Wht %t\\^ pbmjj^ PRESENTED TO TSB COBNEZL UNITBHSIIT, 187^, , -t The Hon. William Kelly Of Rhinebeck. ■Hlil.^a. olin.anx B Cornell University S Library The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031490208 NEW HIGHER ALGEBRA; ANALYTICAL COURSE DESIGNED FOR HIGH SCHOOLS, ACADEMIES, AND COLLEGES. Bt benjamin QREENLEAF, A.M., AUTHOB OF A mIiBBIUIIOIL SSIUES. CORNELL LIBRARY BOSTON: PUBLISHED BT ROBERT S. DAVIS & CO. jrew toek: viluah woob asb oompaht, amb t. i. poomt. PBILASELPHIA : t. S. LtPPIHOOIT AKB OOMFAHT. Bi. Loms : sxhh abb woobs. OINOINNATI : R. W. OARROLL & 00. 1868 6> GEEEHLEAF'S NEW COMPREHENSIVE SERIES. An ENTiEELT NE-yv MATHEMATICAL CouESE, /«% adapted to the best methods of Modem Instruction. GREENLEAF'S NEW PEIMAKY ARITHMETIC. GKEENLEAF'S NEW INTELLECTUAL ARITHMETIC. GREENLEAF'S NEW ELEMENTARY ARITHMETIC. GREENLEAF'S NEW PRACTICAL ARITHMETIC. GREENLEAF'S NEW ELEMENTARY ALGEBRA. GREENLEAF'S NEW HIGHER ALGEBRA. 1^ [AJxJ GREENLEAF'S ELEMENTS OF GEOMETRY. GREENLEAF'S ELEMENTS OF TRIGONOMETRY. GREENLEAFIS GEOMETRY AND TRIGONOMETRY. Other Books of a Oomplete Series, in preparation. \* Keys to the Practical Akithmetic, Algebras, Geometet and TKiGOJSLOMEiBy, in separate volumes. Entered accorcUng to Act of CoagresSf in the year 1864, by Besjauih Geeehlsaf, in the Clerk's Office of the District Court for the District of Massachusetts. BIVEBSISS PRESS : PKIHTED BT H. 0. BOOSHION AND CaUTART. PREFACE. The author's New Elementary Algebra having now, after several years' trial, proved a great success, many teachers, acquainted with the practical results accomplished by the use 'of that book, have been very urgent in their demands for a work on the same plan, for advanced students. The aim of this treatise is to meet, more fully than has been done heretofore, the requirements of the highest stand- ard of mathematical instruction in the best high schools and seminaries. To this end, great care has been taken to in- clude all the more important parts of analysis, to treat each topic with as much conciseness as is consistent with clear- ness a,nd elegance, to introduce valuable original processes, and to secure throughout an arrangement most conducive to a philosophical development of the science. The work, therefore, is very comprehensive in its scope, ample in its range for the curriculum of the college, yet furnishing a course not too extended for the higher order of academies. The Examples have been selected with a view to a full illustration of principles. While complicated and puzzling problems have been excluded, a proper regard has been had to such exercises as promote mental discipline, or tend to make expert practical analysts. Each subject, as far as possible, has been made complete in itself, so that certain parts can be omitted by the student when desirable, at the option of the teacher. IV PREFACE The General Theory of Equations, which is often em- braced in a volume by itself, is necessarily briefly treated. All the essential points, however, receive attention, and are, it is believed, rendered sufficiently intelligible. This part of the work may not always be required for the pur- poses of recitation, and undoubtedly will not be so used, in many casesr yet it will be found of great value for occa- sional reference. Recurring Equations, Binomial Equations, Cardan's Rule for solving Cubic Equations, and other methods seldom called into use by the practical algebraist, have been put ■in the form of an Appendix. The valuable works of Lefebure de Fourcy, Reynaud, Mayer and Choquet, and Todhunter, have been freely con- sulted ; but whatever material these have furnished has been assimilated in language and style. Many eminent teachers have kindly furnished hints and suggestions which have been of great service. But more than a general acknowledgment is due to H. B. Maglathlin, A. M. The greater part of the editorial labor of this vol- ume has been performed by him ; and he, therefore, must share largely the credit of whatever merit it may possess. Bkadfokd, Mass., Sept. 1, 1864. CONTENTS. Definitions and Notation. Page Symbols of Quantity 9 Sjonbols of Operation 10 Symbols of Relation 12 Sjrmbbls of Abbreviation 13 Algebraic Expressions 14 Interpretation of Algebraic Expressions 16 Axioms 17 Algebraic Processes 17 Entire Quantities. Addition 19 Subtraction 24 Multiplication 28 Multiplication by Detached Coefficients 34 Division ........... 36 Division by Detached Coefficients 43 General Principles 45 Useful Formulas 46 Factoring 48 Greatest Common Divisor 54 Least Common Multiple 60 Feactions. General Principles 65 Eeduction 67 Addition 75, Subtraction 77 Multiplication 79 Division 80 VI CONTENTS. Simple Equations. Transformation of Equations 85 Solution of Equations 88 Problems — One Unknown Quantity 93 Equations containing Two Unknown Quantities . . . 102 Elimination . .< 103 Equations containing Three or more Unknown Quantities . 109 Problems — Two or more Unknown Quantities. . . . . 112 General Solution of Problems 119 Discussion of Problems . . . . .• . . .123 Intarpretation of Negative Results 126 Zero and Infinity 129' Pi)oblem- of the Couriers I'Sl InfequaJitiee 135 iNyOLUTioiir.- Bowers of Monomials ........... 14^' Powers of Polynomials 144'- Polynomial Squares 145 Polynomial Cubes 146 Evolution. Siquare Root of Numbers 149 Cube Root of Numbers 155- Koots of Monomials . . , . . . . . .158' Square Root of Polynomials 160 Cube Root of Polynomials . .163 Any Root of Polynomials 166 Radicals. Reduetion jgg Addition J75 subtraction . _ _ -.jg Multiplication ' _ -.yg Division . _ _ joq Powers j,gj^ Roots ■•.■....... ISO Raitionalization ••....... 134 Imaginary Quantities . . _ . . . . . _ _ jag Properties of Quadraitic Surds J92 Radiceil Equations' 196 CONTENTS. vii QuadKatic Equations. Pure Quadratics 200 Problems leading to Pure Equations 203 Affected Quadratics 205 Methods of Competing the Square 206 Problems leading to Affected Quadratic Equations . . .217 Equations in the Quadratic Form'. . . • . . . 221 Simultanepus Equations' involving Quadratics .... 228 Problems leading to Quadratic Equations . . . . 237 Theory of Quadratic Equations 240 Formation of Equations 243 Discussion of the Creneral Equation 243 Discussion of Problems leading to Quadratic Equations . 245 Interpretation of Imaginary Results 247 Problem of the Lights 248 Ratio and Propoetion. Definitions 251 Properties of Proportions 253 Problems in Proportion ' . .257 Variation 258 Peogkessions. Arithmetical Progression 262 Geometrical Progression 266 Harmonical Progression ' . . 271 Problems ' . 273 Permutations and Combinations. Permutations and' Conlbinations 276 Binomial Theorem. — Positive Integral Exponent . , . 280 Series. / Undetermined Coefficients . . ." . -J- • . 289 Decomposition of Rational Fractions ..'... 294 Binomial Theorem. — Any Exponent 296 Reversion of Series 301 Summation of Infinite Series 303 Recurring Series 303 Differential Method 307 Interpolation 311 vm CONTENTS. Logarithms. The Common System 314 Properties of Logarithms 315 Computation of Logarithms 317 The Napierian System 319 Common Logarithms 321 Tables of Common Logarithms 323 Exponential Equations .327 Compound Interest and Annuities . . . •. . 330 General Theory of Equations. Divisibility of Equations 335 Number of Boots 337 Formation of Equations 339 Composition of Coefficients 340 Fractional Eoots 342 Imaginary Roots 343 Descartes' Rule 345 Transformation of Equations 347 Synthetic Division 352 Derived Polynomials 358 Equal Roots 360 Signs of Polynomial Functions 362 Sturin's Theorem 364 Solution op Higher Numerical Equations. Commensurable Roots ........ 372 Incommensurable Roots 376 Homer's Method 376 ■ APPENDIX. Recurring or Reciprocal Equations 381 Binomial Equations 384 Cardan's Rule for Cubics 388 Biquadratic Equations 392 Approximation by Double Position 392 Newton's Method of Approximation 394 ALGEBRA. DEFINITIONS AND NOTATION. 1. Quantity is anything that can be measured ; as distance, time, weight, and number. 2i The Unit of quantity is one of the same kind as the quantity, assumed as a standard, or unit of measure. 3f The Measurement of quantity is accomplished by finding the number of times the quantity contains the assumed unit of quantity. 4t Mathematics is the science of quantities and their relations. 5, Algebra is that branch of mathematics in which the relations of quantities are investigated, and the rea- soning is abridged and generalized by means of symbols. 6i The Symbols employed in Algebra are of four kinds : symbols of quantity, symbols of operation, symbols of relation, and symbols of abbreviation. SYMBOLS OF QUANTITY. 7. The Symbols of Quantity generally used, are the figures of arithmetic and the hUers of the alphabet. The figures are used to represent known quantities and determined values, and the letters, any quantity whatever, known or unknown. 10 ALGEBEA. 8. Known Quantities, or those whose values are given, when not expressed by figures, are represented by the first letters of the alphabet, as a, b, c. 9. Unknown Quantities, or those whose values are not given, are represented by the last letters of the alpha- bet, as X, y, z. 10. Zero, or' the absence of quantity, or that which is less than any assignable quantity, is represented by the symbol 0. 11. Infinity, or that which is greater than any assign- able quantity, is represented by the symbol oo. 12. Quantities occupying similar relations in difierent operations are oftfen represented by the same letter, dis- tinguished by different accents, as a' , a", a"', read, a prinie, a second, a third, etc. ; or by different subscript figures, as Oi, 02, Os, 04, read>. asub one, a sub t\^o, a sub three, etc. SYMBOLS OF OPERATION. 13« The Symbols op Operation are certain signs or characters used to indicate algebraic operations. 14.' The Sign op Addition, -\-, is called plus. Thus, a -\- b, read a plus b, indicates that the quantity S is to be added to the quantity a. 15. The Sign of Subtraction, — , is called otiwms. Thus, a — b, read a minus h, indicates that the quantity b is to be subtracted from the quantity a. The sign — ■ may be written between two quantities, when it is not known which of them is the greater. Thus, a , — ■ b indicates the difference of the two quantities a and b. 16. The Sign of Multiplication, X, is read into, or multiplied by. Thus, a X b indicates that the quantity a is multiplied by the quantity b. A simple point (.)-is sometimes used in the place of DEFINITIONS AND NOTATION. 11 the sign X- The sign of multiplication is, however, usually omitted, except between two arithmetical figures separated by no other sign, and multiplication is there- fore indicated by the> absence ' of any sign. Thus, 2 ab indicates the same as 2 X « X 5, or 2 . a . &. 17t The quantities multiplied are called factors, and the result of the multiplication i&^ called the product. 18. The Sign op Division, -=-, is reaji divided hy, Thus^ a -i- b indicates that the quantity a is divided by the quan- tity b. Division is otherwise often indicated by writing the dividend above and . the divisor below a short horizontal line. Thus, t indicates the same as a -r- J. Also, the sign of division may be replaced in an operation by a straight or curved line. Thus, a\b, or b)a, indicates the same as a -i- b. 19- The Sign op iNVottfTioN, or Exponentiai, Sign, is a figure or letter written at the right and above a quantity, to indicate the number of times the quantity is taken as a factor. Thus, in x^, the ^ indicates that x is to be taken three times as a factor ; and x' is equivalent to XXX. The product obtained by taking a factor one or more times is called a power. Thus, a^, read a square, indicates the second power of a; a', read a cube,- indicates the third power of a; a*, read a fourth, indicates the fourth power of a; a", read a nth, indicates the wth power of a. The figures or letters used to indicate powers are called exponents; and when no exponent is written, the^w^pow- er is understood. Thus, a is equivalent to a^ If the minus sign is prefixed to au exponent, it indi- cates that the quantity is tft be used as a diviaoi'. Thus,' a' 1 «2 J-' C-' is the' same as ^, and a;-" is the same as ^. 12 ALGEBRA. The root of a quantity is one of its equal factors. Thus the root of a^, a', ce*, is a. 20. The Sign of Evolution, or Eadical Sign, V, when prefixed to a quantity, indicates that some root of the quantity is to be extracted. Thus, 4^a indicates the square or second root of a ; 4/a indicates the ciibe or third root of a ; 4/a indicates the fourth root of a ; and so on. The index of the root is the figure or letter written over the radical. Thus, ^ is the index of the square root, ' of the cube root, and so on. . / When the radical sign has no index v^ritten over it, the index ^ is understood. Thus, \/« is the same as i^a. A fractional exponent is also used to indicate a root. Thus, a*, a*, a*, indicate, severally, the square, cube, and fourth root of a. In fractional exponents the numerator denotes a power, and the denominator a root, thus, J* indicates the fifth power of the 'fourth root of h, or the fourth root of the fifth power of h. SYMBOLS OF EELATION. 21. The Symbols of Eelation are signs used to indi- cate the relative magnitude of quantities. 22. The Sign of Equality, ==, read equals, or equal to, indicates that the quantities between which it is written are .equal. Thus, x = y, indicates that the quantity x equals the quantity y. An expression in which quantities are connected by the sign = is an equation, a8a;-|-4 = 2a; — 1, read x plus 4 equals 2x minus 1. 23. The Sign of Katio ( : ), read to, indicates that the two quantities between which it is placed are taken as the terms of a ratio. Thus, a : b indicates the ratio of DEFINITIONS AND NOTATION. 13 the quantity a to the quantity b; and is read the ratio of a to b. An equality of ratios, or a proportion, is expressed by writing the sign ^, or the sign (::), between equal ratios. Thus, 30 : 6 ^ 25 : 5 is read the ratio of 30 to 6 equals the ratio of 25 to 5, or 30 is to 6 as 25 to 5. 24. The Sign of Inequality, > or <, read is greater than, or is less than, when placed between two quantities, indicates that the quantity- toward which the opening of the sign turns is the greater. Thus, a; > y is read X is greater than y, x — 6 < ^^ is read x minus 6 is less than y. 25 1 The Sign of Variation, cx:, read varies as, indi- cates that the two quantities between which it is placed increase or diminish together. Thus, a OC ;^, is read a c "■ vanes as -;. a SYMBOLS OF ABBEEVIATION. 26i The Signs of Deduction (.'. and •.■) stand the one for therefore or hence, and the other for since or be- cause. 2Tt The Signs of Aggregation, the vinculum , the bar \, the parenthesis ( ), the brackets [ ], and the braces ■] (• , indicate that the quantities connected, or inclosed, by them are to be taken as one quantity, and to be operated upon in the same way. Thus, -j-ala; r \ a -(- i X a;, -j- ^ I ' (" -f- ^) ^j [« + ^] '^; ] ci -\- ^ r ^> each indicate that the quantity a -|- 5 is to be multi- plied by X. 28. The Sign of Continuation ( ) stands for and so on, or continued by the same law. Thus, a, a -\- b, a -\- 2 b, a -{- Sb, is read a, a -\- b, a -\- 2 b, a -\- Sb, and so on. 2 14 ALGEBKA. ALGEBRAIC EXPRESSIONS. 29i . An Algebraic Expeession is a quantity written in algebraic language. 30i A Coefficient of a quantity is a figure or letter prefixed to it to show how many times the quantity is to be taken. Thus, in 4 a, 4 is the coeflScient of a, and indicates that a is taken four times ; and in b xy, h may be regarded as the coefficient of xy, or bx as the coef- ficient of y. When no coefficient of a quantity is written, 1 is un- derstood to be the coefficient. Thus, a is the same as 1 a, and xy \a the same as Ixy. 31 • The Terms of an algebraic expression are its parts connected by the signs -|- or — . Thus, a and b are the terms of the expression a -\-b; 2 a, ¥, and — 2ac, of the expression 2a-\-W — 2a e. S2i The Degree of a' term is the number of literal fac- tors which it contains. Thus, 2 a is of the first degree, since it contains but one lit- eral factor. ab is of the second degree, since it contains but two literal factors. 3 a J'^ is of the third degree, since it contains but three literal factors. The degree of any term is determined by adding the exponents of its several letters. Thus, a W c' is of the sixth degree. 33i Positive Terms are those having the ■plus sign ; as, + 2 a, or + a Jl When a term has no sigii written, it is understood to be positive. Thus, a is the same as -\- a. Negative Teems are those having the minus sign ; as, — 3o, or -^bci This sign should never be omitted. DEFINITIONS Ata) NOTATION. 15 M, In a positive terra the coefiBcient indicates how many times the quantity is taken (Mditively, and in a negative term it indicates how many times the quantity is taken svMractively. Thus, -)- 2 a; is the same as -j- a: -|- X, and — 2 a; is the same as — x — x. The signs -j- and — are not only used to indicate dper- ations (Art. 14, 15), but the nature' of the quantities to which they are prefixed ; that is, A quantity affected by the plus sign is positive, and a quantity affected by the iriinus sign is negative. A positive quantity is- additive, While a negative quan- tity is subtractive, and necessarily sustains opposite rela- tions ; consequently, equal terms affected by unlike signs, iij an expression, cancel. Thus, al)-\-cd-^cdis the same as ab. Quantities inde^pendently additive or subtractive, may be regarded as added to, or subtracted from, 0, the neu- tral point, or starting-point, of all positive and negative quantities. 35i Similar or Like Teems are those containing the same letters, affected by the same exponents. Thus, ^xy eioA -^*l xy are similar terms; also, 3 a^ W and Q a^¥ are similar terms. 