'— 1280*'+256i<. 
 
 2. Required the coefficient of x^ in the expansion of (i+i/)". 
 
 Ans. 210. 
 
 3. Find the 5" term of the expansion of (c^ — d')". 
 
 Ans. 495c'« d». 
 
 6o("3ESTioN. — (See Cor. 3; Art. 310.) Instead of a, x, n, and r, sub- 
 MHUte c2, — (P, 12, and 5. 
 
 4 Find the 7" term of (a»4-3a6)». Ans. 61236a"i«. 
 
 5. Find the 5" term of (So'— 7a:')». Ans. 13613670aSx"'. 
 
 6 Find the 6'* term of (oi+iy)". Ans. 25'ia^bVy'. 
 
 7. Find the middle term of (a'"+x'')'^ Ans. 924o«'"a;«". 
 
 8. Find the two middle terms of {a-\-x)". 
 
 Ans. 1716a''a«, and 1716a»x'. 
 
 9. Find the 8'" term of (1+a:)". Ans. 330x'. 
 10. Find the 6" term of {x—y)". Aiis. —142506x"yK 
 
 11 Expand (3ao— 2M)'. Ans. 243bV 
 — ''10a<c''6<i+1080a'c'6'J'— 720a'cWds+240acJ<d'— 32i'<i'. 
 
 12 Expand (a+2i— c)'. Alls. o'+6o'J+12ai'+8i» 
 
 — 3a5(^— 1 2a4c— 1 2i'c+3ac'+6ic'— ««. 
 
 13. Prove that the sum of the coefficients of the odd terms of 
 {a-|-*)", is equal to the sum of the coefficie.ats of the even terms. 
 
INDETERMINATE COEFFICIENTS. 271 
 
 CHAPTER X. 
 
 ISDKTBRMINATE COBITICIENTS: BiNOMlAL THE- 
 OREM, General Demonstration: Summa- 
 tion AND Interpolation of Series. 
 
 INDETERMINATE COEFFICIENTS. 
 
 Art. 314. The method of developing algebraic expressiona 
 into seriss, by assuming a series with unknown coefficients, and 
 then, by equating the coefficients of the like powers of x, finding 
 the values of the assumed coefficients, is termed the method of 
 Indeterminate Coefficients. It depends on the following 
 
 theorem. 
 
 If A+Ba;+Cx'+Da;'4-, &c., =A'+B'x+C'a;=+DV+, &g., 
 for every possible value of a; (A,B, A', B', &c., not containing ar); 
 then shall A=A', B=B', C=C', &c. ; that is, the coefficients of 
 the terms involving the same powers of x in the two series, are respect- 
 ively equal. 
 
 For, by ti-ansposing all the terms into the first member, we 
 have A— A'+(B— B')a;4-(C— C')a;2-|-(D— D')a:'+, &c., =0. 
 
 If A — A' is not equal to 0, let it be equal to some quantity p ; 
 then we have (S—'B')x+{C—C')x^-{-(T)—D')x^+, &c., =—p. 
 
 Now since A and A' are constant quantities, their difference, 
 p, must be constant ; but — p=(B — B')a;-|-(C — C')a;'-{-, &c., a 
 quantity which may evidently have various values, depending on 
 the different values of the variable x ; therefore, p must be varia- 
 ble ; that is, we have proved the same quantity (p) to be both 
 fixed and variable, which is impossible. Therefore, there is no 
 possible quantity (p) which can express the difTerence A — A' ; or 
 [a other words 
 
 A— A'=0 .-. A=A'. 
 
 Hence (B— B')j;+(C— C')a;'+(D— D')a;'+, <fcc., =0. 
 By dividing each side by x, we have 
 
 B— B'+(C— C')x+(D— D')a;=+, &c., =0. 
 
272 RAY'S ALGEBRA, PART SECOND. 
 
 By a process of reasoning, exactly similar to that used in the 
 case of A — A', we may show that B=B'. And so on for the 
 remaining coefficients of the like powers of x. 
 
 Coi — If we have an equation of the form 
 
 A-|-Ba;+Ca:=-{-Da:'+Ea;<+, &c., =0, which is true 
 [or any value whatever of x, then A=0, B=0, C=0, &c. ; that is, 
 lack coefficient is separately equal to zero. 
 
 For the right hand member may evidently be put under tlie 
 form 0-\-0x-\-0x'-\-0x'-\-, &c. ; then comparing the coefficients 
 of the like powers of x, we have A^O, B=0, C=0, &c. 
 
 Remark. — As the values of the coefficients assumed are at first un- 
 known, this method might more properly be termed, the method of 
 undetermined coefficients, or the method of unknown coefficients. 
 
 Aut. 315. Let it be required to dcvelope into a aeries 
 
 a-\-bx 
 without a resort to division. 
 
 It is obvious that the series will consist of the powers of x 
 multiplied by certain undetermined coefficients, depending on 
 either a or b, or both of them, and that x may not enter into the 
 
 first term ; therefore, let us assume — —=A-\-'Bx-\-Cx''-{-Dx^ 
 
 a-\-bx 
 +, &c. 
 
 Multiply both sides by the denominator a-\-bx, and arrange the 
 terms according to the powers of a; ; we thus obtain 
 
 a=Aa-\-'Ba\x-\-Ca\x'-^Da\x'-\-, &c. 
 +Ai| -|-Bi| -t-Ci| 
 
 But by the preceding theorem, the coefficients of the same 
 powers of x in each term are equal to each other ; therefore. 
 
 a=Affi ; hence, A=l; 
 
 Bo+A5=0; " B=— -; 
 a 
 
 Ca+B6=0; " C=+-; 
 a' 
 
 Da-|-Ci=0; " D=—l, &c. 
 a' 
 
 Substituting these values of the coefficients in the assumed 
 
 geries, we find 
 
 ■ =1 — -x-\-—x'' — _a;'-(-_a:<, &c., the same as 
 
 a-\-ox a a' a' a* 
 
 would be obtained by actual division. 
 
INDETERMINATE COEFFICIENTS. 273 
 
 RzMARE. — Lest the learner may not see that the left member a, is of 
 the same forrii as the assumed series on the right, it is proper to observe 
 that a is the same as a-\-^x~\-Qx--\-{ix^-\-t &c. 
 
 Aki. 316. A series with indeterminate coefficients, is gener- 
 ally assumed to proceed according to the ascending integral and 
 positive powers of x, beginning with x" ; but in many series this 
 is not the case . the error in the assumption will then be shown, 
 either by an impossible result, or by the coefficients of those 
 terms which do not exist in the actudt series, being found equal to 
 zero. 
 
 Thus, if it be required to develope , and we assume the 
 
 3a: — a;' 
 
 series to be A+Bi+Cx^+Dar'-I-Ea;''-!-, &c., we have, after 
 
 clearing of fractions, 
 
 l=3Ax+(3B— A)x=+(3C— B)a;'-|-, &c. ; 
 from which, by equating the coefficients of the same powers of x, 
 1=0; 
 3A=0, &c. 
 
 But the first equation, 1=0, is absurd, from which we infer 
 that the expression cannot be developed under the assumed form. 
 
 But ~1 — =- X ^— . and if we put J_=A+ Bx+ Cx^+Dx' 
 3x — x' X 3 — x 3 — X 
 
 ■\-, &.C. ; after clearing of fractions, and equating the coefficients 
 
 of the like powers of x, we find A=|, B=j, 0=:^,^, D=:g'y, &c. 
 
 Therefore, -i — =1 ( 1+?+:^+^ +, &c. ) 
 3x— x= x\ 3^9^27 ^81^ / 
 
 — a, I J . d. .J. I ^^ _ 
 3^9 27^81^' ■' 
 
 that is, the development contains a term affected with a mgatim 
 exponent. Hence, at the outset we ought to have assumed 
 
 -i =Ax-'+B+Ox+Dx»+, &c. 
 
 Again, if we assume 
 
 -iz:^^=A+Bx+Cx'+Dx'+Ex<+Fx'+Gx«+, &c. ; 
 
 we shall find the true series to be 
 
 1— 2a?+3x''— 5x«4-8x8— &c., 
 
274 
 
 RAY'S ALGEBRA, PART SECOND. 
 
 the coefficients B, D, F, &c., of the odd powers of a; becoming 
 zero. We might, therefore, have assumed 
 
 ^~^'-<=A+Cx2+Ei<+Gj:»+, &c. 
 
 Aet. 317. Evolution by ihdeteeminate coefficients. 
 
 Ex. Extract the square root of a'-|-x'. 
 
 Assume J(,a^+3c')=A+Bx+Cx^+l)x'+Ex*-\-, &c. 
 
 Squaring both sides so as to obtain quantities of the same 
 form, we find 
 
 aS+a;'=A'+2ABa;+2AC|j;'+2AD|x»+2AE|a;<+, &c. 
 +B^ +2BC| -1-2BD 
 
 From which, by equating the corresponding coefficients, we 
 get 
 
 A'=o2, 2AB=0, 2AC+B2=1, 2AD+2BC=0, &.c. 
 
 From which we find A=a, B=0, C=JL, D=0, E=— i_ 
 &c. 2" 8a»' 
 
 Hence, V(a=+x=)=a+?!-f!^+. &o. 
 
 2a 8a' 
 
 Akt. 31§. Decomposition op rational fractions. — Frac- 
 tions whose denominators can be separated into certain factors, 
 may often be decomposed into other fractions whose denomina- 
 tors shall consist of one or more of these factors. We shall 
 illustrate the method of operation by an example. 
 
 Decompose — — — into two other fractions whose denom 
 
 x'' — Bx-(-8 
 
 inators shall be the factors of x' — 6a:-|-8. 
 
 Since x'— 6a; + 8=(a;— 2)(a:— 4), (Art. 234, Prop. 2nd), 
 assume 
 
 5x— 14 A , B 
 
 x'— 6x+8 X— 2 X— 4" 
 Reducing the fractions to a common denominator, 
 
 behave ^^=±4 ^ A(x-4)+B(x-2) 
 
 x'— 6x+8 (x— 2)(x— 4) 
 
 or 5x— 14==A(x— 4)+B(x— 2)=(A+B)x--1A— 2B. 
 
 Now since this equation is true for any value whatever of x, 
 
INDETERMINATE COEFFICIENTS. 275 
 
 we may equate the coefficients of the corresponding terms, 
 ^A.rt. 314) ; this gives 
 
 A+B=5; — 4A— 2B=— 14; whence, A=2, and S=3. 
 
 . 5x—U _^ 2 , 3 
 
 x^—Sx-^ 8 x—2 x—i' 
 
 EXAMPLES FOR PRACTICE. 
 
 By the method of Indeterminate Coefficients show that 
 
 1. I±|f'=l-f5a;+15x'+45a;»+, &c. 
 1 — cx 
 
 j_ — — ' 
 
 3. 1— 3J^+2j'_i_4j._L5j.i_^3_4a;4 t &c, 
 l^x + x' ^^ ^ 
 
 4. 3_+?f=3_11^7_ai^,_7-ll^ 7-11^ 
 5+7a! 5 5» ^ 5» S" ^ 5» ^ 
 
 5. J±i-=:l'+2'a>4-32a:'-|-42aJ+5V-f,&c. 
 (l-^)» 
 
 c ,i — -_, X x" 3a;' 3 ■ 6x* . „ 
 
 o. x/1 — x=l — - — — — — ,&c. 
 
 ^ 2 2-4 2-4-6 2-4-6-8 
 
 7. V(l+*+i')=l+|+^-5J+, &c. 
 
 8. V(l+i4.x'+x'+, &c.) =l+^+_l_i=+-^^x'+. 
 
 9. l±^=i+_?_ 
 a; — x' X 1 — ^i' 
 
 jQ 8x-4_ 5 , 3 
 
 x'' — 4 a;-|-2 a- •2' 
 U x+1 _ 5 _ 4 
 
 12. 
 13. 
 
 a;2_7a;4-12 a;— 4 a;— 3' 
 
 x' -^ 4 _ 1 I 1 
 
 (x2_l)(x— 2) 3(1—2) 2(x— l)"''6(x+l)" 
 
 1111 
 
 x^—a* ia\x—a) 4ffl5(x+B) 2a2(x=+a')' 
 14 1 -1 5 1 _ 1 ■!- ^-2 _ ^+2 \ 
 
 ' X6— 1 6 'x— 1 X+1 X=— X+1 X2+X+1V 
 
270 RAY'S ALGEBRA, PART SECOND. 
 
 BINOMIAL THEOREM, 
 
 WHEN THE EXPONENT IS FRACTIONAL OB 
 NEGATIVE. 
 
 Art. S19. We shall now proceed to prove the truth of the 
 Binomial Theorem generally ; that is, to show that 
 
 whether n be integral or fractional, positive or negative. 
 First, a+b=a ( 1+- ) : 
 
 .-. .(o-l-6)-=a° ( 1+- ) "=a\\-\-xY, if x=-, 
 \ a I a 
 
 Hence, if we can find the law of the expansion of (l+a?)", we 
 
 may obtain that of (a-J-i)", by writing - for x, and multiplying 
 
 a 
 
 by o". We shall, therefore, first prove thai 
 
 The proof may be divided into two parts : 
 1st. To show that (l+a;)"=l+rea:+, &c. 
 2nd. To find the general law of the coefficients. 
 
 First. To prove that the coefficient of the second term of the 
 expansion of (l-j-a;)" is n, whether n be integral or fractional, 
 positive or negative. 
 
 Let the index be positive and integral ; then, since by multipli- 
 cation we know that 
 
 (l+a;)2=l+2x+, &c., 
 (1+^)5=1 +3a:+, &c. ; 
 let us assume that (l-l-a;)"'''=l+(w — 1)^+) ^• 
 Multiply both sides of this equality by 1+a;, then 
 (14-i)->(l+a;)=ll-f (n-l)a:4-, &,c.)J(l+a) ; 
 or, Q.-\-xy=\-\-nx-\-, &c., by multiplication. 
 
 Hence, if the proposition is true for any one index » — 1, U 
 it will be true for the next higher index n. Now, by multiplica- 
 tion, it is true for the index 3, It is therefore true for the index 
 3+1=4; and therefore true for the index 4-|-l=5, and so on. 
 
BINOMIAL THEOREM. 277 
 
 Hence, by continued induction, it is always true for n when it ia 
 integral and positive. 
 
 Next let ?i be a fraction =? 
 ? 
 
 Also let (l-^x)<i=l-{-ax-\-,S[,c., =1+Aa;, where Ax is put 
 lo represent all the terms after the first. 
 p 
 Since (l+ar)f=l+Aa!, .. by raising both sides 
 
 to the q power (l+x)'^(l-|-Aa;)' ; 
 
 .-. l+pa+j &c., =l-|-9Aa;-|-, &c., 
 
 =l-\-q(ax-{-, &c.,)-|-, &c., 
 =:l-\-qax-\-, &c. 
 By equating the coefficients of the like powers of a; (Art. 314), 
 
 p=:qa .-. a=?, 
 9 
 
 and (1 -^-x^q =1 +£e-(-, &c. 
 
 ? 
 Lastly, let n be negative ; then, (Art. 81), 
 
 (1 +«)-"=— — — -=— — =1 — nx+, &c., by division. 
 
 (l+a;)" l+nx+,&c. ^ ' 
 
 •. (l-f-a!)"=l-}-nj;+, &c., whatever be the value of n 
 
 . ((H-6)"=a" ( 1+- ) "=fl"(l+»-+, &c.), 
 \ a / a 
 
 and the first two terms of the series are determined. 
 Second. To find the general law of the coefficients. 
 
 Let (l+x)''=l+nx+Bx'+Ci^+T)x*-{-, &c., where B, C, D, 
 iStc., depend upon n. 
 For X, put x-\-z, and consider (a-J-z) as one term, then 
 I l+(,x+z)}"=l+n(,x+z)+B(,x+zy+C(x-{-zy+, &c. 
 But (a+i)"=ffl"+no"-ii+, &c. ; 
 .*. {x-\-zy=x'-\-2xz-\-, &c. ; 
 (a;+2)'=a;54-3a;'z+, &c. ; 
 (x-^zy=x^-\-4:X^z-\-, &o. ; 
 •. ll+(,x+z)\"=l+nx+Bx^+Cx>+J)x*+, &c., 
 +(n+2Bx+3Cx2+4Da:'+, &c.)z+, &c., 
 =(l+a;)»+(re+2Ba;+3Cx2-|-4Z)a;'+,&c.)z+, &c. (A"* 
 
278 RAY'S ALGEBRA, PART SECOND. 
 
 But, considering (1+a;) as one term, (l-|-a;+^)"= J (l+i)+z|"; 
 and J(l+a:)+2|"=(l+j;j''+?!(l+x;''-i2+, &.C. (BV 
 
 Equating the coefficients of z in (A) and (B), 
 
 M+2Ba:+3Ca;=+4Da:'+, &c., =7j(l+x)n ' ; 
 
 multiplying both sides by l-\-x, we have 
 
 n+2Bi+3Ca:=+4Dx'+, &.c.) =re(l-|-a;)» 
 + nx-\-2Bir'-\-SCx^+,&.c.l =n(l-\-ra+Bx'+Cx'-^, Sic). 
 
 By equating the coefficients of the same powers of a:, we hava 
 2B+n=n' .-. 2B=n2-^=n(n— 1). 
 j_ra(n— 1) _ 
 1-2 ' 
 3C+2B=Bn .-. 3C=B(n— 2), 
 P^ B(7i^2) ^ m(TO— 1)(7^-2) . 
 3 1 -2 -3 ' 
 
 also, 4D+3C=nC .. 4D=C(n— 3) ; 
 
 r>=2("~^ )^ "("— 1 )(»— 2)(?i— 3 ) 
 4 1 -2 -3 -4 
 
 Similarly 5E=D(n — 4) ; 
 
 y^^DCro— 4)^ w(n— 1 )(w— 2)(n— 3)(ft— 4) 
 5 1-2-3-4-5 
 
 Hence, generally, if N is any coefficient, M the one which next 
 precedes it, and r— 1 the largest factor in the denominator of M, 
 we have ivr= M? 'J-Cr-l ) | _ M(n+l -r) . 
 
 r r 
 
 1-2 ^ l-2-3-^ 
 and .-. putting _ for x, (a+i)"=a" ( 1+- ) ", 
 
 ^ ^ et 1-2 a^^ 1 ■ 2 • 3 a?^ ' ' 
 
 1-2 ^ 1 • 2 • 3 " -r. "'i- 
 
 If — h be put for 6, then since the odd powers of — J are nega- 
 tive (Art. 193) and the even powers positive, 
 
 1-2 1 ■ 2 • 3 T^> •■ 
 
BINOMIAL THEOREM. 279 
 
 which estahlishes the Binomial Theorem in its most general 
 form. 
 
 Remark. — From the preceding formula and demonstrations, corol- 
 laries, similar to those in Art. 310, may be drawn, but it is not neces- 
 sary to repeat them. The following additional proposition is sometimes 
 aseful. 
 
 Art. 330. To find the greatest-term in the expansion sf 
 
 From Cor. 3, Art. 310, we have seen that the r"* term is 
 
 "^"-^^ ^"-"+^V-+'6^-,hence. 
 
 1-2 (r-1) 
 
 from the general law, the (r-(-l)"' term is 
 
 1-2 r 
 
 Therefore, the (r-|-l)'* term is derived from the r'* by multiplying 
 
 the latter by "-^+\j. 
 r a 
 
 While this multiplier is greater than 1, each term must be 
 greater than th.e preceding. Hence, the r** term will be the great- 
 est when n—r+'^.^ ji^^t <1; 
 r a 
 
 or, (n— r+l)J<ar, (Art. 221) ; 
 
 or, r(a-f i)>(ra+l)i, (Art. 219) ; 
 
 or, r>(n+l)4T- (Art. 221). 
 
 Take r, therefore, the first whole number greater than 
 6 
 
 (n-|-l) - , and the r'* term will be the greatest of the series 
 
 If (ra-|-l) is a whole number, then two terms are equal, 
 
 a-\-b 
 
 each of which is greater than any of the other terms. 
 Ex. 1. Find the greatest term in the expansion of (1+|) • 
 Here (n-fl) A_=(L3)_L=62 ; .-. r>2; hence j-=3. 
 
 2. Find the greatest term in the expansion of (l+y'j)*. 
 
 Ans. 2"*. 
 
 3. Find the greatest term in the expansion of (3-|-5a:)', when 
 e^i . Ans. ■")"'. 
 
280 RAY'S ALGEBRA, PART SECOND. 
 
 Cw. 1 . By finding the greatest term of a series, we determine 
 the point at which the series begins to converge; that is, the point 
 from which the terms become less and less. 
 
 Cor. 2. It is also evident that when " ^'' ia Jirst less than 
 
 r 
 
 1 , that the coefficient of the preceding term is the greatest. 
 
 But when "•~^+^ >l, 
 r 
 
 n— r+l>r, (Art. 221), 
 
 or, 2r>n+l , (Art. 219); 
 
 or, r>^, (Art. 221). 
 
 Hence, the whole number next greater than ?lt- , or next less 
 
 than 2ir_-)-l=^ZI_, denotes the term having the greatest 
 
 coefficient. 
 
 If n is an odd integer, there will b(: two coefficients, the 
 ( 'y^ ) , and the ( VUL- j , each greate r than any other. 
 
 Ex. Find the term having the greatest coefficient in the expan- 
 sion of (o-j-J)"' ; and the two terras having the greatest coeffi- 
 cients in the expansion of (x — yy. Ans. 6'*, and 4" and 5". 
 
 Abt. 321. In the application of the Binomial Theorem, it ia 
 merely necessary to take the general formula (a+i)"=a"-l-7io"-'J 
 -|-, &c., and substitute the given quantities instead of the sym- 
 bols to whicli they correspond in the formula, and then reduce 
 each term to its most simple form. 
 
 Ex. 1. Find the expansion of (l-\-xy. 
 Here o=l, b=x, ^=3. 
 
 ... (l+^)*=l+>:r+i^)*'+MM(iz^^+,&C. 
 
 ^2*^2 -4 ^2 •4-6 2-4-6-8 ^ 
 Ex. 2. Develope (1 — x)~^. Here a=l, 6= — x, 7i= — ^, 
 
BINOMIAL THEOREM. 281 
 
 (l-x)-i=l-.(-a;)+tiX±L) 
 
 1 -2 
 
 (,-xy 
 
 In making these developments it will assist the pupil, to recol- 
 lect that every root and every power of 1, is 1. 
 
 Ex. 3. Develope Ja-\-b into a series. 
 
 Since a+i=a ( 1+^ ) , ••• Va+*=Va ( 1+- ) *. 
 
 Here a=l, J=_, n=X. 
 
 a -* 
 
 ' ^a ' ^--^cL 1 • 2 a^^ 1 • 2 • 3 'oT 
 
 _,,, i 1 6' 1-3 V . 
 -^+^-a-2-Ta-'+2-:T^V-' ^''• 
 
 Hence. ^/«T*= J<l+l-^^,+j^- J1+, &c.). 
 
 EXAMPLES FOR PRACTICE. 
 
 1. -i-=(l -*)-'=! +arH-a;»+a'+a<+,&c. 
 1 — ^x 
 
 2. „ ^ ^ =(l-^)-'=l+2a+33;'+4a^+5a<+, &c. 
 
 Qa+a:)-' a a- a' a* 
 
 4. s/TZ:^=l_^_5!_^'_, &c. 
 ^ 3 9 81 
 
 ^ ^ 2a SoV 16a= 128a'^ 
 
 R / 1 \\ X x' 5i? 10a:'' , „ 
 
 D. ut' — a;)'=a — — — — — , &c. 
 
 3a2 9as 81a« 243a" 
 
 7. (l+2a;)i=14-a^ii'+ia:3_6^4_|_, &c. 
 34 
 
282 RAY'S ALGEBRA, PART SECOND. 
 
 ^ 2a 8a' 16o» 128a' 
 
 10. (-'+^)Ml+343-3^+3^6^-.^c0. 
 
 11 . V9=V8+i=2+? . |-^ . J.,+1? . 1.-, &c. 
 
 12. (os-a;0*=a(l— — +;r^^— ^-^^^^ +, &c.). 
 
 ^ ^ ^ 3o' 3 • 6a8 3 • 6 • Qa' 
 
 , „ a' , 21= 2 • 5a;V2 • 5 • Sa:' . . „ 
 
 13. =a+ — + + +. &c. 
 
 . , ,,S ^3a' 3 ■ 6a'^ 3 ■ 6 • 9a»^ 
 
 Aet. 322. To find the approximate roots of numbers by the 
 Binomial Theorem. 
 
 Let N represent any proposed number whose m** root is 
 required, take a such that a" is the nearest perfect w" power to N, 
 so that N=a"±6, h being small compared with a , ai)d + or — , 
 according as N^ or <^o" ; 
 
 then yN=a| 1±-- )» =, by writing -for J in the general 
 \ a" / a" 
 
 formula ; 
 
 aSl±l *— 1 !^f^)±l ^1 2^^(n'-,&c.|. 
 ra ■ o" n ■ 2re V a" / » " Sra ' 3» \ o" / 
 
 Of this series a few terms only, when b is small with regard 
 uo a", will give the required root to a considerable degree of 
 accuracy. 
 
 Ex. Required the approximate cube root of 128. 
 
 Here X/128=y5^+S=5iJl+-^^; 
 
 =5a-i-l l—l HAT-t-i -(—V— I- 
 
 * """S • 125 3 • 3 \ 125 / "*"3' " 9 \ 125 / '•••»' 
 
 "''S' 5' • 3 ■ 5' ■ ■ ■ "''lO' 10' 3 '10' 
 =5-1-0.04—0.000324-0.0000042— . . . 
 =5.0396842. 
 
 Art. 323. In the preceding example, since the series con- 
 tinues to infinity, we obtain only an approximate value for the 
 
BINOMIAL THEOREM. 28-3 
 
 required root, and as the denominators increase more rapidly than 
 the numerators, a few terms only need be taken for practical pur- 
 poses ; still it may be required to find what is the limit in the 
 error occasioned by neglecting the remaining terms of the series. 
 To do this let R be the true root, and as the terms are alternately 
 positive and negative, let 
 
 R =a — b-\-a — d-{-e—f-]-g — h-\-k — 1-\-, &c., and let 
 
 R' =:fl — b-\-c — d-\-e—f, 
 
 R"=za^b-]-o--d-\-e—f-\-g. 
 
 Then since the terms continually decrease, a — b, c — d, e—f, 
 g — A, &c., are all positive, and therefore R', which contains three 
 only of those differences, will be less than R. For the same rea- 
 son all the pairs of terms after g, aa — h-^-k, — l-\-m, &ic., will be 
 all negative, and R" will be greater than R ; therefore, the true 
 value of the series lies between R' and R", or 
 
 a — b-\-c — d-\-e — -f, 
 and a — b-\-c — d-\-e—f-\-g. 
 
 Hence, the error committed by the omission of any number of the 
 terms of a converging series, is less than the first term of the omitted 
 part of the series. 
 
 Thus, in the preceding example, if we had stopped at the sec ■ 
 ond term, the error would have been less than .0000042. 
 
 15. Find the 5'* root of 35. Ans. 2.0361724-. 
 Here N=35=32+3=25 ( 1+^^ ) . 
 
 16. The student may solve the following examples : 
 
 (1). VIO =>/9+l =3.16227 
 =3.10723 
 =2.88449 
 =4.01553 
 
 (5). iyi08=»/128— 20=1.95204 
 
 (2). 
 
 s/30 =lJ21-\-Z 
 
 (3). 
 
 s/24 =V27— 3 
 
 (4). 
 
 */260=V256+4 
 
 true to 0.00001. 
 true to 0.00001. 
 true to 0.00001. 
 true to 0.00001. 
 true to 0.00001. 
 
 Remark. — Instead of extracting the nth root by the formula in Art 
 392, the operation may be performed by the general formula of the pre- 
 ceding article, the number whose root is to be extracted being divided 
 into any two parts whatever. The advantage of the formula in Art. 
 322 consists in the rapid convergence of its terms. Thus in finding ths 
 4th root of 260 true to five places of decimals, it is only necessary to 
 take two terms of the series. 
 
284 RAY'S ALGEBRA, PART SEUUCTU- 
 
 THE DIFFERENTIAL METHOD OF SERIES. 
 
 Art. 324. A Series consists of a number of terms, each of 
 which is derived from one or more of the preceding terms, accord- 
 ing to some determinate law. (Art. 134.) 
 
 The use of the differential method is, 1 st, to find the successive 
 differences of the terms of a series ; 2nd, any particular term of 
 the series ; or, 3rd, the sum of a finite number of its terms. 
 
 If, in any series, we take the first term from the second, the 
 second from the third, the third from the fourtli, and so on, the 
 new series thus formed is called the First order of differences. 
 
 If we proceed with this new series in the same manner, we 
 shall obtain another series termed the Second order of differences. 
 
 In a similar manner we find the third, fourth, &c., orders of 
 differences. 
 
 Thus, if we have the series 
 
 1 , 8 , 27 . 64 , 125 , 216, . . 
 the 1st order of differences is 7 , 19 , 37 , 61 , 91 , ... 
 
 " 2nd " " " " 12 , 18 , 24 , 30 , 
 
 « 3rd « " " " ,6,6,6 
 
 Art. 325. Problem 1. — To find the first term of any order of 
 differences. 
 
 Let the series be a, h, c, d, e, 
 
 then the respective orders of differences are, 
 1st order, h — a , c — b , d — c , e — d, .... 
 
 2nd order, c — 2b-\-a , d — 2c+6 , e — 2(J-|-c, 
 
 3rd order, d — Sc-\-Sb — a, e — 3d-j-3c — 6, 
 
 4th order, e — 4(i-|-6c — ib-{-a. 
 
 Here each difference pointed off by commas, though a com- 
 pound quantity, is called a term. Thus the first term in the Ist 
 order is 6 — o ; in the second order c — 2b-^a, &c. 
 
 If we denote the first terms in the Ist, 2nd, 3rd, 4th, &c., 
 orders of differences by D,, Dj, Dj, D4, &c., and invert the 
 order of the letters so that a shall stand first, we have 
 D,=— a+i;" 
 Dj=<i— 26+c ; 
 D3=— a+36— 3c+d; 
 Di=o— 4i+6c-4d-l-e ; 
 &c., =2 . . . . &c. 
 
SERIES — DIFFERENTIAL METHOD. 285 
 
 Here the coefficients of a, h, c, d, &.O., in the n" order of differ- 
 ences are evidently the coefBcients of the terms of a binomial 
 raised to the n"" power ; and their signs are alternately positive 
 and negative ; hence, when n is even, the first term of the ra" 
 order of differences is 
 
 a-^fc+"("-l ),_! '("-! )("-S) d+, &c., and 
 ^1-2 1 • 2 ■ 3 ^ 
 
 -ffi4-ni-!^^^)c+^:^!L=lX?=l)i-, ic, when n is odd. 
 
 Cor. It is evident from the coefficients that when re=l, the 
 value of D„ has only two terms, for then n — 1=0; when ?t=2, 
 this value has only three terms, for then n — 2=0, and so on. 
 
 Ex. 1 . Find the first term of the fourth order of differences of 
 the series 1», 2^, 3', 4=, 5' or 1, 8, 27, 64, &c., . . . 
 
 Here re=4, hence take five terms of the first value of D„, and 
 a=l, 6=8, c=27, 3=64, e=125, and 0^= 
 
 . i-4x8+i><gx27-i><l><gx64+^X^X^X^XI2.5= 
 '^ ^1X2"" 1X2X3 ^1X2X3X4 
 
 1_32+162— 256+125=0, Ans. 
 
 Remark. — It is evident the first term of any particular order of dif- 
 ferences may be found by continued subtraction. It is important, how- 
 ever, that the learner should be acquainted with the general law as 
 expressed in the above series. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Find the first term of the second order of differences of the 
 series 1=, 2% 3', 4=, .... or 1,4,9,16,25. . . . Ans. 2. 
 
 3. What is the first terra of the third order of differences of 
 the series 1,3,6, 10, 15, &c. 1 Ans. 0. 
 
 4. Required the first term of the fifth order of differences of 
 the serifs 1, 3, 3', 3=, 3*, &c. Ans. 32. 
 
 5. Find the first term of the fifth order of differences of the 
 »eries 1, jj 5. 5, x'g. &c. Ans — /j. 
 
 Art. 326. Problem II. — To finA the ra'* term of the series 
 a, h, c, d, ty &c. 
 
 From the preceding article we have seen that 
 
286 RAY'S ALGEBRA, PART SECOND. 
 
 D,= — 0+A ; whence J=a+D, ; 
 
 Dj=o— 2J-|-c; " c=:a+2D,+Dj; 
 
 0,=— a+3i— 3c+<i ; " (i=o+3D,+3Dj+Ds ; 
 D4=a-^6+6c— l(H-e ; " e=a44D,+6Dj+4D5+D4. 
 
 It is evident from inspection that the coefficients of the first 
 terms of the different orders of differences, in the value of any 
 term of the series, as of e the fifth term, are the coefficients of the 
 terms of a binomial involved to a power whose exponent is one 
 less than the number denoting the place of the term ; that is, the 
 coefficients of the n"' term of the series, are the coefiicients of 
 the (n — 1) power of a binomial. Hence, writing n — 1 instead 
 of n, in the coefficients of the n"" power of a-\-b, (Art. 319), the 
 re'* term of the series is 
 
 a4-(;,_l)D^4, ('^lX>^-2) D^^.(n-l)(re-2)(re-3)p^_^^ ^^_ 
 
 1 ' <& ± ' -d ' o 
 
 Ex. 1. Find the 12"* term of the series 1, 3, 6, 10, 15, 21, . . 
 
 1 , 3 , 6 , 10 , 15 
 
 2,3,4 , 5, hence D,=3; 
 
 1,1,1, " D,=l; 
 
 , " D,=0; 
 
 snd the succeeding orders of differences are also evidently Oj 
 hence 12" term 
 
 =<^+(;^_l)D,+(!^)^)D,=l+nx2+ll|l°Xl 
 
 =1+22+55=78, Atis. 
 2. Find the n'» term of the series 2, 6, 12, 20, 30 
 
 2 , 6 , 12 , 20 , 30, . ... . 
 4,6,8 10, hence D,=4; 
 
 2,2,2 « Dj=2; 
 
 0,0, « D,=0; 
 
 hence n'" term =2+(>t— l)4+^*'~^^^"~^^x2=re'+?t. Ans. 
 
 From the formula n^-\-n, or re(re+l\ any term of the series 
 is readily found ; thus the 20"' term =20(20+l)=420. 
 
 It is also evident that the re'* term of a series can be found 
 exactly only when some order of differences is zero. 
 
SERIES — DIFFERENTIAL METHOD. 287 
 
 EXAMPLES FOR FRACTICE. 
 
 3. Find tne 15'* term, and the »'* term of the series 1, 2', 3', 
 4», ... or, 1, 4, 9, 16, ... . Ans. 325, and re». 
 
 4. Find the 12" term of the series 1, 5, 15, 35, 70, 126, &c 
 
 Ans. 1365 
 
 5. Find the re" term of the series 1, 3, 6, 10, &c. 
 
 Ans. "("+^> 
 2 • 
 
 6. Find the ji'* term of the series 1, 4, 10, 20, 35, 56, &c. 
 
 ^^ <7t+l)(«+2) 
 2X3 
 
 7. Find the O** term of the series 2 • 5 • 7, 4 • 7 • 9, 6 • 9 • 11, 
 8-11-13, &c. Atw. 8694. 
 
 8. What is the re'* term of the series 1 X2, 3X4, 5x6, &c. 1 
 
 Ans. 4n' — 2re. 
 
 Abt. 327. Peoblem III. — To find the sum of n terms of the 
 series a, b, c, d, e, &c. 
 
 Assume the series 0, a, a^b, a-\-b-\-c, a-\-b-\-c-\-d, .... 
 
 Subtracting each term from the next succeeding, we have 
 
 a, b, c, d, e, &c., 
 
 which is the series whose sum it is proposed to find. Hence, the 
 sum of n terms of the proposed series, which it is now required 
 to find, is the (re+1)'* term of the assumed series. 
 
 It is evident the re'* order of dififerences in the given series, is 
 equal to the (?i-|-l)'* order in the assumed series. Hence, if we 
 compare the quantities in the assumed series, with those of the 
 formula for finding the re'* term of a series (Art. 326), we have 
 
 for a, 
 re+1 for re, 
 a for D,, 
 Dj for Dj, &c. 
 
 Substituting these values in the formula, we have 0+(re+l — l)a 
 
 (^l-l)(;t-|-l-2) ^ _^_ (re+l-l)(re+l -2)(re+l^) T^ 
 
 1-2 '^ 1 -2 -3 '"^" 
 
288 RAY'S ALGEBRA., PART SECOND. 
 
 or, n(7i-l) n(n-l)(w-2_)p ^ ^,^.^^ 
 
 1-2 '^ 1 • 2 • 3 '^ 
 the sum of n terms of the proposed series. 
 
 Ex. 1. Find the sum of n terms of the odd numbers 1, 3, G 
 
 "9 
 
 Here o=l, Di=2, Dj=0; hence, 
 
 Sum =7ta+"^J^=12D , =n X 1 +"^"~^ ^ x 2=?t+w'— »=«'. 
 
 2. Find the sum of re terms of the series 1^ 2', 3', 4', 5', . . . 
 Here 0=1, D,=3, Dj=2, ©3=0; hence, 
 
 ^1-2 '^l-2-3 ' ^ 2 
 
 , w(7t— 1 )(?t— 2) _ n(7H-l )(2?H-1 ) 
 "^ 3 6 • 
 
 EXAMPLES FOR PRACTICE. 
 
 3. Find the sum of n terms of the series 1+3+6+10+16, 
 
 &c. Ans. "("+l)("+2 ) 
 
 6 
 
 4. Find the sum of 20 terms of the series 3+11+31+69 
 +131, &,c. Ans. 44330. 
 
 5. Find the sum of 20 terms of the series 
 
 1 • 2 • 3+2 • 3 • 4+3 • 4 • 5+, &c. Ans. 53130. 
 
 6. Find the sum of n terms of the series of cube numbers 
 13^2'+3'+,&c. Ans. [|n(n+l)]'. 
 
 7. Find the sum of n terms (Jf the series 1+4+10+20 
 
 +35 An.. "("+^)('^^>("+'^) . 
 
 ^ 1X2X3X4 
 
 8. Find the sum of 25 terms of the series whose n" term is 
 1l=(3?^— 2). Ans. 305825. 
 
 Art. 328. Piling op Cahnok Balls and Shells. 
 
 Balls and shells are usually piled by horizontal courses, either 
 in the form of a pyramid or a wedge ; the base being either an 
 equilateral triangle, or a square, or a rectangle. In the triangle 
 and square, the pile terminates in a single ball, but in the rectan- 
 gle it finishes in a ridge, or single row of balls. 
 
SERIES — PILING OF BALLS. 289 
 
 AsT. 329. To find the number of balls in a triangvlar pile. 
 
 V 
 
 A triangular pile, as V — ABC, is formed of 
 successive horizontal courses of the form of an 
 equilateral triangle, such that the number of 
 balls in the sides of these courses, decreases con- 
 tinnally by unity, from the bottom to the single 
 ball at the top. 
 
 If we commence at the top, the number of balls in the 
 respective courses will be as follows : 
 
 1". 2'^. 3"*. 4". 5". 
 
 «•»• ••••• 
 
 • • ••• e«« •••• 
 
 • e ee 
 
 » » » » ® , 
 
 and 60 on. Hence, the number of balls in the respective courses 
 is 1, 1+2, 1+2+3, 1+2+3+4, 1+2+3+4+5, and so on ; 
 orl, 3 6 10 15 
 
 Hence, to find the number of balls in a triangular pile, is to 
 find the sum of the series 1, 3, 6, 10, 15, &c., to as many terms 
 (») as there are balls in one side of the lowest course. 
 
 By applying the formula (Art. 327) to finding the sum of n 
 terms of the series 1,3, 6, 10, &c., we have o=l, D,=2, 
 Dj=l,and D3=0. 
 
 Hence, the formula ;ia+"'^"~^)D.+<"— ^)("-^)d, becomes 
 1-2 '' 1-2-3 * 
 
 r^ rijnr-l) n(n~lXv^2) ^ n'-W+2n 
 
 2 ^ 2X3 ^ '^ "^6 
 
 _n'+3n'+2«_m(w'+3n+2)_n(7H-l)('H-2) /a\ 
 
 _ _ _ _ _ _. (A) 
 
 Aet. 330. To find Oie number of balls in a sqmre pHe. 
 
 A square pile, as V — EFH, is formed of 
 «;ccessive square horizontal courses, such 
 .hat the number of balls in the sides of 
 these courses, decreases continually by unity, 
 from the bottom to the single ball at the 
 
 '"P- 25 
 
 
 H 
 
290 RAY'S ALGEBRA, PART SECOND. 
 
 If we commence at the top, the number of balls in the 
 respective courses will be as follows : 
 
 l->. ^o-i. 3"'. 4'*. 
 
 -. ••• •••• 
 
 . ** ••• •••a 
 
 • •• ••• •••• 
 
 • ••• 
 
 &ad BO on. Hence, the number of balls in the respective courses 
 is 12, 32, 3^ i", 5=, &c., or 1,4, 9, 16, 25, and so on. There- 
 fore, to find the number of balls in a square pile, is to find the 
 sum of the squares of the natural numbers 1,2,3, &c., to as 
 many terms (n) as there are balls in one side of the lowest 
 course. 
 
 But the sum of the series 1,4,9, 16, &c. (see example 2, 
 page 288), is 
 
 n(,n+l)(2n+l) „ 
 
 6 • ^ ' 
 
 Akt. 331. To find the number of balls in a rectangular pUe. 
 
 A rectangular pile, as EFDBCA, is E A 
 
 formed of successive rectangu- 
 lar courses, such that the number 
 of balls in each of the sides of 
 these courses, decreases contin- 
 uously by unity, from the bottom D B 
 to the single row of balls at the top. 
 
 If we commence at the top, the number of balls in the breadth 
 of the first row is 1, of the second 2, of the third 3, and bo on. 
 Also, if m-\-l denotes the number of balls in the top row, the 
 number in the length of the second row will be m.-\-2, in the third 
 row jn-]-3, and so on. Hence, the number of balls in the 
 respective courses, commencing with the top, will be l(m-|-l), 
 3(m-|-2), 3(ni-|-3), and in the m'* course n(m-\-n'). Therefore, 
 the number of balls (S) in a complete rectangular pile of n 
 courses will be 
 
 S=l(m+l)+2(m-f2)+3(m+3)+ +n(m+n) 
 
 =m(l-f2+3-H . . . -|-m)+(12+22+32-f42+ . . . -fjj'); 
 
 but the sum of n terms of the series in the first parenthesis, 
 
 fArt. 327,) is -1-J—J, and the sum of n terms of the series in 
 
 to 
 
 
SERIES — PILING OF BALLS. 291 
 
 iho second parenthesis has just been found (Art. 330) to be 
 
 ^ "^ -^^ — lt_i ; hence, by substitution, we have 
 6 
 
 2 6 6 
 
 Here m-j-re represents the number of balls in the length of the 
 lowest course. If we put m-|-»=2, we have 3TO-|-2n=3Z — n; 
 substituting this for 37n-{-2n, in the preceding formula, it becomes 
 
 6 
 
 It is evident that the number of courses in a triangular or 
 square pile, is equal to the number of balls in one side of the 
 base course, and in the rectangular pile to the number of balls in 
 the breadth of the base course. 
 
 Aet. SS2. Collecting together the results of the three pre- 
 ceding articles, we have for the number of balls 
 
 in a Triangular pile -n(n-\-l)(n-\-2) (A) ; 
 
 6 
 
 in a Square pile -«(»+! )(2ra-|-l) (B) ; 
 
 6 
 
 in a Rectangular pile _ra(7i-|-l )(3Z — tj+I) (C) 
 
 6 
 
 In formula (A) and CB), n denotes the number of courses, or 
 the number of balls in the base course. In formula (C) n denotes 
 the number of balls in the breadth of the base course, and I the 
 number in the length. 
 
 The number of balls in an incomplete pile is evidently found 
 by subtracting the number in the pile which is wanting at the 
 top, from the whole pile considered as complete. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Find the number of balls in a triangular pile of 15 courses. 
 Here n=15, and substituting this value instead of n in formula 
 
 A, (Art. 332), we have the number 
 
 _ 15(15+l)(15+2 )_ 15xl6Xl7 _B«n Ans 
 2X3 6 
 
 2. Find the number of balls in an incomplete triangular oila 
 of 15 courses, having 31 balls in the upper course. 
 
293 RAY'S ALGEBRA, PART SECOND. 
 
 Here we must first find the number of shot in one side of the ■ 
 upper course. From the illustrations in Art. 329, it is evident 
 that the number of balls in any triangular course, is equal to the 
 sum of the natural numbers 1,2,3, &c., to the number (ji) in one 
 side. Now the sum of the numbers 1, 2, 3, &c., to n, is (Art. 
 
 327) "•("+^) ; hence, ^^^±11=21, or ra'+»=42, from which 
 
 (Art. 231) we find 7i=6, and therefore 5 courses have been 
 
 removed from the pile ; hence, by formula A, (Art. 332), the 
 
 nimiber of balls in the pile considered as complete, is 
 
 20v21^22 
 
 — — — i — =1540, and the number in the pile removed is 
 2X3 ^ 
 
 - ^ ='35 . . the number in the incomplete pile is 1540 — 35 
 2X3 
 
 =1505. 
 
 3. Find the number of balls in a square pile of 15 courses. 
 
 Am. 1240. 
 
 4. Find the number of balls in a rectangular pile, the length 
 and breadth of the base containing 52 and 34 balls respectively. 
 
 Ans. 24395 . 
 
 5. Find the number of balls in an incomplete triangular pile, 
 a side of the base course having 25 balls, and a side of the top 
 13. Atw. 2561. 
 
 6. Find the number of balls in an incomplete triangular pile 
 of 15 courses, having 38 balls in a side of the base. 
 
 Ans. 7580. 
 
 7. Find the number of balls in an incomplete square pile, a 
 side of the base course having 44 balls, and a side of the top 22. 
 
 Ans. 26059. 
 
 8. The number of balls in the base and top courses of a square 
 pile are 1521 and 169 respectively ; how many are in the incom- 
 plete pile. Am. 19890. 
 
 9. The number of balls in a complete rectangular pile of 20 
 tjurses is 6440; how many balls are in its base ? Am. 740. 
 
 10. The number of balls in a triangular pile is to the number 
 \n a square pile having the same number of balls in the side of 
 the base, as 6 to 11 ; required the number in each pile. 
 
 Ans. 816, and 1496. 
 
 11. How many balls are in an incomplete rectangular pile of 
 
INTERPOLATION OF SERIES. 293 
 
 8 courses, having 36 balls in the longer side, and 17 in thf 
 shorter side of the upper course. Ans. 6520. 
 
 Art. S33. Interpolation op Series. 
 
 Each of the various tables employed in the different depart 
 ments of science, may be regarded as the terms of a mathemati 
 cal series. These tables are generally calculated from particulai 
 formulsE, but in many cases the computations are so very laborious, 
 that only certain terms at regular intervals, are calculated, and 
 the intermediate ones are derived from these by a process termed 
 Interpolation. Also, in many investigations values of the quanti- 
 ties in the tables are required, intermediate between those given, 
 or extending beyond them. These, likewise, are determined by 
 Interpolation. 
 
 The principle on which Interpolation is founded is that ex- 
 plained in Art. 326; that is, having certain terms of a series 
 given, to find the n"' term. To do this with entire accuracy, re- 
 quires that we should have such a number of terms of the series 
 given, that we can obtain an order of differences equal to zero. 
 In most cases, however, the differences, D,, Dj, Dj, Sic, do not 
 vanish, but become so small that their omission after Dj, or D,, 
 causes no sensible error in the result, and we obtain what is 
 termed, approximate values of the required quantities. 
 
 Art. 334. When the 3rd order of differences of any given 
 series of quantities vanishes, or becomes very small, then (Art. 
 326) we have the equation — o-|-3i — 3c-|-<i=0, and any of the 
 quantities a, b, c, or d, may be found, when the other three are 
 given. Similarly, if the fourth differences vanish, then 
 
 a— 4i-|-6c— 4d+e=0. 
 
 Ex. Given 2/25=2.92401, ^26=2.96249, »/27=3, 
 5/29=3.07231, to find the cube root of 28. 
 
 Here four quantities are given to find a fifth, therefore, sup- 
 posing the fourth order of differences to vanish, we have 
 
 a — 4li-]-6c — id-\-e=0, where d is the term to be 
 interpolated ; hence, 
 
 4<i=ij-|-6c+e—4i=2.92401+18-f3.07231— 11.84996 
 =12.14636, 
 where of, or ^28=3.03659, which is true to .00001. 
 
294 
 
 RArS ALGEBRA, PART SECOND. 
 
 Art. S35. When the terms are equidistant, and it is required 
 to interpolate a term intermediate to any two of them, we may 
 put p to represent the distance of the required term (i) from a, 
 the first term of the series, in which case j>=n — 1 in the formula 
 Art. 326, and the required term is 
 
 The interval between the given numbers is always to be con- 
 sidered as unity, and p is to be reckoned in parts of this interval ; 
 hence, p will be fractional. 
 
 Interpolation is of extensive application in Astronomy ; and in 
 most instances sufficient accuracy is obtained by making use of 
 first and second differences only. The correction to be applied to 
 the first term then is 
 
 In practice, howeveri the method generally adopted is, to take 
 the two terms of the series which precede, and the two terms 
 which follow the term required, and find from them the three first 
 differences, and the two second differences. Then, taking the sec- 
 ond of the three first differences and calling it d, and the mean of 
 the two second differences and calling it d', and denoting the 
 fractional part of the interval by t, the correction to be applied to 
 this second term is 
 
 Ex. Having given the logarimths of 102, 103, 104, and 105 
 let it be required to find the logarithm of 103.55. 
 
 Nos. 
 
 Logarithms. 
 
 Ist Dili, 
 
 2nd Di/r. 
 
 Mean of 
 2Dd Diir. 
 
 102 
 103 
 104 
 105 
 
 2.0086002 
 2.0128372 
 2.0170333 
 2.0211893 
 
 42370 
 41961 
 41560 
 
 —409 
 —401 
 
 —405 
 
 Here /=.55,<i=41961, d'=— 405, and 
 /(d4-tzLj')=.55(41961+-^X405), = 23129 
 
 log. 103 =2.0128372 
 log. 103.55=2.0151501' 
 
INFINITE SERIES. 295 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Find the 2"^ term of the series of which the 4"' differences 
 vanish, the l"", S"*, 4'*, and 5'* terms being 3,15,30,55; and 
 find the 6'*, 7'" and 8"> terms. Ans. 7 ; and 93, 147, and 220. 
 
 2. Find the 5'* term of the series of which the 6"' differences 
 •vanish, and the 1", 2"^ S"', 4», 6«, and 7"' terms are 11, 18, 30, 
 50,132,209. Ans.S2. 
 
 3. Given the logarithms of 101,102, 104, and 105; viz.. 
 2.0043214, 2.0086002, 2.0170333, and 2.0211893, to find the 
 logarithm of 103. Ans. 2.0128372. 
 
 4. Given the cube roots of 60, 62, 64, and 66; viz. : 3.91487, 
 3.95789, 4, and 4.04124, to find the cube root of 63. 
 
 Aws. 3.97905. 
 
 5. Having given the squares of any two consecutive whole 
 numbers, show how the squares of the succeeding whole numbers 
 may be obtained by addition. 
 
 INFINITE SERIES. 
 
 Art. 336. An infinite series is a series consisting of an un- 
 limited number of terms, each of which is derived from the pre- 
 ceding term or terms, according to some law. For examples see 
 Art. 134, and page 253. 
 
 The sum of an infinite series, is the limit to which we approach 
 more nearly by adding together more terms, but which cannot be 
 exceeded by adding together any number of terms whatever. 
 
 A convergent series is one which has a sum or limit. Thus, 
 
 is a convergent series, whose limit is 2, since the sum of any 
 number of terms whatever cannot exceed 2, but will approach it 
 more nearly as the number of terms taken is greater. 
 A divergent series is one which has no sum or limit, as 
 1-1-2+4-1-8+16+32-1-, &.C. 
 
 An ascending series is one in which the powers of the leading 
 quantity continually increase ; and a descending series is one in 
 which the powers of the leading quantity continually diminish. 
 
 Thus, a+6x+cx^+(Zx'+, is an ascending series, and 
 
 a-\-hx~^-\-cx~'-\-dx~'-{-, 
 
296 RAT'S ALGliBRA, PART SECOND. 
 
 or a-\---\-—-\-—-{-, is a descending serieB. 
 X x^ x^ 
 
 Art. 337. There are four general methods of converting an 
 algebraic expression into an infinite series of equivalent valuei 
 each of which has been already exemplified ; viz. : 
 
 Is*. By Division. See Art. 134. 
 
 2nc, By Extraction of Roots. See examples 17, 18, page 
 136. 
 
 3rd By Indeterminate Coefficients. See Art. 314, and exam- 
 ples, page 275. 
 
 4th. By the Binomial Theorem. See Art. 319, and examples, 
 pages 281, 282. 
 
 Akt. 33§. The summation of a series is the finding a finite 
 expression equivalent to the series. 
 
 The general term of a series is an expression from which the 
 several terms of the series may be derived according to some de- 
 terminate law. Thus, in the series _ -|-_ -(- ?-|-"-)- the 
 
 general term is _, because by making x^l, 2, 3, &c., each term 
 
 of the series is found. 
 
 Again, in the series 2 • 2-|-2 • 3-f 2 • 4+2 -5+ .... the 
 general term is 2(a:-|-l). 
 
 As different series are in general governed by different laws, 
 the methods of finding the sum, which are applicable to one class, 
 will not apply universally. 
 
 We shall now explain two of the methods of most general 
 application. 
 
 First Method. — If the series is a regular decreasing geomet- 
 rical series, whose first term is a, and ratio r, its sum is _^ 
 
 (Art. 299.) 
 
 Second Method. — By subtraction To find the sum of a 
 
 •eries whose general term is — i — . 
 "(H-P) 
 
 Since 9— ?_=_Pi_ .-. _?— =1: 'il-A_\ , 
 
 
 n n-\-p nin+p) n(rir\-p) p }n n-j-p^ ' 
 
 or, any fraction of the form — ? — is equal to -", the differ- 
 n{n-\-p) p 
 
INFINITE SERIES. 297 
 
 ence between the two fractions t and _2_ ; that is, any term 
 
 n n-\-p 
 
 of the series whose general term is 2 is equal to the dif- 
 
 n(n+p) 
 
 ference between the corresponding terms of the two series whose 
 
 general terms are 1 and — ?- ; hence, the sum of the former 
 n n-\-p 
 
 series is equal to the difference between the sums of the two 
 
 latter. Therefore, if the sums of the two latter were known, by 
 
 taking their difference the sum of the former series would be 
 
 found. The sums of these two series, however, are not. known, 
 
 but their difference can be found, when, after a certain number 
 
 of terms of the series £, the succeeding terms are identical with 
 
 71 
 
 those of _?_ In general, this certain number is after p terms 
 n+p 
 
 of the ra-|-p former series. 
 Ex. 1 . Required the sum of the series - — -+- — _+_—+, 
 
 I'oo'OO'i 
 
 &c., ad infinitum, that is, to infinity. 
 
 Here q=l, p=2, and n=l, 2, 3, &.c. ; and the two series are 
 
 Jl +1+5+7+' <^o-. ^^ i°f- I . 1= .- 
 
 «-Q+i+4+.'S"=•.^'iinf■)> '"P ' 
 
 The sum of n terms of the same series is found in a manner 
 nearly similar. Thus, 
 
 1+ i+l+ 
 
 ^^' ■ ■ ■ '2^1:1 1=1. 
 
 h+^+^> 2;i-r 2n+l J 
 - ^" . , and - of this sum is '* ^ - — 
 
 2n+l p 2m+l 
 
 2. Find the sum of the series - — _+__-4---_+&c., ad inf 
 
 1'.* -i ' o o'4 
 
 Here 5=1, ^=1, and 7i=l, 2, 3, &c. Ans. 1. 
 
 3. Find the sum of the above series to n terms. 
 
 Am. -?_. 
 n-j-1 
 
298 RAY'S ALGEBRA, PART SECOND. 
 
 4. Find the sum of the series —Ll+J—+——+^—-\-,iic. 
 ad infinitum. 
 
 Here 5=1, and p=3. Aits. fj. 
 
 5. Find the sum of the series + + +, &c., ad 
 
 1 -3 2-4 3 -5 
 
 infinitum. 
 
 Here q=-\, p=:2, and n=l, 2, 3, &c. Am. |. 
 
 6. Find the series whose general term is ; also find iti 
 
 sum continued to infinity. 
 
 Ans. Series = ^ +-j_+ ^ +_j_4-, &c., sum =— 
 1 ■ 5^2 • 6^3 • 7^4 • e 48" 
 
 The sums of series may often be found by reducing them, by 
 multiplication or division, to the forms of other series whose sums 
 are known. 
 
 7. Find the sum of the series l-f-i+e+TS+j ^c., ad infini- 
 tum. Ans. 2. 
 
 SnGGEsTiox. — By dividing by 2 this series becomes the same as that 
 in example 3nd. 
 
 8. Find the sum of the series + + - +, &c., 
 
 3 -8^6 -12^9 -16^' ' 
 
 ad infinitum (Multiply by 3 • 4). Ans. J„. 
 
 Kkhare. — The preceding examples afford an illustration of the man- 
 ner in which the sums of certain classes of infinite series may be found. 
 The sums of a great variety of series may be found by otlier and more 
 complicated methods But the subject is more curious than useful, and 
 Is too complex and extensive for an elementary work. 
 
 RECURRING SERIES. 
 
 Art. 339. A Recurring Series is a series so constituted that 
 every term is connected with one or more of the terms which 
 precede it by an invariable law, usually dependent on the opera- 
 tions of addition, subtraction, &c. Thus, in the series 
 
 14-2a:+3a;=+5a;'-f 8a:<+13a;'-4-21a:»+, &c., 
 
 the sum of the coefficients of any two consecutive terms is equa 
 
RECURRING SERIES. 299 
 
 10 the coefficient of the next following term. If the series be 
 Jiwressed by 
 
 A+B+C+D+E+F+G+H+, &c., then 
 
 the 1" term A= 1 ; 
 
 the 2"'' « B= 2x; 
 
 the 3"" " C= 3a;2=Bx+Aa:"; 
 
 the 4» " D= 5a;'=Car+Ba' ; 
 
 the 5'* " E= 8x*=Dx+Cx^; 
 
 the 6" " F=13a:S=Ea;+Da;S &c. 
 
 That is, each term after the second is equal to the one next 
 preceding, multiplied by x, plus the second next preceding, multi- 
 plied by x^ ; hence, all the terms after the first two recur accord- 
 ing to a definite law. 
 
 Art. 340. The particular expression by means of which any 
 term of the series may be found when the preceding terms are 
 known, is called the scale of the series, and that by means of which 
 the coefficients may be found, the scale of the coefficients. Recur- 
 ring series are said to be of the first order, second order, die, 
 according to the number of terms contained in the scale. Thus 
 
 in the expansion of — ? — , (Art. 315), we find 
 a-\-bx 
 
 " -' JLx+tx^~^lx^+^lx*-^lx^+,&.c., 
 
 a-{-hx a 
 
 where each term after the first is equal to the preceding, multi- 
 plied by — -X. In this case — -x is termed the scale of the 
 a a 
 
 series, — - the scale of the coSfficients, and the series is said to 
 a 
 
 be of the first order. This is the most simple form of a recurring 
 
 series. 
 
 Art. 341. To find ike scale of a series. 
 
 When the series is of the first order, the scale is easily deter- 
 mined, being the ratio of any two consecutive terms. 
 
 When the series is of the seccKid order, the law of the series 
 depends on two terms, and the scale consists of two parts. Let 
 p-j-j represent the scale of the r jcirring series 
 
 A-fB+C-f D-f E+F+, &c. 
 
800 RAY'S ALGEBRA, PART SECOND. 
 
 Then the 3'^ term C =B pi+Aja;' ; 
 
 the 4"* term T)=Cpx+Bqx^ ; 
 
 the 5'* term E —Dpx-]-Cqx' ; &c. 
 
 The values of p and g may be found from any two of these 
 equations. Taking the last two, and making x=l, since the 
 scale of the series is the same, whatever be the value of x, we 
 have 
 
 D=Cp+Bq, 
 
 E=Dp+Cg ; whence, (Art. 158), 
 
 CD— BE „ CE— D' 
 
 ^ C'—BD C— BD" 
 
 Since these formula were obtained by supposing x=l, there- 
 fore, in substituting the values of B, C, D, &.C., x must be con- 
 sidered 1. 
 
 Ex. Find the scale of the series l-^2x-\-2x'-\-ix'-{-5x*-\-, 
 &c. 
 
 Here A=l, B=2ar, C=32=, D=ix>, E=5xS &,c. 
 
 Making a;=:l, and substituting the values of B, C, D, &c., 
 
 3X4—2X5 ^^ 3X5— 1X4 ^_1 
 ^ 3X3—2X4 '^^3x3- 2X4 
 
 Thus, the 4'* term, 4a;'=2Xa:X3x'+2xX— 1 Xa:=. 
 Other exercises will be had in finding the sums of recurring 
 series. 
 
 Akt. 312. In a recurring series of the third order the law of 
 the series depends on three terms. If we let i'+J+r represent 
 the scale of the series 
 
 A+B+C+D+E+F+, &c., 
 then the 4" term D=Cpx+B jx'+Arx' ; 
 the 5'* term E=T)px+C jx'+Brx' ; 
 the 6" term P=E;)X+D5x'+Crx» ; &c. 
 
 Making i=l, the values of p, q and r, are readily found, (Art. 
 158) ; and in a similar manner the scale may be determined in 
 the higher orders of recurring series. 
 
 In finding the scale of a series we may first make trial of two 
 terms. If the results thus obtained do net reproduce the series 
 we may try three terms, four terms, and so on, till a correct result 
 is obtained. If in any case we assume too many terms, the 
 redundant terms will be found equal to zero. 
 
RECURRING SERIES. 301 
 
 Aet. 343. To find tJie sum of an infinite recurring series whose 
 scale of relation is known. 
 
 Let A+B+C+D+E-j-, &.c., be a recurring series whose scale 
 of relation is p-\-q ; then 
 
 the 1'' term A=:A; 
 
 the 2"'' " B =B ; 
 
 theS"* " C—Tipx+Aqxi'; 
 
 the 4" « D=Cpx-\-Bqx'; 
 
 the5'» " E=J)px-\-Cqx^; &c. 
 
 If the series be continued to infinity, the last term may be con- 
 sidered zero. Then if S represent the required sum, I)y adding 
 together the corresponding members of the preceding equalities, 
 and observing that B-|-C-)-D+, &c., =S — A, we have 
 
 S=A+B+pa;(S— A)+?a!=X S ; 
 or, S — pxS — }x^S=A-|-B — Apx; 
 or, S(l — fx — qx'')=A-\-B — Apx; 
 g^ A+B-Ap^ _ 
 1 — px — qx^ 
 
 If we make 5=0, the formula becomes 
 
 g_A4-B— Api^ which is the formula tor 
 1 — px 
 finding the sum of an infinite recurring series of the first order. 
 
 In a manner similar to the preceding, the sum njay be found 
 when. the scale of the series consists of three, four, &c., terms. 
 
 Remark. — Every summable infinite series, of which recurring series 
 are only a particular class, may be supposed to arise from the develop- 
 ment of a rational fraction ; hence, to find the sum of an infinite re- 
 eurring series, is to find the generating fraction of the series. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Find the sum of the infinite recurring series 14-3a;-|-5a' 
 +nx'+9x^+Ux'+, <Sz,c. 
 Here A=l, B='ix, C=5a;2, D=7x', E=9a;<, &c. 
 Making x=l, and substituting in the formula (Art. 341), we 
 
 ^ 5X5—3X7 5X5-3X7 
 
 „_ A+B— A;ia; _ l+3a;— 2j _^ 1+x 
 l—px—qx^ 1— 2a;+i2 il—xY 
 
803 RAY'S ALGEBRA, PART SECOND. 
 
 In each of the following series find the scale of relation, and 
 the sum (S) of an infinite number of terms. 
 
 2. l+6a!+l 21=448x5+1 20a;<+, &c. 
 
 Ans.p=X, 5=6; S=-^+^* 
 
 S. 14-2a:+3ar'+4i'+5a«+6a;'+, &c. 
 
 Ans. p=2, g=— 1 ; S=. ^ 
 
 {l-xt 
 
 4. ?— ?^+^j5!— '*!^+,&c. 
 c c' c' c* 
 
 Aks. ;>=—*, 9=0; S=_f_ 
 c c-l-ia" 
 
 5. x+x'+ar'+i ^c. Ans. p=l, g=0; 8= 
 
 6. !i!—3?->r3c>—<e^-\-,&x,. Ans.p=—\, g=0; S=_^ 
 
 7. l+2a;+8ar'4-28it3+100x«4-356a;S4-, &c. 
 
 A»s. ;)=3, 5=2; S=- ^"~* 
 
 1— 3a^2a!»' 
 
 8. l+3z+5a;2+7a;'4-9a;<+, &c. 
 
 A7«. p=2, 9=-l, S= \+^ 
 
 9. l»+2'a;+3V-l-4V+5V+6V+, &c. 
 
 Ans. p=Z, 5=— 3, r=l; 8=-!+^- 
 
 REVERSION OF SERIES. 
 
 Art. 344. To revert a series is to express the value of the 
 unknown quantity in it by means of another series involving the 
 powers of some other quantity. 
 
 Let X and y represent two indeterminate quantities, and let the 
 value of y be expressed by a series involving the powers of x ; 
 thus, 
 
 ■y=ax-\-hx'-\-cx^-\-dx*-\-, &c., (1). 
 
 in which a, h, c, d, &c., are known quantities ; then to revert this 
 series is to express the value of x in a series containing the 
 known quantities a,b,c, d, &c. , and the powers of y. 
 