36. Dissimilar or Unliee Terms are those containing- different letters or exponents. Thus, a b and a d are dissimilar terms ; also, hify aind bx^ are dissimilar terms. 37. A Monomial or Simple Quantity is an algebraic expression consisting of only one term ; as, 5 a, 7 « b, or 3J^c. 38. A Polynomial or Compound Quantity is an alge- braic expression consisting of more than one term ; as, a -f i, or 3 a^ -f J — 5 W. A polynomial- is sometimes called a multinomial. 16 ALGEBRA. ' 39i A Binomial is a polynomial of two terms ; as, a — h, 1a-\-W, or 3 a c^ — i. A binomial whose second term is negative, as a — J, is sometimes called a residual. 40. A Tkinomial is a polynomial of three terms ; as, a _|_ J _|_ c, or a J + c= — J'. 41, Homogeneous Teems are those of the same degree. Thus, the terms a^, 3Jc, — 4 3;" are homogeneous. 42, A Polynomial is Homogeneous when all its terms are homogeneous. Thus, the polynomial c? -\- abc — W is homogeneous. 43. The Eeciprocal of a quantity is 1 divided by that quantity. Thus, the reciprocal of a is -, and oi x -{■ y x-\-y The reciprocal of a fraction is that fraction inverted. Thus, — is the reciprocal of — . TO '^ n INTEKPEETATION OF ALGEBEAIC EXPRESSIONS. 44. The Inteepeetation of an algebraic expression con- sists in rendering it into Arithmetic, by means of the numerical values assigned to its letters. 45. The NuMEEicAL Value of an algebraic expression is the result obtained by substituting* for its letters their numerical values, and then performing the operations in- dicated. > Thus, the numerical value of 4a-f3Jc — rf, when a = 4, J = 3, c = 5, and rf = 2, is 4 X 4 ■+ 8 X S X 5 — 2 = 59. DEFINITIONS AND NOTATION. 17 AXIOMS. I6i An Axiom is a self-evident truth. Algebraic operations are based upon definitions and the following axioms : — 1. If the same quantity, or equal quantities, be added to equal quantities, the sums will be equal. 2. If the same quantity, or equal quantities, be sub- tracted from equal quantities, the remainders will be equal. 3. If equal quantities be multiplied by the same quan- tity, or equal quantities, the products will be equal. 4. If equal quantities be divided by the same quantity, or equal quantities, the quotients will be equal. 5. If the same quantity be both added to and subtracted from another, the value of the latter will not be changed. 6. If a quantity be both multiplied and divided by an- other, the value of the former will not be changed. "T. Quantities which are equal to the same quantity are equal to each other. 8. Like powers and like roots, of equal quantities, are equal. 9. The whole of a quantity is equal to the sum of all its parts. ALGEBEAIC PROCESSES. 47. The Processes of Algebra, in general, are only those of Arithmetic extended, or rendered more compre- hensive, by the aid of symbols. (Art. T.) Exercises on preceding Definitions and Principles. 48. Put in the form of algebraic expressions : — 1. Five times b, added to two times a. Ans. 2 a -\- 5 b. 2. Two times x, minus i/ second power. 3. The difference of x and y. Ans. x f-- y. 4. a.b multiplied by c square and by d cube. 2* 18 ALGEBRA. 5. a: -|- y divided by of — J. 6. The difference' of a square and J square; T. One third of x, increaised by two, is equal to' three y diminished by eleven. ^^^_ f + 2 = Sjr — 11. 8. The reciprocal of two fifths of a, plus the square of a minus the square of S, is equal to the square root of c square plus d square. 9. The ratio of five a divided by i to 2a + 36 When i = 3, When the terms are dissimilar, or some similar, and others dissimilar. 1. Let it be required to find the sum of 3 a a;, 4?ren, — 6 y^, and *l xy. OPERATION. The given terms 3aa; -\- 4,mn — 6 y^ + 1 xy. being dissimilar, the quantities have no common unit, consequently, we can only indicate their addition, by uniting them by their respective signs, and have Sax -{- 4:mn — Q y^ -\- 1 x y for the sum required. 2. Let it be required to find the sum of 6 a — 1 x, *l a — 4a:, and — a -\- Q x -\- 4:xy. OPERATION. Finding the sum of the 6 a — *l X similar terms, as in Art 52, la — 4 a; • we obtain 12 a — b x, and — a -\- % X -\-Axy there being no term similar 12 a — b X -\- 4: xy to -j- 4 a; y, it is united by its proper sign to those terms, which gives 12 a — bx -{- i^xy for the sum required. Hence, The Algebraic Sum of dissimilar terms can only be indicated, by uniting them by their respective signs. It is also evidently immaterial in what order the terms are united, provided each has its proper sign. Thus, — b -|- a is the same as a — b. 54. From the preceding principles and illustrations is deduced the following 22 ALGEBRA. General Rule. Find the sum of the similar terms, and to the result obtained unite the dissimilar terms, if any, by their proper signs. Examples. (1) (2.) (3.) (i-) (5.) la — 6m 4: ax 13 n *l a — 7np^ Ba m — Sax n a -\- 6 m p^ a — Urn ax — 20n — 11 a — 3 mp^ 5 a 5 m — lax 6 n 8 a + 11 mp^ 11 a — m a x 8n — 9a— 1 mp^ a 20 m 12 ax — n 18 a — 15 mp^ 28 a 8 w 14 a — {) m p^ 6. Add together a -\- x and y — c. Ans. a — c -\- X -\- y. I. Add 3a + J— 10, c — d — a, — 4c + 2« — 3J — t, j,nd 4x^ -|- 5 — 18 m together. Ans. Aa — 2b — 12 — 3c — rf+4a;''— 18m. 8. Add la — 5^, 8 \/x -{-2 a, 5 if — \/^x, and — ■ 9 a -\- 1 \/x together. Ans. 14 \/x. 9. Add imn -\- Bab — 4c, 3x — iab -\- 2mn, and 3 m^ — 4p together. Ans. 6 mn — ab — 4c-|-3x-j-3to^ — 4 p. 10. Find the sum of 3 a^ + 2 a 6 + 4 5^ 5 a^ — 8 a b -\-b\ — a=+5ai — i^ 18a^ — 20 a S— 19 52^ and 14 a^ — Sab -{-20 b\ II. Find the sum of 4 a:' — 5 a' — 5 aa:^ -)- 6 a'' a;, 60^ + 3 Jc' + 4 a a;2 +.2 a^ x, — lY a;' + 19 a x^ — 15 a^ x, 13 a x^ — 21a^ a; + 18a', Sa^x- 20o'+ 12ce', and Sla^^x— 2^3 — Slax'^—1 3?. Ans. — 1 a;' — a". ADDITION. 23 12. Find the sum of 3 a & + 3 (a+ 5), — a J + 2 (a + &), T a 6 — 4 (a 4- J), and — 2 a 6 4- 6 (a + 6). Ans. 1 ab -\-1 (a -\- b). 13. Find the ^um of t vV — 4: (a + S), 6 v^y + 2 (a + i), 2 Vy + (« + S), and s/y — 3 {a -\- b). 14. Find the sum of 3 (c + rf) a;' + 4y — 2 VV. Cx'^y — 2ax -{- 12, y-j- \/y — 5 a a;, 3 ax — x^ i/ -\- (c -\- d) x', and ai^'y -|- 2y — 14. Ads. 4(c4"'^)^+'fy — Vy + ^^y — ^'"^ — 2. S5f When dissimilar terms have a common factor (Art. 17), the addition may be expressed, by prefixing the sum of the coeflBcients, inclosed in a parenthesis, to the com- mon factor. 15. Eequired the sum of a^, b^, and c^. OPKKATIOJJ. a ^ Since a times, b times, and c b if times tf will equal the sum of c y* a -\-b -\- c into if, {a-\-b -\- c) ^ (a -j- 6 -|- c) ^ will indicate the sum required. 16. Add ax^, bx^, 2 c a:^ and Zdx\ Ans. (a + 6 4- 2 c + 3 rf) a;^ It. ^AA. amx -\-2dy, 2 ex — Bdy, and 3 * — ^ — ^m) x. 20. Add a a: 4- 5 3;, §ax — y, and 2 J a: -f- 2 y. Ans. (1 a + 2 J) ar -[- 6 y. 21. Add (2 a -|- c) (ot — \/»»)> ^ (»« — >/»»), and — 3 c (m — \/m). Ans. {2a-{-b — 2 c) (m — v'w)- 22. MAiax-\- dy^, by — dx, and — Sy^+wy. Ans. («'— It follows, from the two preceding cases, that Subtracting a positive quantity is the same as adding an equal negative quantity, and subtracting a negative quantity is the same as adding an equal positive quantity. Hence, the following GENERAL RULE. Conceive the signs of all the terms of the subtrahend to be changed, from -\- to — , or from — to -\-, and then pvoceed as in addition. Examples. (1.) (2.) (3.) (4.) 2T« *l X — 16 a; — %d 13 a 14 a; — bx —lid 14: a —1x —11a: 3d 3 26 ALGEBBA. (5.) (6.) ab -\- cd — ax 1 x -\- by — 3a 4q6 — ^ cd -\- i^cLX X *l y -\- b a _3a6-|-4crf— 5aar & x -\- 12 y — 8a a.) (8.) lahc—llx + by— 48 5\/a— 3y=+ lai — l nahc-\- Zx-\-1y-\- 100 3 \/a + y^ — 5 a^ + 7 — 4aic— 14a: — 2y— 148 2 y/a — 4cy^ -\- I2ai — 2, 9. From 31 a;'^ — 3 y= + a 5 take ITa:* -f- Sy'' — 4aS+ 7. 10. Take a — h from a -{-b. Ans. 2 5. 11. Take6a — 3 &— 5c from 6a + 3 5— 5c + I. 12. From 5a;y + la"^ — %b^ take ^xy — Q a^ ~ b^ -\. ^. 13. From 11a — t6 + ctakea-[- 7 6 — 3c+ 11. Ana. 10a— 14 S + 4c — 11, 14. From jn^ + 3 m» take — 4 m^ — 6 »* + 71 ar. Ans. bm^ -\- 9 w* — 71 ar. 15. From 31 a — 15 a; — 7 take 2 a — 25 a: + /. Ans. 29 a -f 10 a; — 2^2 ^ 7. 16. From a + * take 2 a — 2 5 and — a -f- 5. Ans. 2 5. 17. From a — b — c take — a-j-5-f-canda — 5 + 0. Ans. a — 5 — 3 e. 60. It is sometimes suflScient to indicate the subtrac- tion of a polynomial, by inclosing it in a parenthesis, and prefixing the sign — . Thus, ba^ — (a" -f- 52 _ c) indicates that the entire quantity a^ -j- 5^^ — e is to be taken from 5 a^. Performing the operation indicated, by the rule. Art. 59, we have 6 a" — a» — 5^^ -f c, or 4 a^ — 52 -f c. SUBTRACTION. 2T The first of these expressions may be transformed to its previous equivalent form, by changing the signs of the last three terms and inclosing them in a parenthesis, with the sign — prefixed. Hence., W/ien an expression within a parenthesis is preceded by the sign — , the parenthesis may be removed, if the sign of every inclosed term he changed; and Any number of terms in an expression may be inclosed by a parenthesis and the sign — prefixed, if the sign of every term inclosed be changed. 18. Indicate the subtraction of 6 a -|- J^ from — T a — b. Ans. — T a — i — (6 a + J''). 19. Indicate the subtraction ofJ^c — d — e from a. 20. Reduce the expression Za -\- b — (o — b) to its simplest form. Ans. 2 a -\- 2 b. 21. Reduce the expression 2o' — 3a^b + 4 a 5" — (a' -\-¥ -\- ab^) to its simplest form. Ans. o8_3a2j_|_3aJ2_ js 22. Remove the parenthesis from the expression 6 a ^- 9a; — (4a — 5a;). 23. Remove the parentheses from 4a — 5a: — (a — 4a;) + (a; — 8 a). Ans. — 5 a. 24. Place in a parenthesis, with the negative sign pre- fixed, the last three terms of the expression 2 a' — 'Sa^b -^ 4:ab^ — a^ — ^ — a¥. Ans. 2a» — 3a2J + 4a5''— (a' + J» + aS2). 25. To what is 5a — 4J + 3c + (— 3a + 2 J — c) equal ? Ans. 2a — 2J + 2c. 26. To what is 6 ar^ + 2y» — (3 a;» + y''), — [2 a;^ + 4.f ^ (4 a:« — /)] equal 7 Ans. 5 a;'' — 4 yK 21. Sim plify 1 — [2 — (7 — a; — 4)] + 2 — [3 + (4 -a; — 5)]. Ans. 0. 61 • When dissimilar terms have a common factor, the 28 ALGEBRA. subtraction may be indicated by prefixing the difiference of their coefficients, inclosed in a parenthesis, to the com- mon factor. 28. From ay — h take dy — c. Ans. {a — d)y — h + c. 29. From aa? + dy^ take hx^—cyK Ans. {a — b)a?-\-{dJ^c)y\ 30. From aa; — Ja; + ca; take a: + aa; — hx. Ans. (c — 1) X. 31. From aa? — bx'^ -\- ex — d take ha? ■\- ex — 2d. Ans. aa? ~ 2 bx^ -{- (c — e) x -\- d. 32. Take Shxy — (5 + c) »=' — (3a — 2b) (x + y)i from (4 a — 5 5) \/^x -\- y -\- (a — b)xy — c z\ Ans. (7a — 7 J) ^ai + y + (a -^ 4i) a;y + 5»^ MULTIPLICATION. 62. Multiplication, in Algebra, is the process of taking one quantity as many times as there are units in another quantity. The Multiplicand is the quantity to be multiplied, or taken. The Multiplier is the quantity by which we multiply. The Product is the result of the operation. The multiplicand and multiplier are often called factors. 63i The Product of factors is the same, in whatever order they are taken. For, the product contains one factor as many times as there are units in the other. Thus, the product of a X S, or 6 X «> will be ai units, since h taken a times is the same as a taken b times. Let a = 4 and 6 = 3, we have 4X3, or 3 X 4=, equal to 12. MULTIPLICATION. 29 In like manner, it may be shown that the principle holds when there are any number of factors. 64. To determine the law of the coefficients of the product. Let it be required to find the product of T a by 2 A. Since the factors can be taken in any order (Art. 63), 7 a X 2 J is the same as 7 X 2 X « X * ", then, 2 times 7 = 14 ; J times a=: ab; and 14 times ab= 14 a 6. Hence, The COEFFICIENT of the product is equal to the product of the coefficients of the factors. 65i To determine the law of the exponents of the product. Let it be required to find the product of 2 a' by a'. Since the exponent of a quantity indicates the number of times the quantity is taken as a factor (Art. 19), 2 a* is the same as 2aaa, and a" is the same as a a ; then, a a times 2aaa = laaaaa = Ic^. Hence, The EXPONENT of a letter in the product is equal to the sum of its exponents in the factors. 66> To determine the law of the signs of the product. 1. Let it be required to find the product of + a by Since a is to be taken as many times as there are units in b, and as the sum of any number of positive quantities is positive, the product, ab, must be positive, or +ai. 2. Let it be required to find the product of ^ a by + S. Since — a is to be taken as many times as there are units in b, and as the sum of any number of negative quantities is negative, the product must be negative, or ■ — ab. 3. Let it be required to find the product of + a by — b. The negative multiplier, — b, indicates that a is to be 3* 80 ALGEBRA. taken suUractivelp (Art. 34) as many times as there are units in h; and since a positive quantity becomes negar tive by subtraction (Art. 69), the product must be nega- tive, or — ah. 4. Let it be required to find the product of — a by — 5. Here the negative multiplier, — h, indicates that — a is to be taken subtractively as many times as there are units in h, and since a negative quantity becomes positive, by subtraction, the product must be positive, or ah. From these results it follows, -|- multiplied hy +, and — multiplied hy — , produce -j-, — multiplied hy -\-, and -)- multiplied hy — , produce — . Or, as may be enunciated for the sake of brevity, with regard to the product of any two terms, Like signs produce -\-, and unlike signs produce — . CASE I. 67 • When both factors are monomials. From the preceding examples is derived the following RULE. Multiply the numerical coefficients together; annex to the re- sult the letters of hoth monomials, giving to each letter an expo- nent equal to the sum of its, escponents in the two factors. Make the product positive when the two factors have like signs, and negative when they have different signs. Examples. (1-) (2.) (3.) (4.) 3ah llabc c?¥c 2 a;"' 2ae — 8abc a^bd 4 re" Ga^ho — 136 a^'l^c^ ^¥cd 8 »!"' + •' jfULTIPLICATION. 31 5. Multiply 15ffi*w° by Smn. 6. Multiply 3 a"* 6" by 2a"'i». Ana. ea^™*"*'. 1. Multiply ix^y by — x^yzf'. Ans. —ix^'+'f^., 8. Multiply lY o'c" by 4 a ace. Ans. 68 a* c*. 9. Multiply 3 a'" 4» by — 4 «"• J^ a2 5».-» _|_ a J»-2 -|- J-"-! ; > ^^^ which may be verified by multiplication. In (1) give particular values to m, and we have 50 ALGEBRA. (flU _ i2) _^ (a _ J) — a _|- J, (flS _ 68) -=- (a — 6) = o^ + a 6 + J", (a* _ J*) ^ (a — b)=c^ + an + aP + J', (a* _ js) ^ (a _ J) := „4 ^ ttS J _|. ^2 J2:j_a J3+6<. In (2) let b or a become 1 ; then, (a" — 1) -f- (a — 1) = a + 1, (2) K-l)-(a-l) ^ + a+l. (3) (a4 _ 1) _^ (a _ 1) = aS _j_ „2 _|_ „ _|_ 1 . and (1 _ J2) ^ (1 _ J) ^ 1 _|_ 5, (1 _ y) ^ (1 _ J) = 1 + 6 + J^ (1 _ 64) ^ (1 _ J) ^ 1 _^ J ^ J2 _j_ j?^ and so on. 97. It may also be proved that the difference of any two equal even powers of two quantities is always divisible by the sum of the quantities. Thus, (a2 — 6") -T-(a + J)=a — 5, (a*_6*)-=-(a-j-5) = a= — a^J + aJ^ — J', J-fl) (a6_6») -T- (a-i- i) = a» — a*6-i- «' 4^— «'** + «**— *"■ Also, (a2 _ 1) -^ (a -j_ 1) = a _ 1, (a* _ 1) -f- (a + 1) = o^ — a^ + a — 1, (V—l) -J- (a-i-.'l) = as — a^+a'—a^ + a— l;and (1 _ 6^) ^ (1 + 6) = 1 — 5, (1 _ 6*) ^ (1 4- 6) = 1 — S + 6^ — 6^ (1 _ 6«) H- (1 + *) = 1 — * 4- ^^ — ^ + ** — *"» and so on. 98. It may likewise be proved that the sum of two equal odd powers of two quantities is always divisible by the sum of the quantities. Thus, (c?^l?)-^ {a + h) = a^-ah^W, \ bKi^'-' (2) (c^ + V) -i- (a + J) = a* — 0=6 + a" ' + 6*. FACTORING. 51 Also, (a' +1) -^ (a + 1) = a" — a + 1, and 1 (1 + 48) -5- (1 + i) = 1 — 6 + i^ j (^) and BO on. 99i The foregoing principles may be enunciated thus : — When the two equal powers are odd, their difference is divisible hy the difference of the quantities, and their sdm hy the SUM of the quantities. When the tioo equcd powers are even, their difference is divisible hy both the difference and the sum of the quantities, hut their sum is divisible by neither the difference nor the sum of the quantities. Note. The quantity a^" -\- h^" may always be factored when n contains an odd factor. Thus, a° -(- 6° is divisible by a? -|- V, and a« -|- 612 y^y a*-j-¥; but a" + b' and a* + b* are prime. 100. The factoring of monomials may be performed by inspectioh, since their prime factors are evidently those of their coefficients, and each letter of the expression taken as many times as there are units in its exponent. Thus, 12 a^J = 2 X 2 X Baah. But the decomposition of polynomials often requires a resort to trial, or the law of their formation, and the exercise of much skill. 101. The following observations will aid in factoring some polynomials ; but with regard to others, no specific directions can be given : — 1. When the terms of a polynomial have a common monomial factor, it may be written as one of the factors, with the quotient obtained by dividing the given polyno- mial by this factor as another. Thus, 6a^-+-3a* — 9a; = 3a; (2a:' + ar' — 3). 62 ALGEBEA. 2. Any trinomial can be resolved into two equal bino- mial factors, by 'Formulas 1 and 2, Art. 90, when two of the terms are squares and positive, and the other term is twice the product of their square roots. Thus, a' + 2ai + J^ = (a + 5) (a + i) ; (f — ia¥ + 4:b* = (a — 26") (a — 2J=). 3. Any binomial can be resolved into two binomial factors, by Formula 3, Art. 90, when its terms represent the difference between two squares. Thus, ^ 4.0^ — f= (2a; + y) (2a: — y). 4. Any binomial which consists of the difference of any two equal powers, or the sum of any two equal odd powers, may be factored by Articles 96, 97, and 98 ; for the quotient and divisor are factors of the dividend. Thus, a' — 6» = (a — 6) (a= + a J + J2) ; «* + 1 = (a + 1) (a" - a + 1) ; ^-3,*={^ + f) {x' - f) = {a? + f) {x 4- y) {x - y). Examples. 1. Factor 12 am + 20 ma;. Ans. 4m (3a + 6a;), or 2 X 2m (3a + 5«). 2. Factor o» — 2 a + 1. Ans. (a _ 1) (a — 1). 3. Required the binomial factors of a' -\-2ax •\- a?. 4. ^Actot m^nx -{- m^n^y -\- m^nxy. Ans. m'n (mx -\- ny -^ xy). 5. Resolve Sab* -\- ^a'bd into its factors. 6. Resolve 1 — So? into its factors. Ans. (1 — 2x) (1 + 2a: + 4ar'). 7. Factor 3 Ja;" + 6 6a;y -f- 3 bf. Ans, 3*(a;-f y) (x -^ y). FACTORING. 53 8. Factor 81o»— 1. Ans. (9 a* + 1) (3 a' + 1) (3 a" — 1). 9. Factor 5a:« + 5/. Ans. 5 (a: + y) (at* — jc«y + a;V — ^Y + /)• 10. Factor nx' — 2nxi/ -\-ny^. Ans. n (x — y) (x — y). 11. Factor «-«+ 2a-»6-i -f- *-«. Ans. (a-i-f.^^-1) («-i + 5-1). ► 12. Factor a^ — ^. Ans. (3i*-\-i^) {x' + y) (a^ — y). 13. Factor a? -)- 2ab -\- ^ — e^. a2+2«ft4-^— «« = (« + J)» — c^ = (a + 6 + c) (^+J — c) = (a-|-6 + c) (a-\-b — c). 14. Factor c» — (a=» — 2 a 6 + 5^^. Ans! (e — « + *) ('^ 4* « — ' *)• 15. Kesolve arf+ ay + cifa; -|- a;^ into its factors. ad -\- ay -\- dx -\- xy = a (d -\- y) + x (d -\- y) = {a-\-x) {d-\-y). 16. Factor am — bn -\- bm — an. Ans. (a + ft) (»» — «). It. Eesolvg' a* — J* into its factors. Ans. (a -f i) (o^— a6 -f 4^^ (« _ J) (a^ _|- « J + i^^. I02t A trinomial may be resolved into two binomial • factors, when one of the terms is a square and another the product of the algebraic sum of the two Jaefors of a third term by the square root of the first. For, if we multiply (ar-f- a) by (x -\-V), we have a:»_|_ (ffl 4- J) a; + air . (1) Or, if we multiply (a; — a) by {x — b), we have a^ _ (a 4- 6) a; 4- aJ. (2) Also, if we multiply (a; -|- a) by (a; — fi), or.(a; — a) by (a; -\- b), we have 5* 54 ALGEBRA. aP ^ (a — h) X — ab, (3) aJ! 4- (5 _ a) a: — a 5. ' (i) 18. Factor a^ -\-12x -\- 20. Ans. (x -\- 10) (a; + 2). 19. Factor a'—16a + 64. Ans. (a — 8) (a — 8). 20. Factor a^ -\- a — 2. Ans. (a — 1) (a + 2). 21. Factor b* — ¥ — 20. Ans. (¥ + 4) (S^ — 5). 22. Factor x'^f + 12a;y + 21. Ans. (a;y -j- 9) (xy -(-. 3).« 23. Eesolve a;? -|- ^J a; -|- f into two binomial factors. Here § = 3 X i, and 3^ = 3 + ^ ; .-. x^ + 3^x + I = (a; 4- 3) (a; + J). 24. Factor a;"— V^ + f- -A-HS. (a;_ — 3) (a; — J) . 25. Factor a;^y — -^-^y — -^V- Ans. 2^ (a: — 5) Cb + §). 26. Eesolve 5 a' J — ba^b — 30 a 5 into its factors. Ans. bab{a — 3) (a + 2). GEEATEST COMMON DIVISOE. 103, A Common Divisor or Measure of -two, or more quantities is a quantity that will divide each of them without a remainder. Hence, any factor common to two or more quantities is a common divisor of those quantities. Also, when quantities are prime to each other, they have no common measure greater than unity. 104. The Greatest Common Divisor of two or more quantities is the greatest quantity that will divide each of them without a remainder. Hence, the greatest common divisor of two or more quan- tities is the product of all the prime factors common -to those GREATEST COMMON DIVISOR. 55 By the greatest of two or more algebraic quantities, it may be remarked, is meant. the highest, with reference to the coefficients and exponents of the same letters*' 105i The greatest common divisor of two or more quantities is the same as the greatest common divisor of the least of those quantities, and the remainder, or remainders, if any, after the division of the least into the other, or each of the others. For, let a and 4 be two quantities, of which h is the least, and m some integral number. Suppose, now, that b is not contained in a an exact number of times, but m times, with a remainder, r. Then, since the dividend is equal to the product of the divisor by the quotient, plus the remainder, we have a = »i J -j- r. Also, since the remainder is equal to the dividend minus the product of the divisor by the quotient, Now, any quantity that will exactly divide h will exactly d^ide m times b, or mb ; and any quantity that will exactly divide b and r will exactly divide mh and r, and consequently will exactly divide their sum, mb -\- r, or its equal, a. Hence, any quantity that is a common divisor of b and r is also a common divisor of a and b. Ag'ain, any quantity that will exactly divide a and b will exactly divide a and mb, and consequently will exactly divide their difference, a — mb, or its equal, r. Therefore, any common divisor of a and b must also be a common divisor of 6 "and r. But the converse of this has already been proved ; consequently, the common divisors of a and b, and of b and r, must be identical, and the greatest common divisor of a and b must be also the greatest common divisor of b and r ; which was to be proved. 56 ALGEBRA. In like manner, the same may be proved for more than two quantities. 106. In determining the greatest common divisor of algebraic quantities, it is convenient to distinguish two cases. CASE I. 107i When the quantities can be factored hy in- spection. 1. Find the greatest common divisor of 3 a* — 3 6" and OPERATION. 3 a2 — 3 i= = 3 (a + J) (a — b) 3a'-\-6ab-\-3P = sla-^-b) (o + 5) 3 (a + 6) Factoring the quantities, we find that 3 and (a -\- b) are their only common prime factors ; then, the product of these (Art. 104), or 3 (a + *)> is the required greatest common divisor. Hence the • RULE. Resolve the quantities into their prime factors, and fnd the product of all the factors common to the several quantities. Any factor forming a part of the greatest common divisor will take the lowest exponent with which it occurs in either of the given quantities. Examples. 2. Find the greatest common divisor of 3ax^ — 2a^x and a^x^ — 3abx. Ans. ax. 3. Find the greatest common divisor of 4 a* 5 e, 6 a^ i d, and 2a*¥c. Ans. 2a^b. GREATEST COMMON DIVISOR. 57 i. Find the greatest common divisor of 0a:'y — 12a;'^y'' 4-3ar^ and 4aa;*-j-4aa;y -f- 4a^a;, 5. Eequired the greatest common measure of a;' — a' and x^ — a^ Ans. a; — a. 6. Required the greatest common factor of m^-\-2mn + n" and rn^ — n^. Ans. m -\- n. 1. Required the greatest common divisor of 5a' — lOa^b + 5a42 and Sa'—Sa^b.' Ans. a(a — b). 8. Required the greatest common divisor of b^e-\-1bc + 12c and 3 ab^ -\- 21 ab -\- 36a. Ans. (J + 3) (i + 4). 9. ^ Required the greatest common divisor of a;* — 1, af +V, and as* + 2a:» -f 1. Ans. a;*+ 1. CASE II. M8. When the quantities cannot readily be factored by inspection. 1. Find the greatest common divisor 6f a;'-|-2a; + l and a^ + 2a:»+ 2a;+ 1. OPERATION. a;" + 2a; + 1) a:* + 2a;» + 2a: + 1 (a; a^-|-2ar^+ a; a: + l)a;= + 2a;4-l (a;+I a:^ -|- X x-\- 1 ar-i- 1 It is evident that if ar" -(- 2a: -|- 1 will exactly divide a^ -\- 2x^ -\~2x-\-l, it will be the greatest common di- visor,, since no quantity can have a divisor greater than itself. But we find that the latter is not divisible by the former, there being a remainder, x -j- 1. Now, w* know 58 ALGEBRA. that the greatest common divisor cannot be greater than this remainder ; for the greatest common divisor of two quantities must be a divisor of their remainder after divis- ion (Art. 105). We therefore divide the divisor by the remainder, which it exactly divides. As a; -|- 1 is the greatest common divisor of the remainder and divisor, it must also be that of the divisor and dividend (Art. 105) ; consequently it is the greatest common divisor required. 2. Find the greatest common divisor of 36 a' — 18 a' — 27a^ + 9o' and 27 0=6^ — 18^*5^ — 9a=5?. OPERATION. 36a«— ISa" — 2Ta^ + 9a' = 9a= (4 a' — 2 a^ — 3a + 1) 27 0^5^— 18an2_9„3ja _ g^s J2 (Sa^ _ 2a — 1) 4:0? — 2a^— 3a+l 3 3 a''- -2a- -1) 12 a^ - 12a=- -Ga"- -8 a'' — 9a + 3(4a 4a 2 a'' — 5a-l-3 3 6a^ — Qa"- 15a + 9(2 4a— 2 — Ha+11 a— l)3a2 — 2a— l(3ffl + l Sa" — 3 a a— 1 g- 1 9 a' (a — 1) = the greatest common divisor. Here, 9 a', being common to both the given quantities, is, therefore, a factor of the greatest common divisor (Art. 103), and may be reserved to become a factor of the required greatest common divisor. GREATEST COMMON DIVISOR. 59 The factor h'', being common only to the terms of the second quantity, may be canceled, since it can form no part of the common divisor. In proceeding with the other factors, to avoid a frac- tional quotient, the first is multiplied by 3, which, not being a factor of the second, cannot affect their greatest common divisor. The same is true of the first remainder. The factor — 11 of the remainder — l\a-\-\\ is not found in 3a* — 2a — 1, and can therefore form no part of the common divisor; hence, we cancel it, and as a — 1 divides 3 a* — 2 a — 1 without a remainder, 9 a' (a — 1) is the required greatest common divisor. • From the preceding solutions is deduced the following RULE. Divide the greater quantity hy the less, and if there is no remainder, the less quantity will be the divisor required. If there is a remainder, divide the divisor by it, and con- tinue thus to make the preceding divisor the dividend and the remainder the divisor, until a divisor is obtained which leaves no remainder ; the last divisor will be the greatest common divisor. When the quantities have a common monomial factor, it may be reserved as a factor of the common divisor required. If in the operation a monomial factor is found common to all the terms of one of the quantities and not of the other, it may be canceled ; and, when necessary, the dividend may be multiplied by a factor not common to all the terms of the divisor. When more than two (fnantities are given, find the greatest common divisor of two of them, and then of that common divisor and the third of the quantities, and so on. The last divisor will be the greatest common divi- sor required. 60 ALGEBRA. Examples. 3. Find the greatest common divisor of a* — a" a: + 3aa;"— 3a;= and a^ —5ax -\- ixK Ans. a — x. 4. Find the greatest common divisor of a;'* — 1 ar -j- 10 and ia^ — 25x^-{-20x + 25. Ans. a;— 5. 5. What is the greatest common factor of ^ — x* and / — fx — i/x^-^-x?? Ans. f — x\ 6. Find the greatest common measure of a:* — a?-\-%a? + a; + 3 and a;* + 2 a.'' — a; — 2. ^ Ans. a;^ -|- a: -|- 1. t. Find the greatest common measure of 2 a* -1-3 a' a: — ^a'x^ and Qa^x — 1*1 a^x^ -\- I4.a^a? —.^ax\ 8. Kequired the greatest common factor of a;* -|- 6 a^ -1- 11 a;'' 4- 4a; — 4 and a;* -f- 2a:' — 5 a^^ — 12 a; — 4. Ans. 3? -\- iiX -\- 4:. 9. Find the greatest common divisor of 6a;^y-|-4a;^ — 2^ and ia? -\-4ia?y — 4a;^. Ans. 2 (x -\- y). 10. Find the greatest common divisor of 6 a:" — x — 5, 21a;' — 26 ar" -1-5 a;, and 4 a;' — 12a;2 _|_ g^. _ i_ Ans. X — 1. LEAST COMMON MULTIPLE. 109. A Multiple of a quantity is any quantity that can be divided by it without a remainder. Hence, a multiple of a quantity must contain all the prime factors of that quantity. 110. A Common Multiple gf two or more quantities is one that can be divided by each of them without a re- mainder. . Hence, a common multiple of two or more quantities must centedn all the prime factors of each of the quantities. LEAST COMMON MULTIPLE. 61 111. The Least Common Multiple of two or more quan- tities is the least quantity that can be divided by each of them without a remainder. Hence, the least common multiple of two or more quantities must be the product of all their different prime factors, each taken only the greatest number of times it is found in, any one of titose quantities. 112i If the product of two quantities he divided by their greatest common divisor, the quotient will he their least common multiple. For, since the greatest common divisor of two quan- tities is composed of aU the factors common to those quantities (Art. 104), these factors will enter twice into the product of the quantities. Hence, if the product be divided by the greatest common divisor, the quotient will contain only the factors common to the quantities, and those peculiar to each of them. Now these are the factors of the least common multiple. (Art. 111.) CASE I. 113i When the quantities can be factored by in- spection. 1. Find the least, common multiple of af — a" and a? — c^. OPERATION. se' — a^ = (x — a) (a: + a) s? — c^= (a; — a) (x^ -{-ax-\- «") (as -\- a) (x — a) (a? -\-ax-\- a^) Factoring the quantities, and finding the product of all their different prime factors, w« have (Art. Ill) for their 6 62 ALGEBRA. least common multiple (x -\- a) (x — a) (3? -\- ax -]^ a"). Hence the RULE. Resolve the quantities into their prime factors ; and the prod- uct of these, taking each factor only the greatest number of times it enters into any one of the quantities, will he the least common multiple. When quantities are prime to each other, their product is their least common multiple. Examples. 2. Find the least common multiple of 8 a*, 10 c^b, and 12an\ Ans. 120 a* i^. 3. Find the least common multiple of 2 (a -\- b) and 4. Find the least common multiple of (x -}- 1) (x'' — 1) and a^ — 1. Ans. {x -f 1)= (a:= — 1). 5. Find the least common multiple of (a -(- by, a" — V, (a — by, and c^ -\- 3aH -\- 3ab^ -\- l^. Ans. (a + bf (a — by, or (a + b) (a" — J")^. 6. Find the least common multiple of 4 (1 — x^), 8 (1 — x), 8 (1 + xy, and 4 (1 + a;^). Ans. 8 (1 — a;*). CASE II. 114. When the quantities cannot readily be factored by inspection. 1. Find the least common multiple of a* — 5a^-\-9a' — 7 a + 2 and a* — 6 a" -f 8 a — 3. LEAST -COMMON MXJLTIPLE. 63 OPERATION. (a«-5a'-|-9a'-7a+2)(a«-6a'+8a-3) _ ,_ , „„,„ „. ^,_3^._^3„_^ = (a— 2) (a«— 6a=+8qf— 3) Finding the greatest common divisor of the quantities, a' — 3a^-{-3a — 1, we divide their product by it (Art. 112), and have (a — 2) (a* — 6,0" + 8a — 3), the least common multiple required. Hence the RULE. Divide the product of the quantities by their greatest common divisor, and the quotient will be their least common multiple. When more thaii two quantities are given, find the least common multiple of two of them, and then of that oem- mon multiple and the third quantity, and so on. Examples. 2. Find the least common, multiple of x^ — Zse'y — xy" + 2/ and 3k^ — x^y — 3xy^ -\- y\ Ans. {x — 2y) (Sx — y) {?? — f). 3. Find the least common multiple of a? — ^x^-\-13x — 15 and a:^ — 8 a; + Y. Ans. (a:« — 9a;2+23a;— 15) {x—X). 4. Find the least common multiple of 6 a:^ — x — 1 and 2a^+3a; — 2. Ans. (20^^ + 3a; — 2) (3a: + 1). 5. Find the least common multiple of 7? -^-bx-^- ^, «» + 2 a: — 8, and a;2 4- T a; -}- .12. Ans. a;*-|-6a;' + 3a:'' — 26 a; — 24. 6. Find the least common multiple' of x^ — 4 a^ a? + 2oa;'' + 4a2a: + 8a', and ar' — 2 aa;^ + 4 a'^a; — 8a=. Ans. x^— 1 6 a*. 64 ALGEBRA. FEACTIONS. DEFINITIONS AND NOTATION. 115. A Fractional Unit is one of the equal parts into which a unit has been divided. Thus, |, ^, t are frac- tional units. 116i A Fraction is a fractional unit, or a collection of fractional units. Thus, ^, f , r are fractions. The denominator of a fraction indicates into how many parts the unit has been divided, and the numerator how « a many of those parts are taken. Thus, r indicates that 1 has been divided into b equal parts, and that a such parts are taken. The numerator and denominator are called the terms of a fraction. 117, An Entire Quantity is one which has no frac- tional part ; as, ab^ or a — b. 118. A Mixed Quantity is one having both entire and fractional parts ; as, a , or c ' c' T^r-j-y 119i The Value of a fraction is the quotient of its numerator divided by its denominator. For, a fraction is an expression of division, the nume- rator answering to the dividend, and the denominator to the divisor. (Art. 18.) Thus, the value of the fraction ab . -r- IS a. 9 FEACTIONS. 65 GENEEAL PEINCIPLES OP FRACTIONS. 120. If the numerator _U imiltiplied, or the denominator divided, hji any quantity, the fraction is muhiplied by the same quantity. For, let y denote any fraction ; then, ah -j = a. Now, if we multiply its numerator by any quantity b, we have aV -J- = ab, and, in like manner, if we divide its denominator by h, we obtain also a b. Hence, in both cases the value of the fraction has been multiplied by b. 121. ^ the denominator be multiplied, or the numerator di- vided, by any quantity, the fraction is divided by the same quantity. a V For, let -T- denote any fraction ; then, aV -r— =. a 0. Now, if we multiply its denominator by any quantity b, we have aV -^ = a; and, in like manner, if we divide its numerator by b, we obtain also a. Hence, in both cases the value of the fraction has been divided by h. 122. If th^ numerator and denominator be both multiplied, or both divided, by the same quantity, the value of the fraction will not be changed. 6* 66 ALGEBEA. For, if any quantity be both multiplied and divided by the same quantity, the value of the former will not be changed. (Art. 46, Ax. 6.) I23i A fraction is positive when its numerator and denom- inator have the same sign, and negative when they have dif- ferent signs. For a fraction represents the quotient of its numerator divided by its denominator; consequently its proper sign must be determined as in division. (Art. 67.) 124. The Sign of a fraction, or that prefixed to its dividing line, shows whether the fraction is to be added or subtracted. Thus, in a: -( — r—, the sign -|- denotes that — r— , al- though essentially negative (Art. 123), is to be added to X. The sign written before the dividing line has been termed the apparent sign of the fraction, and that depend- ing upon the value expressed by the fraction itself has been termed the real sign. Thus, in -| .— , the apparent sign is -|-, and the real sign — . 125. If any one of ike signs prefixed to the numerator, denominator, and dividing line of a fraction be changed, the value of the fraction will be changed accordingly. ri„„ "b — ab ah ah . For, -J- — a , -J- =—a;---^ = — a;—^=—a. Also, . — a6 „. ah —ab , — ab — ^rb——"' — :r6 = + «; j- =+«; tzt="- 126. Any two of the signs prefixed to the numerator, de- nominator, and dividing line of a fraction may be changed, without affecting the value of the fraction. FRACTIONS. 67 Pnr "* _ "* _ — a* —ah , Also, _^* = =^ = ili = _:^6 _ _„ 6 6 —b —b~ • I27i -5^ aU the signs prefixed to the terms and the dividing line of a fraction he changed, the value of the fraction 'will he changed accordingly. For, y = a, but — ~ j, = — «! — a6 , . ah -J— = ^°. but ——J = a; ab , , — ah a6 , , — ah ;- = — a, but J- = a. b — h EEDUCTION OF FEACTIONS. 