REVERSION OF SERIES. 303 
 
 To revert this series, assume 
 
 sc=Ay-\-By^+Cf+J)y'', &o. (2), in which the 
 
 coefficients A, B, C . . . . are undetermined. 
 
 Find the values of 2/^ i/S y'' .... from (1), thus, 
 
 y^=aV+2dbx=-\-(b^+2ac)x'+ .... 
 
 f= aV+3a'bx^+ . . . 
 
 y*-= a* x'^-\- . . . &c. 
 
 Substituting these values in (2), and arranging, we have 
 
 I x'+, &c. 
 + Bi' 
 -i- Co3 
 
 0=Ao|i+Ai|a'+ Ac 
 — 1| Bo=| -i-2Bai 
 
 ; x'-}- Ad c 
 + Bi2 
 4- 2Bflc 
 
 4- Da" 
 
 and since this is universally true, whatever be the value of x, the 
 coefficients of x, a', x', &,c., will each =0. (Art. 314, Cor.) 
 Hence, we have 
 
 Aa— 1 =0, . . A=l 
 
 a 
 h 
 
 2i=. 
 
 a' 
 
 A6+Ba' =0, .'. B=— - 
 
 Ac+2Ba6+Ca' =0, .-. 
 
 Ad+-Bd'+2Bac+3Ca^+-Da'=0, . . d=— ^^Zi^f^fifif. 
 
 Hence, x=-y~ -y^+ - _^!/S— ^ y+, &c. (3; 
 
 Art. 345. If the given series has a constant term prefixed, 
 
 thus, y=a'-\-ax-\-bx^-\-cx'-\-dx^-\- 
 
 assume y — a'=z, and we have 
 
 z=ax-\-bx'-\-cx'-\-dx^-\-, &c. 
 
 But this is the same as (1) in the preceding article, except that 
 z stands in the place of y ; hence, if z be substituted for y in 
 [(3), Art. 344], the result will be the required development of x; 
 and then y — a' being substituted for z, the result is 
 
 x=-iy—a')—-(y~a'y+ ~'"^ (y— a')'— ■ &^c. 
 a a' a= 
 
304 RAY'S ALGEBRA, PART SEUUINU. 
 
 AsT. 346. When the given series contains the odd powers of 
 r, assume for a; another series containing the odd powers of y. 
 Thus, if 
 
 y=ax-\-bx'-\-C3f-\-dx'-^. 
 
 to develope x in terms of y, assume 
 
 x=Ay+By'+Cy'+Dy''+ . . . . 
 
 Then by substituting the values of y, y', &,c., derived from the 
 former equation, in the latter, and equating the coefficients to 
 zero, we find 
 
 1 b , , Zb'—ac 5 a'd—Sabc-\-l2b^ , , - 
 
 If both sides of the equation be expressed in a series, as 
 ay-\-by^'{-cy'-\-, &c., =a'a;+i'a;'-|"'^'^'+> ^c., 
 and it be required to find y in terms of x, we must assume, as 
 before, 
 
 y=Ax+Bx2+Ca;'+Da!<+, &c., 
 and substitute the values of y, y^, y', die, derived from this last 
 equation, in the proposed equation ; vye shall then, by equating 
 the coefficients of the like powers of x, determine the values of 
 A, B, C, &c., as before. 
 
 EXAMPLES FOR PRACTICE. 
 
 The following exercises may be solved either by substituting 
 the values of a, b, c, &c., in the equations obtained'in the preced- 
 ing articles, or by proceeding according to the methods by which 
 those equations were obtained. 
 
 1 . Given the series y=x — x'+x' — x''-(- .... to find the 
 ealue of X in terms of y. Ans. x=y-\-y''-\-y^-\-y*-\-, &c. 
 
 Find the value of x, in an infinite series in terms of y : 
 
 2. When y=x-\-x''-\-3i?-\-, &c. 
 
 Ans. x=y — y^+y' — y*-\-y^ — > ^c- 
 
 3. When ?/=2x+3i'+4a;5+5x'+, &c. 
 
 Ans. x=|y— yV+Tssy'— . &c- 
 
 4. When y=l—'2x-\-Zx^. 
 
 Ans. x=-|(y-l)+|(y-l)»_T?g(y-l)'4-. &c. 
 
 5. When y=l+x— 2a:'+a;'. 
 
 Ans. a;=y— l4-2(y— l)=+7(y— l)'4-30(y— 1 •)*. 
 
CONTINUED FRACTIONS. 305 
 
 6. When y=x+ix^+lx^+^\x*+, &c. 
 
 Ans. x=y—y-Jf-iy^—ii/*+, &c. 
 
 7. When y-\-ay^-\-by'-{-cy' . . . =gx-\-h3f'-\-kc'-\-lx* . . . 
 
 A^. ^yj^ (.<^9'-h)y' j ^W-kg-2h(,aff^-h)]f _^ 
 
 CHAPTER XI. 
 
 CoKTiNUED Fbaotionb: Looabithhs: Expo- 
 
 KENTiAL Equations: Inteebst, ahd An- 
 
 NTJI TIES. 
 
 CONTINUED FRACTIONS. 
 
 Art. 347. A continued fraction is one whose denominator is 
 continued by being itself a mixed number, and the denominator 
 ot the fractional part again continued as before, and so on ; thus, 
 
 11 1 
 
 111 
 
 in which a, b, c, d, &c., are positive whole numbers, are called 
 continued fractions. 
 
 Continued fractions are useful in approximating to the values 
 of ratios expressed by large numbers, in resolving exponential 
 equations, in resolving indeterminate equations of the first degree, 
 &c. 
 
 Aet. 34§. To express a rationed fraction in the form of a con- 
 tinued fraction. 
 
 30 
 
 Let it be required to reduce to a continued fraction. 
 
 157 
 
 If we divide both terms of the fraction by the numerator, we 
 
 30 1 
 
 find 
 
 1^'^ 5+1 
 ^30 
 
 36 
 
306 RAf'S ALGEBRA, PART SECOND. 
 
 Since the value of a fraction is the quotient arising from dividing 
 
 the numerator by the denominator (Arith., Part 3rd, Art. 136), if we 
 7 1 
 
 omit -_, the denominator will be too smaU, and consequently _' 
 
 the value of the fraction, will be too large. 
 
 Again, if we divide both terms of the fraction — by the nu- 
 
 30 
 
 30 1 
 
 Aerator, we find =: 
 
 157 ,^1_- 
 
 2 Id 
 
 If we omit -, the value will be expressed by _^_ 
 
 "^ 5+1 ^^' 
 
 2 
 
 By omitting -, the denominator 4 will be kss than tile true 
 
 denominator, and - will be larger than the number wh>bh ought 
 
 to be added to 5: hence, 1 divided by 5+_, or — wil. be les> 
 
 ' ^4 21 
 
 than the true value of the fraction. 
 
 We see from this, that by stopping at the Jirst reduction, and 
 omitting the fractional part, the result is too great ; but by stop- 
 ping at the second reduction and omitting the fractional part, the 
 result is too small. Hence, generally. 
 
 By stopping at an odd reduction and neglecting the fractional 
 part, the result is too great ; but by stopping at an even reduction, and 
 neglecting the fractional part, the result is too smaU. 
 
 2 1 
 
 Since -= , we find 
 
 ^2 
 
 30 1 
 —-S3 1'' reduction, too great ; 
 
 1^'' 5+L._ 2" '• too small; 
 
 44-i 3"* " too great; 
 
 3+1 4" " true value. 
 
 ' 2 
 
 It is evident that the process of reducing a fraction to a con- 
 tinued fraction, is the same as that of finding the greatest common 
 divisor of the two terms of the fraction. (See Arith., Part 3rd, 
 Art. 128.) 
 
CONTINUED FRACTIONS. 307 
 
 Br this process we find 
 
 13_1 _i?.=l 
 
 30 2_^1 204 4_j_l 
 
 3+1' 6+1 
 
 Art. 319. The different quantities 
 1 1 1 
 
 "' a-{l' 0+1 ,&c., 
 
 * b+1 
 
 c 
 
 are called converging fractions, because each one in succession 
 gives a nearer value of- the given expression. 
 
 The fractions _, _, _, &c., are called integral fractions. 
 a b c 
 
 Aet. 3S0. To explain (he mamier in which the converging frac' 
 tioTbs are found from the integral fractions. 
 
 1. - =- l** conv. fraction. 
 
 a a 
 
 1 
 
 2. _ , 1 = 2"^ conv. fraction. 
 
 *+j fli+1 
 
 1_ 
 
 3. a-!-l = — ^?il 3"" conv. fraction. 
 
 ^j_|_l c(ai+l)+a 
 
 By examining the third converging fraction, we find it is formed 
 from the 1", and 2"^, and from the S'"* integral fraction as follows: 
 
 < num. =3"'quot. Xnum.of 2"<'conv.fract.+hum.of l"conv.fract. 
 <denom.=:3"'quot.Xden. of 2"''conv.fract.+den. of l''conv.fract 
 
 p Q P ' 
 
 To prove the general law of formation, let — , _, — be the 
 1-6 ' P'' Q,'' R' 
 
 three converging fractions corresponding to the three integral 
 
 fractions _, _, and _, and, as has already heen shown, 
 a c 
 
 R ^Q, c+P 
 
 R' Q,'c+P" 
 
308 
 
 RAY'S ALGEBRA, PART SECOND. 
 
 Let us now take the next integral fraction -, and let _ ex- 
 
 d S' 
 
 press the 4" converging fraction. Then it is obvious that _ 
 
 R 
 
 S 1 
 
 will become _ by substituting c+-, instead of c ; hence, 
 S d 
 
 s <^(=+-J+P 
 
 .(Q, c+P )d+Q, _R d+Q, 
 
 "^ Q'(c+1)+P' CQ'c+POd+Q.' R'd+Q.'- 
 
 From this -we see that the fourth converging fraction is deduced 
 from the two immediately preceding it, according to the same law 
 by which the third was deduced from the 1'' and 2"'', and it is evi- 
 dent the fifth converging fraction may be deduced in the same 
 manner. Hence, to find the ra" converging fraction, 
 
 Multiply the denominator of the n'* integral fraction by the numer- 
 ator of the (n — 1)" converging fraction, and add to the product the 
 numerator of the (n — '2)"' converging fraction. This will give the 
 numerator of the n" converging fraction. 
 
 Multiply the denominator of the n'* integral fraction Try the denom- 
 inator of the (n — 1)'* converging fraction, and add to the product the 
 denominator of the (n — 2)'* converging fraction. This will give the 
 denominator of the n" converging^ fraction. 
 
 Ex. To find a series of converging fractions for /jly. 
 
 The integral fractions are 5> \i 5. 3. \> \, \. 
 
 The converging fractions are h ^j> |> i\> 2?, /jfj, j^jly. 
 
 Akt. 351. To show that the difference between any two consecutive 
 converging fractions is always a fraction having -|-1, or — \,for 
 its numerator, according as the fraction subtracted is in an even or 
 odd place. 
 
 I b _ab+l—ab 
 a ab-\-\ o(at+l) 
 
 6 ic+1 
 
 _ +1 
 a(aH-l) 
 
 _ fc(aJ+l )+aS— (aJ+1 )(ie+l ) 
 
 oJ+l c(ai+l)4-a 
 
 (fl6+l)[c(a6+l)+a] 
 —1 
 
 (afr-t-l)rc(ai+l)+oJ- 
 
CONTINUED FRACTIONS. 309 
 
 To prove the property in a general manner, let 
 P Q, R 
 
 p'' a" R' 
 
 be three consecutive converging fractions, corresponding to tha 
 
 three integral fractions _, _, _. Then 
 a b c 
 
 P_Q_ PQ'— P'Q , 
 P' Q,' P'Q' ' 
 
 and ^-5:=?^^^:^ ; 
 
 Q' R' R'Q,' 
 
 but R=Qc+P, and R'=Q'c4-P', (Art. 350.) 
 
 Substituting these values" in the last equation, we find 
 
 Q, _R _ (Q'e+P')Q— (Qc+P)Q.' _P'a— PQ' 
 Q,' R' R'Q' R'Q' ■ 
 
 But the numerator of this result P'Q — PQ' is the same with a 
 
 P Q 
 
 contrary sign as the numerator of _ — ~-, which we have before 
 
 shown is -|-1. Hence, the difference between the numerators of 
 any two consecutive approximating fractions, when reduced to a 
 common denominator, is the same with a contrary sign, as that 
 which exists between the last numerator and the numerator of 
 the fraction immediately following. 
 
 But it has been already shown that the difference of the nu- 
 merators of the 1" and 2"'' fractions is -|-1 ; the difference of the 
 numerators of the 2"'' and 3"* fractions is — 1 ; therefore, the dif- 
 ference of the numerators of the 3'''' and 4" is -|-1, and so on. 
 And since (Art. 348) any converging fraction of an even order is 
 less than the true value, and of an odd order greater than the true 
 value ; therefore, if a converging fraction of an even order be 
 subtracted from the consecutive converging fraction of an odd 
 order, the numerator of the difference will he -\-l ; and, con- 
 versely, if a converging fraction of an odd order be subtracted 
 from the consecutive converging fraction of an even order, the 
 numerator of the difference will be — 1. 
 
 Art. 352. To show that every converging fraction is in its lou^M 
 terms ; and to find the limit of error in taking any convergent for 
 
 the true fraction -. 
 
810 RAY'S ALGEBRA, PART SECOND. 
 
 A C 
 
 If — and _ be any two consecutive converging fractions, by 
 
 Art. 351 ^— 2=4.J_, or — L ; that is, AD— BC=+1, or 
 B D ^BD BD ^ 
 
 — 1. Now if A and B have a common divisor greater than I, it 
 
 will divide AD and BC, and consequently their difference 
 
 ±1 ; that is, a quantity greater than 1 is a divisor of 1 
 
 A 
 
 which is impossible ; hence, _ is in its lowest terms. 
 
 "^ A C 
 
 Again, if _ and _ be any two consecutive convergents, as 
 
 A C 
 
 has just been shown, AD — BC=±1; and of _ and -we know 
 
 B D 
 
 that ?< one and > the other (Art. 348) ; therefore, the differ- 
 b 
 
 ence between _, and either of them, is less than the difference 
 b 
 
 between^ and - ; that is, <-i-, since -^.-.^2, or ^.PT^::?? 
 B D ^BD B D .BD 
 
 ~BD" 
 
 But, since D is greater than B, — is greater ~ ; hence, 
 
 BD D' 
 
 since the result is true to within — , it is certainly true to within 
 
 BD 
 
 _ ; that is, the~approximate result which is obtained, is true to 
 
 within unity, divided by the square of tJte denominator of the last 
 converging fraction. 
 
 Thus, in the example, (Art. 348,) i differs from -21. by ■ 
 
 5 157 
 
 quantity less than — ; — differs from by less than — 
 
 o 21 loi 21* 
 
 = , and so on. 
 
 441 
 
 Akt. S53. To express ^N, when N=:o'-j-l, in the form of « 
 tontinued fraction. 
 
 1 
 
 Ja'+l+a 
 
CONTINUED FRACTIONS. 81 1 
 
 1 
 
 1 
 
 =a+. 
 
 2o4-L 
 
 2'H-J^+,&c. 
 
 (•) ^a»H-l-a= 
 
 ^a=+l+a 
 
 because (Va'+l— fl)(V'»'+l +«)=!• 
 
 Ex. V17==^/42^-l=4+ J- 5 
 
 "■"8+, &c. 
 
 1 8 65 
 
 the converging fractions to be added to 4, are -, — , __ fee. 
 
 o 65 528 
 
 Aet. 354. To convert ^N, where N=a'+5, into a contl wd 
 fraction. 
 
 VN=<i+ VN-a=a+ -^^— =0+^-;: 
 
 6 _ 1 
 
 let r, be the nearest integer to -(^/N+a) ; 
 — " -=r, - 
 
 VN+a, '^UvN+a.)' 
 by making a,^r,6 — o ; 6,^_(N — aj). 
 
 
 
 Similarly, UvN+a,)=r»4 
 
 *• J-(VN+<««) 
 
 and we must proceed till we get a quotient 2a, after which tha 
 quotients will recur in the same order 
 
312 RAY'S ALGEBRA, PART SECOND. 
 
 Thus, Vl9=4+1 
 
 2-' ^ 
 
 l+3-f.&c. 
 
 The quotiente are 4, 2, 1, 3, 1, 2, 8; 
 
 .-. fractions are f, |, Y' ". tI. W. V? • 
 
 Aet. 355. To Jind the value of a eontinued fraction, when the 
 denominators q, r, s, ^c, of the integral fractions recur ad irvfinitum 
 in a certain order. 
 
 Ex. 1. Let \ 
 
 '+T1 
 '■+-1 
 
 '~'^-|-, &c., ad infinitum, 
 then — , =x, or J =1 ; 
 
 hence, r-^x=qrx-\-qz'-\-x, and ii;'-|-ra! — -=:0. 
 
 9 
 
 From the ralution of this equation, the value of x is easily 
 found. 
 
 Abt. 356 To find in the form of a continued fraction, the 
 value of X, uhtch satisfies the equation a'=b. 
 
 Substitut* for x the numbers 0, 1, 2, 3, &c., until two consecu- 
 tive numbers are found n, and n-\-l, such that 
 
 a"<^b, and a»+'>6; 
 then it is evident that x<jtr^l, and >n. 
 
 Let x=n-\--, where y>l. 
 
 y 
 
 I I 
 
 then a"+<i==b, or <!".<» =6; 
 
 hence, ai=—, 01 [ ~ |!'=o. 
 a" \ a" / 
 
 Again, since — >1, and <ji, by substituting the numbers be- 
 
 Cween 1 and a for y in the last equation, two consecutive num- 
 bers, p, and p-\-^., will be found, such that y'>j» and <j)+l, so 
 
CONTINUED FRACTIONS. 313 
 
 ■ 1 1 
 
 that y=p-\-- i and .-. x=n-\- r ; and \j continuing the pro- 
 
 P+- 
 z 
 
 cess in the same .nanner, the fraction expressing the value of i 
 may be continued. 
 
 Ex. Required the value of x in the equation 10'=2. 
 
 Bj substituting and 1 for x, it appears that x">0 and <[1 ; 
 
 let a^-, then Wy='i., or 2v=10. 
 
 y 
 
 Since 2'=8, and 2''^16, one of which is less and the othei 
 
 greater than 10, therefore, i/]>3, and <^4; let 2/=3-f— ; 
 
 z 
 
 then 2'+j=10, 
 
 or 2'.2z=10, or 2i'=ig0=1.25; 
 .'. (1.25)'=2. 
 
 Again, it appears that z>3, and <^4; let z^S-j— , then 
 
 u 
 
 a.25)'+S =(1.25)3(1 .25)^ =2 . . (1.25)i=-^A^^=1.024; 
 .-. (1.024)"=1.25, and by trial u>9 and <10. 
 Hence, x= 
 
 3+9+, &c. 
 This gives x=\ — , to+, || — =.30107 nearly, &c. 
 
 EXAMFLES'FOR PRACTICE. 
 
 Reduce each of the foUoviring fractions to a continued fraction, 
 «nd find the successive integral and converging fractions. 
 
 J 130 Ans. Integral fractions ^j \> \> \- 
 421" Converging fractions \i 7*3, |^i \W. 
 
 q 130 Ans. Integral fractions j) \, \i \. 
 
 291" Converging fractions 31 1, |4> 3I?. 
 
 Q 157 Ans. Integral fractions |. \, \, \, \. 
 
 9T2' rjTf Converging fractions \-, -i^, f^, -fi^j, m. 
 
314 RAY'S ALGEBRA, PART SECOND. 
 
 4. The hight of Mt. Etna is 10963 feet, and of Vesuvius 3900 
 feet ; required the approximate ratio of the hight of the former 
 to that of the latter. 
 
 A„, 11 6 !£ 37 90 127 _a,900 
 
 Am. i> 3' T4' 45' ITS?' v;5^' 357> TTiBBS- 
 
 5. ThehightofMt. Perdu, the highest of the Pyrenees, is 11 283 
 feet ; that of Mt. Hecla is 4900 feet ; required the approximate 
 
 atic of the hight of the former to that of the latter. 
 
 /ln« J- -3 '" -3 3 -?6, far 
 ■fins. 2, 7, .,3, 7^, J-!^, etc. 
 
 6. When the diameter of a circle is l,the circumference is 
 found to be greater than 3.1415926, and less than 3.1415927; 
 required the series of fractions converging to the ratio of the 
 circumference to the diameter. Ans. 3> 231 -3I3, and -jjf. 
 
 Show that this last ratio, -J^f , is true to within less than three 
 ten millionths of the circumference. 
 
 Suggestion. — In examples of this kiad the integral fractions, corres- 
 ponding to both fractions, should be found, and then the converging 
 fractions calculated from those integral fractions that are the same in 
 both series. 
 
 7. Express approximately the ratio of 24 hours to 5 hours, 48 
 minutes, 49 seconds, the excess of the solar year above 365 days. 
 
 4^- I 7 8 3] 30 C55_ 694 1349 20^23 
 Ans. :j, 3-5, 3 3, yjg. X5T, }7-fS4> 3b 55' 56'g"9' BSf c8- 
 
 Hence, after every 4 years, we must have had 1 intercalary day, 
 as in leap year ; after every 29 years, we ought to have had 7 in- 
 tercalary days ; after every 33 years we ought to have had 8 inter- 
 calary days. This last was the correction used by the Persian 
 astronomers, who had seven regular leap years, and then deferred 
 the eighth until the fifth year, instead of having it on the fourth. 
 
 8. Find the least fraction with only two figures in each term, 
 approximating to -J|H- ■^'*''- H' 
 
 9. The lunar month, calculated on an average of 100 years, ia 
 27.321661 days. Find a series of common fractions approxi- 
 mating nearer and nearer to this quantity. 
 
 Ans. Y. ¥' W' \W. &c. 
 
 10. Find a series of fractions converging to ^2. 
 
 Ans. i, |, J, jj, li, &c. 
 
 11. Show that ^5 is greater than Iff and less than f|||. 
 
 12. If 8'=32, find x. Ans, |. 
 
 13. If 3»==15, find X. Ajw. 2.465. 
 
LOGARITHMS. 315 
 
 LOGA RITHMS. 
 
 Art. 357. In a system of logarithms, all numbers are con- 
 Bidered as the powers of some one number, arbitrarily asEumed, 
 which is called the base of the system ; and the exponent of that 
 power of the base, which is equal to any given number, is called ikt 
 LosARiTHM of that number. 
 
 Thus, if a is the base of a system of logarithms, N any num- 
 ber, and X such that 
 
 then X is called the logarithm of N, in the system whose base 
 is a. 
 
 For particular examples suppose we have the equations a'=N, 
 and a'=N', then 2 is the logarithm of N, and 3 is the logarithm 
 of N'. 
 
 The base of the common system of logarithms . (called from 
 their inventor " Brigg's Logarithms") is the number 10. If we 
 designate the logarithm of any number in this system by I. or 
 log., we shall have 
 
 (10)»=1 
 
 , hence, is the log. of 1 ; 
 
 (10)1=10 
 
 " 1 " " log. of 10; 
 
 (10)2=100 
 
 " 2 " " log. of 100; 
 
 (10)5=1000 
 
 « 3 " " log. of 1000; 
 
 (10)''=10000 
 
 « 4" " log. of 10000; 
 
 &c.. 
 
 &c. 
 
 From this it appears that, in the common system, the logarithm 
 of every number between 1 and 10 is some number between 
 and 1 ; that is, a proper fraction. The logarithm of every num- 
 ber between 10 and 100 is some number between 1 and 2; that 
 is, 1 plus a fraction. The logarithm of every number between 
 100 and 1000 is some number between 2 and 3; that is, 2 pirns 
 a fraction ; and so on. 
 
 Akt. 358. The integral part of a logarithm is called the index 
 or characteristic of the logarithm. 
 
 Since the logarithm of 1 is 0, of 10 is 1, of 100 is 2, of 1000 
 is 3, and so on ; therefore, 
 
 The characteristic of the logarithm of any number greater than 
 unity, is one less than the number of integral figures m the given 
 nurnber. 
 
316 
 
 RAY'S ALGEBRA, PART SECOND. 
 
 Thus, the logarithm of 123 is 2 plus a fraction ; the logarithm 
 of 1234 is 3 plus a fractiou, and so on. 
 
 Art. 359. The computation of the logarithms of numhers in 
 the common system, consists in finding the values of x in the 
 equation 
 
 10*=:N, when N is successively 1, 2, 3, &c. 
 
 One method of finding an approximate value of x has been ex- 
 olained in Art. 356, but other methods more expeditious will be 
 given hereafter. 
 
 T^e following table contains the logarithms of numbers from 
 1 to 100 in the common system : 
 
 N. 
 
 r,oa. 
 
 N. 
 
 Loa. 
 
 N. 
 
 Loa. 
 
 N. 
 
 Loa. 
 
 1 
 
 O.OUODOO 
 
 26 
 
 1.414973 
 
 51 
 
 1.707670 
 
 76 
 
 1.880814 
 
 a 
 
 0.301030 
 
 27 
 
 1.431364 
 
 52 
 
 1.716003 
 
 77 
 
 1.886491 
 
 3 
 
 0.477121 
 
 28 
 
 1.447168 
 
 63 
 
 1.724276 
 
 78 
 
 l;89209B 
 
 4 
 
 0.602060 
 
 29 
 
 1.462398 
 
 64 
 
 1.732394 
 
 79 
 
 1.897627 
 
 B 
 
 0.698970 
 
 30 
 
 1.477121 
 
 56 
 
 1.740363 
 
 80 
 
 1.903090 
 
 6 
 
 0.7781B1 
 
 31 
 
 1.491362 
 
 66 
 
 1.748188 
 
 81 
 
 1.908485 
 
 7 
 
 0.84B098 
 
 32 
 
 1.6051B0 
 
 67 
 
 1.7B687B 
 
 82 
 
 1.913814 
 
 8 
 
 0.903090 
 
 33 
 
 1.B18514 
 
 58 
 
 1.763428 
 
 83 
 
 1.919078 
 
 9 
 
 0.9S4243 
 
 34 
 
 1.531479 
 
 69 
 
 1.770862 
 
 84 
 
 1.924279 
 
 10 
 11 
 
 1.000000 
 
 35 
 
 1.644068 
 
 60 
 
 1.778161 
 
 85 
 
 1.929419 
 
 1.041393 
 
 36 
 
 1.656303 
 
 61 
 
 1.785330 
 
 86 
 
 1.934498 
 
 12 
 
 1.079181 
 
 37 
 
 1.568202 
 
 62 
 
 1.792392 
 
 87 
 
 1.939519 
 
 13 
 
 1.113943 
 
 38 
 
 1.579784 
 
 63 
 
 1.799341 
 
 88 
 
 1.944483 
 
 14 
 
 1.146128 
 
 39 
 
 1.691065 
 
 64 
 
 1.806180 
 
 89 
 
 1.949390 
 
 IB 
 
 1.176091 
 
 40 
 
 1.602060 
 
 66 
 
 1.812913 
 
 90 
 
 1.964243 
 
 16 
 
 1.204120 
 
 41 
 
 1.612784 
 
 66 
 
 1.819544 
 
 91 
 
 1.969041 
 
 17 
 
 1.230449 
 
 42 
 
 1.623249 
 
 67 
 
 1.826075 
 
 92 
 
 1,963788 
 
 18 
 
 1.25B273 
 
 43 
 
 1.633468 
 
 68 
 
 1.832609 
 
 93 
 
 1.968483 
 
 19 
 
 1.278754 
 
 44 
 
 1.643463 
 
 69 
 
 1.838849 
 
 94 
 
 1.973128 
 
 20 
 
 1.301030 
 
 45 
 
 1.653213 
 
 70 
 
 1.846098 
 
 95 
 
 1.977724 
 
 21 
 
 1.322219 
 
 46 
 
 1.662758 
 
 71 
 
 1.851258 
 
 96 
 
 1.982271 
 
 23 
 
 1.342423 
 
 47 
 
 1.672098 
 
 72 
 
 1.857333 
 
 97 
 
 1.986772 
 
 23 
 
 1.361728 
 
 48 
 
 1.681241 
 
 73 
 
 1.863323 
 
 98 
 
 1.991226 
 
 24 
 
 1.380211 
 
 49 
 
 1.690196 
 
 74 
 
 1.869232 
 
 99 
 
 1.995636 
 
 3B 
 
 1.397940 
 
 50 
 
 1,698970 
 
 75 
 
 1.875061 
 
 100 
 
 2.000000 
 
 In the common tables, only the fractional part of the logarithm 
 is given. Thus, in searching for the logarithm of such a number 
 as 3530 we find in the table opposite to 3530 the number 
 647775; but since 3530 is expressed by four figures the charac- 
 teristic of the logarithm is 3; hence, 
 
 log. 3530=3.547775 
 
LOGARITHMS. 317 
 
 GENERAL PROPERTIES OF LOGARITHMS. 
 
 Art. 360. Let N and N' be any two numbers, x and x' theii 
 respective logarithms, and a the base of the system. Then, by 
 the definition of logarithms (Art. 357), 
 
 a'=N. . . . a). 
 
 a''=N'. . . . (2). 
 
 Multiplying equations (1) and (2) together, we find 
 
 a'X«^=o*+''=NN'. 
 
 But, by the definition of logarithms, x-\-x', the exponent of a 
 is the logarithm of NN' ; hence, we have 
 
 Property I. — The sum of (he logarithms of two numbers is 
 equal to the logarithm of their product. 
 
 It may be shown similarly that the sum of the logarithms of 
 three or more factors, is equal to the logarithm of their product. 
 Hence, to multiply two or more numbers together, add their logarithms 
 together, and the product will be the nuniber corresponding to this 
 sum. 
 
 Art. 361. Taking the same equations, (Art. 360), we have 
 
 a'=N. . . . (1), 
 
 a«'=N'. . . . (2). 
 Dividing equation (1) by equation (2), we find 
 
 ^=a"'= — 
 a'' " N" 
 
 But, by the definition of logarithms, x — a', the exponent of a 
 
 N 
 is the logarithm of — ; hence. 
 
 Property II. — The logarithm of the dividend, minus the loga- 
 rithm of the divisor, is equal to the logarithm of the quotient. 
 
 The same principle may be expressed otherwise thus, the log- 
 arithm of a fraction is equal to the logarithm of the numerator 
 minus the logarithm of the denominator. 
 
 From this article, and the preceding, we see that by means of 
 logarithms, the operation of Multiplication is performed by Addi- 
 tion, and of Division by Subtraction. 
 
 Ex. 1 Find the product of 9 and 6 by means of logarithms. 
 
'18 RAY'S ALGEBRA, PART SECOND. 
 
 By the table (page 316) the log. of 9 is ... . 0.954243 
 
 the log. of 6 is ... . 0.778151 
 
 The sum of these logarithms is 1 .732394 
 
 and the number corresponding in the table is 54. 
 
 2. Find the quotient of 63, divided by 9, by means of log 
 arithms. 
 
 The log. of 63 is 1.799341 
 
 " log. of 9 is 0.954243 
 
 The difference is •-.... 0.845098 
 
 and the number corresponding to this log. is 7. 
 
 By means of logarithms 
 
 3. Find the product of 7 and 8. 
 
 4. Find the continued product of 2, 3, and 7. 
 
 5. Find the quotient of 85 divided by 17. 
 
 6. Find the quotient of 91 divided by 13. 
 
 Akt. 362. Resuming equation (1), (Art. 360), we have 
 
 Raising both sides to the m"' power, we find 
 
 But, by the definition (Art. 357), mx is the logarithm of N" ; 
 that is, m times log. N= log. N". Hence, 
 
 Peopeett III. — If we multiply the logarithm of a number by 
 any expmient, the product wiU he the logarithm of that fewer of the 
 given nvmber. 
 
 Abt. S63. Taking the same equation 
 o'=N, 
 and extracting the n"' root of both sides, we have 
 aS =N'i' 
 
 But, by the definition, (Art. 357), - is the logarithm of N» ; 
 
 n 
 1 1 
 
 that is, _ of log. N= log. N" . Hence, 
 n 
 
 Pkopektt IV. — If vie divide the logarithm of a nuniber by amy 
 index, the quotient will be the logarithm of that root of the given 
 number. 
 
 From this article and the preceding, we see that by means of 
 logarithms, the operation of raisinfr a number to any given powei 
 
LOGARITHMS. 319 
 
 IB performed by a simple muUiplication, and the extraction of any 
 root,' by a simple division. 
 Ex. 1 . Find the third power of 4 by means of logarithms. 
 
 The logarithm of 4 is 0.602060 
 
 Multiply by the exponent 3 3 
 
 The product is 1.806180 
 
 which is the logarithm of 64. 
 
 2. Extract the fifth root of 32 by means of logarithms. 
 
 The logarithm of 32 is 1.605150 
 
 Dividing by the index 5, the quotient is ... . 0.301030 
 
 which is the logarithm of 2, the required root. 
 
 Solve the following examples by means of logarithms • 
 
 3 . Find the square of 7. 
 