128t Reduction op Fractions is the process of changing their forms without altering their values. CASE I. 129i To reduce a fraction to its lowest- terms. A fraction is in its lowest terms, when its terms are prime to each other. Since dividing both numerator and denominator by the same quantity, or canceling equal factors in each, does not alter the value of the fraction (Art. 122), we have the following 68 ALGEBRA. ETTLB. Resolve loth terms of the fraction into their prime factors, and cancel all that are common to both. Or, Divide both terms by their greater common divisor. Examples. 1. Reduce . ~ . to its lowest terms. a* — ar a' a; — a? x(a-\-x)(fl — x) x . 0*^ — 3* ~ (a + k) (a — a;) (a« + ^) ~ cF+^' ' 2. Keduce ^ . ~ — -j-ra to its lowest terms. 3/ ~4~ A X 7^~ i:>?-Vx)^{x-^V) _ x'-bx (x'-[-2bx + V)^(x + b) ~ x + b' "''■ Here the reduction is performed by using the greatest common divisor of the terms of the fraction. 3. Eeduce ^rrr — a to its lowest terms. Ans. -^rz — . Tbojnar 19 19 ab^cd e^ h 4. Reduce „„ „, . , to its lowest terms. Ans. 5. Reduce - , , ,„ r-rzii to its lowest terms. 6 a' -\- IS ax -\- 6 3^ . 3a — 2x 6. Reduce -j -j to its lowest terms. Ans. -r~, — 5. t a — ar a' -\- x' 1. It is required to reduce i_\ to its lowest terms. Ans. ^+/^±t. „ , 2k« — 16a;— 6 ^ ., , 2 8. Reduce 3 ^ _ 24 ^ _ 9 to its lowest terms. Ans. 5. FRACTIONS. 69 9. Reduce " ■ „!_ to an entire quantity. Ans. bn? — 2x — 3. CASE III. 132t To reduce a mixed quantity to a fractional form. This is the converse of Case II. Hence the following EtTLB. Multiply the entire part by the denominator of the fraction ; add the numerator to the product when the sign of the fraction is plus, and subtract it when the sign is minus, and write the result over the denominator. Examples. q2 J2 5 1. Reduce a-\-b — ■ j — to a fractional form. a-\-h.- a — J a^ — V — 5 _ d? — V—(a? — W — 5) a — 6 a — 6 • _ a" — 6" — a» + 6' + 5 5 . .^ =- — — 7, Ans. o — a — 72 ALGEBRA. 2. Change a •\ ^^^^ to a fractional form. an + 6^ — cd Ans. ' . n 3. Beduce 1x — ^ "^ y ^? to a fractional form. O , 56 a; — 4 n^ — 5 a Ans. g . 4. Reduce x ^^ to the form of a fraction. m < 5. Reduce a-\^b ^|^ to the forifl of a fraction. ' a 4"" 6 . 2a6 Ans. — r—r. a-\-b 6, Reduce 1 A ^^^-n to the form of a fraction. ' 2 DC Ans (Hif)!^:^' 3 d^ — ■ 2 6^ T. Reduce 2 a to the form of a fraction. 2 a Ans. — i . 2a a; + 1 8. Reduce a; + 1 -I ^^^— to a fractional form. ' ' , a; Ans. (^+^. 9. Reduce a — h- ~^ to a fractional form. a -\- o IV Ans. — 10. Reduce a:' — 3 a; ^-i — -^ to a fractional form. X — 2 a + 6' al fori X (a? — 2 X — 3) Ans. „ x — i CASE IV. 133i To reduce fractions to equivalent ones having a common denominator. FRACTIONS. 73 Fractions have a common denominator when they have the same quantity for a denominator. 1. Let it be required to reduce — and — to fractions mx my having a common denominator. FIRST OPERATION. Multiplying both terms of each fraction by the de- « __ g X wy __ amy nominator of the other, mx mxxmy . m'xy which does not change the b ly,mx hmx value expressed (Art. 122), , my~~myXm:x rn? x y we have fractions equiva- lent to those given, and with a common denominator, m?xy. But this is not the least common denominator, for all the terms of the frabtions. found have a common fac- tor, m. Now, the common denominator n^xy is a common, multipk of the lowest denominators ; hence, their kast common mul- tiple is their least common dehominatDr. SECOND OPERATION. mxi/-7-mx=:y', " " ' mx X y mxy mxy-r-my'=.x', , a X y __ ay I X y, y mxy h X X hx my X X mxy The least common multiple of the denominators is mxy. Dividing this by each of those denominators, we ascertain that the multipliers required to change the given fractious to others having the least common denominator, are, re- spectively, y and x. Multiplying both terms of the first fraction by y, and both terms of the second byar, we have' fractions equivalent to- those given, and with the least common denominator. 7 T4 ALGEBEA. RtTLE. Multiply each numerator hy all the denominators except its own, for new numerators, and all the denominators together for a COMMON denominator. Or, Find the least common multiple of all the denominators for the LEAST COMMON denominator. Divide this multiple by each denominator, separately, and multiply the corresponding numera- tors by the quotients for new numerators. Before applying the rule, each fraction should be in its lowest terms, and entire and mixed quantities should be changed to a fractional form. Examples. 2. Eeduce — , — , and j to fractions having a y' X ' b — c ° ' common denominator. . 36^^ — 3c a^ 4b my — 4 c my Ans. -7 , — = , axy bxy — cxy' bxy — cxy ' bxy — cxy' 3. Eeduce a, h, c, d, and t- to fractions having a com- mon denominator. Ans. b ab V be bd a T' b' T' T' 5' 2z ~\- 1 5 X 4. Eeduce — and — to fractions having the least common denominator. Ans. ^(^^+^) l^ 12a ' 12a' 5. Eeduce ^^^ , -^, — -^ to fractions having a com- mon denominator. ^jjg 25rf y —J 05 6c a; 286(far — %hbdx' — Uhdx' —35bTx' 6. Keduce ^-j-^, ^-^, and ^--^ to fractions having the least common denominator. Ans "'(" — '') (« + by a'V a? — fia ' a' — V a^ 6^" FRACTIONS. 75 Q, ti (1 St or t? T. Eeduce j-^, > ^ _ ^y and . J_ to fractions haying the least common denominator. . gy(l— a:)' aar'(l— a;) xf ■^ ^" (X — xf ' (l—xf > (l—xy 134i A fraction may be reduced to an equivalent one having a given denominator, by dividing the given denomi- nator by the denominator of the fraction, and multiplying both terms hy the quotient. 8. Change — into a fraction having by for a denomi- nator. , , ai? X h abx? . by -i-y = ; — — ^- = ^^ — , Ans. * " yXb hy ' 9. Change -^—^ — , , „ into a fraction having «' -f" ^^ for a denominator. Ans. a' + V 5 ah 10. Change -^ and — r- into fractions having 10 a" S , .. . ia'hd SOac' for a common denommator. Ans. ^- „, , ,- ,.. 10 a ' 10 o o 11. Convert ab, — ;— £1 and "^ , into fractions having a -f- a — ° o' — J^ for a common denominator. a6(a'-6') (a - 6)' (a + 6)' ADDITION OF FEACTIONS. 135, When fractions have a common denominator, they may be added by finding the sum of their numerators. For the sum of - and - is evidently the same as a di- vided by c, added to b divided by c, or the sum of a and b divided by c. 76 ALGEBRA. Also, since -= ac-^, and - = bc-^ (Art. 131), c c ^+ j= ac-^ + hc-^ = (« + i) c- = ^. Hence the RULE. Seduce the fractions, if necessary,, to equivalent ones having a common denominator. Add the numerators, and write the sum over the common denominator. The final result should be reduced to its lowest terns, whenever such reduction is possible. ExA^MPLBS. 1. Find the sum of and X — 3 a;-)- 3" X 3? -\- 3 X 1^ aP' — 3x a; — 3'^a;+3 a^ — 9 ^^ x' — 9 a^-f Sr + z' — 3a: 2^° . 2. Find the sum of -z- and . Ans. ^ -. 3. Find the sum of 4 m, — ^ — > and ""' . 2- 3 6 4. Find the sum of ■■ — = — , — J — , and — . 5 ' 3 ' 2 3 S ft /» 5. Find Uie sum of — j— r and r. Ans. ' . °j.> ill 3a:'.,,2aa; 6. Add a =- to J -4 . b ' c Ans. a + & + ^^^Ir:l£^. BKACTIONS. T7 1. Add 2-\ ^ to — -r-r. Ans. -= j=. 'a — a -\-o or — &' 8. Add , , I — 3 to -^,. Ads. 1+2 + 0? "" l_a; + a:»* "^°- l + a? + a:*' 9. Find the sum of -, — ^±i and — ^^. , 36^ — 2a'6— 2a — 3 Ans. js . a¥ — a . , 2 ^ Q « 10. Find the Etum of 2 a, ,3a+^^r-, and a —. Ans. Q a -^. 45 n Add ° + ^ ■^ + '' and " + " Ans. 0. 12. Add — 7 V'v / ' • jf > tT"?! \—ni j> aiid Ans. ^7— c (c — tr) (c — *)' " o6'c* SUBTEACTION OF FEACTIONS. 136i When two fractions have a common denominator, the one may be suhtraclted from the other by finding the difference of their numerators. For - subtracted from - is evidently the same as b di- c c vided by c, subtracted from a dirided by *, or a minus b divided by c. ft h Also, since - = ac''^, and — = bc~^, c c = a c-^ — ftc * = (a — 6) c * = — - — • c c ^ ' c Hence the 7* 78 ALGEBRA. RULE. Seduce the fractions, if necessary, to equivalent ones having a common denominator. Subtract the numerator of the subtra- hend from that of the minuend, and write the difference over the common denominator. Examples. CL C 1. From r subtract -j. h d a c ad he ad- — 6c ,. b~d~ hd~hd~ id ' ■ 2. From — subtract — . Ans. — . 2 ' 2 '' 3. From subtract — ;— :• Ans. a+ 1' o"— 1' ^ a ui. i. 2a — 45 „ 3a — 26 4. Subtract — r-r — from — r . 56 8c =L-y r^^rr. ^ + y A„„ 4a;y 5. Subtract — r- ^ from ^ . Aiis. ^ ». X -\- y X — y ar — y' 6. Subtract a- 3— from 3a -I- °T • d ' d Ans. 2 a -| j-. t. Subtract a; — i^^ from T a: — ^^zzl^_ 2 3 Ans. 6 a: -|- a -|- fi • 8. From.^^i^' take ^^^t Ans. 4. 10. From ^r-r^,-^ — ,» , ^ .. take ^* 2 (a; + 1) 10 (a — 1) "^ 5 (2a; + 3)" 2a; — 3 Ans. (a;^ — 1) (2 a; + 3)* FRACTIONS. 79 MULTIPLICATION OP FRACTIONS. 137i It has been shown that a fraction may be multi- plied by an entire quantity, either by multiplying its nu- merator, or by dividing its denominator (Art. 120). If, however, both multiplicand and multiplier are frac- tions, their product equals the product of their numerators divided by the product of their denominators. For, let it be required to multiply r- hy -;. Then, since ^=aJ~\ and -3=cd~^, o a e a Hence the RULE. MedvLce entire and mixed quantities, if any, to. fractional forms. Mvkiply the numerators together for a new numerator, and the denominators for a new denominator. When there are common factors in the numerators and denominators, they may be canceled before performing the multiplication. EXAMFEES. g ^ -J. JQ " 1. Multiply ^ — by 6 •' 2a:^ — 4a:" 23? — X 10 Z3? — X 5 2x — 1 1 tx — 1 . ~~f~'^2x'—ix~~'J~^¥^^~~^'^x^~T^' ' 2. Multiply --„ by -. — > Ans. . „ . . ^ •' mr? •' Trend, m^n'd _ ,, ,.. , 3a'x , iab . 12a*hx 3. Multiply -^ by ^^. Ans. j^j^. 4. Required the product of - — j- by „ ,^ . 80 ALGEBRA. 5. Multiply ^-^-- by -^-^y Ans. -^,:^--^:^-^ . 6. Multiply a-f^ by ™. ara 1. »^P^y S by ^■ Ans. J. 8. Multiply i^^ by ^-^. Ans. 3 . 9. ,, ,^. , 2a;— 1 , 3K + 1 Multiply ^^2 by ^j;^^. Ba? — a;— 1 Ans. "^-v-e- 10. Multiply ^ by p ■Ana. '-— -• ^m + Tt 11. M4«p..&^'b,,-^ 6)' . ? (a — &) Ans. -y-^.—K- x(a + b) 12. Multiply x^ — x-\-l by ^ + 5+>- Ans. a;^ + 1 + i 13. Find the product of - by -. Ans. ^^^^j. 14. Multiply together \~, ^^, and 1 + ^-^. Ans. 2. X 15. Multiply l^j by a- - h\ ^^^_ 3 ^ j, ^^ _^ ^^^ 16. Eequired the value oi b Ij- \ \j--\--)- Ans. — 2^-. DIVISION OF FRACTIONS. 138i It has been shown that a fractioTri may be divided by an entire quantity, either by dividing its numerator or by multiplying its denominator (Art. 121). FEACTIONS. 81 If, however, both diTicJehd and divisor are fractions, their quotient equals the product of the dividend by the recip- rocal of the divisor, or by the divisor inverted (Art. 43). For, let it be required to divide r by d" Then, since j-=ab~^, and -5 :=/; The Solution of an Equation is the process of finding the value, or values, of the unknown quantity, or the roots of the equation. 154i To solve an equation, such transformations are required as shall make the unknown quantity stand alone as one member of the equation, and all the known quan- tities as the other member. SIMPLE EQUATIONS. 89 I55i To solve simple equations of one unknown quan- tity. 1. Let it be required to find the value of x in the equation, ¥ + J = 22 + |. (1) Clearing (1) of fractions, gives VU 4- 15 X ■= 660 -\-bx. (2) Transposing in (2), we have 12a; + 15a: — 5a: = 669. (3) Uniting similar terms in (3), 22 a; = 660. (4) Dividing both members of (4) t)y the coefficient of x, we obtain a; = 30. This value of x we may verify (Art. 148). Thus, sub- stituting the value of x in (1), the original equation becomes £^ + 5^= 22 +i^; or, 12 + 15 = 22 + 5 ; whence 2t == 2t, in which the members are the same ; and the value of x is verified. Hence the following KULE. (Jlear the equation of fractions, if it has any. Transpose the unknown terms to the first mender, and the known terms to the second member, and reduce each member to its simplest form. Divide both members by the coefficient of the unknown quan- tity, and the second member of the resmUing equation will be the value of the unknown quantity. 8* 90 ALGEBRA. In some cases the operation may be abbreviated by some artifice, such as transposing and uniting similar terms be- fore clearing of fractions, or by partially clearing at first, and then making reductions. Examples. 2. Solve the equation. ■ + 6 Clearing of fractions, 10 a; +180 = 12 Transposing and uniting, — 2a; = — Dividing by — 2, a; = 45 a; + 90 90 3. Solve the equation, 1 — hx 1 — ax a ~ h Clearing of fractions, h — l^x = a — a^x Transposing, a^x — V^x = a — h Factoring, (c? — Jf) x :^ a — h Dividing by a^ — 6', x a? — V~a + b 4. Solve the equation, , ix-\-3 , Ix — 29 : ,8a;+19 S ^ + -i^ + 5^^12 = S^ + -^8— Suppressing 5 a; in each member (Art. 151), 4a; + 3 . 7a;— 29 8 3; + 19 9 "*" 5x— 12 18 Multiplying by 18, Subtracting 8 a; + 6 from each member (Art. 150), 126 a: — 522 SIMPLE EQUATIONS. 91 Clearing of fractions, 126a; — 522 = 65a:— 156 Transposing and uniting, 61' a; = 366 Dividing by the coefficient of x, X = 6 Solve each of the following equations. 5. 12 a: — 3 a: — 2 a; = 63. Ans. a; = 9. 6. a; — J + 13 = I + 40. Ans. x = 108. 1. Ta; + 2a; = 12a; — 36. 8- I + S = fo + 22- Ans. a;=120. 9. - = 6 4- c. Ans. x = a b-\-c- 10. a: — I + 20 = I + I + 26. Ans. x = 56. 11. 3a; + ^+15 = 1 + 41. Ans. a; = 8. 12. X i^±? = 8. Ans. X = 28. 13. 21 + i^----il=^-^^+^--^. U. i>j.-.l^-'JL±A=20x-'^p-5.- Ans. a; = 2. 15. a3^-4-bx = mx''-\-nx. . n — 6 ' ' Ans. X = . o — m 16. a + a;-^ = 5 + c + dx-\ j^^^ ^ ^ ^lzrJ_ a — b — c x-^3 ^ x-\-l 6 92 AL6EBBA. 18. - — 4: =: -. Ans. X = — 2. 19. aa; + 5 = ? + 5-i. Ans. x = t9rV. - a ' (V — 1) 20. 1.2 a: — •18^--Q5 -^ .4 a; -^ 8.9. Ans. a: = 20. 21. (a -1- x) (i 4- «) — ,« (J 4- c) = ^ + x\ Ans. x^=^. 22. i5 _ ? = ^_ ,. Ans. ;« = »40!iz^)_ c 3c — 4k + 3 7a;— 29 8a; + 19 23. — 3^ + ^-_^ = _^_. Ans. x=Q. 24 -^^^^ — /'^c — ?-^^^^ ) = T. Ans. a; = 5. 25 -('-^'^) = (g — 6) a ; ■ x _ ab ■ 3a(S + 4) • 2~~ "+■ 3 — 4 ■+ "• _ ■*°^- 6(a-6)+4- 26. 4.8 a; — :^^^^=^= 1.6a: + 8.9. Ans. a; = 5. 21. — J = a — b. c i 4 ah — ac -\- ahc — V c Aas. X = =— ' — -• — — . — c ^i. X X , X 2 .. 2 6 28. -+-=—. Ans. X = : a 6 ' c ac ' be — ac-^ab' 29. ^r^ + ' ab — ax ^^ h c — hx ac — a a?' _hja-b±e) Ans. X : a 30. {x — f) (x + I) — (a: — 6) (« + 3) — ^^ = 0. Ans, a;==-l^. SIMPLE EQUATIOKS. 93. PROBLEMS LEADING TO SIMPLE EQUATIONS CONTAINING ONE UNKNOWN QUANTITY. 156i A Pkoblem is a (question proposed for solution, or soinethirig to be done. 157i The Solution of a Problem by Algebra consists of two distinct parts : — 1. The Statement, or the process of expressing the con- ditions of the problem in one or more equations. 2. The Solution of thie resulting equation or equations, or the process of determinjijag from them "the value of the unknown quantities. 158. The statement of a problem often includes, a con- sideration of ratio and proportion (Art. 23). Ratio is the relation, in respect to magnitude, which one quantity bears to anotkeic of the same kind, and is the result arising from the division of one quantity by the other. PfeopoKTloN is an equality of ratios; Thusj a : h, or r- , indicates the ratio of a to 6 ; a : h = c : d indicates, a proportioB. 159. In a proptortioa, the relation of the terms iS' such/ that the product of the first and fourth is equal to the product of the second and third. d c ' For, a •.h=zc : d is the same as r = .^ , which gives ad ■= he. 160. Much must depend upon the skill and ingenuity of the operator j. in the statement oft problems, as will bei seen from the foHowiug examples. 94 ALGEBRA. 1. What number is that to which if four sevenths of itself be added, the sum will equal 99 ? SOLUTION. Let X = the number. Then — = four . sevenths of it. By the conditions, x -\ — =- = 99 Clearing of fractions, 7 as -|- 4 a; = 693 Or, 11a; = 693 Whence, x = 63, the number. Verification, f of 63 = 36 ; and 63 + 36 = 99. 2. The sum of two numbers is 144, and their difference is -30; required the numbers. SOLUTION. Let X = the snialler number. Then a; -|- 30 = the larger number. By the conditions, a; -j- a; -|- 30 = 144 Transposing and uniting, 2x = 114 Whence, ^ x = 67, the smaller number, and a; -|- 30 = 87, the larger number. Verification. 57 -}- 57 + 30 = 144. 3. A workman engaged for 48 days, at the rate of $2 per day and his board, which is estimated at $ 1 per day. At the end of the time he receives $ 42 only, his employer having deducted the cost of the board for every day he was idle. How many days did he work ? SIMPLE EQUATIONS. 95 SOLUTION. Let X = number of days he worked, and 48 — x = number of days he was idle. Then 2 a: = amount received for labor, and 48 — x = amount deducted for board. By the conditions, 2x — (48 — x) = 42 Or, 2ar — 48 + a; = 42 Transposing and uniting, 3 a; = 90 Whence, x = 30, number of days [he worked. 4. A can do a piece of work in 8 days, which it re- quires B 10 days to perform ; in how many days, can it be done by both working, together ? SOLUTION. Let X = number of days required. Then - = what both can do in one day. X Also, z = what A can do in one day, cS and T^ = what B can do in one day. By the conditions, - -|- — = - Clearing of fractions, bx-\-4tx =. 40 Or, 9a; = 40 Whence, x = 4|, no. of days required. VeBIPICATION. I- -f- t^ = i'iT' 5. Two pieces of cloth were purchased at the same price per yard, but as they were of different lengths, the one cost $ 5 and the other $ 6.50. If each had been 10 yards longer, their lengths would have been as 5 to 6. Bequired the length of each piece. 96 ALGEBRA. soLtmoN. Since the price of each per yard is the same, the lengths of the two pieces must be in the ratio of their prices, that is, as 5 to 6|, or 10 to 13. Therefore, Let 10 a; = length of first piece, in yards, and 13 a: = length of second piece, in yards. By the conditions, 10 a: -j- 10 : 13 a; + 10 :: 5 : 6 Or, by Art. 159, 60 a; + 60 = 65 a; + 50 Transposing and uniting, — 5 a; = — 10 Whence, x ^ 2 Then 10 a; = 20, length of first, and 13 a; = 26, length of second. Veeification. 