 4. Find the fourth power of 3. 
 
 5. Extract the cube root of 27. 
 
 6. Extract the sixth root of 64. 
 
 The preceding properties and examples will suffice to show the 
 great utility of logarithms in mathematical calculations. It is, 
 however, rather the province of algebra, to explain the principles 
 of logarithms, than their use in actual calculations, as the latter 
 requires a set of logarithmic tables, which are usually inserted 
 in works on Trigonometry, Surveying, &,c. 
 
 Art. 3G4. By means of negative exponents, we can also ex- 
 press the logarithm of fractions less than 1. Thus, in the com- 
 mon system, since 
 
 (10)-' =^5 =.1 , therefore —1 is the log. of .1 
 (10)-2=yj5 =.01 , " —2 " log. " .01 
 (10)-^=T(5'o<J =-001 , « -3 " log. « .001 
 (10)-'=T^f(,^=.0001 « —4 " log. " .0001 ; 
 &c., &c. 
 
 The logarithm of any fraction between one and one-tenth, for 
 tiample, seven-tenths, may be expressed thus, 
 
 log- (t'ij)= log- (toX7)= log. tV+ log- 7=-l+ log. 7. 
 In like manner the logarithm of any fraction between one-teath 
 and one-hundredth, may be expressed thus, 
 
 log- {ilii)= log. (-145X3)= log. ■j-Jb+ log. 3=— 3+ log. 3. 
 
320 RAY'S ALGEBRA, PART SECOND. 
 
 Similarly, for fractions between -, Jj and j^jsts> thus, 
 log- (TTi^T!)= log- Td'5U+ log. 1=— 3+ log. 4. 
 
 It is customary not to perform the subtraction thus indicated 
 but to unite the logarithm of the numerator of the decimal con- 
 sidered as a whole number, to the negative characteristic. Thus, 
 
 log. 0.7 =— 1+ log. 7=— 1.845098 
 
 log. 0.03 =—2+ log. 3 =—2.4771 21, 
 
 log. 0.004=— 3+ log. 4=— 3.602060. 
 
 Since the logarithm of .1 is — 1, of .01 is — 2, of .001 is — 3, 
 Knd so on ; therefore. 
 
 The characteristic of the logarithm of a decimal fraction is a 
 negative number, and is one more than the number of zeros immedi- 
 ately following the decimal point. 
 
 Art. 365. To explain the principle generally, by means of which 
 the logarithms of decimals are represented. 
 
 Let a represent a decimal fraction containing m zeros immedi- 
 ately following the decimal point, and n other places of figures ; 
 then the number of zeros in the denominator will be m-\-n, and 
 by the nature of decimals the fraction will be represented by 
 
 log. < — ^L — i = log. a—(m-\-n) log. 10= log. a— (m+re), 
 
 since log. 10t=l. 
 
 But, by supposition, a contains n figures ; hence, the character- 
 istic of its logarithm (Art. 358,) is n — 1 ; and if d represent the 
 decimal part of the log., the entire log. of a will be n — l-\-d. 
 Substituting this instead of log. a, we have 
 
 log. \ 1 =re— 1+d— (m+7i)=— (OT+l)+d. 
 
 Hence, to find the logarithm of any decimal fraction, find the 
 decimal part of the logarithm from the tables, as if the fraction were 
 a whole number, and unite to it u negative characteristic, greater by 
 unity than the number of zeros immMiately following the decimal 
 point. 
 
 Akt. 366. It is of the highest importance to the student to 
 make himself familiar with the application of the properties of 
 
LOGARITHMS. 321 
 
 logarithms (Arts. 360 to 304) to algebraic calculations. The 
 following examples will afford a useful exercise : 
 
 1. log. (a .1 . c .d. . )= log. a-\- log. h-\- log. c-\- log. d . . 
 
 2. log. /?-!?)= log. a+ log. l-\- log. c — log. d — log. s. 
 
 3. log. (a™ . 6" . c^ . )=m log. a-\-n log. l-\-p log. o. 
 
 — :: — I z=m log. a+n log. h — -p. log. c. 
 
 5. log. (a^ — x^)=\og. {(a-\-x){a — a;)]= 1. (a-|-a;)4-l (<> — ^). 
 
 6. log. ^a' — J!^=3 log. (a+a;)+2 log. (a — a;). , 
 
 7. log. {a? X V^)=3| log. a. 
 
 8. log. ^^^^=^{log. (a-a;)-3 log. (a+x)|. 
 Art. 367. Let us resume the equation 
 
 in which x is the logarithm of N. 
 1st. If we make x=l, we have 
 
 a'=N=ffl, hence log. (Z=l ; 
 that is, whatever he the base of the system, its logarithm in that system 
 isl. 
 2nd. If we make a;=0, in the equation (r*=N, we have 
 a°=N=l , hence log. 1=0 ; 
 that is, in any system the logarithm of 1 is 0. 
 
 Art. 368. In the equation o'^=N, consider o^l, as in the 
 
 common system, and suppose x negative, we then have 
 
 As X increases the value of the fraction _ will diminish ; and 
 
 a" 
 
 when X is infinite, the value of the fraction becomes 0; that is, 
 
 JL=o-«= =0; or, log. 0=— 00 . 
 a* 
 
 Hence, the logarithm of in a system whose base is greater than 
 1 is an infinite number and negative. 
 
 In the Naperian, as well as the common system of logarithms, 
 the base is greater than ] ; but it may be shown that in a system 
 whose base is less than 1 , the logarithm of is infinite and posiliim 
 
322 RAY'S ALGEBRA, PART SECOND. 
 
 Art. 369. In the equation o'=N, every positive value of a 
 gives a corresponding positive value of N. 
 
 If X is negative, we have o~'i=_=N. Hence, for every nega- 
 
 a' 
 
 live value of x the corresponding value of N is also positive. 
 Therefore, whether x is positive or negative, the corresponding 
 value of N is positive; hence, Negative numbers have no real loga- 
 rithms. 
 
 COMPTJTATIOK OF L O G A K I T H M S . 
 
 Art. S70. Before proceeding to explain the methods of com- 
 puting Idgarithms, we may observe that it is only necessary to 
 compute the logarithms of the prime numbers. 
 
 This is obvious when we consider that every composite number 
 is the product of two or more prime numbers, and that the loga- 
 rithm of any product is equal to the sum of the logarithms of its 
 factors. (Art. 360.) 
 
 For example, if we have the logarithms of 1, 2, 3, 5, 7, we can 
 find the logarithms of all composite numbers produced by the mul- 
 tiplication of two or more of these numbers together. Thus, 
 
 4=22 . hence, log. 4=2 log. 2, (Art. 362) ; 
 
 6=2X3; " log. 6= log. 2+ log. 3; 
 
 8=2' ; " log. 8=3 log. 2 ; 
 
 9=3= ; " log. 9=2 log. 3 ; 
 
 10=2x5; " log. 10= log. 2+ log. 5, 
 
 12=3X4; " log. 12= log. 3+ log. 4 ; 
 
 We can proceed in a similar manner to find the logarithms of 
 14, 15, 16, 18, 20, 21, 24, 25, 27, 28, 30, and so on. 
 
 Exercise 1. Suppose the logarithms of the numbers 2,3,5 
 and 7 to be known ; show how the logarithms of the numbers 
 just named may be found. 
 
 2. Of what numbers between 30 and 100, may the logarithms 
 be found from those of 2, 3, 5, and 7; and why ■? 
 
 Ans. Of 23 different numbers, from 32 to 98. 
 
 Aet. 371. In the common system the equation o'^N (Art 
 357) becomes 10''=N. 
 
 If we multiply both sides by 10, we have 
 10»X10=10'+>=10N; 
 also, ]0'X100=10'X10'=10'+«=100N. 
 
LOGARITHMS. 323 
 
 Hence, in the common system, the logarithm of any number 
 will become the logarithm of 10 times that number, by increasing 
 the characteristic by 1; of 100 times by increasing the charac- 
 cBristic by 2, and so on. 
 
 Thus, the log. ot 3 is 0.477121, 
 
 30 " 1.477121, 
 
 300 " 2.477121. 
 
 Also, the log. of .2583 is —1.412124, 
 
 " 2.683 " 0.412124, 
 
 25.83 " 1.412124. 
 
 Akt. 372. If we compare the different powers of 10 with 
 their logantlims in the common system, we have 
 
 numbers 1 , 10 , 100, 1000, 10000, 
 logarithmii fi , 1 , 2,3, 4 , and so on. 
 
 Hence, in the common system, while the numbers are in geo- 
 metrical progresiwn, their logarithms are in arithmetical progres- 
 sion. Therefore, if we take a geometrical mean between two 
 numbers, and an Arithmetical mean between their logarithms, the 
 latter number wiii be the logarithm of the former. Thus, the 
 geometrical mean between 10 and 1000 is ^10x1000=100, 
 and the arithmetical mean between their logarithms, 1 and 3, is 
 (l+3)-i.2=2. 
 
 In general, if N and N' are two numbers, and x and a' their 
 logarithms in the common system, then the 
 
 log. of JNTT is ^?±i. 
 
 By means of this principle, the common, or Briggean, system 
 of logarithms was originally calculated. To exemplify the 
 method of operation, let it be required to calculate the logarithm 
 of 5. 
 
 First. — The proposed number lies between 1 and 10; hence, 
 its logarithm will lie between and 1. 
 
 The geometrical mean between 1 and 10 is ,^(1X10) 
 =3.162277; the arithmetical mean between and 1 is (0+1 
 ^2=0.5. 
 
 Hence, the log. of 3.162277 is 0.5. ' 
 
 Secondly. — Take the numbers 3.162277 and 10, and their 
 logarithms .5 and 1 , we find 
 
324 RAY'S ALGEBRA, PART SECOND. 
 
 the geometrical mean is V(3-162277xl0)=5.623413; 
 the arithmetical mean is (.5-l-l)-r-2:=0.75. 
 
 Hence, the log. of 5.623413 is 0.75. 
 
 Thirdly.— Take the numbers 3.162277 and 5.623413, and 
 their logarithms 0.5 and 0.75, we find 
 
 the geometrical mean is ^(3.162277x5.623413)=4.216964 ; 
 the arithmetical mean is (.5+.75)-r-2=0.625. 
 
 Hence, the logarithm of 4.216964 is 0.625. 
 
 Fourthly.— Take the numbers 4.216964 and 5.623413, and 
 their logarithms 0.625 and 0.75, we find 
 
 the geometrical mean is V(4.216964x5.623413)=4.869674; 
 the arithmetical mean is (.625+.75)-s-2=0.6875. 
 
 Hence, the logarithm of 4.869674 is 0.6875. 
 
 By continuing this process, observing always to take the two 
 numbers nearest to 5, one of which is less and the other greater, 
 and finding their ffeomeirical mean, and the corresponding arith- 
 metical mean of their logarithms, at each step we shall obtain a 
 number nearer to 5 than either of the preceding, with its corres. 
 ponding logarithm. And after twenty-two operations we ob- 
 tain the number 5.000000+, and its corresponding logarithm 
 0.698970+. 
 
 Having the logarithm of 5 we readily find that of 2, 
 
 since 2=V, and log. 2= log. 10— log. 5 =1—0.698970 
 =0.301030. 
 
 We might now proceed to find the logarithm of 3 by taking 
 the numbers 2 and 3.162277, and their logarithms 0.301030, 
 and 0.5, and pursuing a process similar to that used in finding 
 the logarithm of 5. But the method of series is much shorter, 
 and is the one now generally used. 
 
 Art. Sya. Logarithmic Series. — The most convenient 
 method of computing logarithms is by means of Series, which we 
 shall now proceed to explain. 
 
 Let a; be a number whose logarithm is to be expressed in a series, 
 and let us apply the method of Indeterminate Coefficients (Art. 
 314^. If we assume 
 
 log. x=A+Ba;+Cx'+Da:3+, &,c., 
 and make a;=0, we have 
 
 log. 0=A. Butlog. 0=QO (Art. 368); hence, 
 00=A, which is absurd. 
 
LOGARITHMS. 325 
 
 If we assume log. x=Ax-\-Bx'-\-Cx'-\-, &.c., and make x=0, 
 we have log. 0=0 ; that is, (Art. 368), 
 00=0, which is also absurd. 
 Hence, it is impossible to develop the logarithm of a number in 
 powers of that number. 
 
 But if we assume 
 
 log. (l+x)=Aa4-Bj;'+Cr'+Dar'+, &o. . . (1) 
 and make a;=0, we have 
 
 log. 1=0, which is correct, (Art. 367). 
 In like manner, also assume 
 
 log. il+z-)—Az+'Bz'+Cz^-\-Dz*+, &c. . . (2) 
 Subtracting equation (2) from (1) we get 
 
 log. (1+a;)— log. (l+2)=A(j;— 0)+B(a;'— z') 
 +C(x'—z>)+, Sic. . . (3). 
 
 The second member of this equation is divisible by x — z (Art. 
 83) ; let us reduce the first member to a form in which it shall 
 also be divisible by the same factor. 
 
 Since the logarithm of a fraction is equal to the logarithm of 
 the numerator, minus the logarithm of the denominator (Art. 
 361), therefore, 
 
 log. (1+a:)- log. (1+^)= log. ( J^ ) . 
 But, by division, we find '"'''= 1-}-^ ^; therefore, 
 
 1+2 1+Z 
 
 H.(yj)=,.,.(,+^j. 
 
 Now regarding ■ as a single quantity, we may assume 
 
 1+z 
 
 "'^- ( ^+fS ) =-fS+« ( wz ) +' ( fS ) +' ^^ 
 
 Substituting this development in the place of log. (l+ar) 
 — log. (1+z), in equation (3), and dividing both sides by x — z, 
 Ko obtain 
 
 A. J_+B.-^=:!_+c.^f=:!.)J+, &o., 
 
 1+z^ (l+z)=^ (1+^)'^ 
 =A+B(a;+2)+C(x'+a;2+2')+, &c.. 
 
326 RAY'S ALGEBRA, PART SECOND. 
 
 Since this equation, like the preceding, is true for all values of 
 X and z, it must be true when x^z. 'Making this supposition, we 
 have 
 
 A.-l_=A+2Bar+3Ca;'44Da:'+5Ei<+, &c. ; 
 
 \-\-x 
 
 or, performing the division of 1 by 1+x, we have 
 
 A(l— x+x^— ajs+a;"— . . . )=A+2Ba:^3C.T'+4Dx'+ . 
 
 Equating the coefficients of the like powers of x (Art. 314), 
 we obtain 
 
 A=A, — A=2B, A=3C, — A=4D. . . . 
 whence, 
 
 A=A, B=-:^, C=4, D=-4. . . 
 2 3 4 
 
 The law of this series is obvious, the coefficient of the »"* term 
 
 A 
 
 being rt_ , according as n it odd or even. 
 n 
 
 Hence, log. (l-|-a;)=Aa; — ^x^-\--x^ — _a:^-|-- • • • 
 
 ■*i2 iflB 'vi4 mS 1^0 . 
 
 =A(a;— i+^— ^+1— ?-+. . . . ) (4) 
 2 3 4 5 6 
 
 There still remains one quantity, A, undetermined. This is as 
 it should be, for the question to find the logarithm of a givet 
 number is indeterminate, unless the base of the system be given. 
 The value of the quantity A may be considered as dependent or. 
 the base of the system, so that when A is given the base may be 
 determined ; or, when the base is known, A may be determined 
 
 If we denote the series in the parenthesis in equation (4) by 
 x', we may write 
 
 log. (l-\-x)=Ax' . 
 
 Hence, the logarithm of a number consists of two factors, one 
 of which depends on the number itself, and the other on the base 
 of the system in which the logarithm is taken. That factor 
 which depends on the base is called the MoEUUrs of the system of 
 logarithms. 
 
 Lord Napier, the inventor of logarithms, assumed the modulus 
 equal to unity, and the system resulting from such a modulus, U 
 called the Naperian system. 
 
 Designating the logarithms in this system by log'., we have 
 
 log'. (l+x)=*-^+^-J+, &c. (5) 
 
LOGARITHMS. 32T 
 
 By making is=Q, 1, S, 3, &c., we may obtain from this equa- 
 tion the Naperian logarithms of all numbers. 
 
 Thus, if a;=0, we find log'. 1=0, as in Art. 367. 
 
 If we make x=l , we have 
 
 log'. 2=l-^+i-l+J_, &c. 
 
 Akt. Sf*. The preceding series converges so slowly that it 
 would be necessary to take a great number of terms to obtain a 
 near approximation. But we may obtain a more converging 
 series in the following manner : 
 
 Resuming equation (5), 
 
 nr* 'V«2 'v«3 'yi4 fr%S 
 
 log'- (l.+x)=-— -+-—-+-— , &,c. . . (5). 
 Substituting — x for x, in this equation, we obtain 
 
 log'. (1-^)=—-—^— -—-—-—, &c. . (6). 
 
 Subtracting equation (6) from (5), and observing that 
 
 log'. (1+x)— log'. (1-^)= log'. ( llt^ ) , we have 
 
 \ 1 — x/ 
 
 log'. L+?=2/'?4-^+^+^4-^'+. . . 1 
 
 Since 1+^=1+-^^, let 1±?=1+1, .-. x=_l_, 
 1 — X 1 — X 1 — X z 2a-|-l 
 
 and log'. J+^= log'. ( 1+J ) = log'. ( !±1 ) 
 
 = log'- (2+1)— log'. Z. 
 
 By substitution, the preceding series becomes 
 
 log'. (2+1)— log'. 2=2 J_Jl_ 4- I + ^ 4- . \ . 
 
 ^ ^^ ^ s 122+1^3(22-1-1)3^5(224-1)=^ ) ' 
 
 or, log.' 
 
 (2-1-1)= log'. 2-1-2 I _1_+ L__-l- L__-|- . . I n\. 
 
 ' ^ -^ ^ ^ 1 22-f 1^3(22-1-1)5^5(224-1 )s^P'' 
 
 Aet. 375. By means of this series the Naperian logarithm 
 of any number may be computed, when the logarithm of the 
 preceding number'is known. But the log', of 1 is 0, (Art. 367) ; 
 therefore, making 2=1, 2, 4, 6, &c., we obtain the following 
 
328 
 
 RAV'S ALGEBRA, PART SECOND. 
 
 Naferian or Htferbolic Logarithms. 
 log'. 2=log'. 14-2 ^]:+_Ji_+J_H — L-4-..i =0.693147 
 
 log'. 3=log'. 2+2 H+J— + ^ + ^ +..1 =1.098612 
 
 log'. 4=2. log. 2 =1.386294 
 
 log'. 5=log'. 4+2 ^?:+_J_+J_+_J_+..i =1.609438 
 
 log'. 6=log'.2+ log'. 3 =1.791759 
 
 log'. 7=log'. 6+2 ^-L+_i_+_i__+. . . i =1.945910 
 ^ ^ (13^3 • 13'^5 • 13'^ J 
 
 log'. 8=3 log'. 2, or log'. 2+ log'. 4 =2.079442 
 
 log'. 9=2 log'. 3 =2.197225 
 
 log'. 10= log'. 2+ log'. 5 =2.302585 
 
 In this manner the Naperian logarithms of all numbers may 
 be computed. 
 
 When the numbers are large their logarithms are computed 
 more easily than in the case of small numbers. Thus, ii) calcu- 
 lating the logarithm of 101, the first term of the series gives the 
 result true to seven places of decimals. 
 
 Art. 376. To explain, the method of computing common logo- 
 rithms from Naperian logarithms. 
 
 We have already found (Art. 373, Equation 4), 
 
 (M >*<2 lyiS •!/•* V^ T* \ 
 
 i 2+1-1+56+- • • )■ 
 
 Denoting the Naperian logarithm by an accent, we have 
 
 log'. (l+a;)=A'/'5— ?!+^— '^+?!— ?!+. . . ) 
 ^ ^ "^ U 2^ 3 4^ 5 6^ / • 
 
 Since the series in the second members are the same, vye have 
 
 log. (l+a:):Iog'. (l+ar)::A:A'. 
 
 Therefore, the logarithms of the same number, in two differemi 
 lystems, are to each other as the moduli of those systems. 
 
 But in Napier's system the modulus A'=l, Therefore, 
 log. (l+a:)=A log'. (1+r). 
 
LOGARITHMS. 329 
 
 Hence, to find the common logarithm, of any number, multiply the 
 Naperian logarithm of the number by the modulus of the common 
 lystem. 
 
 It now remains to find the Modulus of the common system. 
 From the equation, log. (l+a;)=A. log'. (l-\-x), 
 
 we find A='°g- (^+^) 
 log'. (1+x)- 
 
 Hence, the modulus of the common system is equal to the common 
 logarithm of any. number divided by the Naperian logarithm of the 
 same number. 
 
 But the common logarithm of 10 is 1, and we have calculated 
 the Naperian logarithm of 10, (Art. 375) ; therefore, 
 
 A =1%I^= \ =.4342944, 
 
 log.' 10 2.302585 
 
 which is the modulus of the common system. 
 
 Hence, if N is any number, we have 
 
 com. log. N=.4342944x Nap. log. N. 
 
 On account of the importance of the number A, its value has 
 been calculated with great exactness. It is 
 
 A=.43429448190325182765. 
 
 Art. Syy. To calculate the common logarithms of number), 
 directly. 
 
 Having found the modulus of the common system, if we multi- 
 ply both members of equation (7), Art. 374, by A, and recollect 
 that Ax Nap. log. N= com. log. N, the series becomes 
 
 log. (0+1)= log. 0+2A j 1 + 1 
 i2z4-l d(2z4 
 
 1 
 
 2z+l 3(2z+l)' ' 5(22+1)5 ' S ■ 
 
 Or, by changing z into P, for the sake of distinction, and put- 
 ting B, C, D, &c., to represent the terms immediately preceding 
 tliose in which they are used, we have 
 
 log. (P+l)= log. P+ ^^ + ? +__i2 
 
 6 V -r ; s ^2P+1^3(2P+1)2 5(2P+1V 
 
 5D , 7E , 9F 
 
 -, &.C. 
 
 ' 7(2P+1)2 9(2P+1)= ' 11(2P+1)' 
 We sliall now exemplify its use in finding the logarithm of 2. 
 
 Hare P=l, and 2P+1=3. 
 
 28 
 
330 RAY'S ALGEBRA, PART SECOND. 
 log. P ' =log:l =.00000000; 
 
 '^^ .86858896 =.28952965; (B.) 
 
 2P+1 3 ■' 
 
 ? ^ -28952965 =.01072332; (C.) 
 
 3(2P+1)» 3X3' 
 
 ^^ 3 X -01072332 =.00071489; (D.) 
 
 l.(2P+iy 5X3' 
 
 5D - 5X.00071489 _ 00005674; (E.) 
 
 7(2P+1)' ~ 7X3' 
 
 _J1 ^X -00005674 _ _ _ =.00000490; (F.) 
 
 9(2P+1)' ~ 9X3' 
 
 9F 9 X -00000490 ._ =.00000045; (G.) 
 
 11(2P+1)'~ 11X3' 
 
 IIG llX-00000045 =.00000004; (H.) 
 
 13(2P+1)'~ 13X3' 
 .-. common logarithm of 2 =.30102999. 
 
 Exercise. In a similar manner let the pupil calculate the com- 
 mon logarithms of 3, 5, 7, and 11 . 
 For the results to 6 places of decimals, see the Table, page 316. 
 
 Akt. 378. To find the base of the Naperian system of logo- 
 Hthms. 
 
 If we designate the base by e, we have, (Art. 376), 
 
 log. e : log', e : : A : A'. 
 But A=.4342944, A'=l, and log'. e=l, (Art. 367) ; 
 
 hence, log. e : 1 : : .4342944 : 1, 
 
 whence log. e=.4342944, 
 
 But since we have explained the method of calculating common 
 logarithms, they are supposed to be known, and we may use them 
 to obtain the number of which the logarithm is .4342944, which 
 we shall find to be 
 
 e=2.71828128. 
 
 We thus see that in both the common and the Napierian sys- 
 tems of logarithms, the base is greater than unity. 
 
 Brigg's logarithms are used in the ordinary operations of multi« 
 plication, division, &c., and hence are called common logarithms. 
 Napier's logarithms are used in the applications of the Calculus. 
 These are the only systems much used. 
 
POSITION. 331 
 
 Akt. 3'3'9. The student may prove the following theorems : 
 
 1. No system of logarithms can have a negative base, or have 
 unity for its base. 
 
 2. The logarithms of the same numbers in two different sys- 
 tems have the same ratio to each other. 
 
 3. The difference of the logarithms of two consecutive num- 
 bers is less as the numbers themselves are greater. 
 
 SINGLE AND DOUBLE POSITION. 
 
 Note. — On account of the use made of Double Position in the solu- 
 tion of exponential and other equations, it becomes necessary to explain 
 the principles on which it is founded. We shall also explain Single 
 Position. 
 
 Abt. 3§0. Single Position. — The Rule of Single Position 
 is applied to the solution of those questions in which there is a 
 result which is increased or diminished in the same ratio with 
 some unknown quantity which it is required to find. Of this 
 class are all questions which give rise to an equation of the form 
 ax=m (1). 
 If we assume x' to be the value of x, and denote by m' the 
 result of the substitution of x' for x, we have 
 ax'—m' (2). 
 Comparing equations (1) and (2), we have 
 m' : m : : ax' : ax : x' : X ; 
 that is. As the result of the supposition ts to the result in the question, 
 so is the supposed number to the number required. 
 
 Art. 381. Double Position. — The Rule of Double Position 
 is applied to those questions in which the result, although it is 
 dependent on the unknown quantity, does not increase or dimin- 
 ish in the same ratio with it. The class of questions to which it 
 is particularly applicable, gives rise to an equation of the form 
 ax-\-b=m (1). 
 If we suppose x' and x" to be near values of x, and e' and e'' 
 to be the errors, or the differences between the true result and 
 tlie results obtained by substituting x' and x" for x, we have 
 ax' -\-b=m-{-e' (2), 
 aif'-\-b=m-\-e" (3). 
 If we subtract equation (1) from (2), and (3) from (2), we have 
 a(x' — X )=e' (4). 
 
 a(a/— a!")=e'— e" (5\ 
 
333 RAY'S ALGEBRA, PART SECOND. 
 
 From these equatione, we easily obtain 
 
 x'^-x" X — X ^f>^ 
 
 e' — e" e' 
 By subtracting equation (1) from (3) we also find 
 a(x" — x')^e", and thence, 
 
 X X X X t^\ 
 
 e' — e" e" 
 
 Hence (Art. 263), The difference of the errors is to the differ 
 ence of the two assumed numbers, as the error of either result is to 
 the difference letween the true result and the corresponding assumed 
 number. 
 
 When the question gives rise to ah equation of the form 
 ax-{-b=m, this rule gives a. result absolutely correct ; but when 
 the equation is of a less simple form, as in exponential equations 
 (Art. 383), the result obtained is only approximately true. 
 
 Cor. The value of x, found either from equation (6) or (7), is 
 x=^~ — - — . This, expressed in ordinary 
 
 language, furnishes the common arithmetical rule. 
 
 EXPONENTIAL EftUATIONS. 
 
 Aet. 382. An exponential equation is an equation in which the 
 unknown quantity appears in the form of an exponent or index, aa 
 a'=b, xi'=a, a'^=c, &c. 
 Such equations are most easily solved by means of logtirithms 
 Thus, in the equation 
 
 a'=b 
 
 if we take the logarithms of both members, 
 we have a; log. o= log. b, 
 
 or, xJ^S:! 
 log. a 
 
 Ex. 1 . What is the value of x in the equation 2''=64:'t 
 
 Here x log. 2= log. 64. 
 
 , log. 64 1.806180 R . 
 
 whence, a;= -s = =6. Ans. 
 
 log. 2 .301030 
 
 Aet. 383. If the equation is of the form ai':=a, the value of 
 r may be found by Double Position as follows : 
 
 Find by trial two numbers nearly equal to the value of x ; sub« 
 
EXPONENTIAL EQUATIONS. 333 
 
 atitute them for x in the given equation, and note the results. 
 Then, 
 
 As the difference of the errors ; 
 
 Is to the difference of the two assumed numbers ; 
 
 So is the error of either result ; 
 
 To the correction to he applied to the corresponding assumed num- 
 ber. 
 
 Ex. 1. Given afi=100, to find the value of x. 
 
 The value of x is evidently between 3 and 4, since 3'=27, and 
 4''=256 ; hence, taking the logarithms of both sides of the equa- 
 tion, vre have 
 
 x\og.x= log. 100=2. 
 
 By trial, we readily find that x is greater than 3.5, and less 
 than 3.6; then let us assume 3.5 and 3.6 for the tviro numbers. 
 
 Second Supposition. 
 
 2=3.6; log. ^=.556303 
 multiply by 3 .6 we find 
 K. log. a; =2.002690 
 
 true no. =2.000000 
 
 error + .002690 
 
 Diff. results : Diff. assumed nos. : : Error 2nd result : Its cor. 
 .098452 : 0.1 : : .002690 : .00273 
 
 Hence, a;=3.6— .00273=3.59727 nearly. 
 
 By trial we find that 3.5972 is less, and 3.5973 greater than 
 the true value ; and by repeating the operation with these num- 
 bers we would find a;=3 .5972849 nearly. 
 
 EXAMPLES FOR PRACTICE. 
 
 2. Given 20»^=100, to find x. A?^. 1=1.53724. 
 
 3. Given 100»'=250, to find ar. Ans. a;=l. 198 97. 
 
 4 Given af=5, to find x. Ans. x=2. 129372. 
 
 5. Given k*=42.8454, to find a;. Aras. x=3.2164. 
 
 G. How many places of figures will there be in the number 
 expressing the 64" power of 2? Am. 20. 
 
 7. Given a"+'^c, to find x. Ans. x= !?Si=^:i?S:_'' 
 
 b. log. a 
 
 First 
 
 Supposition. 
 
 a:=3.5; 
 
 log. a:= .544068 
 
 multiply by 3.5 we find 
 
 X. log. X 
 
 =1.904238 
 
 true no. 
 
 -2.000000 
 
 error 
 
 =—.095762 
 
834 RAY'S AiGEBRA, PART SECOND. 
 
 8 . Given a^.}f-=:c, to find x 
 
 Ans. x=. ^ — 
 
 m. log. a-}" "■ ^°S- ^' 
 
 9. Given c'"=a.i"^"', to find x. 
 
 . log. a — log. b 
 
 Ans. x= & 5 _ 
 
 m. log. c — 71. log. i 
 
 10. Given a;-(-y=a, and m'~i=n, to find a; and y. 
 
 Aju. x=h(.a-\- log. Ji-:- log. m), y=l(a — log. n-j- log. m). 
 
 ] 1 . Given a'.bi^c, and my=nx, to find x and y. 
 
 , TO. log. c. ji. log. c 
 
 m. log. a-\-n. log. 6 m. log. a-\-n. log. J' 
 
 12. Given 2^3»=2000,and3^=5a;,tofind the values ofaranda, 
 
 Ans — 3(3+ log. 2) ._ 5(3+ log. 2) 
 
 3 log.' 2+5 log. 3' 3 log. 2+5 log. 3' 
 
 13. Given a='— 2a'=8, to find a. Ans, x= ^ '°^" ^ . 
 
 log. a 
 
 Suggestion. — This is a quadratic form, therefore let a*=!/ and com 
 plete the square. 
 
 14. Given 2=='+2'=12,tofinda;. Ans. as=l.r,8496. 
 
 15. Given 2a<'+o»=a«', to find x. 
 
 Ans. x J°S- (V2+I) 
 2 log. a 
 
 16. Given 0^+1=6, to find *. 
 
 log. a 
 
 17. Given 3fl=y', and x'=y', to find a; and y. 
 
 Ans. x=2\, y=3| 
 
 18. Given (a'— 6')«^-"=(a— 6)'', to find ar. 
 
 Ans. x=l+^°g- ("-^^ 
 log. (o+t) 
 
 19. Given (a<— 2a'P+M)'-'=(a— i)»(a+i)-=', to find a;. 
 
 Atw. x= '°g- (°-^> 
 log. (a^-i) 
 
 20. Given 3!S:=y', and a?'=y«, to find x and j/. 
 
 Ans. a:=f ^ \^„y={t ji£^ 
 
 21. Given S^^"**"!^ W2OO, to find x. 
 
 Ans. x=4.33, or —0.38 
 
INTEREST AND ANNUITIES. 33a 
 
 INTEREST AND ANNUITIES. 
 
 Akt. 384. The solution of all questions connected with inter- 
 est and annuities, may be simplified, and also generalized, by means 
 of algebraical formul«e. 
 
 We shall employ the following notation : 
 Let P= the principal, or sum at interest in dollars. 
 r= the interest of 1 $ for cme year. 
 /= the time in years that P draws interest. 
 A= the amount of principal and interest, at the end of I 
 years. 
 
 Note. — It must be recollected that r is not the rate per cent.,bxit only 
 the hundredth part of it. Thus, at 5 per cent., r=.05$, at 6 per cent. 
 r^.06$ ; and so on. 
 
 Akt. 3§5. Simple Interest. — Since the interest of the same 
 sum for 2 years, is twice the interest for 1 year ; for 3 years, three, 
 times the interest for 1 year, and so on ; therefore, if 
 
 r= the interest of 1 $ for one year, 
 tr= the interest of 1$ for t years, 
 Pir= the interest of P$ for t years, 
 .-. A=P+Pir=P(l+i?-) (1). 
 
 From this equation, any three of the quantities P, r, t, A, being 
 given, the fourth may be found. Thus, 
 
 P=^^, t=^=?, r=^=P 
 1+tr Pr Pi 
 
 Examples may be taken from any treatise on arithmetic to 
 illustrate these formulae. 
 
 Art. 386. Compound Interest. — Let R=l-|-r, the amount 
 of 1$ for one year ; then at the end of the first year, R may be 
 considered as the principal or sum due, and since the amount is 
 proportional to the principal, that is, the amount of R$ for 1 
 year is R times the amount of 1 $ for the same time ; therefore. 
 
 1 : R : : R : R^, the amount of 1$ in 2 years. 
 
 1 : R . : R2 : R', the amount of 1$ in 3 years. 
 And in like manner R' is the amount of 1$ in < years. 
 Then, since for the same time the amount is proportional la 
 
336 RAT'S ALGEBRA, PART SECOND. 
 
 the principal, the amount of P§ will be P times the amount of 
 1$. Hence, 
 
 A=P.R'=P(1+)-)' ; whence, 
 
 log. A= log. P+t. log. (1+r) (1). 
 
 log. P= log. A—t. log. (1+r) (2). 
 
 log. A— log. P ,g. 
 
 log. (1+r) 
 
 log.(l+r)= ^°g-A-'°g-P (4). 
 
 Cor. 1.— The interest =A— P=PR'— P=P(R'— 1). 
 
 Cor. 2. — If the interest is paid half-yearly, then 24 will be the 
 number of payments, and _ the rate of interest ; hence, in this 
 case we have 
 
 ''=H'+iy' ^'>- 
 
 paid quarterly, A=P I 1+^ J (6). 
 
 Cor. 3. — From the equation A=P.R', we can readily find the 
 time in which any sum at compound interest, will amount to 
 twice, thrice, or m times itself. 
 
 Thus, if A=2P ; then 2P=PR' .-. R'=2, and 4= }^^. 
 
 log. R 
 
 if A=3P ; then R' =3, and fc= log. 3 -i- log. R ; 
 
 if A=mP ; then R' =m, and t= log. m-i- log. R. 
 
 Ex. Let it be required to find the time in which any sum will 
 double itself at 10 per cent, compound interest. 
 Here r=.10, R=l+r=l+.10=1.10; 
 
 hence, 4=l2gLg-= -g0103° =7.272yrB. Av.. 
 log. R .041393 ^ 
 
 Aet. SSV. The increase of the population of a country may 
 be computed on the same principles as compound interest. Thus 
 if we know the population at two difierent periods, we may find 
 the rate of increase ; or, if we know the population at any given 
 period, with the rate of increase, we may determine lie popular 
 tion at any future period. 
 
INT^IREST AND ANNUITIES. 337 
 
 Ex. The population of the United States in 1790 waa 
 3900000, and in 1840, 17000000. Required the average rate 
 of increase for each 10 years. 
 
 Hero there are 5 periods of 1 years each. Hence, by com- 
 paring the quantities given, with those in equation (4), Art. 380, 
 «B have A=17000000, P=3900000, and t=5. 
 
 log. A, (see table, page 316), 7.230449 
 
 log. P 6.591065 
 
 Divide by 5 5)0.639384 
 
 log. (1+r) 1.342 0.127877 
 
 Hence r=1.342—l=.342=:34^ per cent. Ans. 
 
 Abt. 38§. Compound Discount.— The preseni value of a 
 sum P, due t years hence, reckoning compound interest, is easily 
 obtained Irom Art. 386. 
 
 Let P'= the present worth, then in t years, P' at compound 
 'nterest, will amount to P, .-. 
 
 P=P'(l+r)', .-. P'=^^^,. (1). 
 
 (1+r)' 
 
 Let D= Comp. Discount, then D=P— P'=P— —^ (2). 
 
 (1+r)'- 
 
 From equation (1), log. P'= log. P—t log. (1+r) (3). 
 
 Akt. 389. Annuities Certain. — An Annuity is a sum of 
 money which is payable at equal intervals of time. 
 
 When the annuity has already commenced, it is said to be in 
 possession ; but should it not begin until some particular event 
 has happened, or a certain number of years has elapsed, it is 
 then called a deferred annuity, or an annuity in reversion. 
 
 An annuity certain is one which is limited to a certain number 
 of years ; a life annuity is one which terminates with the life of 
 any person, and s. perpetuity, or perpetual annuity, is one which is 
 •ntirely unlimited in its duration. 
 
 All the computations relating to annuities are made according 
 ts compound interest. 
 
 Art. 390. To find the amount of an annuity in any number of 
 years, at compound interest. 
 
 Let a denote the annuity, p the present value, m the amount : 
 and r, R, /, the same as in the preceding articles. 
 29 
 
S38 
 
 RAT'S ALGEBRA, PART SECOND. 
 
 The first annuity a, becomes due at the end of the year, ana 
 thus, in i — 1 years, will amount to aR'' (Art. 386). The sec- 
 ond annuity becomes due at the end of two years, and in t — 2 
 years it will amount to oR'-. In like manner, the third annuity 
 will amount to aR'"', and so on to the last annuity, which is sim- 
 ply a. Hence, the entire amount is the sum of a geometrical 
 series, whose first term =aR'~', common ratio =R, and last term 
 ■^fl ; therefore, by reversing the order of the terms, we have 
 OT=o+aR+aR2+aR'-|-. . . . -(-aR'-'+aR'"'. 
 
 .-. (Art.297),m==.g!=i=a ^^+'-)'-l 
 R — 1 r 
 
 If the annuity is to be received in hilf-yearly installments, 
 
 then we have m=_ / '^- ' =o. '■ ^ - ^ 
 
 2 )r 
 
 xr . 1 a (1+^'-)'"— 1 (l+jr)<'— 1 
 If quarterly, m=z_ '■ ^ '■ -' — .^a. ^ '^ ' 
 
 4' ir r ■ 
 
 4 
 
 Cor. If d dollars are placed out annually for n successive years, 
 and the whole be allowed to accumulate at compound interest, 
 then will the amount A=dR-|-dR2-(-dR3-f .... -^iK". 
 
 A=dR(l+R+R2+. . . +R''-')=dR.?lZll 
 
 R — 1 ' 
 
 Art. 391. To find the present valve of an annuity to he paid t 
 years, at compound interest. 
 
 Let p denote the present value of the annuity a ; then the 
 amount of jd$ in t years ==pR' (Art. 386), and the amount of the 
 
 annuity a in the same time is (Art. 390) a. ; but these two 
 
 R — 1 
 
 amounts must be equal to each other ; hence, we get 
 
 T,, R'— 1 , R'— 1 a I , 1 \ 
 
 pR'=o , and B=a = I 1 — _ I 
 
 ^ R— 1 ^ R'(R— 1) R— iV R'/- 
 
 Cor. If the annuity is to continue forever, t is infinite, and 
 
 therefore R' is infinitely great, and _ vanishes ; 
 
 , a a 
 
 hence, p= =_ 
 
 ■^ R— 1 r 
 
 Aet. 392. To find the present value of an annuity in reversion ; 
 that is, an annuity which is to comment at the end of n years, and la 
 continue t years. 
 
mi ^iREST AND ANNUITIES. 339 
 
 By Art. 391, the present value of the annuity for Tir-\-t years, is 
 
 — — I 1 — I , and the present value 
 
 of the annuity for n years is | 1 — — | ; and the difference 
 
 ■^ •^ R— 1 ^ R" y 
 
 f these two sums is obviously the value in reversion, 
 
 R— 1 \ R" R»+' / rR" \ Wr 
 
 If the annuity is payable /orerer after the expiration of wears, 
 then the value of the reversion of the perpetuity is (since t is 
 
 Infinite), p= — . 
 
 EIAMPLES IN INTEREST AND ANNUITIES. 
 
 1. What is the amount of 1$ for 100 years, at 6 per cent, per 
 annum, compound interest 1 Ans. $339.30. 
 
 2. How many figures will it require to express the amount of 
 1$ for 1000 years, at 6 per cent, per annum, compound interest i 
 
 Ans. 26. 
 
 3. How many years will it require for any sum of money to 
 double itself at compound interest, at the rates of 5, 6, 7, and 8 
 per cent, per annum respectively 1 
 
 Atis. 14.2066, 11.8956, 10.2447, and 9.0064 yrs. 
 
 4. Find in what time, at compound interest, reckoning 5 per 
 cent, per annum, $10 will amount to $100. Ans. 47.14 yrs. 
 
 5. If P$, at compound interest, amount to M$ in t years, what 
 Bum must be paid down to receive P$ at the end of t years ? 
 
 Ans.^ 
 M 
 
 6. Three children. A, B, C, who come of age at the end of 
 a, b, c, years, are to have a sum of money $P divided among them, 
 so that their shares being placed at compound interest, each shall 
 receive at coming of age the same sum. Find the share of A, 
 
 p 
 
 the youngest. Ans. _ . 
 
 ' =■ l-j-R^-'+R"-' 
 
 7. To what sum will an annuity of $120 for 20 years amount 
 to at 6 per cent, per annum 7 Ans. $4414.27. 
 
340 RAY'S ALGEBRA, PART SECOND. 
 
 8. What is the present worth of an annuity of $250, payable 
 yearly for 30 years, at 5 per cent, per annum 'i 
 
 Ans. $3843.1135. 
 
 9. What is the present value of an annuity of $112.50, to 
 commence at the end of 10 years, and to continue 20 years, at 4 
 per cent. 1 Ans. $1032.877. 
 
 10. A debt of a$, accumulating at compound interest, is dis- 
 charged in n years, by equal annual payments of b$ ; find the 
 value of n. Ans. ^ ^og- ^ l°g- (t-^a) 
 
 log. (1+r) • 
 
 CHAPTER XII . 
 
 GENERAL THEORY OF EQUATIONS. 
 
 Akt. 393. An equation is the statement of equality between 
 two algebraic expressions. Equations are of different degrees. 
 
 From what has been already shown (Art. 113), it is obvious 
 that 
 
 ax-{-b=0, is an equation of the 1st degree. 
 
 x''-\-bx-{-c=Q, is an equation of the 2nd degree. 
 
 x'-\-bx^-\-cx-\-d^O, is an equation of the 3rd degree , 
 
 and in general, 
 
 a;"+Ax»-i+Ba?'-=+Ca»-'+. . . . 4-Ta;+V=0, 
 is an equation of the n"' degree.- The coefficients. A, B, C, &.C., 
 may be positive or negative, integral or fractional ; and either of 
 them may be equal to zero. The coefficient of the highest power 
 of X is represented by unity, because if it is not unity, the equa- 
 tion may be reduced to this form by dividing by such coefficient. 
 
 Akt. 394. The root of an equation is such a number, or 
 quantity, that being substituted for the unknown quantity, the 
 equation will be verified. Thus, in the cubic equation a^-\-2x' 
 — 14 r — 3=0, the root is 3, because when this number is substi- 
 tuted for X, the first member becomes equal to the second. 
 
 Every equation must have at least one root, for if there is no quan- 
 tity whatever that will satisfy the equation when substituted for 
 the unknown quantity, then is the equation itself not true. 
 
GENERAL THEORY OF EQUATIONS. 341 
 
 K function of a quantity is any expression dependent on that 
 quantity. Thus, Sx-J-S is a Smction of x, 
 
 5x', is a function of x, 
 
 Ix — 3y', is a function of x and y. 
 
 In a series, when the signs of two successive terms are aUke^ 
 they constitute a permaneTUX, when they are unlike, a variatioTi. 
 Thus, ill the polynomial, / 
 
 — r — s-\-t-\-u, 
 the signs of the first and second terras constitute a permanence, 
 of the second and third a variation, and of the third and fourth a 
 permanence. 
 
 Aet. 393. Peop. I. — If a is a root of any equation, 
 ;c"+Aa»-'+Ba?-2+Ci»-'+. . . . +Ta;+V=0, (n), 
 then will the equation ie divisible by x — a. 
 
 For if a is one value of x, the equation will be verified when a 
 is substituted for ar. This gives 
 
 a"+Aa»-'+Ba"-=+Ca»-»+. . . . +Ta+V=0; 
 or, V=— a"— Ao"-'— Bo"-'— Ca"-»— . . . . — Ta. 
 
 Substituting this value of V in the given equation, and arrang- 
 ing the terms according to the same powers of x and a, we have 
 
 (3C"— a")+A(a?'-'— a"-')+B(af-'— o"--')+. . . +T(a;— a)=0. 
 
 Now, (Art. 83), each of the expressions (j^ — o"), (a;""' — a"""')i 
 &c., is divisible by x — a, therefore the given equation is divisible 
 by X — a. 
 
 Cor. Conversely, if the equation 
 
 ;c"+Aa:"-'+Ba?'--'+. . . . +Ta;-fV=0, (n) is divisible 
 by X — a, then a is a root of the equation. 
 
 Tor if the equation (n) is divisible by x — a, if we call the quo- 
 tient Q,, we have (a; — a)Q,=0 (re), 
 which may be satisfied by making x — o=0, whence x=a. 
 
 D'Alembeut's pkoof of Prop. I. — If said division leave a re- 
 mainder, let it be called R, and the quotient Q, ; then the equation 
 (n) becomes 
 
 (i— a)Q,4-R=0. 
 
 But a; — 0=0, .•. R=0; that is, there is no remainder on divid- 
 ing equation (n) by x — a. 
 
 Illustration 1 . In the equation a:* — 9x'+26i — 24=0, the 
 
842 RAY'S ALGEBRA, PART SECOND. 
 
 roots are 2, 3, and 4 ; and the equation is divisible by x — 3 
 X — 3 , and x — 4 . 
 
 2. In the equation x^-\-x^ — 14a: — 24=0, the roots are — 2, 
 — 3, and 4; and the equation is divisible by 1+2, a;+3, and 
 X— 4. 
 
 Akt. 396. Peop. II. — Every equation containing hit arm un- 
 known quantity, has as many roots as there are units in the number 
 denoting its degree; that is, an equation of the n" degree has n roots. 
 
 Let a be a root of the equation 
 
 a;»+Aa?-'+Bx»-»+Ci'-=+. . . . +Ti+V=0 (n) 
 By Art. 395 this equation is divisible by x — a. If we perform 
 the division, and denote by A,, B,, &c., the coefficients of the 
 powers of x in the quotient after the highest, equation (w) 
 becomes 
 
 (I— a)(a;^'H-A,af^'+B,a?'-»+. . . . +Tja;+V,)=0. 
 This equation will be satisfied by making 
 
 a:"-'+A,a:^=+B,i^=+. . . . +T,a;+V,=0. 
 
 Now this equation must also have a root, which may be denoted 
 by i ; it is therefore (Art. 395) divisible by x — h, and may be 
 placed under the form 
 
 (a;-fi)(x'-='+A,x'-'+B,af-<+. . . . +Tjx+V,)=0. 
 
 This equation will be satisfied by placing the second member 
 equal to zero, which gives another equation of a degree still 
 lower by a unit, and as x must here also have some value, as c, 
 this equation must be divisible by x — c ; and if the division be 
 performed we shall have an equation of a degree still lower by a 
 unit. 
 
 It is evident that if this operation be continued, the exponent 
 n will be exhausted, and the last quotient will be unity ; hence, 
 calling the last root I, we shall have 
 
 (jo — a)(x — J)(x — c)(x — d),. . . . {x — Q=0, which is satisfied 
 
 by making x=a,h, c,d,. . . . or I; that is, there are n 
 
 quantities, either of which, when substituted for x, will satisfy 
 
 the conditions of the equation ; or, in other words, the equation 
 
 has n roots, a, b, c, d, &c. 
 
 Cor. 1. From this theorem it follows that if we know one roo 
 of an equation jce may, by dividing (Art. 395), find the equation 
 containing the remaining roots. Hence, when all the roots of an 
 equation but two are known, it may be reduced to a quadratic by 
 division, and the remaining roots found by methods already given. 
 
GENERAL THEORY OF EQUATIONS. 343 
 
 Thus, one root of the equation x'- — \2x'-\-41x — 60=0, is 5, and 
 by dividing it by x — 5, the quotient is x' — 7a;-)-12=0, of which 
 the roots are found to be -|-3 and -f-4. 
 
 Cor. 2. From the preceding, it is obvious that when any equa- 
 tion, whose right hand member is zero, can be separated into fac- 
 tors, the roots of the equation may be found by placing each of 
 the factors equal to zero. Thus, in the equation x' — 1=0, by 
 factoring we have {x-\-l)(x — 1)=J), .•. a:-|-l=0, and x — 1=0, 
 whence x= — 1, and x=-\-l. 
 
 Again if x^-\-ix^O, we have a:(a;-|-4)=0, whence x=0, and 
 x=—4 . (See Art. 254.) 
 
 EXAMPLES FOR PRACTICE. 
 
 1. One root of the equation x' — lla;'+23a:+35=0 is — 1; 
 find the equation containing the remaining roots. 
 
 Ans. a'— ■12a;+35=0. 
 
 2. One root of the equation a;' — 9x'-\-26x — 24=0 is 3; find 
 the remaining roots. Atis. 2 and 4. 
 
 3. One root of the equation x' — '7x-{-6=0 is 2; find the 
 remaining roots. Ans. 1 and — 3. 
 
 . 4. Two roots of the equation x^-\-2x> — 41a;= — 42a;+360=0, 
 are 3 and — 4; required the remaining roots. A?is. 5 and — 6. 
 
 5. Two roots of the equation x'' — 3a;' — 5a:'-|-9a: — 2=0, are 
 -|-1, and — 2 ; find the remaining roots. 
 
 Ans. 2-{-JS, and 2—^3". 
 
 Remarks. 1. When it is stated, for example, that 1=4 and x^3, in 
 the same equation, it is not to be understood tliat x is equal to 4 and 3 at 
 the same time, but that x is equal either to 4 or 3. 
 
 2. This proposition proves that an equation of the nth degree is com- 
 posed of n binomial factors, but these are not necessarily unequal. Two 
 or more of Iheni may be equal to each other. Thus, the equation 
 2;3_6s2_|-]2i— 8=0, is the same as (x— 2)(i— 2)(x— 2)=0, or (x—2)f 
 =0, from which, by placing each factor equal to zero, we find the three 
 rcott to be x=2, z=2, and a:=2. 
 
 Art. 397. Prop. III. — No equation can have a greater number 
 of roots than there are units in the number denoting its degree, that 
 is an equation of the n"' degree can have only n roots. 
 
 If it be possible let the equation 
 
 *"+Aa:»-i+Ba;"-3+Ca:"-'+. . . , -|-Ta;+V=0. 
 
344 
 
 RAY'S ALGEBRA, PART SECOND. 
 
 Besides the n roots a, b, c, d, &c., have another root, r, not iden. 
 tical with either of the roots a, b, c, d, &.c. ; then since r is q root 
 of the equation it must be divisible by i — r (Art. 395) ; this 
 gives 
 
 x''+Ax"-'+Bx"-2+, &.C., =(a;— r)(a;'-'+A'a;"-2-|-, &c., ■ 
 « >— o)(a:— 6)(x— c). . (i— /)=(a:— r)(a;"-'+A'a;"-'+, &c.) 
 But since r is a value of it, we have, by substitution, 
 (r~a)(7-— *)()•— c). . . (r— Z)=(r— r)(j;"-i+A'x"-2+, &c.) 
 
 Now the second member of this equation is =0, because 
 (r — r)=0; but the other side cannot be 0, since r is not equal to 
 any of the quantities a, h, c, &lc. ; hence the supposition is absurd 
 that X can have any value other than a, b, c, d,. . I. 
 
 Art. 398. Pkop. IV. — To discover the relations between the 
 coefficients of an equation, and its roots. 
 
 Let x=a, 1 Then, t'x—a=0, 
 
 x=b, I 
 
 x=c, 
 x=d, &c. 
 
 Jx — a=0, 
 x—b=0, 
 \ x—c=0, 
 V X — d=0 , Sic. 
 
 By multiplying together the corresponding terms of the last 
 set of equations, we have (x — a)(x — l)(x — c)(x — (i)=OxOxO 
 
 xo=o. 
 
 If we perform the actual multiplication of the factors, we find 
 
 '—a 
 
 x'+ab 
 
 x' — ale x-\-abcd 
 
 —b 
 
 -\-ac 
 
 —abd 
 
 — c 
 
 -\-ad 
 
 —acd 
 
 —d 
 
 +bc 
 
 —bed 
 
 +bd 
 
 
 
 +cd 
 
 
 Similarly, in the equation of the ra" degree, 
 
 a?'+Ax^'+Bai"-'+, &c., =(x— aXx— i)(x— c). . (a^— Z)=0. 
 
 If we perform the multiplication of the n factors, we shall have 
 
 —a — b — c. . . . — k — l=A ; 
 
 ab+ac+. . . . +A:Z=B ; 
 
 — dbc — abd, . . . atfc=C ; 
 
 ±abcd. 
 
 M=y. 
 
 The double sign is placed before the last term, because the 
 product — aX — *X — c. . . . X — 'j will be plus or minus, ar^ 
 cording as the degree of the equation is even or odd. Hence, 
 
GENERAL THEORY OF EQUATIONS. 345 
 
 1 . The coefficient of the second term of any equation, is equal to 
 the sum of all the roots, with their siffns cha7iged. 
 
 2. Tlie coefficient of the third term is equal to the sum of the pro- 
 ducts of all the roots taken two and two. 
 
 3 . Tlie coefficient of the fourth term is equal to the sum of the pro- 
 ducts of all the roots taken three and three, with their signs changed. 
 
 And so on, and 
 
 4. The last, or absolute term, is equal to the product of all the roots. 
 
 Cor. 1 . If any term of an equation is wanting it is hecause ill 
 coefficient is 0. 
 
 2. If the 2'"' term of any equation is wanting, the sum of the 
 roots is equal to . 
 
 3. If the 3"' term of any equation is wanting, the sum of the pro- 
 ducts of the roots, taken two and two in a product, is equal to . 
 
 4. If the absolute term is wanting, the product of the roots must 
 be , and hence one of the roots must be . 
 
 5 . Since the last term is ihe product of all the roots, therefore it 
 must be divisible by each of them ; that is, every rational root of an 
 equation is a divisor of the last term. 
 
 EXAMPLES ILLTTSTRATING THE PRECEDING PKINCIFLES. 
 
 1. Form the equation whose roots are 3, 4, and — 5. 
 
 The equations x =3, x=4, and x= — 5, give x — 3=0, x — 4=0 
 and a:-)-5=:0; 
 hence, (a^— 3)(a;— 4)(a;+5)=a;3— 2a;'— 23a;+60=0. 
 Here 34-4 — 5=-f-2, the coefficient of the 2"'' term with a 
 contrary sign. 
 3X44-3X— 5+4X— 5=— 23, the coefficient of the 3''' term. 
 3X4X — 5=: — 60, the last term, with the minus sign, because 
 tlie degree of the equation is odd. 
 
 2. What is the equation whose roots are 2, 3, and — 5 "! (Sea 
 Cor. 2.) Ans. a;'— 19a;+30=0. 
 
 3. Find the equation whose roots are 3, — 2, and 7. 
 
 Ans. x'— 8a;2-j-a;-|-42=0. 
 
 ^ 4. Form the equation with roots 0, — 1, 2, and — 5. 
 
 Ans. a;''+4x' — 7a:' — 10a;=0 
 
 5. Form (he equation whose rocts are — 2, 4^4, and -]-4 
 (See Cor. 3.) Ans. a:'— fia;'4-32=0. 
 
3^^ RAY'S ALGEBRA, PART SECOND. 
 
 6. Find the equation whose roots are 14-<s/3, and 1 — ^3. 
 
 Ans. x'—2x—2=Q. 
 
 7. Find the equation whose roots are l±^"-i and 2±^3. 
 
 Ans. a;"— 6x3+8x2+2^—1=0. 
 S. What is the 4" term of the equation whose roots are — 2, 
 —1,1,3,47 Ar!5. 29a;'. 
 
 9. Find the middle term of the equation whose roots are 5, 3, 
 1,-1,— 2, —4. Atw. 28aJ. 
 
 10. Form an equation of the 4'* degree, when two of the rooti 
 are —J2, and +^—3. Ans. x^+a;'— 6=0. 
 
 Akt. 399. Pkop. V. — No equation having unity for the coeffi- 
 cient of the first term, and all the other coe^ients integers, can have 
 a root equal to a ratiojndl fraction. 
 
 Talce the general equation of the n"' degree, and suppose all 
 its coefficients integers, 
 
 xn+Aa?'-'+Bx"-2+ . . . . +Tx+V=0. 
 
 If possible, let -, a fraction in its lowest terms, be a root of 
 
 
 
 this equation ; then by substituting it for x we have 
 
 ^•+a^'+b:^+. . . . +T?+v=o. 
 
 J" ' 6"-' ' 6»-2 ' h 
 
 Reducing all the terms to a common denominator, 
 
 a" , Aa"-'i , T,a"-2i2 TaJ»-' , Vi" „ 
 
 — -4- -T--t> + • . • » + + =U. 
 
 i" 6" ' i» ' ' 6" ' J" 
 
 Transposing all the terms to the second member, except the 
 first, and omitting the common denominator, 
 
 a"=— Aa"-'i— Ba"-2J''— . . . . — Tai—'— V6". 
 Dividing both members by h, 
 
 -=— Aa"r'— Ba"-'^— . . . . — ToJ"-'— V6"-'. 
 b 
 
 But, by hypothesis, a and h contain no common factor, therefore 
 
 1 is an irreducible fraction, and the right member is a series of 
 
 integral quantities ; therefore, an irreducible fraction is equal to 
 a series of integers, which is absurd. Hence, the supposition 
 which leads to this conclusion is absurd, namely that the equation 
 has a fractional root. 
 
GENERAL THEORY OF EQUATIONS. 347 
 
 Remark. — This proposition only proves that in an equation of the 
 form described, the real roots must be integers, otherwise they cannot 
 be exactly expressed in numbers. It often happens that the roots of an 
 equation can be expressed approximately by fractions. Thus, in the, 
 equation x^ — 3^2 — 5j;-|-10=0, one of the roots is — 2, and the other 
 two are expressed nearly by 1.382, and 3.618. 
 
 When a real root cannot be expressed exactly in numbers it is 
 termed incommensurable. 
 
 Art. 400. Pkop. VI. — If the signs of the alternate terms of an 
 equation be changed, the signs of all the roots will be changed. 
 
 Let a be a root of the equation 
 
 a:"+Aa:'-i+Bj;"-2+Caf-'+. . . . +V=0, (1) 
 then a"+Aa''-i+Bo"-'+Ca''-34-. . . . +V=0, (2) 
 
 By changing the signs of the alternate terms of equation (1 ) 
 It becomes 
 
 a;"— Ax"-'+B2»-2— Ca;"-'4-. . . . =hV=0. (3) 
 By substituting — a for x In this equation, we have 
 
 a"— Aa"-'+Ba"-^— Ca"-= ±V=0. (4) 
 
 Now if n be even, the 2"'', 4'*, &c., terms will contain odd 
 powers of a, which will be negative (Art. 193), and the signs of the 
 terms being negative, the results of each term will be positive ; 
 hence, the whole result will be the same as that produced by the 
 substitution of a for x in equation (1). 
 
 But if n be odd, the odd powers of a will be negative, and the 
 even powers positive ; and the signs Sf the same terms being 
 negative, these terms will be negative, which will render all the 
 terms of (4) negative. 
 
 But this result is the same as that which would be produced by 
 multiplying all the terms of (2) by — 1 . Hence, if a is a root 
 of equation (1), — a is a root of (3), whether n be odd or even. 
 
 Remare. — If the signs of all the terms be changed, the signs of the 
 roots will remain unchanged, because this is the same as multiplying 
 both members by — 1. (Art. 148.) 
 
 Ex. 1. The roots of the equation x''-\-2x — 24=0, are 4 and 
 — 6 ; what are the roots of the equation x' — 2x — 24=:0 1 
 
 Atis. — 4 and 6. 
 
 2. The roots of the equation k' — Zic' — 10x+24=0, are 2, 
 — 3, and 4; what are the roots of the equation x'-\-2x' — 10a 
 —24=0. 
 
348 
 
 RAY'S ALGEBRA, PART SECOND. 
 
 Art. 401. Prop. VII. — When IJie coefficients of an equation 
 are real, if it conUins imaginary roots, the numler of these roots 
 
 mxist be even. 
 
 If a-\-b,J — 1 be a root of the equation 
 
 a;-— Ax"-'+Ba:''-2— Ca^-3+, &c., =0; 
 then a — bij — 1 is also a root. 
 
 In the equation substitute a-\-bjJ — 1 for x, and the result will 
 consist of two parts :.!'', possible quantities which involve the 
 odd and even powers of a, and the even powers of b^ — 1; and 
 2"'', impossible quantities which involve the odd powers ofbJ — 1; 
 call the sum of the possible quantities P, and of the impossible 
 Q,iJ — 1, then P+QV — 1 is the whole result ; hence, 
 P+QV^=0. 
 
 But the first quantity being real, and the second imaginary, in 
 order to satisfy the equation, each of the quantities must be ; 
 this gives P=0, and 0.^^=0. 
 
 Again, let a — bj — 1 be substituted for x, and the 1" part of 
 the result will be the same as before, and the 2'"' part, which 
 arises from the odd powers of bj — 1 , will differ from the former 
 imaginary part only in its sign ; therefore, the result will be 
 P — QV — 1; but since P=0, and Q;^ — 1=0, we must have 
 
 Hence a — 6^ — 1 is a root of the equation, since its substitu- 
 tion for X gives a result equal to . 
 
 Cor 1. — If for b^ — 1 we put ^b, it is evident that in the re- 
 sult we must put ;^Q, instead of Q.,^ — 1, so that P-j-,^Q=0 
 and therefore P — ;yQ,=0; hence, surd roots of the form a±Jb, 
 enter an equation by pairs. 
 
 Cor. 2. — In the same manner it may be proved that roots of 
 the form dzb^ — 1, or dzijb enter equations by pairs, for in both 
 cases we have only to make o=0. 
 
 Cor. 3. — Since irrational and imaginary roots always occur ii 
 pairs where the coefficients are real, it follows that every equa 
 tion of an odd degree must have at least one real root. 
 
 Cor. 4. — Corresponding to any pair of imaginary roots 
 orhJ.^/ — 1, we shall have in the equation, the quadratic factor 
 
GENERAL THEORY OF EQUATIONS. 849 
 
 therefore every equation of an even order, with real coefficients, 
 is composed of real factors of the second degree. 
 
 Ex. 1. One root of the equation a;' — 26a;-)-60=:0 is — 6 ; 
 required the other roots. Ans. SdrV — !• 
 
 Ex. 2. One root of the equation a' — 15a;-|-4=0 is — 4 ; re- 
 jjuired the other roots. Ans. 2±^J'i■ 
 
 Ex. 3. Two of the roots of the equation x'' — 4a;' — ^7x^4-26^ 
 . -14=0 are 3+^2, and 3 — J2; required the remaining roots. 
 
 Ans. — 1±V3. 
 
 Ex. 4. One root of x'— 1 3^+12 x—S=0, is 2—JS; find the 
 other roots. Ans. 2-\-^'d and 3. 
 
 Ex. 5. One root of a;<—3a!=— 42a;— 40=0 is —1(3+^—31); 
 find the other roots. Ans. — 3(8 — ^ — 31), 4, and — 1. 
 
 Ex. 6. Two roots of a;'— 10a;''+29a;'— 10a;=— 62a;+60=0 are 
 3 and ^^2; find the other roots. Ans. — ,J2, 2, and .5. 
 
 Art. 402. Pkop. VIII. — Descartes' Rule of the Signs. — 
 JVo equation can have a greater number of positive roots than there 
 are variations of sign ; nor a greater number of negative roots 
 than there are permanences of sign. 
 
 In the equation a; — a=0, where the value of x is -\-a, there is 
 one variation, and one positive root. 
 
 In the equation x-\-a=0, where the value of x is — a, there is 
 one permanence, and one negative root. 
 
 In the equation x' — (o-f-i)a;+ai=0 , v\rhere the values of x are 
 -\-a and -\-h, there are two variations and two positive roots. 
 
 In the equation x''^{a-\-b)x-\-ab=Q , where the values of x are 
 — a, and — b, there are two permanences, and two negative roots. 
 
 In the equation x^ — x — 12=0, where a;=:4-4, and — 3, there 
 is one variation, and one positive root, and 07te permanence, and one 
 negative root. 
 
 Ifwe form an equation of the third degree, (Art. 397), whose 
 roots are +2, +3, +4, we shall have x'— 9a;24-26a;— 24=0 
 where there are three variations, and three positive roots. 
 
 But if we form an equation whose roots are — 2, — 3, -|-4, we 
 shall have x'-f-x' — 14x — 24=0, where there is one variation, and 
 oTix positive root, and two fermanences, and tiuo negative roots. 
 
850 RAY'S ALGEBRA, PART SECOND. 
 
 To prove the proposition generally, let the signs of the terms 
 in their order, in any compkle equation, be 
 
 -| 1 1 \- -\ \-, and let a new factor 
 
 X — 0=0, corresponding to a new positive root be introduced, the 
 •igns in the partial and final products will be 
 +.^ !__ + + + 
 
 + - 
 
 + + - + - + + + 
 1-- + 
 
 + ± — H h±±— . 
 