20 + 10 : 26 + 10 = 5 : 6. 6. A general arranging his troops in thp form of a solid square, finds he has 21 men over, but attempting to add 1 man to each side of the square, finds" he wants 200 men to fill up the square 4 required the number of men on a side at first, and the whole number of troops. SOLUTION. Let X =^ the number on a side at first. Then a;^ -(" 21 = th6 whole number of troops. By the conditions, (x + 1)=' = x^ + 21 + 200 Or,/ ' :E2 + 2a;+l = a:2 + 221 Transposing and uniting, 2a; = 220 Whence, x= 110, no. men on a side at first. Sq'uaring,. ar' = 12100 and ar' + ^l = 12121, the whole number. From the preceding examples and illustrations is de- duced the following SIMPLE EQUATIONS. 97 RULE. Denote the unknown quantity or quantities hj some of the final letters of the alphabet. Form an equation, hy indicating the operations required to verify Ike answer, were it already obtained. Determine the value of the unknown quantity in the equation thus formed. Problems. I. My horse and chaise are worth $336, but the horse is wo"rth two times as much as the chaise. Eequired the value of each. Ans. Horse, $224; chaise, $112. 8. A drover has a lot of oxen and cows, for which he gave $ 1428. For the oxen he gave $ 65 each, and for the cows $ 32 each, and he had twice as many cows as oxen. Required the number of each. Ans. 12 oxen ; 24 cows. 9. What two numbers are those whose diiference is 3, and the difference of whose squares is 51 ? Ans. *I and 10. 10. A gentleman, at his decease, left an estate of $ 1872 for his wife, three sons, and two daughters. His wife was .to receive three times as much as either of her daughters, and each son to receive one half as much as each of the daughters. Eequired the sum that each received. Ans. Wife, $ 864 ; daughters, $ 288 each ; sons, $ 144 each. II. A laborer agreed to serve for 36 days on these con- ditions, that for every day he worked he was to receive $1.25, but for every day he was absent he was to forfeit $0.50. At the end of the time he received $ IT. It is required to find how many days he labored, and how many days he was absent. Ans. He labored 20 days, and was absent 16 days. 12. A man being asked the value of his horse and sad- dle, replied that his hofse was worth $ 114 more than his 9 9g ALGEBKA. saddle, and that f the value of the horse was 1 times the Talue of his saddle. What was the value of each ? Ans. The saddle, $ 12 ; the horse, $ 126. 13. What number is that from which if 1 be subtracted, i of the rejnainder will be 5 f Ans. 31. 14. In a garrison of 2744 men, there are 2 cavalry sol- diers to 25 infantry, and half as many artillery as cavalry. Required the number of each. Ans. Infantry, 2450 ; cavalry, 196 ; artillery, 98. 15. The stones which pave a square court would just cover a rectangular area, whose length is 6 yards longer', and breadth 4 yards shorter, than the side of the square. Find .the area of the court. Ans. 144 square yards. 16. A person has traveled altogether 3036 miles, of which he has gone 1 miles by water to 4 on foot, and 5 fey water to 2 on horseback. How many miles did he travel in each manner ? Ans. By water, 1540 ; on foot, 880 ; horseback, 616. It. A certain man added to his estate | its value, and then lost $ T60 ; but afterwards having gained 1 600, his property then amounted to $ 2000. What was the value of his estate at first? Ans. $1728. 18. A capitalist invested § of a certain sura of money in government bonds paying 5 per cent, interest, and the remainder in bonds paying 6 per cent. ; and found the interest of the whole per annum to be $ 180. Eequired the amount of each kind of bonds. Ans. 5 per cent, bonds, 1 1200 ; 6 per cent., | 2000, 19. A woman sells half an egg more than half her eggs. Again she sells half an egg more than half her remaining eggs. A third time she does the same ; and now she has sold all her eggs. How many had she at first ? Ans. 1. 20. A merchant has grain worth 9 shillings per bushel, and other grain worth 13 shillings per bushel. In what SIMPLE EQUATIONS. 99 proportion must he mix 44 bushels so that he may sell the mixture at 10 shillings per bushel ? Ans. 30 bushels at 9 shillings, and 10 at 13 shillings. 21. What number is that, the treble of which, increased by 12, shall as much exceed 54 as that treble is less than 144? Ans. 31. 22. A asked B how much money he had. He replied, if I had 5 times the sum I now possess, I could lend you $ 60, and then |- of the remainder would be equal to J- the dollars I now have. Eequired the sum which B had. Ans. $24. 23. A, B, and C found a purse of money, and it was mutually agreed that A should receive $ 15 less than one half, that B should have $ 13 more than one quarter, and that should have the remainder, which was $ 27. How many dollars did the purse contain ? Ans. $ 100. 24. Two persons, A and B, 120 miles apart, set out at the same time to meet each other. A goes 3 miles an hour, and B 5 miles. What distance will each have traveled when they meet ? Ans. A, 45 miles ; B, 75 miles. 25. The first digit of a certain number exceeds the second by 4, and when the number is divided by the sum of the digits, the quotient is 1. What is the number ? Ans. 84. 26. A person bought a certain number of acres of land for $ 180. After reserving two of them, he sold the re- mainder for $ 180. Now he found that he had gained on the cost price of each acre one third more per cent, than that cost price. How many acres did he buy ? Ans. 12. 27. A prize of $1000 is to -be divided between A and B, so that their shares may be in the ratio of 7 to 8. Ee- quired the share of each. Ans. A's share, $466|, and B's, $533^. 100 ALGEBRA. 28. A gentleman let a certain sum of money for 3 years, at 5 per cent, compound interest ; that is, at the end of each year there was added ^ to the sum due. At the close of the third year there was due him $ 2315.25. Ee- quired the sum let. Ans. $2000. 29. Bought a picture at a certain price, and paid the same price for a frame ; if the frame had cost $ 1 less, and the picture $0.'75 more, the price of the frame would have been only half that of the picture. Eequired the cost of the picture. Ans. $2.75. 30. A and B can do a piece of work together in *l days, which A alone could do in 10. days. In what time could B alone do it? Ans. 23^ days. 31. A gentleman gave in charity $ 46 ; a part in equal portions to 5 poor men, and the rest in equal portions to t poor women. Now, a man and a woman had be- tween them $ 8. What was given to the men, and what to the women ? Ans. The men received $ 25, and the women $ 21. 32. A man has two farms, and his stock is worth $ 183. Now, the stock and his first farm are worth once and two sevenths the value of the second farm ; and the stock and the second farm are worth once and five eighths the value of the first farm. What is the value of each farm ? Ans. First farm, $384; second farm, $441. 33. At 6 o'clock the hands of a watch are opposite the one to the other. When will they next be in like position ? Ans. At 5/y minutes after Y o'clock. 34. A vessel can be emptied by three taps ; by the first alone it could be emptied in 80 minutes, by the second in 200 minutes, and by the third in 5 hours. In what time will it be emptied if all the taps be opened ? Ans. 48 minutes. 35. A fox is pursued by a greyhound, and is 60 of her own leaps before him. The fox makes 9 leaps while the SIMPLE EQUATIONS. 101 greyhound makes but 6 ; but the latter in 3 leaps goes as far as the former in T. How many leaps does each make before the greyhound catches the fox ? Ans. The greyhound, 72 leaps ; the fox, 108. 36. It is found that T men and 3 boys pulling together produce just the same effect as 9 men pulling against 9 boys. What is the relative strength of one of the men to one of the boys ? Ans. 6 to 1. 37. It is required to divide the number 43 into two such parts that one of them shall be 3 times as much above 20 as the other wants of IT. Eequired the num- bers. Ans. 29 and 14. 38. Two horses ran over a mile course ; the winner completed the distance in 2 minutes 54 seconds, win- ning by 2 seconds. How many yards start might have been allowed to the other without risk of losing, supposing the same rates be kept ? Ans. 20. 39. Gold is 19 J times as heavy as water, and silver 10 J times. A mixed mass weighs 4160 ounces, and .displaces 250 ounces of water. What proportions of gold and silver does it contain ? Ans. Gold, 33VT ounces ; silver, 783 ounces. 40. A alone could perform a piece of work in 12 hours ; A and together could do it in 5 hours"; and C's work is I of B's. Now, the work has to be completed by noon. A begins work at 5 o'clock in the morning ; at what hour can he.be relieved by B and C, and the work be just fin- ished in time ? Ans. At 10 o'clock. 41. A merchant possesses $ 5120, but at the beginning of each year he sets aside a fixed sum for family expenses. His business in*eases his capital employed therein annually at the rate of 25 per cent. At the end of four years he finds that his capital is reduced to $ 3275. What are his annual expenses 1 Ans. % 1280. 9* •••"^ ALGEBBA. SIMPLE EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES. 161. Independent Equations are such as camiot be made to assume the same form. If they relate to the same problem, they must, there- fore, express essentially different conditions of that prob- lem. 162. When the conditions of a problem require the- unknown quantities to be denoted by different letters, in order to determine their values, as many independent equa- tions are necessary as there are unknown quantities. For, if we have an equation containing two unknown quantities, x and y, as a: — y = 1, transposing y, we have X = y-]- 1. (1) But the value of y is not known ; consequently, from thia equation alone the value of x cannot be determined. If, however, we have a second equation, as ^+y = 'J; or, . a; = 1 — y, (2) in which the value of x and y are the same as in the first, the second members of (1) and (2) being equal to the same quantity, x, and consequently equal to each other (Art. 46, Ax. 7), give y+1 = 1 —y, or, 2y = 6. . Whence, y = 3. Substituting 3, the value of y, for y in either equation (1) or equation (2), we obtain 4 as the value of x ; and SIMPLE EQUATIONS. 103 the values obtained for the two unknown quantities satisfy the two equations. 163. Simultaneous Equatmns are those in which the unknown quantities are satisfied by the same values. Two unknown quantities, denoted by different letters, require for their determination, as has been shown (Art. 162), at least two indepeadent, simultaneous equations. ELIMINATION. 164i Elimination is the process of corabining simulta- neous equations, in such a manner as to cause one or more of their unknown quantities to disappear. There are four principal methods of elimination ; "-^ by Comparison, by Substitution, by Addition or jSubtraction, and by Undetermined Multipliers. CASE I. 165i Eliimnation by comparison. i. Given 5x — 3^=9, and 2a;-l-5^=16, to find the values of x and y. OPERATION. 5a; — 3y = 9 (1) 2x+5i/=16 (2) From (1), by transposition and division, x = -J^ — - (3) From (2), by transposition, etc. x = — - (4) Combining (3) and (4), by Ax. 1, ?J^ = i^-^ (5) Clearing of fractions, 18 -{- 6y = 80 — 25 y Or, Sly = 62 Whence, y = 2 Substituting the value of ^ in (3), x ^ 3 1*^4 ALGEBRA. Hence the EULE, Find the value of the same unknown quantity, in terms of the other, from each of the equations ; and form a new equation by placing one of these two values equal to the other. It is usually most convenient to reduce each equation, at first, to its simplest form. CASE II. 166. Elimination by substitution. 2. Given ^ — !/ = 1> 3,nd x — | := 8, to find the values of x and y. OPERATION. i-y = i (1) ^-1=8 (2) Clearing (1) of fractions, x — 2y = 2 Or, X = 2-{-2y (3) Substituting the value of a; in (2) , 2 + 2t/— | = 8 Clearing of fractions, 4-|-4y — !/= 16 Or, 3y = 12 Whence, y ^ 4 Substituting the value of y in (3), a; = 10 Hence the KULE. Find the value of one of the unlcnown quantities, in terms of the other, from either of the equations ; and substitute this for the same quantity in the other equation. The rule applies very advantageously when either. of the unknown quantities has 1 for a coefiBcient. SIMPLE EQUATIONS. 105 CASE III. 167. Elimination by addition or subtraction. 3. Given 6a;4-4y = 32, and 4a; — 2^=12, to find the values of x and y. OPEKATION. 6a; + 4y = 32 4a; — 2y = 12 Dividing (1) by 2, 3 a: -f- 2y = 16 (1) (2) (3) Adding (2) and (3), Ta; = 28 Whence, a; = 4 Substituting the value of a; in (3), 12 + 2y = 16 Or, ■ 2^/ = 4 Whence, y = 2 4. Given 6 a; + 4:^=56, and 4 a; — 3^=9, to find the values of x and y. OPERATION. 6x-\-iy = 56 (1) 4:X — 3y = 9 (2) Multiplying (1) by 2, 12a; + 8y = 112 (3) Multiplying (2) by 3, 12x—9y = 21 (4) Subtracting (4) from (3), H y =^ 85 Or, yz=5 Substituting the value of ^ in (2), 4 a: — 15 = 9 Or, 4a; = 24 Whence, a: = 6 Hence "the RULE. Multiply or divide one or loth of the equations, if necessary, by such a number or quantity that one of the unknown quanti- ties shall have the same coefficient in both. Then, if tKe signs of the terms having the same coefficients are alike, subtract one equation from the other ; or, if unlike, add the two equations. 106 ALGEBRA. When the coefficients of the qtiantity to be eliminated are prime to each other, each equation must be multiplied by the coefficient found in the other. In general, such a multiplier may be. used for each as will produce the least common multiple of the coefficients. CASE IV. 168. Elimination, by undetermined multipliers. An UNDETERMINED MULTIPLIER is a factor, at first undeter- mined, but to which a convenient value is assigned in the course of the operation. 5. Given ix-{-2i/ = 2G, and 6x-\-ii/^S2, to find the values of x and y. OPERATION. 4a; + 22^ = 20 (1) Qx-^iy = 32 (2) Multiplying (I) by »wy 4,mx-\-2my^20m (3) Subtracting (2) from (3), and factoring with reference to x and y, (im — 6)x-{-(2m — 4:)y:=20m — 32 (4) Assuming 2m — 4 = 0^ the. second term of (4) disappears, and (4w — 6) x — 20 w — 32 20 m — 32 Or, - X = -; ^ ' Am — 6 Whence, since 2m — 4 = 0, m = 2, a; ^ 4 Substituting the: value of x in (1), y = 2 Instead of subtracting (2)- from (3), we might have added them, and obtained' like results. In (4),, X might have been eliminated by assuming 4 m — 6 = 0; but the value of m would have been J. Now, if we multiply the first of the given equations by 2, the first value of »», the coefficdent of y~ in it. will SIMPLE fiQtJATIONS. 107 become the! same, as iu the second equation"; and, if we divide the second of the given equations by |, the second value of m, the coefficient of x in it will become the same as in the first. Hence, this may be regarded hb' a generalization of the method by addition or subtraction. RULE. Multiply one of the given equations hy the undetermined quantity m, and add to, or' suUraci from, the i'esttki tJie other given equation. In the remitting equation, factored with reference to the un- known quanUiies, place the eo^cient of the qtmntity first to be elimindted equal to zero, and find the value of m. Substitute this value in the equation containing the uneMminated unknown quantity and m,. and reduce the equaiion: 169t Solve by whichever met&od- tOj be most advan- tageous, the following ExAltPLESl Find the values of the urikii^wBE qiiantities in each of the following equations. 6. Given iJ"+I^ = S[- ^^B-j^^t n. Given \Y^]y=^^l]. ^ne.|-=^- 13a; — 4^= 7) 1^ = 2. 8. Given {^^-^^ = '^,1. Ans.j-=J- 9. Givetx {^" + ^^ = 2|. Ans.^=^»- l5x — 2y — 3Q) (.y = 2. 10. Given \'l^'^=''l\. Ans.{-=^- 11. Given Pl- + ;^= ^f\. Ans.|-= «• 108 12. Given ALGEBEA. 13. Given 14. Given 15. Given (9a;-}-4y=58) L12") 80) by: .2y: (6a:+5y=112"J (8a; — 2v = n^-2y = — 6> |2a;+2y= 24>' (6a;+lly= 115| l8a; — 22y = — 30) IT. Given 18. Given 19. Given 20. Given 21. Given 22. Given \ T-f= H. (3a:+3y=126) (l4. + V^= 38j ( a;+12v=146) --^=-20 7 10 [ + 3y= 134 Ans. Ans. Ans. = 2. J.= 12 \y= 8 (.= 16. Given \ 2-+ 3y= 4^ Ans.|-= "^ llOa;— 12« = — 62) (3^=11 Ans. 2. 10. a; =10. Ans. = 12. 18. Ans. ty = = 35. 10. Ans. Ans. Ans. |.= |.= (y = |. <-3/ = 28. 21. = 24. [y=18. :a:= 2. 12. Ans. (.« = = 56. y = 40. SIMPLE EQUATIONS. 109 23. Giyen|«»'tt^ = ''l. (a' x-\-o'y = c' ) 24. Given 25. Given X a X y.—. X , v X Ans. Ans. - 2-12 = 1 + 8 x+j,x g 2y— g -_8- ■3 ^• ■21 h'c — be' ^==-57 t:- ao — a ac' — a' c .•' ao — a abcm-\-acdn ad -\- be bcdn — abdm ad-\-bc Ans. = 60. = 40. SIMPLE EQUATIONS CONTAINING THREE OR MORE UNKNOWN QUANTITIES. I70# The methods given for the solution of simxilta- neous equations containing two unknown quantities may be extended to those containing three or more unknown quantities. (10a: — 6y4-43=30j 1. Given •< 2a;-|- y — z^ 9[-to find x, y, and z. ( 5a;-|-4y — 22 = 28) OPEEATION. 30 10a; — 6y4-4z 2a;4- y— z = 9 5a;-|-.4y— 2z = 28 — 5a: + 3y — 2« = — 15 4a:-|-2y — 2« = 18 10 a; + 2? — 9a; + y Dividing (1) by —2, Multiplying (2) by 2, Subtracting (4) from (3), Subtracting (5) from (4), Subtracting (7) from (6), Or, Substituting the value of x in (6), Substituting the values of x and y in (2), 10 (1) (2) (3) (4) . (5) 43 (6) — 33 (T) 19a; y = 76 4 3 2 110 ALGEBRA. ^+ y-\- « = 53 jj Given -j a; + ^^^ + 3 2 = 107 > to find x, y, and z. OPERATION. a; + 2^-|-3« == lOT x-\-Sy-\-iz = 13T (i)- (2) (3) y-{.2z == 54 • 2/ + z = 30 (4) (5) Subtracting (1) from (2), Subtracting (2) from (3),- Subtracting (5) from (4), z = 24 Substituting the value of z in (5), y = 6 Substituting the values of y amd z in (1), x = 23 Froni these examples we deduce the following RHLB. €i)mUne the gia^ equations, s^ as td form a new set of equations, with a less numhef of unknown quantities ; drid con- tinue the process with the new equations ; and so on, ilrifil an equation is obtained containing but one unknown quantity. Find the value of the unknown quantity in this equation. By substituting this value in either of the equations containing only two unknown quantities, find the i>alue of a second uriknown quantity. Then, by suh^tvting these values in either of the equations lehich- contain thfee unknown quantities, find the value of a third;, and so on, till the vkilues of all are found. Examples. 3. Given 3a;— y~2z= Ov - 6a;+2y+3z = 45L Ax-\-By— z = 3l) /a; =4. Ans. -| 3/ = 6. (z = 3. 4. Given r 8x — 9y — 1z=—36i jl2a;— y—3z= 36 [. . ( 6a: — 23r— 2!= 10 > (a; =4. Aiis. y W ^^^-(^)'2^^ = i (5) 20y = 24/ (6) y = 5 a) 20 20 x + y 6y 4, no. hrs. going; 20 20 x — y iy 6, no. of hours in 114 By the conditions, and From (1), From (2), Substituting value ( Or, Whence, From (4) and (7), Also, [returning. 3. A and B can perform a piece of worfe in 6 days, A and in 8 days, and B and C in 12 days. In how' many days can each of them alone perform it 1 SOLUTION, Let X = no. days in which A alone- can perform it, y = no. days in which B alone can perform it, and z = no. days in which C alone can perform it. Then - = x~^ = the part which A can perform in one day, - = y~^ =; the part which B can perform in one day, and - = 2~' = the part which C can perform in one day. Also, ^ = the part A -and B can pCEform in one day, i = the part A and C can perform in one day, and xV ^^ ^^^ P*''* ^ ^^^ ^ '^^^ perform in one day. SIMPLE EQUATIONS. 115 By the conditions, x-i + 2^-' = * (1) x-^ -^ z- ' = \ (2) and y-* + «- = tV (3) Adding (1), (2), and (3), 2 a:- -i+2y-i + 20-' ■-%. (4) Dividing (4) by 2, x-^^y-' + '-' = T% (5) Subtracting (3) from (5), X ' ' = i^ (6) Subtracting (2) from (5), w' ^^z (7) Subtracting. (1) from (5), a~' = is (8) From (6), x = 9f , no. of days in which A can perform it. From (7), y = 16, no. of days in which B can perform it. From (8), z = 48, no. of days in which can perform it. 4. A Says to B, if ^ of my age were added to | of yours, the sum would be 19J years. But, says B, if | of mine were subtracted fi-om ^ of yours, the remainder would be 18|- years. Eequired their ages. Ans. A's age, 30 years ; B's, 20 years. 5. If 1 be added to the numerator of a certain fraction, its value is 4 i tiut if 1 be added to its denominator, its value is ^. What is the fraction ? Ans. -j*^. 6. A farmer had 89 oxen and cows ; . but,, having sold 4 oxen and 20 cows, found he then had 1 more oxen than cows. Required the number of each at firs!*. Ans. 40 oxen ; 49 cows. T. A says to B, if 1 times my property were added to f of yours, the sum wauld be $990. B replied, if 1 times my property were added to ^ of yours, the sum wonld be $510. Eequired the property of each. Ans. A's, $ 140 ; B's,. $ TO. 8. If } of A's age were subtracted from B's age, and 5 years added to the remainder,, the sum would be 6 years ; and if 4 years were added to -J of B's age, it would be equal to j^^ of A's age. Required their ages. AnSf A's, 9S years ; B's, 15 years. 116 ALGEBRA. 9. It is required to divide 50 into two such parts that f of the larger shall be equal to § of the smaller. Ans. 32 and 18. 10. A gentleman, at the time of his marriage, found that his wife's age was to his as 3 to 4 ; but, after they had been married 12 years, her age was to his as 5 to 6. Eequired their ages at the time of their marriage. ' Ans. The man's age, 24 ; his wife's, 18 years. 11. A farmer hired a laborer for ten days, and agreed to pay him $12 for every day he labored, and he was to forfeit $ 8 for every day he was absent. He received at the end of his time $40. How many days did he labor, and how many days was he absent ? Ans. He labored 6 days ; was absent, 4. 12. A gentleman bought a horse and chaise for $208, and ^ of the cost of the chaise was equal to | the price of the horse. What was the price of each ? •Ans. Chaise, $112; horse, $96. 13. A and B engaged in trade, A with $240, and B with $ 96. A lost twice as much as B ; and, upon settling their accounts, it appeared that A had three times as much remaining as B. How much did each lose ? Ans. A lost $96 ; B lost $48. 14. Two men, A and B, agreed to dig a well in 10 days, but, having labored together 4 days, B agreed to finish the job, which he did in 16 days. How long would it have required A to dig the whole weU.? Ans. 16 days. 15. A merchant has two kinds of grain, one at 60 cents per bushel, and the other at 90 cents per bushel, of which he wishes to make a mixture of 40 bushels that may be worth 80 cents per bushel. How many bushels of each must he use ? Ans. 13^ bushels at 60 cents ; 26§ at 90 cents. SIMPLE EQUATIONS. 117 16. A farmer has a large box, filled with wheat and rye ; seven times the bushels of wheat are 3 bushels more than four times the bushels of rye ;. and the quantity of wheat is to the quantity of rye as 3 to 5. Required the bushels of wheat and the bushels of rye. Ans. Wheat, 9 bushels ; rye, 15 bushels. IT. My income and assessed taxes together amount to $ 50. But if the income tax be increased 50 per cent., and assessed tax diminished 25 per cent., the taxes will to- gether amount to $ 52.50. Eequired the amount of each tax. Ans. Income tax, $ 20 ; assessed tax, $ 30. 18. A and B entered into partnership, and gained $200. Now, 6 times A's accumulated stock (capital and profit) was equal to 5 times B's original stock ; and 6 times B's profit exceeded A's original stock by $200. Eequired the original stock of each. Ans. A's stock, $ 500 ; B's stock, $ TOO. 19. A boy at a fair spent his money for oranges. If he had got five more for his meney, they would have averaged a half-cent each less ; and if three less, a half- cent each more. How many cents did he spend, and how many oranges did he get ? Ans. 30 cents ; 15 oranges. 20. A merchant has three kinds of sugar. He can sell 3 lbs. of the first quality, 4 lbs. of the second quality, and 2 lbs. of the third quality, for 60 cents ; or, he can sell 4 lbs. of the first quality, 1 lb. of the second quality, and 5 lbs. of the third quality, for 59 cents ; or, he can sell 1 lb. of the first quality, 10 lbs. of the second quality, and 3 lbs. of the third quality, for 90 cents. Eequired the price of each quality. Ans. First quality, 8 cents per lb. ; second, T cents ; third, 4 cents. 21. A gentleman's two horses, with their harness, cost him $ 120. The value of the poorer horse, with the har- 118 ALGEBRA. ness, was double that of the better horse ; and the value of the better horse, with the harness, was triple that of the poorer horse. What was the value of each ? Ans. Harness, $50 ; better horse, $40 ; poorer, $30. 22. Find three numbers, so that the first with half the other two, the second with one third of the other two, and the third with one fourth of the other two, shall each be equal to .34. Ans. 10, 22, and 26. 23. Find a number of three places, of which the digits have equal differences in their order ; and, if the number be divided by half the sum of the digits, the quotient will be 41 ; and, if 396 be added to the number, the digits will be inverted. Ans. 246. 24. There are 4 men, A, B, C, and D, the value of whose estates is $ 14,000 ; twice A's, three times B's, half of C's, and one fifth of D's, is $16,000 ; A's, twice B's, twice C's, and two fifths of D's, is $18,000 ; and half of A's, with one third of B's, one fourth of C's, and one fifth of D's, is $4000. Required the property of each. Ans. A's, $2000; B's, $3000; C's, $4000; D's, $5000. 25. A and B driving their turkeys to market, A says to B, give me 5 of your turkeys, and I shall have as many as you. B replies, but give me 15 of yours, and then yours will be f of mine. What number of turkeys had each ? Ans. A, 45 turkeys ; B, 55 turkeys. 26. A person possesses a capital of $30,000, on which he gains a certain rate of interest ; but he owes $ 20,000, for which he pays interest at another rate. The interest which he receives is greater than that which he pays by $ 800. A second person has $ 35,000, on which he gains- the second rate of interest ; but he owes $ 24,000, for which he pays the first rate of interest. The sum which he receives is greater than that which he pays by $310. What are the two rates of interest ? Ans. First rate, 6 per cent. ; second rate, 5 per cent. SIMPLE EQUATIONS. 119 2'!. A says to B and 0, give me half of your money, and I shall have $ 55. B replies, if you two will give me one third of yoijrs, I shall have $ 50. But G says to A and B, if I had one fifth of your money, I should have $ 50. Eequired the sum that each possessed. Ans. A, $20; B, $30; C, $40, 28. A gentleman left a sum of money to be divided among his four sons, so that the share of the oldest was ^ of the sum of the shares of the other three, the share of the second ^ of the sum of the other three, and the share of the third ^ of the sum of the other three ; and it was found that the share of the oldest exceeded that of the youngest by $ 14. What was the whole sum, and what was the share of each person ? Ans. Whole sum, $120 ; oldest son's share, $40 ; second son's, $30 ; third son's, $24 ; youngest son's, $26. GENEEAL SOLUTION OP PEOBLEMS. 171, In the general solution of a problem, all the quantities are represented by letters, or the symbols of general values. The unknown thus found in terms of the known quantities is a general expression, or formula, which can be used for the solution of any similar problem. 172. A problem is said to be generalized when letters are used to represent its known quantities. The algebraic solution of general problems discloses many interesting truths and useful practical rules, as may be seen from the consideration of the following, among other 120 ALGEBRA. General Problems. 1. The sum of two numbers is a, and their difference is h ; what are the two numbers ? SOLUTION. Let X = the greater number, and y = the less number. By the conditions, x -\- y = a (1) Also, X — y z=. b (2) Adding (1) and (2), 2 a; = a + 6 (3) Subtracting (2) from (1), 2y = a — 6 (4) Prom (3), a; = "T" , the greater number. Prom (4), y =■ — 5 — ) tli6 less number. Hence, since a and h may have any value whatever, the values of x and y are general, and may be expressed as rules for the numerical calculations in any like case ; thus. To find two numbers, when their sum and difference are given, — Add the sum and difference, and divide hy 2, for the. greater of the two numbers ; subtract the difference from the sum, arid divide by 2, for the less number. 2. A can do a piece of work in a days, which it requires b days for' B to perform. In how many days can it be done if A and B work together ? SOLUTION. Let X = the number of days required, and 1 = the entire work ; then, in 1 day A can do - of the work, and B T ; therefore, in x days, they can do - and ^ of the a Work. Hence, SIMPLE EQUATIONS. 121 By the conditions, - -|- - =: 1 Clearing of fractions, ax -\-hx = ah Or, (a -|- J) a; = a i ■Whence, x = . , no. days required. Hence, To find the time required for tWo agencies conjointly to accomplish a certain result, when the times are given in which each separately can accomplish the same, — ^ Divide the prodtict of the given times hy their sum. 3. A cistern can be filled by three pipes ; by the first in a hours, by the second in 5 hours, and by the third in c hours. In what time can it be filled by all the pipes running together ? . ahc , " ° Ans. — =— i i— iT^ hours. ah -\- ac -\-oc Here it will- be seen that, when three agencies are employed, the required time is the product of the given times, divided h) the sum of their 'products, taken two and two. 4. In the last example, what will be the time required, if a == 2, J = 5, and c = 10 ? Ans. 1^ hours. 5. Three men. A, B, and 0, enter into partnership for a certain time. Of the capital stock A furnishes m dol- lars ; B, n dollars ; and C, p, dollars. They gain a dollars. What is each man's share of the gain ? SOLUTION. Let a: = A's share. Then — = B's share, m and P— = C's share. m 11 122 ALGEBRA. By the conditions, a; + - + ^ = « Clearing of fractions, mx-^-n-x-^^px =^ ma Or, (jn -\- n -\- p) X = ma Whence, x = ,"'", , A's sharp. Then — = — ; r—, B's share. m m-\-n -\-p Also, ' — == — r- — f-zj C's share. Hence, To find each man's gain, when each man's stock and the whole gain are given, — Multiply the whole gain ly each man's stock, and divide the product hy the whole stock. 6. In the last example, if ?» :== $ 300, m = $ 500, ^ = $ 800, and a ^ $ 820, what is each man's share of the gain ? Ans. A's share, $60 ; B's share, $100 ; C's share, $ 160. "T. Divide the number a into two p^rta which shall have to each other the ratio pf m to n, Ans. — I — and i-\- n m-\-n' 8. A courier left this place n days ago, and goes a miles each daj'. He is pursued by another, starting to-day and going b miles daily. How many days will the second require to overtake the first ? . na , ^ Ans. , days. 9. Eequired what principal at interest at r per cent. wUl amount to the sum a, in t years. . lOOa ^°»- 100 + rf 10. A gentleman distributing some money among beg- gars, found that in order to give them a cents each, he should want b cents more ; he therefore gave them c cents each, and he had d cents left. Eequired the number of beggars. ^^^ h + d a — c' SIMPLE EQUATIONS. 123 11. Required the number of years in which p dollars, at r per CQut. interest, will amount to a dollars. Ans. }^l^L=LPl. rp 12. A banker has two kinds of money ; it takes a pieces of the first to make a dollar, and b pieces of the second to make the same sum. If he is offered a dollar for c pieces, how many of each kind must he give ? Ans. First kind, ° ^'' - . ^ ; second kind, — ^^ j-^. a — a — 13. In the last example, if a = 10, J = 20, and c^l5, how many of each kind must he give ? Ans. First kind, 5 ; second kind, 10. 14. A mixture is made of a pounds of coffee at m cents a pound, b pounds at n cents, and c pounds at p cents. Required the cost per pound oif 4he mixture. Ans. '"° + f + ^" . 15. A, B, and C hire a pasture together for a dollars. A puts in m horses for * months, B puts in n horses for t' months, and puts in p horses for t" months. What part of the expense should each pay ? Ans. A,$ ,,^:'° ,,. ; B, $- ^"^ ■ mt-\- nt' -{'pf' DISCUSSION OF PROBLEMS I/EAI?INa TO SIMPLE EQUATIONS. 173i The Discussiow of a problem, or of an equation, is the process of attributing any reasonable values and rela- tions to the arbitrary quantities which enter the equation, and interpreting the results. 174i An Arbitrary Quantity is one to which any rea- sonable value may be given at pleasure. 124 ALGEBRA. 175. A Determinate Problem is one in which the given conditions furnish the means of finding the required quan- tities. A determinate problem leads to as many independent equations as there are required quantities (Art. 162). 176. An Indeterminate Problem is one in which there are fewer imposed conditions than there are required quan- tities, and, consequently, an insufficient number of inde- pendent equations to determine definitely the values of the required quantities. 177. An Impossible Problem is one in which the con- ditions are incompatible or contradictory, and consequently cannot be fulfilled. 178. A determinate problem, leading to a simple equa- tion involving only one unknown quantity, can be satisfied by but one value of that unknown quantity. . For, after transposing and uniting, if the coeflScient of X is represented by a, and the second member by b, the equation will take the general form, ax = b. (1) Suppose now, if possible, this equation has two roots, r and r', then, by substituting successively these values in (1), ar = b, (2) rar'= b. (3) Therefore, by subtracting (3) from (2), we have ar — ar' = 0, Or, a (r — r') = 0. But the last equation is impossible, since by supposition r — r' is not zero, and a is not zero. That is, A determinate simple equation can have but one root. SIMPLfi EQUATIONS. 125 179t An- indeterminate problem, or one leading to a less number of independent equations than it has unknown quantities, may be satisfied by any number of values. For example, suppose a problem involving three un- known quantities leads only to two equations, which, on combining, give x — z=z 10, or, X = 10 -f- «. Now, if we make IS = 1, then a; = 11 ; 8 = 2, then a; = 12 ; « = 3, then x = 13. Thus, we may find sets of values without limit that will satisfy the equation. Hence, An indeterminate equation may have an infinite number of 180t When a problem leads to more independent equa- tions than it has unknown quantities to be determined, it is impossible. For, suppose we have a problem furnishing three inde- pendent equations, as,. a: = 2, + 1 (1) y=n-x (2) xy = 16 (3) From (1) and (2), we have X = 4o, and y = 3. But (3) requires their product to be 16, which can not be satisfied ; hence the problem is impossible. If, however, the third equation had not been indepen- dent, but derived from the other two ; as, xy=12; 11* 126 ALGEBRA. then the problem would have been possible, but the last equation, not being required for the solution (Art. 162), would have been redundant. INTERPRETATION OE NEGATIVE RESULTS. 181. In a NEGATIVE RESULT, Or a result preceded by the sign — , the negative sign is regarded as a symbol of interpretation. Its significance when thus used it is now proposed to investigate. 1. Let it be required to find what number must be added to the number a, that the sum may be h. Let X = the required number. Then, a -{- x = b. Whence, x = b — a. Here, the value of x corresponds to any assigned values of a and b. Thus, for example, Let a = 12 and b = 25. Then, a; = 25 — 12 = 13, which satisfies the conditions of the problem ; for if 13 be added to 12, or a, the sum will be 25, or b.