 Now in this product, it is obvious, that each permanence is 
 changed into an ambiguity ; hence, the permanences, take the am- 
 biguous sign as you will, are not increased in the final product by 
 the introduction of the positive root -\-a ; but the number of signs 
 is increased by one, and there.ore the number of variations must 
 be increased by OTie. Hence, the introduction of any positive 
 root introduces, at least, one additional variation of sign. 
 
 Now the equation x — a=0, contains one positive root, and has 
 one variation of sign. Therefore, since every additional positive 
 root introduces, at least, one additional variation of sign, the num- 
 ber of positive roots can never exceed the number of variations of 
 sign. 
 
 Again, if we change the signs of the alternate terms, the roots 
 will be changed from positive to negative, and conversely (Art. 
 400). Hence, the permanences in, the proposed equation will be 
 replaced by variations in the changed equation, and the variations 
 in the' former by permanences in the latter ; and since the 
 changed equation cannot have a greater number of positive r lots 
 than there are variations of sign, the proposed equation ccnnol 
 have a greater number of negative roots than there are perman^ices 
 of sign. 
 
 Cor. 1 . Since the whole number of variations and permanences 
 is evidently equal to the degree of the equation, (the equation if 
 not complete being rendered so by the introduction of ciphers) 
 Therefore, if the roots of an equation be all real, the immber of 
 positive roots must be equal to the number of variations, and the 
 number of negative roots to the number of permanences. (See 
 examples, pages 343, 345.) 
 
 2. By means of this theorem we can often determine whether 
 there are imaginary roots in an equation. 
 
GENERAL THEORY OF EQUATIONS 351 
 
 For example, the equation 
 
 a;2-|-I6=0, 
 may be written a;2±0a:-|-16=0. 
 
 Now, if we take the upper sign there are n. variations, hence 
 there is no positive root ; and if we take the luwer sign there are 
 no permanences, hence there is no negative root. But since the 
 equation has two roots (Art. 396), they must, therefore, both be 
 imaginary. 
 
 In like manner the cubic equation 
 
 a;3_|_Bx+C=0, 
 may be written a;'±Oa;'-f-Ba;+C=0. 
 
 Now if we take the upper sign there are no variations, and 
 consequently no positive root. But if we take the lower sign, 
 there is one permanence, hence there can be hut one negative 
 root. Therefore, the other two roots must be imaginary. 
 
 Akt. 4®3. Prop. IX. — If two numbers, when substiluted for 
 the unknown quantity in an equation, give results affected with differ- 
 ent signs, one root at least of this equation lies between tiiese numbers. 
 
 Let the equation, for example, be 
 
 a'— x'+a;— 8=0. 
 
 If we substitute 2 for x in this ee(uation, the result is — '2; and 
 if we substitute 3 for x, the result is -|-13. These results have 
 different signs, and it is required tO'show that there must be one 
 real root, at least, between 2 and 3. 
 
 The equation may evidently be written thus, 
 
 (l3_|_^-)_(^2_|_8-)_0. 
 
 Now in substituting 2 for x, x^-\-x=10, and a;'-|-8=12, 
 .-. a?'+a:<a:2-|-8 ; 
 also, in substituting 3 for x, x^-\-x='AO, and a:^+8=17, 
 ,-. x^+xyx^+9. 
 
 Now both these quantities increase while x increases, but the 
 first increases more rapidly than the second, since when x=2, il 
 is less than the second, but when a;=3 it is greater. Consequently, 
 for some value of x between 2 and 3, we must have x'-\-x=x^-{-S , 
 and this value of x is, therefore, a reil root of the equation. 
 
 In general, suppose we have an equation X=0, where X rep- 
 resents a polynomial involving x, and that two numbers, p ard y, 
 when substituted for x, give results with contrary signs. Let P 
 be the sum of the positive, and N the sum of the negative terms ; 
 
352 RAY'S ALGEBRA., PART SECOND. 
 
 also suppose that when x=p, P — N is negative, or P<]^f, and 
 that when ar=}, P — N is positive, or P]>N. 
 
 Suppose X to change by imperceptible degrees from p to q, 
 then P and N must also change by imperceptible degrees, and 
 both increase, but P must increase faster than N, otherwise from 
 having been less it could never become greater ; there must, 
 therefore, be some value of x between p and 5, which renders 
 P=N, or satisfies the equation X=0, and this value of a; is, there- 
 fore, a real root of the equation. 
 
 Cor. If the difference of the two numbers,^ and 5, which give 
 results with contrary signs, is equal to unity, it is evident that we 
 have found the integral part of one of the roots. 
 
 Ex. 1 . Find the integral part of one value of x in the equation 
 
 If 1=3, the value of the equation is — 2, but if a;=4, the 
 value is 47. Hence, a root lies between 3 and 4; that is, 3 is 
 the first figure of one of the roots. 
 
 2. Required the first figure of one of the roots of the equation 
 x^—bx''—x-\-\=f). Am. 5. 
 
 TRANSFORMATION OF EQUATIONS. 
 
 Akt. 404. The transformation of an equation is the changing 
 it into another of the same degree, whose roots shall have a speci- 
 fied relation to the roots of the given equation. 
 
 Thus, in the general equation of the n'* degree 
 
 K»+Aa;»-i+Ba?'-=. . . . +Ta;+V=0; (1) 
 
 f — y be substituted for x, the equation will be transformed into 
 another who?e roots are the same as those in (1), but with con- 
 trary signs, for y= — x. 
 
 If - be substituted for x, the roots of the new equation in y 
 
 y 
 
 will be the reciprocals of those of equation (1), for y=-. 
 
 Art. 405. Prop. I. — To transform an equation into one whose 
 roots are the roots of the given equation multiplied or divided ly any 
 given quantity. 
 
 Let a, b, c, &c., be the roots of the equation 
 
 ar"+Aaf^i-fBaf^=. . . . +Tx+V=Q; 
 
TRANSFORMATION OF EQUATIONS. 353 
 
 assume y=hc, or x=M ; then substitute this value for x, and the 
 k 
 
 proposed equation becomes 
 
 . k"^ i"-'^ 4"-' ^k^ 
 
 then, multiplying by k", we have 
 
 y"+A%"-'4-B4'y»-'. . . . +Tk''-^y+k''V=0 . 
 
 Since y=Jcx, the roots of this equation are ka, kb, kc, &c. 
 
 It is evident this equation may be derived from the proposed 
 equation, by multiplying the successive terms by 1, i, k', k', &c., 
 and changing x into y. 
 
 In the case of division, assume y=-, or x=ky, and substitute. 
 
 Car. By this transformation an equation may be cleared of 
 fractions, or if the first term be afiected with a coefBcient, that 
 coefficient may be removed. 
 
 Ex. Let it be required to transform the equation 
 
 into ona which is clear of fractions, and which has unity for the 
 coefficient of the highest term. 
 
 By multiplying by 6, we have 
 
 6x>+3px^+2qx-\-6r=0 . 
 
 Let y=Qx, or x=\y, and the equation becomes 
 
 and multiplying by 6', we have 
 
 y'+3?i/'+1292/+216r=0, 
 
 an equation of the required form. 
 
 Ex. 1 . Find the equation whose roots are those of the equation 
 j;<_|-7a;2__4a;-)-3=0 multiplied by 3. 
 
 Alls. y'+63y'— 1083/+243=0. 
 
 2. Find the equation whose roots are each 5 times those of 
 the equation x*-{-'i,x^ — 7x — 1=0. 
 
 Am. !/<+103/3—87.5y— 625=0. 
 
 3. What is the equation whose roots are each \ of those of 
 jj_3a;54-4a:+10=0'! Ans. 4y'— 6u.2-f4y4-5=0. 
 
 30 
 
454 
 
 RAY'S ALGEBRA, PART SECOND 
 
 4. What is the equation whose roots are each J of those of 
 a:3+18x'+99a:+81=0 taken negatively 1 
 
 Ans. y'— 6^2+11^^3=0. 
 
 5. Transform the equation x' — 1x^-\-\x — 10=:0, into one hav- 
 ing integral coefficients. Arts, y' — Cy'+Sj/ — 270=0 . 
 
 Art. 406. Peop. II. — To transform an equation into one whose 
 foots are greater or less by any given quantity than the corresponding 
 roots of the proposed equation. 
 
 Let the proposed equation be 
 
 a?'4-Ai»-'+Ba?^'. . . . +Ta;+V=0, 
 whose roots are a, b, c, &.c. 
 
 The relation between x and y will be expressed by the equation 
 y=;x±r. As the principle is the same, whether x is increased or 
 diminished, we will consider the case where y=x — r. This gives 
 x=y-\-r, and by substituting this for a; in the proposed equation, 
 we have 
 
 (j-+r)''+A(y+r)'-'+B(7/+r)»->, . . . +T(y+7-)+ V=0 . 
 
 Developing the different powers of y+r by the Binomial the- 
 orem, and arranging the terms according to the powers of y, we 
 have 
 
 ,»(n— 1), 
 
 y+nr 
 
 ' 1-2 
 +(n-l)Ar 
 
 r-' 
 
 +r» 
 
 +Ar»-' 
 4-Br'-=' 
 
 ^=0. 
 
 +Tr 
 
 -t-v 
 
 Now since y=x — r, the values of y in this equation are 
 b — r, c — r, &c. 
 
 Aet. 407. Cor. — By means of the preceding transformation 
 we may remove any intermediate term of an equation. Thus, 
 to transform an equation into one which shall want the second 
 
 term, r must be assumed, so that «r-l-A=0, or r^ — _. 
 
 To take 
 
 away the third term, jnCn — l)r'-\-(n — l)Ar4-B must be put =0, 
 and the value of r derived from the solution of this equation. 
 
 The pupil may solve the following examples b> *he preceding 
 principles : 
 
TRANSFORMATION OF EQUATIONS. 355 
 
 1 . Transform the equation x^ — Tx-\-7=:0 into another whose 
 roots shall be less by 1 than the corresponding roots of this 
 equation. Arts. y^-\-Zy^—^-\-l—0. 
 
 2. Find the equation whose roots are less by 3 than those of 
 the equation x' — 3x' — 1 5^2+4 9a>— 12=0. 
 
 Ans. y'-\-9y'+l2y^— 14:^=0. 
 
 3. Transform the equation i' — 6a:2_|_8a: — ^2=:0 into anothei 
 hose second term shall be absent. 
 
 Here A= — 6, n=3, .-. r:=2; hence, x=y-{-2. 
 
 Ans. f — 4j/— 2=0. 
 
 4. Transform the equation x^-\-2px — 5=0 into another want- 
 ing the second term. Atis. y^ — p' — q=0. 
 
 Art. 40S. There is a more easy and elegant method of per- 
 forming the operation of transformation, so as to increase or 
 diminish the roots of an equation, than by direct substitution, 
 which we will now proceed to explain. 
 
 Let the proposed equation be 
 
 Aa;''+Bj:'+Ca;'+Dx4-E=0, (1) 
 
 and let it be required to transform it into another, whose roots 
 shall be less by r ; then y=x — r and x=y-\-r. 
 By substituting y-\-r, instead of x, we have 
 
 A(y+ry4-B(.y+ry+C(y+ry+-D(y+r)+B=0. 
 By developing the powers of y-\-r, and arranging the terms ac- 
 cording to the powers of y, as in Art. 406, the transformed equa- 
 tion will take the form 
 
 As,^+B,j/'+C,2/^+D,j/+E,=0. (2) 
 
 where the coefficient A must evidently be the same as in equation 
 (1), while the coefficients B,, Cj, D,, and Ei, are unknown 
 quantities, whose values are now to be determined. For y, sub- 
 stitute its value x — r, and equation (2) becomes 
 
 A(x— r)<4-B , (ar_r)3+C , (,x—ry+D , (x— r)+E , =0 (3) 
 
 Now since the values of x are the same in equations (1) and 
 (3), it is evident these equations are identical. Hence, whatever 
 operation is performed on equation (1), the result will be the 
 same as if this process had been applied to equation (3). Now 
 as the object is to obtain the values of the coefficients Bj, C,, 
 &c., let equation (3) or (1 ) be divided by x — r, and it ia evident 
 that the quotient will be 
 
 ACx— r)»-l-B , {x—ry+C , (a^-r)+D , , 
 
356 HAY'S ALGEBRA, PART SECOND. 
 
 and the remainder will be the last coefficient E, ; hence, E, is 
 determined. ^ 
 
 Again, divide this quotient by a? — r, and the next quotient will 
 be 
 
 A(x — ^)*-|-B,(a^ — r)+C, with a remainder D, ; 
 hence, D, is determined. Dividing again by x — r we get the re- 
 mainder C, ; and lastly, by another division, we obtain the 
 remainder B, ; and thus we find all the coefficients of equation 
 (2)- 
 
 To illustrate this method we will now proceed to solve Ex. 1 , 
 Art. 407; that is, to find the equation whose roots are less by 1, 
 than those of the equation x' — ^7a-|-7=0. 
 
 Here y=x — 1, and we proceed to divide the proposed equation 
 and the successive quotients, by x — 1. The successive remain- 
 ders will be the coefficients of y in the transformed equation, ex- 
 cept that of the highest power, which will have the same coeffi- 
 cie/it as the highest power of x in the proposed equation. 
 
 x—l)x'—1x+lix'+x-^ I— l)a:2+x— 6(a;+2 
 
 a;3 X'^ ^st- lliot. x^ — X 2nd. qnot. 
 
 -\-x'—lx -|-2x— 6 
 
 x'~- X 2j— 2 
 
 — 6x-{-7 2nd. rem. = — 4 
 
 — 6a+6 
 
 Ist rem. =+1 . X — 1 )a;-f-2(l , 3rd quot. 
 
 X—1 
 Srd. rem. =-1-3 
 
 Here the last quotient is 1, and the successive remainders are 
 4-3, — 4, and-|-l. Comparing these with the general equation, 
 we have A=l, Bj=-)-3, C,= — 4, and D,=-|-l. Placing these 
 as coefficients to the respective powers of y, the transformed equa- 
 tion is 2/'-f 33/2 — iy+l=0. 
 
 This method of transforming an equation, would not, in gen- 
 eral, be shorter than direct substitution, were it not that the suc- 
 cessive divisions may be greatly shortened by a process, called 
 from its discoverer, Hornek's Synthetic Method or Divisios, 
 which we shall now proceed to explain. 
 
 Aet. 409. Sththetic Division. — This may be considered as 
 an abridgment of the method of division by Detached Coefficients 
 (Art. 77). To explain the process, we shall first divide Sx* — 12a:' 
 ~\-2x^-\-^x — 5 by X — 2, by detached coefficients. 
 
TRANSFORMATION OF EQUATIONS. 357 
 
 DlTieor 1_2)5— 12+3+4— 5(5— 3— 1+2 Quotient 
 
 5—10 oi- 5x^—2x^—x-{-2. 
 
 -3 
 
 — 2- 
 —2-1-4 
 
 -1+4 
 
 —1+2 
 
 +2—5 
 2—4 
 
 I Kem- 
 
 By changing the signs of the terms of the divisor, except thai 
 of the first term, which must not be changed, as by means of 
 that we determine the signs of the respective terms of the quo- 
 tient, and adding each partial product instead of subtracting it, 
 except the Jirst term, which being always the same as the first 
 term of each dividend, may be omitted, the operation may be 
 represented thus : 
 
 1+2)5—12+3+4—5(5—2—1+2 
 *+10 
 
 —2+3 
 
 * 4 
 
 -1+4 
 *— 2 
 
 *+4 
 —1 
 
 Let it be observed that the figures over the stars are the coefii- 
 cients of the several terms of the quotient. It will also be seen 
 that it is unnecessary to bring down the several terms of the 
 dividend. Hence, the last operation may be represented as fol- 
 lows : 
 
 +2)5—12+3+4—5 
 +10—4—2+4 
 — 2—1+2—1 
 
 In this operation 5 is the first term of the quotient, +10 is the 
 product of +2, the divisor, by 5; the sum of +10 and — 12 
 gives — 2, the second term of the quotient, — 4 is the product of 
 +2, the divisor,by — 2, the second term of the quotient, and the 
 sum of — 4 and +3 gives — l,the third term of the quotient, 
 and so on. The last term, — 1 , is the remainder. 
 
 Supplying the powers of x, the quotient is Sje^ — 2i^ — s+S, 
 with a remainder — 1 . 
 
858 RAY'S ALGEBRit, PART SECOND. 
 
 A similar method may be used when the divisor contains three 
 terms ; and if the coefficient of the first term of the divisor ia 
 not unity, it may be made unity by dividing both dividend and 
 divisor by the coefficient of the first term of the divisor. The 
 method, however, is rarely used except when the divisor is a bino- 
 mia., the coefficient of whose first term is 1 . 
 
 In the application of Synthetic division, when any term of an 
 equation is absent, its place must be supplied with a zero. 
 
 Art. 41©. We shall now illustrate the use of Synthetic 
 division in the transformation of equations, by the metJiod 
 described in Art. 408. 
 
 1 . Let it be required to find the equation whose roots ar* less 
 by 1 than those of the equation x' — 7j:-|-7. 
 
 To effect this transformation, it is required to find the suc- 
 cessive remainders which arise from dividing a;' — 7a;+7, and the 
 successive quotients, by x — 1. 
 
 Since the second term is wanting, its place must be supp'ied 
 with 0. Also, in arranging the operation, it is customary to 
 place the second term of the changed divisor on the right, as in 
 division. 
 
 OPERATION BY SYNTHETIC BIVISIOH. 
 
 +1 
 
 —7 
 +1 
 
 —6 
 
 (+1 
 
 '.: +1= 1" 
 
 
 +1 
 +1 
 
 —6 
 
 +2 
 
 +1 
 
 R. 
 
 +2 
 +1 
 
 —4 
 
 .-. —4= S"* R. 
 
 
 +3 . . +3= S'" R. 
 
 Hence, the required coefficients are 1, -|-3, — 4, and +1 . 
 
 •■• y'+Sy' — 4j/+1=0 is the transformed equation required. 
 
 2. Transform the equation 5a:<4-28x'+.'51a;'+32a;— 1=0, in- 
 to another having its roots greater by 2 than those of the given 
 equation. 
 
 Here, y=a;+2; hence, to find the coefficients of the trans 
 formed equation, we must find the successive remainders arising 
 from dividing the proposed equation and the successive quotients, 
 by a:-4-2. Changing the sign of 4-2, th« operation is 6S follows 
 
TRANSFORMATION OF EQUATIONS. 350 
 
 5 +28 
 —10 
 
 +51 
 —36 
 
 +32 
 —30 
 
 -1 (- 
 — 1 
 
 -2 
 
 +18 
 —10 
 
 +15 
 —16 
 
 + 2 
 + 2 
 
 —5 .-. 
 
 -5= 1- R. 
 
 —10 
 
 — 1 
 
 + 4 
 
 + 4 
 
 .-.+4= 
 
 : S"" R. 
 
 — 2 
 —10 
 
 + 3 
 
 . . +3= 
 
 = 3'" R. 
 
 
 —13 .-. —12= 4'* R. 
 
 Comparing this with the general equation (Art. 408,) we find 
 A=5, B,=— 12, C,=+3, D,=+4, and E,=— 5. 
 
 .-. 5y* — 12y"+3y*+4y — 5=0 is the transformed equation 
 required. 
 
 3. Find the equation whose roots are less by 1.7 than those 
 of the equation x' — ^2a;'+3a:— 4=0. 
 
 If we transform this equation into another whose roots are 
 less by 1, the resulting equation is y'-\-y'-\-2y — ^2=0. We may 
 then transform this into another whose roots are less by .7, and 
 the result will be the equation required, or, the whole operation 
 may be performed at once as follows : 
 
 —2 
 
 +3 -4 (+1.7 
 
 +1.7 
 
 — .51 +4.233 
 
 — .3 
 
 +2.49 + .233 .-. +.233= 1" R. 
 
 +1.7 
 
 +2.38 
 
 +1.4 
 
 +4.87 .-. +4.87— 2«'R. 
 
 +1.7 
 
 
 +3.1 . 
 
 . 3.1= 3'" R. 
 
 Hence, the required equation is y'+3.1y'+4.87y+.233 
 =0. 
 
 It is generally preferable to perform the operation by successive 
 transformations, using only one figure at a time, as there is less 
 liability to error. It is not necessary, however, after each trans- 
 formation to arrange the coefficients in a horizontal line. 
 
 Thus, the two transformations necessary in the preceding 
 example mny be combined as follows : 
 
360 
 
 RAY'S ALGEBRA, PART SECOND 
 
 1—2 +3 —4 (1.7 
 
 +1 
 
 —1 
 
 +2 
 
 — 1 
 
 +2 
 
 —2.* 
 
 +1 
 
 
 
 +2.233 
 
 
 
 +2.* 
 
 ■ .233* 
 
 +1 
 
 +1.19 
 
 
 +1* 
 
 3.19 
 
 
 .7 
 
 1.68 
 
 
 +1.7 
 
 4.87* 
 
 
 .7 
 
 
 
 +2.4 
 
 
 
 .7 
 
 
 
 3.1* 
 
 The stars indicate the coefficients after each transformation 
 
 EXAMPLES FOR PRACTICE. 
 
 4. Find the equation whose roots are each less by 3 than the 
 loots of a'— 27ic— 36=0. Am. y'-\-9y^—90=0 . 
 
 5 . Find the equation whose roots are each less by 3 than thfc 
 roots of a;<— 27a;2— 14x+120=0. 
 
 Ans. y<+12y'+27j(2— 683^— 84=0. 
 
 6. Required the equation whose roots are less by 5 than those 
 of the equation a;*— 18a;'— 32ap2+17a;+9=0. 
 
 Atis. y<+23/'— 1 52y2— 1 153^-233 1=0. 
 
 7. Required the equation whose roots are less by 1.2 than 
 those of the equation a* — 6a;*+7.4a;' + 7.92a:= — 17.872ar 
 —.79232=0. Am. y>—^y^+2y—8=0. 
 
 Transform the following equations into others wanting the 
 2nd term. (See Art. 407.) 
 
 8. a;'— 6a2+7a:_2=0. Ans. y'— 5y— 4=0. 
 
 9. a:'— 6a;'+5=0. Ans. y'— 12y— 11=0. 
 .0. a;'— 6a:'+12a:+19=0. Atis. y'+27=0. 
 
 11. 33:'+15a;'+25a^-3=0. Ans. 27y»— 162=0. 
 
 Transform the following equations into others wanting the 3rd 
 term. 
 
 12. a'— 6a:'+9a>-20=0. 
 
 Ans. j/'+3y'— 20=0, or y»— 3y»— 16=0. 
 
 13. !»— 4i2+5a;_2=0. Ans. y— y»=0, or y>+y^—^^=zO 
 
TRANSFORMATION OF EQUATIONS. 361 
 
 Art. 411. Pkop. III. — To determine the law of Derived Poly- 
 nomials. 
 
 Let X represent the general equation of the n"' degree ; that is 
 
 X=x"+Aaf-'+Ba;"-^ . . . +Tx+V=0. 
 
 If we substitute x-\-h for x, and put X, to represent the new 
 value of X, we have 
 
 X , =(a;+A)"+A(a;+A)"-'+B(a:+A)"-2+, &,c., 
 
 and if we expand the different powers of x-\-h by the Binomia 
 theorem, we have X| = 
 
 + Ai"-'+ (n— 1 ) Aa;"-2 
 4-Ba;"--4-(ra— 2)Bx"-3 
 
 -j-, &.O. 
 
 +(»— l)(re— 2) Aa:»- 
 +(ra— 2)(n— 3) Ba?-- 
 
 JiL+, &c. 
 1 -2^ 
 
 But the first vertical column is the same as the original equa- 
 tion, and if we put X', X", X'", &c., to represent the succeeding 
 columns, we have 
 
 X =x"+Aa;"-'+Ba;"-2-|-, &c., 
 
 X' =nar-i+(n— l)Aa!»-2+(re— 2)Ba;"-'+, &c., 
 
 X"=7t(n— l)a?'-='+(ra— l)(7i— 2)Aaf^'4-, &c., 
 
 dicr, &c. 
 
 By substituting these in the development of Xj, we have 
 
 X.=X+X'A+— A''+ ^"' /j'+, &c. 
 ' ^ ^1 -2 ^1 •2-3 ^ 
 
 The expressions X', X", X'", &c., are called derived polynomu 
 als of X, or derived functions of X. X' is called the first derived 
 polynomial of X, or first derived function of X ; X" is called the 
 iecoTid, X'" the third, and so on. 
 
 It is easily seen that X' may be derived from X, iy multiplying 
 ach term by the exponent of x in that term, and diminishing the ex- 
 ponent hy unity. And each succeeding polynomial may be derived 
 from that which precedes it by the same law. 
 
 Aet. 412. Cor. If we transpose X we have X, — X=X'A 
 
 X" 
 -f-. i'+, &c. Now it is evident that h may be taken so small 
 
 31 
 
362 
 
 RAVS ALGEBRA, PART SECOND. 
 
 ihai the sign of the sum X'A+ i'+, &c., will he the same at 
 
 the sign of the first term X'h. For, since 
 
 X'A+JX"A2+, &c., =A(X'+iX' h+, &c.), 
 
 if A be taken so small, that iX"h-{-iX"'h'-\-, &c., becomes leso 
 than X' (their magnitudes alone being considered), the sign of 
 the sum of these two expressions must be the same as the sign 
 of the greater X'. 
 
 Art. 413. By comparing the transformed equation in Art. 
 406, with the development of X, in Art. 411, it is easily seen 
 that Xj may be considered the transformed equation, y corres- 
 ponding to X, and r to h. Hence, the transformed equation may 
 be obtained by substituting the values of X, X', &.C., in the devel- 
 opment of X,. As an example, let it be required to find the 
 equation whose roots are less by 1 than those of the equation 
 a;3— 7a;+7=0. 
 
 Here X = x'—lx+T, 
 X' =3a;=— 7, 
 X" =6x, 
 X"-'=6, 
 X" =0. 
 Observing that h=l, and substituting these values m tJis 
 equation X^=X+X'h+—h'+ ^"' h^+, &.C., we have 
 
 1 ' ^ 1 * -ft * o 
 
 X,=(a;'-7x+7)+(3i'-7)l+(6i-)j^+j-:|-^3. 
 
 ^x^-^Zx' — 4a;+l, in which the value of x is equal 
 to that of X in the given equation diminished by 1. 
 
 By this method the learner may solve the examples in Art. 
 410. 
 
 BttUAL ROOTS. 
 
 Akt. 414. To determine the equal roots of an equation. 
 
 We have already seen (Art. 396, Rem. 2,) that an equation 
 may have two or more of its roots equal to each other. Thus 
 the equation ii?—6x' + l2x—8=0, or (,x—2Xx—2){x—2) 
 =(a: — 2)'=0, has three roots, each of which is 2. We now pro 
 
EQUAL ROOTS. 363 
 
 poee to determine when an eqnation has equal roots, and how to 
 tind them. 
 
 If we take the equation (x — ^2)^=0 (1) 
 
 Its first derived polynomial is 3(a: — 2y=0. 
 
 Hence, we see that if any equation contains the same factoi 
 •taken three times, its first derived polynomial will contain the 
 same factor taken twice ; this last factor is, therefore, u, common 
 divisor of the given equation, and its first derived polynomial. 
 
 In general, if we have an equation X=0, containing the fac" 
 tors {x — a)'"(a; — i)", its first derived polynomial will contain the 
 factors mQc — a)'"~'»t(af — i)"~' ; that is, the greatest common divisor 
 of the given equation, and its first derived polynomial will be 
 {x — a)'""-'(x — b)"-\ and the given equation will have m roots, 
 each equal to o, and n roots, each equal to b. 
 
 Therefore, to determine whether an equation has equal roots, 
 find the greatest common divisor between the equation and its first 
 derived polynomial. If tliere is no common divisor the equation has 
 no equal roots. 
 
 If the greatest common divisor contains a factor of the form 
 X — a, then it has two roots equal to a ; if it contains a factor of 
 the form (x—ay it has three roots equal to o, and so on. 
 
 If it has a factor of the form {x — a)(x — 6) it has two roots 
 equal to o, and two equal to b ; and so on. 
 
 Ex. 1. Given the equation x' — x^ — 8a;-|-12^0, to determine 
 whether it has equal roots, and if so, to find them. 
 
 We have for the first derived polynomial (Art. 411), 
 3i;=— 2a;— 8. 
 
 The greatest common divisor of this and the given equation 
 (Art. 108) is x—2. 
 
 Hence x — 2^0, and x=-\-2. 
 
 . Therefore, the equation has two roots equal to 2. 
 
 Now since the equation has two roots equal to 3, it must be 
 divisible by (a— 2)(x— 2), or (2—2)'. (Art. 395.) 
 Whence, x^—x-'—Sx-\-12=(x—2)\x+Z)=Q, 
 and x-\-Z=0, or a;= — 3. 
 
 It is evident that when an equation contains other roots be- 
 sides the equal roots, that these may be found, and the degree of 
 the equation depressed by division (Art. 39-5), after which the 
 unequal roots may be found by other methods. 
 
364 RA.Y'S ALGEBRA, PART SECOND. 
 
 The following equations have equal roots ; find all the roots. 
 
 2. a;'— 2a;'— 15x+36=0. An«. 3, 3,— 4. 
 
 3. !<— 9x'+ 4a:+12=0. Ans. 2, 2, —1, —3. 
 
 4 . x'—6x>+l 2a;'— 1 0a;-|-3 =0. Ans. 1,1,1, ^■ 
 5 a;<— 7a;'+9a;2+27a;— 54=0. Ans. a:=3, 3, 3, — 2. 
 
 5. a;<+2a;'— 3a;=— 4i+4=0. Ans. —2, —2, +1, +1. 
 7. a:<— 12aJ4-50a;»— 84a;4-49=0. Ans. 3±V2, 3±^2 
 8 x=— 2a;<+3a!'— 7a;'+8a>-3=0. 
 
 Ans. 1,1,1,— ^±W— 11- 
 9. a;«+3x'— 6a;<— 6aJ+9ir'+3a;— 4=0. 
 
 Ans. 1,1,1,— 1,-1, —4. 
 
 Sdggestiox. — When the greatest common divisor of the given equa- 
 tion and its first derived polynomial, contains a factor of the form 
 (x — ay, or of any higher degree than the first, it is evident that the 
 first and second derived polynomials will also contain a common divisor, 
 of which the first or some higher power of x — a is also a factor. This 
 principle may be sometimes used, as in the last example, to simplify the 
 solution. 
 
 LIMITS OF THE ROOTS OF EaUATIONS. 
 
 Art. 41S. Limits to a root of an equation are any two num- 
 bers between which that root lies. A superior limit to the pos- 
 itive roots is a number numerically greater than the greatest posi- 
 tive root ; and an inferior limit to the negative roots, is a number 
 greater without regard to its sign, than the greatest negative root. 
 
 The characteristic of a superior limit is, that when it or any 
 number greater than it, is substituted for x in the equation, the 
 result is positive. 
 
 The characteristic of an inferior limit is, that its substitution 
 for a; produces a negative result, as likewise do all negative num- 
 bers numerically greater, provided the equation is of an odd degree. 
 
 The object of ascertaining the limits of the roots is to diminish 
 the labor necessary in finding them. 
 
 Aki. 41<>. Pkop. I. — The greatest negative coefficient, increased 
 by unity, is greater than the greatest root of the equation. 
 
 Take the general equation 
 
 a;"+Aar'-'-|-Ba;»-'. . . . +Ta;+V=0, 
 and let us suppose A to be the greatest negative coefficient. 
 
LIMITS OF THE ROOTS OF EQUATIONS. 363 
 
 .>A(^),ora.>45-_ 
 
 The reasoning will not be affected if we suppose all the coeffi. 
 tients to bo negative, and each equal to A. 
 It is required to find what number substituted for x, will make 
 a">A(x"-'+a;"-2+a?-^ . . . +x+l). 
 
 By Art. 297, the sum of the series in the parenthesis is *" ; 
 
 X — 1 
 enoe, we must have 
 
 A 
 x—V 
 
 But if x"=. , we find a=A+l ; therefore, A+l substituted 
 
 a; — 1 
 
 for X will render a^:= , and consequently 
 
 X — 1 
 
 „^ Ax" A 
 
 X 1 X 1 
 
 It is evident that by considering all the coefBcients after the 
 first negative, we have taken the most unfavorable case ; if either 
 of them, as B, were positive, the sum of the terms in the paren- 
 
 thesis would be less than 
 
 ar— 1" 
 
 Art. 417. Peop. II. — If we increase by unity that root, of the 
 greatest negative coefficient, whose index is equal to the number of 
 terms preceding the first negative term, the result will be greater than 
 the greatest positive root of the equation. 
 