- But suppose a == 30 and b = 24. Then a; = 24 — 30=— 6, which indicates that, nnder the latter hypothesis, the prob- lem is impossible in an arithmetical sense, though it is pos- sible in the algebraic sense of the words " number," " added," and "sum." . The negative result, — 6, points out, therefore, either an error or an But, taking the value of x with a contrary sign, we see that it will satisfy the enunciation of the problem, in an arithmetical sense, when modified so as to read : siMriiBi jiyuAilONS. 127 What number must be taken from 30, that the difference may be 24? 2. Let it be required to find the epoch at which A's age is twice as great as B's, A's age at present being 35 years, and B's 20 years. Let us suppose the required epoch to be after the present date. Then x ■=. tbe number of years after the present date, and 85 -j- a; = 2 (20 + «) ; whence, x =. — 5, a negative result. On recurring to the problem, we find it is so worded as to admit also of the supposition that the epoch is before the present date, and taking the value of x .obtained, with the contrary sign, we find it will satisfy that enun- ciation. Hence, a negative result here indicates that a wrong choice was made of two possible suppositions which the problem allowed. Prom the discussion of these problems we may infer : — 1. That negative results indicate either an erroneous enuncia- tion of a problem, or a wrong supposition respecting the quality of some quantity belonging to it. 2. That we may form, when attainable, a possible problem analogous to that which involved the impossibility, or correct the vjrong supposition, by attributing to the unknown quantity in the equation a quality directly opposite to thai whfich had been attributed to it. In general, it is not necessary to form a new equation, but simply to change in the old one the sign of each quantity which is to have its quality changed. Interpret the negative results obtained, and modify the enunciation accordingly, for the following 128 ALGEBRA. Problems. 3. If the length of a field be 10 rods, and the breadth 8 rods, what quantity must be added to its breadth that the contents be 60 square rods ? Ana. — 2 rods. 4 If 1 be added to the numerator of a certain fraction, its value becomes | ; but if 1 be added to the denominator, it becomes |. What is the fraction ? » — 5 5. The sum of two numbers is 90, and their difference is 120 ; what are the numbers ? Ans. 105 and — 15. 6. A is 50 years old, and B 40 ; required the time when- A will be twice as old as B. Ans. — 30 years. t. A and B were in partnership, and A had 3 times as much capital in the firm as B. When A had gained $2000, and B $ ^50, A had twice as much capital as B. What was the capital of each at first ? Ans. A was in debt $1500, and B $500. 8. A man worked 14 days, his son being with him 6 days, and received $ 39, besides the subsistence of himself and son while at work. At another time be worked 10 days, and had his son with him 4 days, and received $ 28. What were the daily wages of each ? Ans. The father's wages, $ 3 ; the son's, — 50 cts. That is, the father earned $ 3 a day, and was at the expense of 50 cents a day for his son's subsistence. 9. A worked 10 days, B 4 days, and C 3 days, and their wages amounted to $ 23 ; at another time, A worked 9 days, B 8 days, and C 6 days, and their wages amounted to $ 24 ; a third time, A worked T days, B 6 days, and C 4 days, and their wages amounted to $ 18. What were the daily wages of each ? Ans. A's, $2.00 ; B's, ; and C's, $ 1.00. SIMPLE EQUATIONS. 129 ZEEO AND INFINITY. 182i Zero, or the symbol 0, not only denotes absence of value, or nothing,, but may, in Algebra, stand for a quantity less than any assignable value. Infinity, or the symbol oo , denotes a quantity greater than any assignable value. 183. Hence, zero and infinity constitute the limits of all absolute or finite values ; and, in comparison with infinities, finite values may be considered as all equal. INTERPRETATION OF d, ^, ^, AND 1 oo' A 184. In order to explain the meaning of these symbols, let us take the fraction t. 6 1. Suppose the numerator, a, to remain constant, while the denominator, h, continually decreases. Then, since the value of a fraction depends upon the relative value of its terms (Art. 119), the fraction must increase as the denomi- nator decreases ; consequently, when b decreases below any determinate limits, the value of the fraction must exceed any determinate or assignable quantity. Hence, representing any finite quantity by A, we have A That is, If a finite quantity is divided by zero, the quotient is infinity. 2. Suppose the numerator, a, to remain constant, while the denominator, J, constantly increases. Then, the value of the fraction must decrease as the denominator increases ; consequently, when b increases beyond any determinate limits, the value of the fraction must be less than any determinate or assignable quantity. Hence we have 130 ALGEBEA. 00 That is, If a finite quantity is divided hy infinity, the quotient w zero. 3. Suppose, now, the denominator, h, to remain con- stant, while the numerator, a, constantly decreases. Then the value of the fraction must decrease as the numerator decreases ; consequently, when a decreases below any determinate limits, the vaiue of the fraction must be less than any determinate or assignable quantity. Hence we have ! = »• That is, If zero is divided hy a finite quantity, the quotient is zero. 4. Suppose, next, a and h both to decrease, at the same time and in the same ratio. Then, the value of the fraction will not be changed ; but when a and h decrease below any determinate limits, the terms of the fractions each become zero, and the fraction itself becomes ^. As - may have any value, - will represent any finite quan- tity. Hence, If zero is divided hy zero, the quotient may he any finite quantity. It is not true, however, that ^ is always a symbol of an indeterminate finite quantity. For, if j: is the result of an expression whose numerator contained more zero factors than its denominator, its value is ; or if its denominator contained more zero factors than its numerator, its value is oo . Thus, when a =: i, (« — 6)'' _ _ a — h _ ^ _ c^ — W g^-f aS-f-S' 3 a'' (a — by ~~ ~ (o — 6)' "^ "0~ — °° • SIMPLE EQUATIONS. 131 Also, -, by canceling a common factor in the terms of the fraction from which it originates, is sometimes found to have a definite finite value. Thus, when a=.b. The expressions X <» and ^ may also hav6 the same four values that have just been attributed to -^. Note. — A proper understanding of the theory of indetermination, and of the relations of to finite quantities, will lead to the detection of the fallacy in some apparently remarkable results: For instance, Let a = h (1) Multiplying both members by a, and subtracting ¥ from both, a^ — V = ah — ¥ (2) Factoring, (a -f 6) (o — 6) = 5 (o — 6) (3) Dividing by a — 6, a-\-b ^ b (4) Substituting & for a, 2i = 6 Whence, 2=1 Again, take the identical equation, 20 + 12 = 20 + 12 (1) Transposing, 20 — 20 = 12 — 12 (2) Factoring, 5 (4 — 4) = 3 (4 — 4) (3) Whence, 5 = 3 (4) The members of equation (3) in each of the alaove fexamples are equal, because each member becomes 0, through the influence of a factor ; but it does not follow that the other factors are equal, as assumed in (4). In other woi-ds, the result of the division of each member ©f (3) is _, which may represent any finite quantity. PROBLEM OF THE COURIERS. 185i The discussion of the following general problem, commonly known as that of Clairaut, will serve to further illustrate the significance of the anomalous forms consid- ered in the last Article. 182 ALGEBRA. PROBLEM. Two couriers, A and B, are traveling along the same road, in the same direction, R' R, at the rate of m and n miles respectively. If at any time, say 12 o'clock, A is at the point P, and B a miles from him at Q, when and where are they together ? Let t = the required time in hours, X = the distance B travels in time (, and a -{- X ^ the distance A travels in time t. ' But the distance each travels is equal to the rate of each per hour multiplied by the number denoting the time. Therefore, mt = a -\- X ■ (1) Also, nt = X (2) Subtracting (2) from (1), mt — nt = a (3) Whence, t- " (4) Substituting the value of i in (2), x = (5) It is proposed, now, to discuss these values on different' suppositions ; — 1. m '> n. This hypothesis makes the common denominator, m — n, in (4) and (5), positive ; hence, the values of both t and x are positive. That is, the couriers are together after 12 o'clock, and at the right of P. This interpretation corresponds with the conditions of the problem. For, A will eventually overtake B, and in advance of their position at 12 o'clock. 2. m <_ n. This hypothesis makes the common denominator, m — n, negative ; hence, both t and x are negative. Now, from what is known of negative results (Art. 181), these values SIMPLE EQUATIONS. 133 of t and X indicate that the couriers were together before 12 o'clock, and at the hft of P. This interpretation corresponds with the supposition made. For, A will never overtake B, but as they are traveling along the same road, they were together before they could have advanced as far as P. 3. j» = n. This hypothesis makes m — « = ; then, we have a . na t= - = CO , and a; = — ^ CO . Hence, the values of both t and x are infinity. That is they will never be together. This interpretation corresponds with the supposition made. For, if they travel at the same rate, and are a miles apart at 12 o'clock, they always have preserved, and always will preserve, the same distance. 4. = 0, and m > w or «j < n. Then, we have t = = 0, and X = = 0, m — n m — n or, both the time and distance are nothing. That is, the couriers are together at 12 o'clock, at the point P, but at no other time or place. This interpretation corresponds with the supposition made. For, since the distance between them at 12 o'clock is nothing, they are together then at P, but as their rates are unequal, they cannot be together after 12 o'clock, nor could they have been together before that time. 5. a = 0, and m — m = 0. Then we have t = Q, and a? = ^, or the values of t and x indeterminate. That is, the cou- 12 134 ALGEBRA. riers are together at aM times, and at any distance whatever from P. This interpretation corresponds with the supposition made. For, being together, and advancing at the same rate, they can never separate. 186. It may be remarked that the formula, a t = , m — n in the foregoing problem, is general, and may be applied to the soluti-oit of all similar PeoblemS. 1. A train of cars leaves Boston at 1 o'clock, P. M., at the rate of 24 miles an hour, and is followed by an express train at 3 o'clock, at the rate of 36 miles an hour. When will the latter overtake the former ? Here a = 48, m = 36, and m = 24 ; whence, t = — — — = 4 hours ; oo -^ i4 and 3 o'clock -^ 4 hours = 7 o'clock^ P. M., Ans. 2. At 12 o'clock both hands of a clock are together ; when will they next be together ? Here o = 12, ot^12, and w = l, hour spaces. 12 12 — 1 12 Whence, t = == 1 h. 5/y m., Ans. 3. At what time between 3 and 4 o'clock will the two hands of a clock be together? Ans. 3h. 16^*^™- 4. Two men, A and B, set out from two places, distant from each other 144 miles, and travel in the same direction, along the same road. A goes 12 miles an hour, and B 4 miles an hour. In how many hours will they be . together, and how far will each have traveled ? Ans. 18 hours ; A 216 miles, and B 12 miles. SIMPLE EQUATIONS. 135 5. At what time between 12 and 1 o'clock do the hour and minute hands of a watch point in exactly opposite directions? Ans. 32/x minutes after 12. INEQtJALITIES. 187i An Inequality is an expression indicating that one of two quantities is greater or less than the other ; as, a > 5, and m <. n. The quantity on the left of the sign is called the left member, and that on the right, the riffht member of the inequality. I88t Two inequalities are said to subsist in the same sense, when the first member is the greater or less in both. Thus, a > b and 6 > d, or 3 < 4 and 2 < 3, are inequalities •W'hich subsist in the same sense. 189. Two inequalities are said to subsist in a contrary/ sense, when the first member is the greater in the one, and the second in the other. Thus, a > J and c <. d, or a: < y and « > «, are inequalities which subsist in a contrary sense. 190i In the discussion of inequalities, the terms greater and less must be taken as having an extended, or alge- braic meaning. That is. Of any two quantities, & and b, a is the gficUet- when & — b is positive, and a is the less when a — b is negative. Hence, a negative quantity may be considered as less than nothing, and, of two negative quantities, that is the greatest which has the least number of units. Thus, > —2, and —2 > —3. 136 ALGEBRA. 191. An inequality vnU continue in the same sense, after the same quantity has been added to, or subtracted from, each member. For, suppose a > b ; then, by Art. 190, a — 6 is positive ; consequently, (a .+ c) — (J + c) and (a — c) — (6 — c) are positive. Therefore, a -\- c > J -j- c, and a — c > J — c. Hence, it follows that a term may be transposed from one member of an inequality to the other, with its sign changed. 192. If the signs of all the terms of an inequality be changed, the sign of inequality must be reversed. For, to change all the signs is equivalent to transposing each term of the first member to the second, and each term of the second member to the first. 193. If two or more inequalities subsisting in the same sense, be added, member to member, the resulting inequality will also subsist in the same sense. For, let a> b, a' > b>, a" > b", ; then, by Art. 190, a — b, a' — b', a" — b", , are all positive ; consequently their sum, a — 5 + a' — 5' + a" — b", -f , or (a + a' + a" + ) — (b -]- b' J^ b" -\- ), is positive. Hence, a _1_ a' + a" + > b -\- b' -{- b" -\- 194. If two inequalities subsisting in the same sense be subtracted, the one from the other, member from member, the resulting inequality will not always svhsist in the same sense. SIMPLE EQUATIONS. 137 For, let a "> h, and a' > J' ; then a — b and a' — b' are positive, but a — h — (a' — J'), or a — a' — (J — b'), may be either positive, negative, or ; that is, a — a' "^ b — b', a — a' <. b — 5', or a — a'=:b — b'. 195. An inequality will continue in the same sense, after each member has been multiplied or divided by the same posi- tive quantity. For, suppose a > J ; then, since a — b is positive, if m is positive, m (a — b) and — (a — b) are positive. Hence, ma '> mb and — > — . mm 196i ^ each member of an inequality be mtdtiplied or divided by the same negative quantity, the sign of inequality must be reversed. For, since multiplying or dividing by a negative quan- tity must change the signs of all the terms, the sign of inequality must be reversed (Art. 192). 197. The solution of an inequality consists in deter- mining the limit to the value of its unknown quantity. This may be done by the application of the preceding principles. 198. When, however, an inequality and an equatibn are given, containing two unknown quantities, the process of elimination will be required in the solution. In verifying an inequality, if the symbols of the unknown quantities be taken equal to their respective limits, the inequality becomes an equation. 12* 138 ALGEBRA. Examples. 1. Find the limit of x in the inequality, Clearing of fractions, 21 a: — 23 > 2 a; -j- 15 Transposing and uniting, 19 a; '> 38 Whence, a: > 2 2. Find the limits of x in the inequalities, ax -\- bhx — h ah > c?, hx — *l ax -\- lah < V^. Adding 5a6 to both members of (1), and factorifify (a -j- 5 J) a; > {a -\- bh) a\ whence, a; > a. Subtracting *l ah from both members of (2), and factoring, (h — *la)x< (h—1a)h; whence, a; < J. 3. Find the limits of x and y in the following inequality and equation, (4a: + 6y > 52 (1) l4a;-j-2y = 32 (2) Subtracting (2) from (I), 4y > 20 Whence, y > 5 Dividing (2) by 2, and transposing, y = 16 — 2a; Subst. value of y in (1), 4a; -)- 96 — 12a; > 52 Or, _8a; > — 44 Whence, a; < 51- 4. Given 5 a; — 6 > 19, to find the limit of x. Ans. a; > 5. ( 2x - 5 ■> 25 ) 5- ^'^«° i3a;-7< 2^+13}'*° ^"'^*'^«^^'"^*«°f^- Ans. a; > 15 and a; < 20. SIMPLE EQUATIONS. 133 of X. 6. Given -5, '^^ "*''''5-,to find an integral value lix — 1 < 2a;4-3j ^ Ans. a; = 4. 1. Given { 5a: + 3y > 46 -y) ^^ ^^^ ^^^ li^j^^ ^^ (2y — 2a; = — 8 j' and y. Ans. a: > 6| and y > 2f . 8. Given -j-{-dx — cd > |- I T-''^+'''^< 8 J , to find the limits of a;. Ans. X > c and x 9f and ^^ < 12J. 10. A teacher being asked the number of his pupils, replied that twice their number diminished by *l is greater than 29, and three times their number diminished by 5 is less than twice their number increased by 16. Eequired the number of his pupils. Ans. 19 or 20. Here the inequalities gi\?e a; > 18 and a;< 21. That is, all numbers between 18 and 21 will satisfy the inequalities ; but as, from the nature of the problem, the answer should be a whole number, the solutions are limited to a; = 19, or a; = 20. 11. Three times a Certain nuaibel* plus 16 is greater than twice that number plus 24, and two fifths of the number plus 6 is less than 11. Eequired the number. Ans. as > 8 and x < 15. That is,, any number, whple or fractional, between 8 and 15. 140 ALGEBRA. 12. A shepherd has a number of sheep such that three times the number increased, by 2 exceeds twice the number increased by 61 ; and 5 times the number dimin- ished by 70 is less than 4 times the number diminished by 9. How many sheep has he ? Ans. 60. INVOLUTION. 199. Involution is the process of raising a quantity to any required power. This may be effected, as is evident from the definition of a power (Art. 19), by taking the given quantity as a factor as many times as there are units in the expo- nent of the required power. That is. If ike exponent is n, the power is the product of n factors, when n is any entire quantity whatever. 