 Let Caf^ be the first negative term, C being the greatest neg. 
 ative coefficient, then any value of x which makes 
 
 x">C(a»-'+a?"-' . . . . -\-x-\-\) (1) 
 will evidently render the first member of the equation )>0, or 
 positive ; because this supposes all the coefficients after C nega- 
 tive, and each equal to the greatest, which is evidently the most 
 unfavorable case. 
 
 jpft— r-)-l 1 
 
 By Art. 297, the series in the parenthesis is equal to . 
 
 x—\ 
 
 hence, by substitution, the inequality (1) becomes 
 
 .>o('_Z-l),„.>^' 
 
 x—V 
 
 But this inequality will be true if 
 
 a?'= __, or > 
 
 X — 1 x- 
 
366 RAY'S ALGEBRA, PART SECOND. 
 
 or, by multiplying both members by x — 1, and dividing by af^'+^, 
 when (I— l)a?-'=C, or >C (2). 
 
 But X — 1 is <^x, and . . (x — ly-'-C^a?'-' ; 
 therefore, (2) will be true, if we have 
 
 (x—l)(i— !)■-', or (a>— 1)'=C, or >C ; 
 
 or X — 1=^C, or >^C ; 
 or x=l+VC. or >1 + ^C. 
 Find superior limits of the roots of tlie foUowrlng equations : 
 
 1. a<— 5a^+37a;'— 3x+39=0. 
 
 Here C=5, and r=l .-. l+;/"C=l+5T=6. Ans 
 
 2. a«+7a;<— 12i»— 49ar'-l-52a>-13=0. 
 Here 1+^/0=1+^19=1+7=8. Ans. 
 
 3. x*+llx'—25x'-S^=(i. 
 
 By supposing the second term +0a^, we have r=3; hencci 
 the limit is 1+^67, or 6. 
 
 4. 3a:s—2a;2— 112+4=0. 
 Dividing by 3, a;'— fa;'— yx+|=0. 
 Here the limit is l+V , or 5. 
 
 Akt. 418. To determine the inferior limit to the negative 
 roots, change the signs of the alternate terms; this will change 
 the signs of the roots (Art. 400) ; then the superior limit of the 
 roots of this equation, by changing its sign, will be the inferior 
 limit of the roots of the proposed equation. 
 
 Art. 419. Prop. III. — If the real roots of an equation, taken 
 in the order of their magnitudes, be a, b, c, d, (fee, a being greater 
 than h, b greater than c, and so on; then if a series of numbers, a', 
 b', c', d', (Src, in which a' is greater than u.,h' a number between a 
 and b, c' a number between b and c, and so on, be substituted for x 
 in the proposed equation, the results will be alternately positive and 
 negative. 
 
 The first member of the proposed equation is equivalent to 
 (x — d)(x — b)(x — c)(a — d). . . . =0. 
 
 Substituting for x the proposed series of numbers a', b', d, &c., 
 we obtain the following results : 
 
LIMITS OF THE ROOTS OF EQUATIONS. 367 
 
 (o' — a){a' — 6)(a' — c)(o' — d), &c. . =-|- product, since all the 
 factors are -\-. 
 
 (p' — a)(b' — b)(b' — c)(b' — cI),Sl,c. . = — product, since only 
 one factor is — . 
 
 (c' — a)(c' — b)(c' — c)(c' — d), &o. . =-|- product, since two fac- 
 tors are — , and the rest -|-- 
 
 {i' — a){d' — b)(d' — o)(d' — d), &c. . = — product, since an odd 
 number of factors is — , and so on. 
 
 Cor. 1 . If two numbers be successively substituted for x, in 
 Bny equation, and give results with contrary signs, then between 
 these numbers there must be one, three,five, or some odd number 
 of roots. 
 
 Ccyr. 2. If two numbers, when substituted successively for x, 
 give results affected with the same sign, then between these num- 
 bers there must be two, four, or some even number of real roots, 
 or no roots at all. 
 
 Cor. 3. If a quantity q, and every quantity greater than q, ren- 
 der the results continually positive, q is greater than the gi-eatest 
 root of the equation. 
 
 Cor. 4. Hence, if the signs of the alternate terms be changed, 
 and if p, and every quantity greater than p, renders the result 
 positive, then — p is less than the least root of the equation. 
 
 Illustration. — If we form the equation whose roots are 5, 2, 
 and — 3, the result is x^ — 4i' — lla;-t-30=:0. Now if we substi- 
 tute any number whatever for x, greater than 5, the result is 
 positive. When we substitute 5 for x, the result is zero, as it 
 should be. 
 
 If we substitute for u:, any number less than 5, and greater 
 than 2, the result is negative. When we substitute 2 the result 
 is zero. 
 
 When we substitute for x, any number less than 2, and greater 
 than — 3, the result is positive. When We substitute ^-3, the 
 resull is zero. 
 
 If we substitute for x, any number less than — 3, the result is 
 negative. 
 
 By means of Corollaries 3 and 4, it is easy to find when we 
 have passed all the real roots, either in the ascending or descending 
 scale. 
 
368 
 
 RAY'S ALGEBRA, PART SECOND. 
 
 STUB MS THEOREM. 
 
 Akt. 430. To find the number of real and imaginary roots oj 
 an eqiialion. 
 
 In 1834 M. Sturm gained the mathematical prize of the Prencli 
 Academy of Sciences, by the discovery of a beautiful theorem, 
 Dy means of which the number and situation of all the real roots 
 of an equation can, with certainty, be determined. This theorem 
 we shall now proceed to explain. 
 
 Let X=a;"+Aa^-'+Bx"-2. . . . +Ta;4-V=0, 
 be any equation of the n" degree, which we may suppose con- 
 tains no equal roots ; for if the given equation contains equal 
 roots these may be found (.Art. 414), and the degree of the equa- 
 tion diminished by division. 
 
 Let the first derived function of X (Art. 411) be denoted bj X,. 
 
 Divide X by Xj until the remain- 
 der is of a lower degree with respect ' 'ij ^ 
 
 to X than the divisor, and call this ^j*"^' 
 
 remainder — Xj ; that is, let the re- X — X,Q,,= — Xj 
 
 mainder with its sign changed, be ^ s-^ /q 
 denoted by Xj. Divide X] by Xj X Q, 
 
 in the same manner, and so on, as -^ „ 
 
 in the margin, denoting the succes- ' ^ * ' 
 
 sive remainders with their signs X3)X, (Q,g 
 changed by X,, X^, &c., until we XjQj 
 
 arrive at a remainder which does ^ X Q, =— X 
 
 not contain x, which must always 
 
 happen, since the equation having no equal roots, there can be no 
 factor containing I, common to the equation and its first derived func- 
 tion. Let this remainder, having its sign changed, be called X, + 1 . 
 
 In making these successive divisions, we may either multiply 
 or divide the dividends and divisors by any positive number, for 
 the purpose of avoiding fractions, as this will not afiect the sigm 
 of the functions X, Xj, Xj, &c. 
 
 By this operation we obtain the series of quantities 
 
 X, X„ X„ X,. . . . X,+ , (1), 
 
 which, for the convenience of reference, we shall call series (1 ). 
 Each member of this series is of a lower degree witli respect to 
 X than the preceding, and the last does not contain x. We shall 
 also call X the primitive function, and Xi,X2, &c., auxiliary 
 functions. 
 
THEOREM OF STURM. 309 
 
 The following, then, is Sturm's Theorem : 
 
 If p and q he any two numbers, of which p is less than q, and if 
 these nwmhers be substituted for x in the functions X, X„ X,,<ic., in 
 the series (l),we shall have two series of signs, the one resulting from 
 the substitution of p for x, giving k variatioTis of signs, and the other 
 resulting from tJte substitution of q for x, giving k' variations of 
 signs; then the exact number of real roots of the given equation 
 between the limits p and q will be k — k'. 
 
 To simplify the demonstration of this theorem we shall employ 
 the following Lemmas : 
 
 Akt. 421. Lemma L — TVco consecutive functions, Xj, Xj, for 
 example, cannot both vanish for the same value of x. 
 
 From the process by which X,, Xj, &c., are obtained, we haTe 
 the following equations : 
 
 X =X,Q — Xj (1) 
 
 X, =X,Q,-X3 (2) 
 
 Xj =X,Ql,-X^ (3) 
 
 X_, =X,Q,— X, + ,. . . . (r— 1). 
 
 If possible let X]=0, and Xj=0, then by eq. (2) we have 
 X3=0; hence, since Xj=0, and X3=0, then by eq. (3) we have 
 Xj=0; and proceeding in the same way we shall find Xj=0, 
 X5=0, and finally X,+ i=0. But this is impossible, since X,+ i 
 does not contain a;, and 'therefore cannot vanish for any value 
 of X. 
 
 Akt. 422. Lemma II. — If one of the auxiliary functions van- 
 ishes for any particular value of x, the two adjacent functions must 
 have contrary signs for the same value of x. 
 
 Let us suppose that X5=0, when x^a ; then because 
 Xj=X3Q3— Xj, and Xs=0, therefore Xj=— X^ ; fhat is, X^ 
 and X^ have contrary signs. 
 
 Art. 42S. Lemma III. — If any of the auxiliary fi •Kiions vaji- 
 ishes when x^a, and h be taken so small that no root ii, any of the 
 other functions in series (1 ) lies between a — h, and a.-'r'tt, then will 
 the number of variations and permanences when a — h Cifi* a+h are 
 substituted for x in this series, be precisely the same. 
 
 Suppose, for example, the substitution of a for jc causes the 
 'unction X3 to vanish, then by Art. 421 neither o" the functions 
 
370 
 
 RAY'S ALGEBRA, PART SECOND. 
 
 Xj or X'4 can vanish for the same value of x ; and since when 
 Xj vanishes, Xj and X^ have contrary signs, (Art. 422), there- 
 fore, the substitution of a for x in the three functions Xj, Xj, X4 
 must give 
 
 X2 9 Xg , X^ , or X2, Xg, X^. 
 + - , - + 
 
 Ajid since A is talcen so small that no root either of Xj=0, or 
 X4=0, lies between a — h and a-f-A, the signs of these functions 
 will continue tne sane wnether we substitute a — A, or a-\-h for t 
 (Art. 419). Hence, whether we suppose X, to be -(^ or — by 
 the substitution of a — h and a-|-A for x, there will be (me varia- 
 tion and one permanence. Thus we shall have either 
 
 Xj , Xj , X^, or Xj , Xj, X. 
 
 + ± - - ± + 
 
 So that no alteration in the number of variations and perma- 
 nences can be made in passing from a — A to a-\-h. 
 
 Art. 124. Lemma IV.— Tf a. is a root of the equatim X=0, 
 then the series of fimctions X, X„ X., &c., vdll lose one variation 
 of signs in passing from a— h to a+h ; h being taken so small that 
 no root of the function X,=0, lies between a — h and a+h. 
 
 For X substitute a-\-h in the equation X=:0, and denote the re- 
 sult by H. Also put A, A', A" for the values of X and its derived 
 functions, when a-\-h is substituted for x ; we shall then have 
 (Art. 411) 
 
 H=AH-A'A+^A"A'+, &c. 
 
 But since o is a root of the equation X=0, we shall have 
 A=0, while A' cannot be 0, since the equation X=0 has no 
 equal roots. Therefore, 
 
 H=A'A+iA"A=-|-, &c., =A(A'+^A"A+, &c). 
 
 Now A may be taken so small that the quantity within the 
 parenthesis shall have the same sign as its first term A', (since 
 A' expresses the first derived function of X, corresponding to X. 
 in Art. 412), therefore, the sign of X, when x=a-\-h, will be the 
 sam£ as the sign of X,. 
 
 If we substitute a — A for x in the equation X=0, and denote 
 the result by H', we then have, by changing A into — A, in the 
 expression for H, 
 
 H'=— ft(A'— iA"A-|-, &c). 
 
THEOREM OF STURM. 371 
 
 Now it is evident that for very small values of h, the sign of 
 H will depend upon the first term — A'A, and consequently wil' 
 be contrary to that of A'. Hence, when ar=o — h, there is a varia- 
 tion of signs in the first two terms of the series X, Xj ; and 
 when x^a-\-h, there is a continuation of the same sign. There- 
 fore, one variation of signs is lost in passing from x=a — h to 
 a-\-h. 
 
 [f any of the auxiliary functions should vanish at the same 
 ti/ne by making x=a, the number of variations will not be 
 ifFected on this account (Art. 423), and therefore, one variation 
 if signs will still be lost in passing from a — h to a-\-h. 
 
 Art. 495. SttJiIm's Theorem. — If any two numlicrs, p and q, 
 (p being less than q) be^substiliUud for x in the series of functions 
 X, X,, X-j, (I'c, the substitution of p for x giving k variations, and 
 that of q for x, giving l<! variations, I 'len k— k' wilt be the exact num- 
 ber of real roots of the equation X=0, which lies between p and q. 
 
 Let us suppose that — oo is substituted for x, (by which sign is 
 meant any quantity so great that the signs of the different func- 
 tions X, X,, Xj, &c., depend on the sign of the first term only), 
 and suppose that x continually increases and passes through all 
 degrees of magnitude till it becomes 0, and after this let it con- 
 tinually increase till it becomes -)-x. . 
 
 Now it is evident, that so long as x, with its minus sign, is 
 less than any of the roots of X=0, X,=0, &c., no alteration 
 will take place in the signs of any of these functions (Art. 
 419); but when x becomes equal to the least root (with its 
 sign) of any of the auxiliary functions, although a change may 
 occur in the sign of this function, yet we have seen (Art. 423) 
 that it is the order only, and not the number of variations which 
 is affected. But when x becomes equal to any of the roots of the 
 primitive function, then one variation of signs is always lost. 
 Since, then, a variation is always lost whenever the value of x 
 passes through a root of the primitive function X=0, and since 
 a variation cannot be lost in any other way, nor can one be ever 
 introduced, it follows that the excess of the number of variations 
 given by 1=^, above that given by x^q, {,p<Cjf) is exactly equal to 
 the number of real roots of X==0, which lie between^ and q. 
 
 Cor. If the equation is of the n"" degree, and m represents the 
 number of real roots, then (Art. 396) the number of imaginary 
 roots will be n — m. 
 
873 RAY'S ALGEBRA, PART SECOND. 
 
 Art. 426. To determine merely the number of real roots, 
 we may substitute — oc- and -\- aa for x in the several functions. 
 In this case the sign of each function will be that of its first 
 term. 
 
 If we substitute for x, the number of variations lost from 
 — 00 to 0, will be the number of neffative roots ; and from to 
 + 00, the number of positive roots. 
 
 Aet. 437. To determiyie the situation of each real root ; that is, 
 the figures between which it lies. 
 
 Substitute the numbers 0, — 1, — 2, — 3, &c., for x, in series 
 (1), till we find a number which produces as many variations as 
 x= — CD produced. This number will be the limit of the greatest 
 negative root. 
 
 We then substitute the numbers 1,2,3, &c., till we find a 
 positive number which gives the same number of variations that 
 x=-\- 00 does. This will be the limit of the greatest positive 
 root. By observing where one or more variations is lost, we find 
 the situation of the roots. If two or more variations are lost be- 
 tween two of the substitutions, we must substitute smaller num- 
 bers, until only one variation is lost between two substitutions. 
 This operation is termed the separation of the roots. 
 
 Ex. 1 . Find the number and situation of the real roots of the 
 equation 4a;'— 12a:2+lla;— 3=0. 
 
 Here we have X = 4.x'—12x^+ll3>S, 
 and (Art. 411) X,=12a;=— 24a; +11. 
 
 Multiplying X by 3, to render the first term divisible by the ftrst 
 term of X,, and proceeding as in the method of finding the 
 greatest common divisor (Art. 108), we have for a remainder 
 — 2a;-|-2. Canceling the factor -{-2, and changing the signs 
 (Art. 420) we have Xj=a: — 1. Dividing Xj by Xj we have for 
 a remainder — 1 ; hence, Xj=-1-] . Therefore, the series of 
 functions are 
 
 X = 4a;3— 12a;'+llx— 3. 
 
 X,=12a;2— 24a!+ll. 
 
 X,= a:-l. 
 
 X,=+l. 
 
 Put — OD and + ™ for a; in the leading terms of these func- 
 tions, and the signs of the results are 
 
THEOREM OF STURM. 373 
 
 For x= — CD, 1 \. three variations, .•. k =3 
 
 x=-\- 00, -^ — 1 — I 1- no variation, .•. k'^0 
 
 .'. k — k'=3 — 0=3, the number of real roots. 
 
 Next, to find the situation of the roots, we must employ nar- 
 rower limits than — oo and -|- ao. But if we substitute in each 
 of the functions, we find three variations, the same number aa 
 Tor — OD ; hence, there is no real root between and — oo. This 
 we might also have learned fi:om Art. 402, Cor. 1, since there is 
 no permanence in the proposed equation. 
 
 In practice it is customary to substitute integral numbers first, 
 and afterward fractional ; particularly where two or more roots 
 lie between two whole numbers. In the present example, how- 
 ever, for the sake of illustration, we shall at once substitute frac- 
 tions. 
 
 A. JL, ^2 ^ij 
 
 For x^ — 00 the signs are — + — -\- giving 3 var 
 .■<;= _ _f- _ -|- « 3 '■ 
 
 ^=+i - + - + " 3 ■• 
 
 x=+i OL + — + 
 
 a=+|- + — — + " 2 « 
 
 x=+l — -f 
 
 ^=+li - - + + « 1 •' 
 
 a;=+li + + -f 
 
 a:=+l-| 4- + _|_ _|- « « 
 
 iE=+aD -J- -]_ 4_ ^ 11 « 
 
 Here we see that the roots are J, 1, and Ij, but if these num- 
 bers had not been substituted, we would have noticed that om 
 variation was lost in passing from | to | ; one in passing from | 
 to Ij, and lastly, OTie in passing from Ij to 1|, which would 
 have given the situation of the roots. 
 
 A careful study of this example will serve to illustrate the 
 theorem. Thus we see that there are three changes of sign of 
 the primitive function, two of the first auxiliary function, and one 
 of the second. We observe, however, that no variation is los* 
 by the change of sign of either of the auxiliary functions, while 
 each change of sign of the primitive function occasions a loss of 
 one t'ariation. 
 
8'5'4 RAY'S ALGEBRA, PART SECOND. 
 
 2. How many real roots has the equation 
 a;3_3a;2-|-a;— 3=0. 
 Here, X = ar^— Sai'-f-x— 3 
 X,=3a;2— 6a+l 
 Xj= x+2 
 X,=-25. 
 
 For x= — OD the signs are \ ,2 variations, .•. k =2 
 
 x=-\- 00 the signs are + + -| > 1 variation, . . k'=l 
 
 .-. k — k'=2 — 1=1, the number of real roots. 
 
 The root is 4-3, and by substitution it will be found that one 
 Tariation is lost in passing from 2 to 4. 
 
 Find the number and situafon of the real roots in each of the 
 following equations : 
 
 3 . a:'— 2a;'— a:+ 2=0 . Ans. Three, — 1 , + 1 , +2 . 
 
 4. 8a:'— 36a:'i+46x— 15=0. Ans. Three. One between and 
 1, one between 1 and 2, and one between 2 and 3. 
 
 5. 3? — 3a;' — 4a;+11^0. Ans. Three, one between — 2 and 
 — 1, one between 1 and 2, and one between 3 and 4. 
 
 6. X? — 2a; — 5=0. Ans. One between 2 and 3. 
 
 7. a?— 15a;— 22=0. Am. Three. One root is —2, one be- 
 tween — 2J and — 2 J, and one between 4 and 5. 
 
 8. a;*+a;' — x' — 2a;+4=0. Ajis. No real roots. 
 
 9. x*-43f—3x+23=0. Ans. Two. One between 2 and 3 
 and one between 3 and 4. 
 
 10. a;*— 2a;'— 7a;2+10a;+10=0. Ans. Four. The limits are 
 (-3,-2); (0,-1); (2,3); (2,3). 
 
 11. !i?—10x'+6x+l=0. Ans. Five. The limits are (—4. 
 -3); (-1,0); (-1,0); (0,1); (3,4). 
 
 CHAPTER XIII. 
 
 RESOLUTION OF NUMERICAL EQUATIONS 
 
 Aet. 428. In the preceding Articles we have demonstrated 
 the most important propositions in the theory of equations, and 
 in some cases have shown how to find their roots. The general 
 solution of an equation higher than the fourth degree has never 
 yet been effected, but the class of equations which most frequently 
 
RESOLUTION OF NUMERICAL EQUATIONS. 375 
 
 occurs in philosophical investigations is numerical ; that is, those 
 that have numerical coefficients. When the roots of these are 
 real, we can find them either exactly, or approximately, as near 
 as we please. The way for doing this has been prepared m the 
 preceding articles, by finding the limits of the roots, and separat- 
 ing them from each other. 
 
 RATIONAL ROOTS. 
 
 Art. 429. Prop. I. — To determine tlie integral roots of an 
 equation. 
 
 If a be an integral root of the equation 
 
 Aa;''4-Bx3+Cx'+Dx+E=0, 
 we shall have Aa*+Ba^+Ca^+Da-\-E=0; 
 
 .: ^— Aa'— Bo2— Ca— D. 
 a 
 
 Now since the second member of the last equation is evidently 
 
 P 
 a whole number, E is divisible by a. Put _=E' ; transpose D 
 
 a 
 
 to the first member, and divide'by a ; this gives 
 
 5^=— Aa'—Bo— C ; 
 a 
 
 therefore, a is also a divisor of E'-(-D. 
 
 Put E'-1-D=D', transpose C, and divide by a; this gives 
 
 ?-:i:5=— Affl— B ; . . a is a divisor of D'+C. 
 a 
 
 Again put , — JI—=C', transpose B, and divide by a, we find 
 a 
 
 a 
 Lastly, making C'+B=B', and transposing A, we have 
 
 B'+A=0. 
 If then, all these conditions are satisfied, a is a root of the pro- 
 posed equation ; but if any one of them fails, u is not a root. 
 Hence, we have the following 
 
 Rule for finding the integral roots of an equation. — 
 Divide the last term of the equation by any of its divisors a, and 
 add to the quotient the coefficient of the term containing x. 
 
 Divide thin swm by a, and add to the quotient the coefficient of x^. 
 
3T6 RAY'S ALGEBRA, PART SECOND. 
 
 Proceed in this manner unto the first term, and if a. be a root of 
 the equation, all these quotients will be whole numbers, and the result 
 will be . 
 
 Cor. 1 . It will be more easy to substitute the divisors +1 and 
 -l,at once in the given equation, and therefore they may be 
 omitted in the operation. Also, by ascertaining the limits to the 
 positive and negative roots (Art. 417), we shall frequently find 
 chat several of the divisors fall beyond the limits ; and therefore, 
 these may be omitted. 
 
 Cor. 2. If the coefficient of the first term be not unity, the 
 equation may have a fractional root. To determine if this be the 
 case, transform the equation into one in which the coefficient of 
 the first term shall be unity (Art. 405, Cor. 1), and then all the 
 rational roots will be integers (Art. 399). 
 
 Cor. 3. When all the roots except two are integral, the inte- 
 gral roots may be found by the rule, and then the proposed equa- 
 tion reduced to one of the second degree by division (Art. 396, 
 Cor. 1 ), and solved as a quadratic. 
 
 Ex. 1. Find the rational roots of the equation 
 a;34-3a:»— 4a!— 12=0. 
 
 Here, by Art. 417, no positive root can exceed 1+^12, or 4, 
 and the limit of the negative roots is l-j-3=4. 
 
 It is also found, by trial, that -|-1, and — 1 are not roots. 
 
 We then proceed to arrange the divisors of — 12, among which 
 'f, is possible to find the roots, and proceed with the operation as 
 'oilows : 
 
 Last term — 12 
 
 Divisors. , 
 Quotients . 
 4,dd —^ . 
 Quotients , 
 Add +3 , 
 Quotients , 
 Add +1 , 
 
 + 2 , +3 , +4 , — 2 , -3 , -4 
 _ 6 , — 4 , — 3 , +6 , -j-4 , +3 
 _10 , —8 , — 7 , +2 , -^ , — 1 
 
 — 5 , * , • , — 1 , * 
 
 — 2 , +2 , +3 , 
 
 — 1 —1,-1, 
 , 0,0. 
 
 Since — 8 is not divisible by -j-3, we proceed no further with 
 this divisor, as it is evident that it is not- a root of the equation ; 
 tn like manner +4 and — 4 cannot be roots. But we find that 
 -f-2, — 2, ap^ — 3 are roots of Ihe proposed equation. 
 
HORNER'S METHOD OF APPROXIMATION. 377 
 
 Find the roots of the following equations : 
 
 2. ics— 7a;2+36=0. Am. 3, 6, and —2. 
 
 SnGGESTioN, — When any term is wanting, as the third term In thi« 
 example its place must be supplied with 0. 
 
 3. x'— 6x2+lla^6=0. Ans. 1, 2, 3. 
 
 4. ar'+aj^— 4x— 4=0. Ans. 2, —1, —2. 
 
 5. x'— 3a'— 46j;— 72=0. Am. 9, —2, —4. 
 
 6. a;'— 5a;2— 18a;+72=0. Ani. 3, 6, —4. 
 
 7. a:<—10a;'+35a;=— 501+24=0. ' Ans. 1,2,3,4. 
 
 8. a;<+4x3—x'— 16a;— 12=0. Ans. 2, —1, —2, —3. 
 
 9. a;<— 4a;'— 19x2+46x+120=0. Ans. 4, 5, —2, —3. 
 
 1 0. x«— 27x'4-14x+120=0. Ans. 3, 4, —2, —5 
 
 11. x<4-x3— 29x2— 9x+180=0. Ans. 3, 4, —3, —5. 
 
 12. x'— 2x2— 4x+8=0. Ans. 2, 2, —2. 
 
 13. x'+3x=— 8x+10=0. (See Cor. 3.) Ans. —J>,l±J—i 
 
 14. x"— 9x3+17x2+27x— 60=0. Ans. 4, 5, dr^/S". 
 
 15. 2^ — 3x2+2x— 3=0. (See Cor. 2.) Ans. |, ±V— 1- 
 
 16. 3x=— 2x'— 6x+4=0. Ans. |, ±V2- 
 
 17. Sx'— 26x=+llx+10=0. Ans. |, J(3d=V41)- 
 
 18. 6x<— 25x'+26x2-|-4x— 8=0. Ans. 2, 2, f , — i. 
 
 19. x<— 9x'4-\^x2+yx— V=0. Ans. |, |, 3±3 ^2. 
 
 IRRATIONAL ROOTS METHODS OF APPROXIMATION. 
 
 After we have found all the integral roots of an equation, we 
 must have recourse to the methods of approximation, the best of 
 which is Horner's, by which we can always obtain the numerical 
 values of the real roots, to any required degree of accuracy. 
 
 Art. 430. Horner's Method of Approximation. 
 
 The principle of this method depends on the successive trans- 
 formation of the given equation, so as to diminish its roots at each 
 step, and the operation is performed by Synthetic Division, af 
 explained in Art. 410. 
 
 Let the equation, one of whose roots is to be found, be 
 
 Px"+Q«"-'. ■ ■ • +Tx+V=0. 
 32 
 
378 RAY'S ALGEBRA, PART SECOND. 
 
 Suppose a to be the integral part of the root required, and r, j, 
 t. . the decimal digits taken in order, BO that a;=a+r+s+'- ■ • 
 Let a be found by trial, or by Sturm's theorem (Art. 427), and 
 transform the equation into one whose roots shall be diminished 
 by o, by the method explained in Art. 410. 
 
 Let Pjfn + Q.y-'. . . . -fT'y+V'=0 be the transformed 
 equation, then the value of y is the decimal r-\-s-\-l. . . ; and 
 since this root is con's'ned between and 1, we may easily find 
 its first digit r. Again, let the roots of this equation be dimin- 
 ished by r, and let the transformed equation be 
 
 P2»-f Q'V-i . . . . +T"2+V"=0. 
 
 Now the value of z in this equation is s-j-i. . . , and the value 
 of s lies between .00 and .1; that is, it is either .00, .01, .02, 
 ... or .09. But since the figure s is in the second place of 
 decimals, the terms containing z^, 2'. . . will be small, and we 
 may generally find s, the next figure of the root, from the equa- 
 tion T"2-l-V"=0; that is, s is nearly equal to the quotient of 
 -V" divided by T". 
 
 Having found «, we next proceed to diminish the roots of the 
 last equation by s, and then from the last two terms, T"'2'-|-V"', 
 of the resulting equation, find t the next decimal figure, and 
 GO on. 
 
 Art. 431.. The absolute number, or last tertn of the equation, 
 is sometimes called the dividend, and the coefficient of the first 
 power of the unknown quantity,(for example T") the incomplete 
 or trial divisor. 
 
 The correctness of the values of the figures s, t, &c., obtained 
 by means of the trial divisor, will always be verified in the next 
 operation. For when we multiply by s, in the operation of 
 transformation, to obtain the product to be subtracted from V", 
 the number multiplied by s (sometimes called the complete divisor) 
 ought to be contained in V" only » times. But if it should be 
 contained a greater or less number of times, then s must be 
 increased or diminished. 
 
 In some cases, where it is small, and when the equation does 
 not exceed the third degree, r, the first decimal figure of the root, 
 may he found by dividing V by T'. The accuracy with whicii 
 each succeeding decimal figure may be found, increases as the 
 value of the figure decreases. In general, after threeor four dec- 
 imal figures have been found, the next three or four figures may 
 be obtained accurately by division, as in the method of finding 
 each previous figure. 
 
HOKNERS METHOD OF APPROXIMATION. 379 
 
 Aet. 432. To find the negative roots of the proposed equa- 
 tion, change the signs of the alternate terms (Art. 400) and find 
 the positive roots of the resulting equation ; these will be the 
 negative roots of the proposed equation.' 
 
 Remark. — Instead of finding the first decimal figure of the root by 
 trial, or by division, it may be found from the transformed equation by 
 Sturm's theorem ; but in general it can be obtained more easily by 
 trial. 
 
 Akt. 433. To illustrate this method, let it be required to find 
 the positive root of the equation x' — ix — 10.708649=0. 
 
 We readily find that a; must be greater than 5 and less than 6 
 therefore a=5. We then proceed to transform this equation 
 into another whose roots shall be less by 5. (See Art. 410.) 
 
 a 
 1—4 —10.768649 (5 
 
 +5 +5 
 
 +1 — 5.768649 
 
 +5 
 
 +6 
 
 Ist Trans, eq. ^"+6^ —5.768649=0. 
 
 Here we may find the value of y nearly, by dividing 5.7 by 6, 
 which gives .9 ; but this is too great because we neglected the 
 square of y. If we assume y=.8, and deduct y'=.64 from 5.7, 
 and then divide by 6, we see that y must be .8. Let us now 
 transform tne equation into another whose roots shall be less 
 by .8. 
 
 s 
 (.8 
 
 +6 
 
 —5.768649 
 
 .8 
 
 +5.44 
 
 +6.8 
 
 — .328649 
 
 .8 
 
 
 7.6 
 
 
 2nd Trans, eq. 2'.+7 .6*— .328649=0. 
 
 The approximate value of z in this equation is the second deci- 
 mal figure of the root. This is readily found by dividing the 
 absolute term by the coefficient of z, the first term, z", being now 
 60 small that it may be neglected. Thus, .328-;-7.6=.04^s. 
 
 We next prcceed ^ diminish the roots of the last equation 
 by .04. 
 
380 RAY'S ALGEBRA, PART SECOND. 
 
 +7.6 
 .04 
 
 s 
 —.328649 (.04 
 .3056 
 
 +7.64 
 .04 
 
 .023049 
 
 +7.68 
 +7.68/ 
 
 —.023049=0. 
 
 3rd Trans, eq. z'»+7.68/ 
 ' Here z' is nearly .023-=-7.68=.003=t. 
 By diminishing the roots of the last equation by .003 we have 
 
 < 
 1 +7.68 —.023049 (.003 
 
 .003 .023049 
 
 +7.683 .0 
 
 The remainder being zero, shows that we have obtained the 
 exact root, which is 5.843. 
 
 By changing the sign of the second term of the proposed equa- 
 tion, we have x'+4a; — 10.768649=0. The root of this equa- 
 tion may be found in a similar manner ; it is 1.843. Hence, the 
 two roots are +5.843 and — 1.843. 
 
 Ex. 2 . To illustrate this method further, let us form the equation 
 whose roots are 3, +^2, — ^2, which gives n? — 3x' — ^2a;+6 
 =0. Let it now be required to find, by Horner's method, the 
 root which lies between 1 and 2; that is, ^2. 
 
 By examination, we readily see that one root lies between 1 
 and 2; hence, a=l, and the first step is to transform the equa- 
 tion so as to diminish its roots by 1 . 
 
 a 
 +6 (1 
 —4 
 
 —3 
 
 —2 
 
 +1 
 
 —2 
 
 —2 
 
 —4 
 
 +1 
 
 — 1 
 
 — 1 
 
 —5 
 
 +1 
 
 
 
 
 
 +2 
 
 V 
 
 r==-=l =.4 
 
 T' 
 
 s 
 
 Hence, y'rtOy' — 5y+2=0, is the first transformed equation, 
 By dividing the absolute term 2, by 5, the trial divisor or coeffi- 
 cient of y, the second figure, r, of the root is readily found =.4, 
 and we proceed to transform the equation so as to diminish iti 
 roots by .4. 
 
HORNER'S METHOD OF APPROXIMATION. 381 
 
 
 
 r 
 
 ±0 
 
 —5 
 
 +2 (.4 
 
 .4 
 
 .16 
 
 —1.936 
 
 .4 
 
 —4.84 
 
 + .064 
 
 .4 
 
 .8 
 
 + .32 
 —4.52 
 
 ,=V"^.064^01 
 T" 4.52 
 
 .4 
 
 
 
 1.2 
 
 
 
 This gives «» + 1.2z2-^.522;+.064=0, for the 2iid trans- 
 formed equation ; and for s the next figure of the root .01. The 
 next step is to transform this equation so as to diminish its roots 
 by .01. 
 
 s 
 1 -1-1.2 —4.52 +.064 (.01 
 .01 .0121 —.045079 
 1.21 —1.5079 -1-.018921 
 .01 .0122 
 
 1.22 —4.4957 
 .01 i=Y:=:^l^=.004 
 
 1.23 T'" 4.495 
 
 This gives i!;''+1.23z'=—4.4957z'+ .018921=0, for the 3rd 
 transformed equation ; and for the next figure of the root i=.004. 
 The next step is to transform this equation so as to diminish its 
 roots by .004. 
 
 +1.23 
 
 —4.4957 
 
 +.018921 (.004 
 
 .004 
 
 + .004936 
 
 —.017963056 
 
 1.234 
 
 —4.490764 
 
 .000957944 
 
 .004 
 
 + .004952 
 
 
 1.238 
 
 —4.485812 
 
 
 .004 
 
 
 
 1.242 
 
 Having obtained three decimal places in the root, we may ob- 
 tain several of the succeeding figures accurately by division ; thus; 
 .0009o7944-7-4.485812=.0002135, which is true to the last 
 decimal place, as will be found by extracting the square root of 2. 
 Hence, 1=1.4142135. 
 
 To illustrate the process fully, the preceding operation has been 
 presented in the most extended form. In practice it is customary 
 
382 
 
 RAY'S ALGEBRA, PART SECOND. 
 
 to make some abridgments. Thus, by marking with a * the to- 
 efficients of the unknown quantity in each transformed equation, 
 it is not necessary to rewrite it. Also, when the root is required 
 only to five or six places of decimals we need not use more than 
 this number in the operation. 
 
 We shall now give the solution of an equation of the ith 
 degree, presenting the operation in a concise form. 
 
 3. Given x* — 8a;'-j-14a;'-f-4x — 8=0, to find a value of at. 
 
 
 OPEHATIOH. 
 
 
 —8 
 
 +14 . 
 
 +4 
 
 —8 (5.236068 
 
 5 
 
 —15 
 
 —5 
 
 —5 
 
 —3 
 
 — 1 
 
 —1 
 
 ♦—13 
 
 5 
 
 10 
 
 45 
 
 10.6576 
 
 2 
 
 9 
 
 *44 
 
 *— 2.3424 
 
 5 
 
 35 
 
 9.288 
 
 1.93880241 
 
 7 
 
 *44 
 
 53.288 
 
 *— .40359759 
 
 5 
 
 2.44 
 
 9.784 
 
 .39905490 
 
 *12 
 
 46.44 
 
 *63.072 
 
 *— .00454269 
 
 .2 
 
 2.48 
 
 1.5.'54747 
 
 .00400954 
 
 12.2 
 
 48.92 
 
 64.626747 
 
 '—.00053315 
 
 .2 
 12.4 
 
 2.52 
 *51.44 
 
 1.566321 
 *66. 193068 
 
 
 .2 
 
 .3849 
 
 .31608 
 
 
 12.6 
 
 51.8249 
 
 66.50915 
 
 
 .2 
 
 3858 
 
 .31656 
 
 
 *12.8 
 
 52.2107 
 
 *66.82571 
 
 
 .03 
 
 3867 
 
 
 
 12.83 
 
 *52.5974 
 
 
 
 03 
 
 08 
 
 
 
 12.86 
 
 52.68 
 
 
 
 .03 
 
 .08 
 
 
 
 12.89 
 
 52.76 
 
 
 
 03 
 
 
 
 
 1 *12.92 
 
 As the root is found only to six decimal places, it is not neces- 
 •ary to carry the true divisor for the third figure (6) to more than 
 
HORNER'S METHOD OF,APPROXIMATION. 383 
 
 five decimal places; this divisor is 66.50915, which multiplied 
 by .006, gives eight decimal places, and the dividend ought to be 
 carried to seven or eight decimal places, in order that the figure 
 in the sixth decimal place of the root may be correct. So the 
 divisor, 66.825, for the fifth figure of the root, requires to be car- 
 ried only to three decimal places, for the product of this number 
 by .00006 gives eight decimal places as it ought to do. So the 
 divisor for the last figure (8) of the root would require to be caiv 
 ried only to two decimal places. The numbers in the vertical 
 columns preceding the divisors, require to be carried to still fewer 
 places, as the pupil will readily perceive. 
 
 After obtaining the third figure of the root, the next three may 
 be obtained merely by division; thus, .00454269-^-66.82571 
 =.000068 nearly. 
 
 The pupil must observe that where decimals are omitted, we 
 always take the figure next to the omitted places, to the nearest 
 unit. Thus, .07752 is nearer .08 than .07; therefore, the former 
 is taken. 
 
 Art. 434. The process illustrated in the preceding examples 
 may be extended to equations of any degree, and is justly re- 
 garded as the most elegant method of approximating to the roots 
 of equations yet discovered. It may be briefly expressed by the 
 following 
 
 Rule. — 1 . Find hy trial, or by Sturm's theorem, the integral part 
 of the required root. 
 
 2. Transform tlie equation (Art. 410) into another whose roots shall 
 be those of the proposed equation, diminished hy the part of the 
 root already found. 
 
 3 . With the absolute term in the first transformed equation for a 
 dividend, and the coefficient of x for a divisor, find tlie first deci- 
 mal figure of the root. 
 
 4. Transform the last equation into another whose roots shall be di- 
 minished by the part of the root already found, and from the first 
 two terms of this equation,find the second figure of the root. 
 
 5 . Continue this process, till the root is found to the required degree 
 of accuracy. 
 
 6. To find the negative roots, change the signs of the alternate terms, 
 and proceed as for a positive root. 
 
 Remarks. — 1. If any figure, found by trial, is either too great or too 
 small, it will be made manifest in the next transformatian. (See Art 
 «31.) 
 
384 
 
 RAY'S ALGEBRA, PART SECOND. 
 
 2. In general, after three figures of the root have been found accu- 
 rately, the next three may be obtained by dividing the absolute term by 
 the coefficient of x. 
 
 EXAMPLES FOR PRACTICE. 
 
 Let the pupil find at least one valua of 
 ing equations : 
 
 1. a;2+5x— 12.24=0. 
 
 2. I'+l 2a;— 35.4025=0. 
 
 3. 4i'—28x— 61.25=0. 
 
 4. 8x2—1201+394.875=0. 
 
 5. 5x'—7.4x— 16.08=0. 
 
 6. x'+x— 1=0. 
 
 7. x=— 6i+6=0. 
 
 8. x=+4x2—9x— 57.623625=0. 
 
 9. 2x'— 50x+32.994306=0. 
 
 10. x»+x=+x— 1=0. 
 
 11. x'+4x2— 5x— 20=0. 
 
 12. x'— 2x— 5=0. 
 
 13. x=+10x=—24i— 240=0, 
 
 14. x'+12x=— 18x=216. 
 
 15. X*— 8x3+20x=— 15x4-.5=0. 
 
 16. x<+x'— 8x— 15=0. 
 
 17. x<— 59x=+840=0. 
 
 18. 2x<4-5x'+4x5+3x=8002. 
 
 19. x'+4x<— Sx^+lOx'- 2x=962. 
 
 20. x'+2x<+3x3+4x'+5x=54321. 
 
 X in each of the follow- 
 
 Ans. x=1.8. 
 
 Atis. x=2A5. 
 
 Ans. x=8.75 
 
 Ans. x=10.125 
 
 Atis. x=2.68. 
 
 Ans. x=.618034. 
 
 Ans. x=4.73205. 
 
 Ans. x=3.45. 
 
 Ans. x^4.63. 
 
 Ans. x=.543689. 
 
 Ans. x=2.23608. 
 
 Atis. x=2.0945515. 
 
 Atis. x=4.8989795 
 
 Atis. x=4.2426407. 
 
 Ans. x=l .284724. 
 
 Ans. x=2.302775. 
 
 Ans. x=4.8989795. 
 
 Atzs. x=7.335554. 
 
 Atis. x=3 .385777. 
 
 Atis. x=8.414455. 
 
 Art. 433. To extract ike roots of nurribers by Horner's Method. 
 
 The extraction of any root of a number is only a particular 
 case of the solution of an equation of the same degree ; for if we 
 call the number N, the root x, and the index of the root n, we 
 shall have x"=N, or x" — N^O; an equation of the n"' degree 
 in which all the terms are wanting except the first and last. 
 
 In performing the operation we may find the successive integral 
 figures in the same manner as the successive decimal places were 
 
APPROXIMATION BY DOUBLE POSITION. 
 
 385 
 
 found in the preceding article. It is only necessary to bear in 
 mind that any two figures in consecutive integer places, have the 
 same relation to each other as if they were in consecutive deci- 
 mal plr.ces. In extracting any root, the cube root for example, 
 it is necessary to point off the given number into periods, as in 
 the operation by the common rule. We shall now illustrate the 
 method of operation by finding the cube root of 12977875; tha* 
 is, by nndiag one root of the equation x'. — 12977875=0. 
 
 
 2 
 
 
 
 4 
 4 
 
 8 
 
 12977875 
 8 
 
 2 
 2 
 
 4977 
 4167 
 
 4 
 
 2 
 
 *12 
 189 
 
 810875 
 810875 
 
 *6 
 
 3 
 63 
 
 3 
 66 
 
 3 
 
 1389 
 198 
 ♦1587 
 3475 
 162175 
 
 
 *69 
 5 
 
 
 
 695 
 
 
 
 Should 056 learner not readily understand the reason for the 
 manner ih which the figures are placed in the successive columns, 
 let him perform the operation, using the numbers 200, and 30, 
 instead of 2 aBiiS, and all the difficulties will vanish. 
 
 By the same method find 
 
 2. The cube root of 34012224. Ans. 324. 
 
 3. The cube root of 9. Ans. 2.080084. 
 
 4. The cube root of 30. Ans. 3.107233. 
 
 5. The fifth root of 68641485507. Ans. 147 
 
 APPROXIMATrOir BY DOUBLE POSITION 
 
 Art. 436. Double Position furnishes one of the most usefu. 
 methods of approximating to the roots of equations. It has th 
 advantage of being applicable, whether the equation is fractional 
 radical, or exponential, or to any other form of function. 
 
 Let X=0, represent any equation ; and suppose that a and ft, 
 33 
 
386 RAY'S ALGEBRA, PART SECOND. 
 
 when substituted for x, give results, the one too small, and the 
 other too great, so that one root of this equation lies between a 
 and h. (Art. 403.) 
 
 Let A and B be the results arising from the substitution of a 
 and 6 for x,m the equation X^O. Let a;=a+A, and b=a-\-k . 
 then if we substitute a-{-h, and a-\-lc for x, in the equation X=0 
 we shall have 
 
 X=A4-A'A+iA"A=+, &c. 
 
 B=A4-A'i+iA"A:2+, &c. 
 
 Here A', A", &c., are the derived functions of A (Art. 411). 
 Now if A and k be so small that their second and higher powers 
 may be neglected without much error, we shall have 
 X — A=A'A nearly ; 
 B— A=A'A: " . 
 Whence, B— A : X— A : : A'k : All : k : h ; 
 
 or B— A -.k : : X— A : h, (Art. 270) , 
 or B — A : b — a : : X — A : h, since k=b — a. 
 Hence we have the following 
 RtiLE. — Find by trial, two numbers which substituted for x in the 
 proposed equation, give one a result too small, and the other 
 too great. Then say, 
 As the difference of the results; 
 Is to the differeTice of the suppositions ; 
 So is the difference between the true result and eillier of the former 
 
 results ; 
 To the correction of the corresponding supposition. 
 
 This correction is to be added to the corresponding supposition 
 when it is too little, and subtracted when it is too great, and the 
 result will be the first approximation. 
 
 Substitute this root for the unknown quantity, and the result 
 will show whether the supposition is too small or too great ; then 
 take another number such that the true root may lie between it 
 and the last supposition, and proceed, as before, to obtain a 
 second approximate value of the required root ; and so on. 
 
 It is generally best to begin with two integers which differ 
 from each other by unity, and to carry the first approximation 
 only to OTie place of decimals. In the next operation the differ- 
 ence of the suppositions may be 0.1, and the second quotient 
 may be carried to two places, and so on, doubling the number of 
 places of decimals at each approximation. 
 
NEWTON'S METHOD OF APPROXIMATIO^. 38T 
 
 Ex, 1. Given x'+x'-\-x=l()0, to find x. 
 
 It is easily found tliat x lies between 4 and 5. We then sub- 
 stitute these two numbers for x in the given equation, and the re- 
 sult is as follows : 
 
 4 X 5 
 
 64 i' 125 
 
 16 a' 25 
 
 4. X 5 
 
 84 results 155 
 
 155 5 100 
 
 84 _4 _84 
 
 71 : 1 : : 16 : 0.22; 
 
 therefore, a;=4.2, the first approximation. 
 
 Again, substituting 4.2 and 4.3 for x in the given equation, 
 and proceeding as before, we get for a second approximation 
 a;^4.264. liy assuming 1=4.204, and x=4.265,and repeating 
 the operation, we obtain for a third approximation x=:4.2644299 
 nearly. 
 
 Find one root of each of the following equations : 
 
 2. a:3_i_30j;=420. Ans. x=Q. 11 OlOZ. 
 
 3. 144x3— 973i-=319. ^„s x=:2.75, 
 
 4. x'+10x=+5x=2600. Ans. x=l 1.00679. 
 
 5. 2x'+3x2— 4x=10. Ans.x=l. 62482. 
 
 6. i'— x'+2x'+x=4. Ans.x=l. 14699. 
 
 7. x'4-.-'+2x2— iE=4. Ans. x=l .09059. 
 
 8. x<— 12x+7=0. Atis. x=2.04727. 
 
 9. 2x^— 13x'+10x— 19=0.. Ares. x=2.4573. 
 
 10 %/7x'-\-ix''+JWx(^2x—l)=2a. Ans. x=4.51066. 
 
 Art. 437. Newton's Method of ArPKOxiMATioN. — This 
 method of approximation is but little used, yet it is so often re- 
 ferred to, that it is desirable the learner should be acquainted with 
 .he principle on which it is founded. 
 
 Find by trial, two numbers which, substituted for the unknown 
 quantity, give results with different signs ; then (Art. 403) one 
 real root, at least, lies between these two numbers. Now by in- 
 creasing one of the limits, and diminishing the other, an approxi- 
 Eiation may be made to the root. When the quantity a thus 
 found is within 0,1 of the value of the root, we may substitute 
 a-f-y for x in the given equation, and it will be of this form 
 
388 RAVS ALGEBRA, PART SECOND. 
 
 A+A'y-\-^A'y+lA"y+, &c., =0 (Art. 411). 
 where A, A', A", &c., are known quantities dependent on a. 
 From this equation, by transposing and dividing, we find 
 
 y A' -A' •=' 6a"^ 
 and since y is <^0.1,y^ will be <^0.01,^'<^0.001, and so on. 
 Therefore, if the sum of the terms containing y', y', &c., be less 
 than .01, we shall, in neglecting them, obtain a value of y within 
 
 A 
 
 .01 of the truth. Let, then, y= — — , which gives for x the value 
 
 A' 
 
 A 
 
 a . This will differ from the true value of a: by less than .01 
 
 A' ^ 
 
 [t is not necessary, however, to carry on the division of A by A 
 beyond the second place of decimals, as the accuracy of the fig- 
 ure in the third place, could not be relied on. 
 
 Now, put 5 for this approximate value of x, and let a=5+i • 
 we have then as before 
 
 B+B'2+AB'V+jB"'z'+, &c., =0; 
 
 and as z is supposed to be less than .01, z' will be <[.0001. If 
 
 then we neglect the terms containing 2', z', &c., we shall obtain 
 
 a probable value of z within .0001; and so on. 
 
 Since A is what the proposed equation becomes when x=a, 
 
 and A' what the first derived function becomes when x=a, there- 
 
 A B 
 fore the corrections — — ;, —,, ^c., are easily found. 
 
 Newton gave but a single example, viz. : to find the value of x 
 in the equation ar^— 2a; — 5=0. Ans. 1=2.09455149. 
 
 The pupil desirous of additional exercises may solve the exam- 
 ples in the preceding article by this method. 
 
 cardan's rule for solving cubic EaUATIONS. 
 
 Art. 438. In its most general form, a cubic equation may bo 
 represented by 
 
 x'-\-px^-{-gx-\-r=0 ; 
 but as we can always take away the second term by the method 
 described in Art. 407, we will suppose, in order to avoid fractions 
 that it is reduced to the form 
 
 a:'+35a;+2r=0. 
 
 Assume x=:7/-\-z, and the equation becomes 
 
 y'+z'+3y2(y+2)+3?(y+z)+2r=0. 
 
CARDAN'S SOLUTION OF CUBIC EQUATIONS. 389 
 
 Now since we have two unknown quantities in tliis equation, 
 y and z, and have made only one supposition respectjng them, 
 namely, that y-\-z-=x, we are at liberty to make another. Let, 
 therefore, yz= — q ; then, by substituting this in the equation, it 
 
 becomes y*-|-z'-(-2r=0 ; but since yz=—q,-we have a'^ — ^ ; 
 hence y' — ^?_-J-2r=0, 
 or y^-\-2ry'=q^. 
 
 whence y'= — r+^r'+9'=A', 
 and simuarlv z'= — r — ^ r'-^q'==B^ ; 
 
 the radical quantity being taken positive in one of these expres- 
 sions, and negative in the other, to render them different. And 
 since x=^y-\-z, we have 
 
 This formula would appear to give but one of the roots. But 
 since the values of y and z are found by extracting the cube roots 
 of A' and B', it will now be shown that each of them must have 
 three values. 
 
 Since 2/'=A^ we have j/' — A'=0, or, by factoring (Art. 83), 
 (y — A)(y'+A!/+A')=0 ; putting each of these 
 factors equal to 0, and solving the resulting equations, we have 
 
 y=A, y== ~^+y~^ A, and y=Z±:^t:±A. 
 
 Similarly, from the equation 2'=B', we find 
 
 2 2 • 
 
 By combining each value of y with the three values of z, it 
 might appear that x had 9 values ; that is, that an equation of the 
 third degree has nine roots, which is impossible (Art. 397). 
 
 That this is impossible from the solution is thus shown : 
 We supposed that y z = — q. 
 
 But six of the products of the values of yfc give imaginary 
 values, and since yz=: — q, a real negative quantity, therefore, 
 the combinations giving imaginary products must be rejected, and 
 the three values of x are 
 
 1« A+B, 
 
 2-- -^+^/-^A+ -1-V-3 b, 
 
 83* ^ ^ 
 
390 RAY'S ALGEBRA, PART SECOND. 
 
 3" — 1— V— a .1 — 1 + V— 3 ^^ 
 
 Aet. 439. If r=+9' be negative, that is, if r'+j3<0, the 
 values of x become apparently imaginary Vi^hen they are actually 
 real, and we shall now show that 
 
 Cardan's Method of Solution does not extend to those cases in which 
 Ike equation has three real and UTiequal roots. 
 
 Since every cubic equation has at least one real root (Art. 401, 
 Cor. 3), we may suppose this to be a ; and the other two roots 
 arising from the solution of a quadratic, may be represented by 
 b-\-,JSc, and b — ;^3c, in which, if 3c be positive, the roots are 
 real, and if 3c be negative they are imaginary ; and because the 
 second term of the equation is 0, we have (Art. 395, Corollaries) 
 
 0=a+(i4-^3^)+(6— 73"c)=a+26 ; 
 39=0X26+6'— 3c=—36='— 3c ; 
 2r=— o(62— 3c)=2J'— 66c. 
 Hence we have 
 
 r2+53=(6'— 3 6c)'— (62+c)' 
 
 =— 96<c+662c'— c'=— c(365— c)' . 
 ... ^7qY=(35'-c)V=^. 
 
 Now this expression la real when c is negative, and imaginary 
 when c is positive, or when the equation has three real roots. 
 
 If we suppose c to become zero, the value of «/'■'+?' ceases 
 to be imaginary. But this supposition reduces the roots to a, 6, 
 and 6 ; hence, Cardan's Rule is applicable to equations of the 
 third degree containing two equal roots. 
 
 Aet. 440. From the last article we see that when the roots 
 of the quadratic equation in Art. 438 are imaginary, the roots of 
 the cubic equation are all real. As this appears paradoxical, we 
 will show by the direct solution of a particular exampla, that 'i« 
 vilue of X, in this case, is a real quantity. 
 
 Ex. To find the three roots of the equation x' — 15a? 4-rf 
 By substituting y-j-z for x, we have 
 
 uid, therefore, as in Art. 488, 
 
 Zyz=15, y'+z' — 4=0. 
 
CARDAN'S SOLUTION OF CUBIC EQUATIONS. 39J 
 
 Prom the solution of these equations we obtain y'=2-\- 
 11 aJ — 1; and it may be proved, by actual multiplication, that 
 y=2 + V^; likewise, z^=2—ll^~—i, and z=2 — J~l. 
 Hence, a:=s^+z=(2+V— 1)+(2— V— 1)=4- 
 
 By dividing the given equation by x — 4, and placing the result 
 equal to 0, we find the other two roots are as= — 2-\-^'d, and 
 -2—J3. 
 
 As no means have yet been discovered for reducing the imag- 
 inary forms to real values, Cardan's rule fails when all the roots 
 are real. 
 
 This is termed the irreducible case of cubic equations, and has 
 been a subject of great perplexity to mathematicians. 
 
 Art. 441. We shall present some examples for solution in 
 the case to which Cardan's rule applies ; that is, when the pro- 
 posed equation contains one real and two imaginary roots. 
 When the proposed equation contains the second term it must 
 first be removed (See Art. 407), and the equation reduced to the 
 form x'-|-3ga;-l-2r=0. 
 
 Then r=V(— ^+V'-'+9')+VC— »•— V'-'+9'). will be the 
 real root of the proposed equation. 
 
 Having the real root, the imaginary roots may be found from 
 the formula in Art. 438, or by reducing the proposed equation to 
 a quadratic. 
 
 Ex. 1. Solve the equation «'+3D=-f-9u— 13=0. 
 
 If we transform this equation into another which shall want its 
 second term, by substituting x — 1 for v (Art. 407), we have 
 x'-l-6x— 20=0. 
 
 Comparing this with the equation x^-{-2qx-{-2r^) , we find 
 j=2, ?■= — 10; hence, 
 
 x=5/(10+V108)+V(10—^/108)=2.732— .732=2. 
 Whence v=a: — 1^2 — 1=1. 
 
 The other two roots are easily found to be — 1±3 ^ — 1. 
 Solve the following equations by Cardan's Rule : 
 
 2. x'— 9x+28=0. Am. x=— 4, 2±V^. 
 
 3. x'+6x— 2=0. Ans. x=5/4— V2=.32748. 
 
 4. x'— 6x'+13x— 10=0. Am. x=2, 2±V^- 
 
 5. x»+6x2— 32=0. Ana. x=2, —4, —4 
 
802 RAY'S ALGEBRA, PART SECOND. 
 
 6. i>+8a;'4-27x— 26=0. Ans. a:=.801245. 
 
 1. x'—9x'+Gx—2=0 A7w.a;=8. 306674. 
 
 Remark. — It is not deemed necessary to introduce the Rules of 
 I'errari, Euler, Descartes, or Simpson, for the Solution of Equations 
 of the fourth degree, since they are applicable only to special cases, 
 and, together with Cardan's Rule for the solution of cubic equations, are 
 regarded, since the discovery of Horner's method, and Sturm's theorem, 
 ts little more than analytical curiosities. 
 
 RECIPROCAL OR RECURRING EQUATIONS. 
 
 Art. 442. A recurring or reciprocal equation is one such that 
 
 if a be one of its roots, the reciprocal of a, that is _ , will be 
 
 a 
 
 another. 
 
 Prop. I. In a recurring equation the coefficients, when taken in a 
 direct and in an inverse order, are the same. 
 
 Let j;-+Aa;"-'4-Bj^-=. . . . +Sa;'+Ta;+V=0, 
 be a recurring equation ; that is, one that is satisfied by the sub- 
 stitution of - for I ; this gives 
 
 X 
 
 L+4_+? . . . +?+T+v=o, 
 
 X" a;"-'^a!»-' ^x' x^ 
 
 and multiplying by if. 
 
 l+Ai+Car'. . . . +S*"-'+Ts'^'+Vx"=0, 
 which proves the proposition. 
 
 Note. — Equations of this kind are called Recurring equations from the 
 forms of their coefficients, and Reciprocal equations from the forms of their 
 roots. 
 
 Prop. II. A recurring equation of an odd degree has one of its 
 roots equal to -\-l , when the signs of the like coefficients are different, 
 iut equal to — 1 , when their signs are alike. 
 
 Since every power of -|-1 is positive ; when the signs of the 
 like coefficients are different, if we substitute +1 for x, the cor- 
 responding terms will be equal, but of different signs ; hence, 
 they will destroy each other. But when the signs of the lilce 
 coefficients are the same, then since one of them will belong to an 
 odd power, and the other to an even power, if we substitute — 1 
 for X, the corresponding terms will be equal, but of different signs , 
 and, therefore, they will destroy each other. Hence, in either 
 
RECURRING EQUATIONS. 393 
 
 case the equation will be satisfied, and may be reduced one degree 
 lower by dividing by x — 1 , or x+l . 
 
 Prop. III. A recurring equation of an even degree, in whtch the 
 like coefficients have opposite signs, and whose middle term is wanting, 
 is divisible by x' — l,and therefore, two of its roots are-\-\,and 
 —1. 
 
 Let a;'"+Ax"-'+Bx2°-». . . . — Bx'— Ax— 1=0, 
 
 be an equation of the kind specified. It may evidently be 
 arranged thus 
 
 (x2"_l)4-Ax(x»»-2— l)+Bx'(x="-<— 1)+, &c.. . . =0, 
 which is divisible by x'— 1 (Art. 83). 
 
 Cor. An equation of this form may therefore be reduced two 
 
 degrees lower by division. 
 
 The most convenient method of reduction, either in this case 
 or the preceding, is by means of Synthetic Division (Art. 
 109). 
 
 Prop. IV. Every recurring equation of an even degree above tlte 
 second, may he reduced to an equation of half that degree. 
 
 For, x'"— Ax'^-'+Bx'"-'— , &c. . . +Bx'— Ax-f 1=0, by 
 dividing by x", and collecting the pairs of terms equi-distant from 
 the extremes, becomes of the form 
 
 (x-.+L)-a(x«-+L_)+b(x"-=+L_)-.&c.,=0. 
 
 Let x+-=z, then x^-\-—=z' — 2, by squaring ; also, 
 X x' 
 
 (x3+l^) = (x^+l,)-(.+^)=(.'-2)-.; 
 
 and generally ( x»+L ) = ( x'-+ L_ ) ^- ( x»-=+ L_ ) . 
 
 Hence, each of the binomials may be expressed in terms of z, 
 and the resulting equation will be of the n"' degree. 
 
 If the signs of the terms from the beginning and end be diffe? 
 ent, let z=x — -, and a similar result will be obtained. 
 
 X 
 
 Ex. 1. Given x*— 5x'+6x= — 5x+l=0, to find x. 
 Here x»-5x+6— 5+1=0, or ( x'+L ) -5 ( x+l ] +6=0. 
 
894 RATS ALGEBllA, PART SECOND. 
 
 Let x-f— =*i then «' — 52+4=0, and z=4 or Ij 
 
 X 
 
 also ar+_=4, gives x=2zkJS; 
 
 X 
 
 and x+_=l, gives x=^(}±^—3). 
 
 Hence, 1=2+^3", 2— ^"3, 
 
 ^+^/^ 1— V— 3 
 2 2~- 
 
 The learner may not see readily that the second of these valnea 
 is the reciprocal of the first, and the fourth of the third, we will, 
 therefore, explain. 
 
 2+V3 2+V3 2— V3 4-3 
 
 2 1 — V^^ 
 
 In like manner — — J=== s ' 
 
 1+V — 3 
 
 EXAMPLES IN RECUREING EQUATIONS. 
 
 1. a:*— 10a;'4-26a:2— 101+1=0. 
 
 Ans. a;=3±2 ^2", 2±V3. 
 
 2. a;<+5i'+2i'+5i+l=0. 
 
 Avs. a;=^(— 5± V21) ; ±-s/^ • 
 
 3. «<—|a:'+2a;'— ^1+1=0. 
 
 Ans. x=2, 3 ; ±^ — 1 
 
 4. I*— 3a:'+3a:— 1=0. Ans. a;=±l, 2(3±V5). 
 
 5. x>—lli<+17i'+17x=— 111+1=0. 
 
 A«s x—l ^±M ^-J^ 3+V5 3-V5 
 
 6. 4j;»— 24a;s+57x<— 73a;'+57a'— 24a:+4=0 . 
 
 A».. :r=2, .,2, 1, ll^^. ^=^. 
 
 BINOMIAL EQUATIONS. 
 
 Art. 443. Binomial equations are those of tlie form 
 jr±A=0. 
 
BINOMIAL EQUATIONS. 395 
 
 Let ^A=a ; that is, A=fl". 
 Then i/"±a"=0. 
 Let y^=ax, then ffi"a;"±o"=0, 
 or i»±l=0, 
 which is a recurring equation. 
 
 Aet. 444. I. — The roots of the equation 
 
 iC"±1^0, are all unequal; for the first 
 derived polynomial 7W:"~', evidently has no divisor in common 
 with x"±l, and therefore there are no equal roots (Art. 414). 
 
 II. — If n be even the equation x" — 1 ^0, or a;"=l , has two real 
 roots, 4-1 and — 1, and no more. That it has these two roots is 
 evident from Art. 442, Prop. Ill ; and that it has no other real 
 root is evident because no other number can by its involution pro- 
 duce 1. 
 
 By dividing a?' — 1=0 by (ir-|-l)C^ — l)=a;' — 1, we have 
 x— 2_|-a;"-<+. . . . +x*+x'-\-l=0, 
 a recurring equation in which all the n — 2 roots must be imag- 
 inary. 
 
 For example, the equation x^=l, or x' — 1=0 divided by a;' — i 
 gives x^-{-x^-}-l=0 ; 
 
 whence x=d=J ^~^^'^~H . 
 
 This gives for the six roots of 1 
 
 +1, _ —1, 
 
 , / -l+V-3 / -1+/-3 
 
 "'"^/ 2 ' V 2 ' 
 
 +. 
 
 V— 3 _ /— 1— V— 3 
 
 4=i^^-4 
 
 III. — If n be odd, the equation x" — 1=0 has only one real 
 root, viz. . -f-l ; for +1 is the only real number of which the odd 
 powers are -|-1 . Divi(fing X" — 1=0 by x — 1, we have the recur- 
 ring equation 
 
 K"-'+a:"-=>-|-a:"-34.. . . . +x^+x+l=0, 
 of which the n — 1 roots are imaijinary. 
 
 For example, the equation a;>=l, cr a;' — 1=0, divided by x — 1 
 gives a;'+a;-j-l=;0; 
 
 whence x= T' V — 
 2 
 
398 RAT'S ALGEBRA, PART SECOND. 
 
 Hence the three third roots of 1 ore 
 
 ' 2 ' 2 • 
 
 IV. — If n be even, the equation a!"+l=0, or x»= — 1, has no 
 real root, since y — 1 is then impossible. Hence, all the roots 
 of this equation are imaginary. For example, the four roots uf 
 the equation a:''-|-l=0, as determined by the method explained in 
 Art. 442, are 
 
 — 1+V=1 —\-J—i 1-W^, IjT^O 
 V2 ' V2 ' V^ ' V2 ■ 
 
 v. — If n be odd, the equation x"+l=0, or a^= — 1, has one 
 real root, viz. : — 1, and no more, because this is the only real 
 number of which an odd power is — 1. For example, the 
 equation 
 
 3?-\-l =:0, divided by a;-|-l gives 
 a;2__j;-l-l=0, 
 
 whence x= ^ .•. the three third roots 
 
 2 
 
 of —1, are —1 , ^+>^~^, and Iz^Zz^. 
 
 Binomial equations have other properties, but some of themcan- 
 not be discussed without a knowledge of Analytical Trigonometry 
 For exercises the pupil may find 
 1 . The four fourth roots of unity. 
 
 Ans. +1, —1, +V— 1. — V— 1 
 2 The five fifth roots of unity. 
 
 Ans. 1, 
 
 |jVl-l+>/(-10-2V5)|. 
 
 4{VB-1-J(-10-2V5)}, 
 -\ \ ^5+1+ V(-10+2 V5) ( , 
 _J| ^5+l_V(-10+2 V5)r 
 
 THE END. 
 
u^ 
 
 
 j^^^iSiSJlt 
 
 -,^','!