200. Any power of the product of two or more factors is equal to the product of the same power of each of the factors. Thus, (a 5)» = a" 6". For, {ahy = al y, ah y, ah to n factors, is the same as a X « X « to n factors X ^ X i X 5 to n factors, or a" X *" i and the same reasoning may be employed when there are more than two factors. That is. The nth power of the product of two or more factors is equal to the product of the nth power of each of the factors. 201. The power of any power of a quantity is equal to that expressed by the product of the exponents of the two powers. Thus, (a")™ = a"". INVOLUTION. 141 For, if we form the product of m factors of which each is a", the result, by the rule of multiplication (Art. 67), is a"". That is, The mth power of the nth power of any quantify is equal to the mnth power of that quantity. 202t If the quantity to be involved is positive, the sign of all its powers will be positive ; but if the quan- tity is negative, all its even powers will be positive, and all its odd powers negative. , Thus, a X « = «° « X o X a = «^ X « = «'j and so on. For, any positive factor taken any number of times must give a positive result (Art. 66). Again, (— a) X (— «) = +«' {■—o) X ( — a) X ( — a) = +«'' X ( — «) = — «', and so on. For, a negative multiplier causes the sign of the product- to be the oppos;ite of that of the multiplicand (Art. 66), and therefore each additional negative factor changes the sign of the result. So that, since n may be any quantity whatever, in the expression, (— a)» = ± a", by the double sign ±, is to be understood, if the wth power of — a is even, it has the positive sign, but if odd, the negative sign. Hence, Every even power is positive, and 'every odd power has the same sign as its root. POWERS OF MONOMIALS. 203i From the preceding principles and the rule for multiplication, we have, for raising a monomial to any power, the following 142 ALGEBRA. EULB. liaise ike numerical coefficient to the required power, and muUiply the exponent of each letter hy the exponent of the re- quired power, making the sign of every even power positive, and the sign of every odd power (he same as that of its root. Examples. 1. Required the third power of — 3a'i. {—^aHf = ^2,^(^¥ = —Qna^W. 2. Eaise c^x to the second power. Ans. a*a;^ 3. Kaise — ah^ It magr likewise be shown by actual multiplication, that {a-{-h-\-ey = a2-|-2a(J + c> + Ji' + 2ie4-c2 (a + i + c + ^O*^ = a^-{-2a(b-\-e.-{-d)-\-V-\-2b{e-{-d) -\-e^-\-2ccl-\-d', and so on. These results, fbr convenience of enunciation, may take another form, {a-\-by = »= + &'+ 2a5 (a — by = a^-^¥ — 2ab (a_|_5 + c)'' = a<'-{-b^-{-c^-\-2ab-\-2ac-\-2bc (a_}_J-|_e-j_«?)2 = aa_|_52_|_g2^_(;2_[,2a6 + 2ac + 2arf -\-2be-\-2bd-{-2cd, and so on. H6nce the' following ^ RULE. Write the square of each term, together with twice its product by each of ike terms falhwing it. Examples. 1, Square a — b-\- c. Ans. a^—2ab-\-2ac-\-V' — 2bc-\-(?. 2. Square 2(c=+3:k + 4. Ans. 4a;*+12a!3 + 25a;2 + 24a:4-16. 13 146 ALGEBRA. 3. Square 2x^ — Zx-\-i. Ans. 4a;*— 12 a? + 11 a?— So; +f 4. Square a — 6 — c -\- d. Ans. a= — 2a6 — 2ac + 2arf+ J^ + 2Jc — 2&(f + c2 — 2crf+rf^. 6. Square a:^ + 2 a;^ + a; -f 2. Ans. «»4-4«*+6a;< + 8a:»+9a:2_|_4a;_|_4_ 6. Expand (1 — 2 a: + 3 x'f. Ans. 1— 4a; + 10a;'>— 12a:' + 9a;*. 7. Expand (1 + a; + a;^ + a;»)2. Ans. l+2a; + 3ar' + 4a;» + 3a;* + 2a« + a;*. 8. Required the square of a-\-h-\-c-\-d-\-e. Ans. a^-\-2ah-\-2ac-lf-2ad-\-2ae-^b'^-\-2hc-\-2hd -\.2he-\-e^-\-2cd-{-2ce^d^-\-2de-\-e\ POLYNOMIAL CUBES. 208. The cube oi a-{-h, and of a — h, readily obtained by multiplication, give the formulas, (« + J)« = a= + 3a2J + 3a5»+5», (a _ 5/ = a^ _ 3a2J -f- 3a62 — W, which, for convenience of enunciation, may be put under the form, (a + 5)3 = a» + 5» + 3o J (a + J), la — by = cP — V — iabla^l). That is, The CUBE of a binomial is equal to the cube of the two terms, together with three times the product of the two terms multiplied hy each of the terms. But it may be also shown by actual multiplication, that mvoLxmoN. 147 -i-3c''(a4-i)-i-6aJc, (a-|-6+c4-c?)» = a» + 3a'' (J + c + rf) 4-i3 -1-36^ (a+c+rf) -\-6abc-\-Gabd-\-6aed-\-6bcd, and BO on. Hence, for finding the cube of expressions containing three or more terms, the following RULE. Write the cube of each term, together with three times the prodiicf of its square by each of the other terms, and also six times the product of every three different terms. Examples. 1. Find the third power of a -\-b — c. Ans. a» + 6^ — c' + Sa^J— 3a»c + 3a6= — 3c6= -\- Zac^ -\- ^hc^ — Qabc. 2. Find the third power of a? -^ x — 1. Ans. a^ — 3a^-\-53^—3x—l. 3. Eequired the third power of a — i -f- 1. Ans. a^—San+Sa^+Sab^—Qab-^-Sa — b' -f 3 J'' — 3 i + 1. ■ 4. Eequired the third power of 1 -{- x -\- x" -\- a?. Ans. l + 3a; + 6a;^ + 10a^4-12a:*+12a^+ lOa;^ -|-6a;' + 3a^ + a^. 5. What is the third power of a — b~2e^ — d^? Ans. a»— S*— 8c«— A Pekfect Square is any number or quantity that can be resolved into two equal factors (Art. 199) ; and, consequently, has an exact square root (Art. 210). 215> If a number be separated into as many groups, or periods, of two figures each, as is possible, from right to left, these groups will correspond, respectively, to the units, tens, hundreds, etc., in its square root. 13* 150 ALGEBRA. For, the first ten numbers are 1, 2, 3, 4, 5, 6, 1, S, 9, 10; and their squares are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100. Also, the square of 99 is 9801, of 100 is 10000, of 999 is 998001, of 1000 is 1000000, and so on. Hence, If a point be placed over every second figure in any integral number, beginning with the units? place, the points will show the number and order of the figures in its square rqot. 216. Every number, expressed by more than one place of figures, can be separated into two parts, tens and units. Therefore, if we denote the tens of a number by a, and the units by 6, the number will be denoted by a-\-b, and its square by (a 4- by = a^ -\-2ab-{-,V = a^ + (2 a + J) J. Then, by this formula, if a :^ 3 tens, or 30, and S = 6, we have 3 tens + 6 units = 30 -f 6 = 36 ; and 36^ = (30 + 6)^ = 30^ + (60 + 6) X 6 = 1296. Hence, Every number expressed by more than two places of figures is equal to the square of the tens in its root, plus the product of twice the tens plus the units, multiplied by the units. 1. Let it be required to find the square root of 4356. 120 + 6 4356 60 + 6] 3600 T56 156 OPERATION. or 4356 36 66 126 756 •756 EVOLUTION. 151 The root of the given number, according to Art. 215, consists of two places of figures. Let, now, a -j- i denote the root, where a is the value of the figure in the tens' place, and J of that in the -units' place. Then a must be the greatest multiple of ten whose square is less than 4300 ; this we find to be 60. Sub- tracting c?, that is, the square of 60, from the given num- ber, we have the remainder 756. Dividing this remainder by 2 a, that is, by 120, gives- 6, which is the value of h. Then {^a-\-h) h, that is, 126 X 6, or 756, is the quantity to be subtracted ; and as there is now no remainder, we conclude that 60 -|- 6, or 66, is the required square root. Had the root consisted of three places of figures, we could have let a represent the hundreds, and h the tens ; then, having obtained a and h as before, we might let the hundreds and tens together be considered as a new value of a, and find a new value of h for the units. Hence the following RULE. Separate the given number into periods, hy pointing every second figure, beginning with the units' place. Find the greatest sqvuire in the left-hand period, and place its root on the right ; suMract the square of this root from the first period, and to the remainder bring down the next period fr)r a dividend. Divide this quantity, omitting tJie last figure, by double the part of the root already found, and annex the result to the root, and also to the divisor. Multiply the divisor as it now stands by the part of the root last obtained, and subtract the product from the dividend. If there are more periods to be brought down, continue the operation in the same manner as before. If there be a final remainder, the given number has not an exact root ; and, since the rule applies equally to 152 ALGEBRA. decimals, we may continue the operation, by annexing periods of decimal ciphers, and thus obtain a decimal part, to be added to the integral part of the root already found. It will be observed that decimals require to be pointed to the right ; and if they have not an exact root, we may continue to form decimal periods, and obtain decimal fig- ures in the root to any desirable extent. As the trial divisor is necessarily an incomplete divisor, it is sometimes found that after completion it gives a product larger than the dividend. In such a case, the figure found for the root is too large, and one less must be substituted for it. The root of a common fraction may be found, it is evi- dent from Art. 204, by taking the root of both the numer- ator and denominator. Examples. 2. Extract the square root of 213529. 3. Extract the square root of 45196. 4. Extract the square root of 106929. 5. Extract the square root of 33.1776. 6. Extract the square root of .9409. 7. Extract the square root of f^f^. 8. Extract the square root of 5^|-. 9. Extract the square root of 2 to three decimal places. Ans. 1.414 -f. Here, the sign + is used to denote that the approxi- mate root is less than the true root. 217. To find the approximate root of an imperfect square, within a given fractional unit, we may Multiply the given number hy the square of thet denominator of the fraction determining the degree of exactness ; then extract Ans. 523. Ans. 214. Ans. 327. Ans. 5.76. Ans. .97. Ans, • li- Ans. • n- EVOLUTION. 153 the square root of the product to the nearest unit, and divide the result b>/ the denomitiator of the fraction. For, let a be any imperfect square, of which it is re- quired to find the square root to withia a fraction - ; then, since a may be put under the form of -, by multi- plying both terms of the expression by n", we have tan' a = — =-. n" Denote the entire part of the square root of a w* by r, and an" will be oomprised between r" and (/--fl)'; consequently — j- will be comprised between ?! and <^^+^ hence the square root of -rj;, that is of a, will be com- prised between - and "^ n n r »■ -I- 1 1 But the difference between - and ■ ' is -^ whence n n n r . . 1 - will be the root of a to within -. n n 10. Find the square root of 59 to within y^. 59 X (12)* = 8496 ; «/ 8496 ^ 92-f ; 92 -j- 12 = 1^%, Ans. 11. Find the eijuare root of 7 to witiiin y^. Ans. 2^6^. 12. Find the square root of 11 to within ^V- Ans. 3^. 13. Find the square toot of 12 to within M. Ans. 3.46. 154: ALGEBEA. 14. Find the square root of 5 to within .00001. Ans. 2.2360'r. 218. To find' the approximate square root of a common fraction, whose denominator, only, is a perfect square, we may, obviously. Take the approximate square root of the numerator, and divide it hy the square root of the denominator. 15. Find the square root of ^ to within \. ^-^ = T, Ans. V' 25 5 16. Find the square root of f to three decimal places. ' )/!, 2.236+ ^._ . i= = —s-^ = .U5, Ans. y/ 9 ^ IT. Find the square root of -^^ to five places of deci- mals. Ans. .20203. 219i The fraction may be changed to a decimal when its denominator is not a perfect square ; or we may mul- tiply both terms of the fraction by any number which will make its denominator a perfect square, and proceed as in the last Article. 18. Find the square root of -f^ to within ■^. 9 15 135 _ v' 135 12 15 ^ 15 225 ' y/ 225 15' 19. Find the square root of f to within -J, and to within hV- 3 5 15 _ v^lS 4, 5 X 5 — 25 ' y/-25 ~ 5 ' 15 /12y 2160 v' 2160 46 25 ^ \12) 3600' ^/36oo 60" J - Ans. 20. Find the square root of f to within j'g-* Ans. ||. 21. Find the square root of -^f to four decimal places. A ns. .8044. EVOLUTION. 155 CUBE EOOT , OF NUMBEES. 220. A Perfect Cube is any number or quantity that can be resolved into three equal factors (Art. 199) ; and, consequently, has an exact cube root (Art. 210). 221. If a number be separated into as many groups, or periods, of three figures each, as is possible, from right to left, these groups will correspond, respectively, to the units, tens, hundreds, etc., in its cube root. For, the first ten numbers are 1, 2, 3, 4, 5, 6, % 8, 9, 10; and their cubes are 1, .8, 2r, 64, 125, 216, 343, 512, Y29, 1000. Also, the cube of 99 is 970299, of 100 is 1000000, of 999 is 991002999, of 1000 is 1000000000, and so on. Hence, If a point he placed aver every third figure in any integral number, beginning with the unitS place, the points will show the number and order of the figures in its cube root. 222. Since every number expressed by more than one place of figures, can be separated into two parts, tens and units, If we denote the tens of a number by a, and the units by b, the number will be denoted by a-\-h, and its cube by (a + 5)3 = o3 _|_ 3 ^2 J _|_ 3 a 52 _|_ j3 _ „s _j_ (3 ^2 _|_ 3 „ 5 _j_ }2jj_ Then, by this formula, if a = T tens, or TO, and 5 = 4, we have T tens + 4 units = fO -|- 4 =; U, and t4» = (10 +4)» = 10'+ (14100 + 840 + 16) X 4 = 405224. Hence, 156 ALGEBRA. Eoery number expressed hy more than three places of figures is equal to the cube of the tens in its roof, plus the sum of three times the square of the tens, three times the product of the tens and units, and the square of the units, multiplied hy the units. 1. Let it be required to find the cube root of 405224. OPERATION. 405224 343 14T00 840 16 15556X4 = T4 62 2 24 62224 The root of the given number, according to Article 221, consists of two places of figures. Let, now, a-\-b denote the root, where a is the value of the figure in the tens' place, and b of that in the units' place. Then a must be the greatest multiple of ten whose cube is less than 405000 ; this we find to be "TO, or 1 tens. Subtracting a', that is the cube of TO, or 343000, from the given number, we have the remainder 62224. Dividing this remainder by 3 a^, that is, by three times the' square of 10, or 14700, we obtain the value of b, or 4. Then (3 a'' + 3 a 6 + 52) j^ that is, (14700 + 840 + 16) 4, or 15556 X 4 = 62224, is the quantity to be subtracted ; and as there is now no remainder, we conclude that 70 + 4, or 74, is the required root. For brevity, instead of writing 70 in the root, we- may simply write 7 in the tens' place ; and for the cube of 70 write the cube of 7, or 343, without the ciphers, under the proper period. EVOLUTION. 157 Had the root consisted of three figures, we could have let a represent the hundreds, and h the tens ; then, having obtained a and h as before, we might let the hundreds and tens together be considered as a new value of a, and find a new value of h for the units. Hence the following RULE. Sepairate the given number into periods, ly poirtting every third figure, beginning with the units' place. Find the greatest cube in the left-hand period, and place its root on the right ; subtract the cube of this root from the first period, and to the remainder bring down the next period for a dividend. At the left of the dividend write three times the square of the root already found, for a tried, divisor; divide the dividend, omitting the last two figures, by it, and write the quotient for the next figure of the root. Add together the trial divisor, with two ciphers annexed ; three times the product of the last root figure by the rest of the root, with one cipher annexed ; and the square of the last root figure; and the sum will be the complete divisor. Multiply the Somplete divisor by the last root figure, and sub- tract the product from the dividend. If there are more periods to bring down, continvie the opera- tion in the same manner as before. The observatioiis made under the rule for the extraction of the square root (Art. 216) are equally applicable to the extriaction of the cube root. EXAUFLBS. 2. Extract the cube root of 8.144865728. 14 158 ALGEBRA. OPERATION. 8 2.01 2 120000 600 1 120601 X 1 = 12120300 12060 4 12132364 X 2 = 144865 120601 24264T28 24264728 Here, it will be observed that, in consequence of the in the root, we annex two additional ciphers to the trial divisor, 1200, and bring down to the corresponding divi- dend another period. 3. Extract the cube root of 186086T. Ans. 123. 4. Extract the cube root of .'724150792. Ans. .898. 5. Extract "the cube root of 1481.544. Ans. 11.4. 6. Extract the cube root of ■^^ViVs'' Ans. f^. 1. Extract the cube root of -^ to four decimal places. Ans. .5609+. Here, it is convenient to change the given fraction to a decimal, and then take the required root. EOOTS OF MONOMIALS. 223. From the principles contained in Articles 209-213, we obtain the following ETJLE. Extract the required root of the numerical coefficient, and divide the eiyaonent of each letter ly the index of the root, making EVOLUTION. 159 the sign of every £ven root of a positive quantity ±, and the sign of every odd root of any quantity the same as that of the quantity. If the given quantity is a fraction, it follows, from Art. 204, that we may take the square root of both its terms for the required root. * Examples. 1. Find the square root of 9a*¥<^. OPERATION. A^QaH^c^ = 9^ X a^ X b^ X c^ = ± Sa^ic', Ans. 2. Find the cube root of — 125 s^i^. Ans. — 5xy^. 3. Find the fourth root of 81 a* J^. Ans. ±3a¥. 4. Find the fifth root of — -— — . Ans. — - — . 24o o 5. Find the square root of 9a*i^e-^^. Ans. ± SaHc-". 6. Find the square root of 625 a^ e. Ans. ± 25 a« c*. Here, as we can not extract the root of c, its root is indicated in the result by the fractional exponent. J. 1. Find the with root of a h""". Ans. a"* J-". 8. Find the cube root of —Sa-n^x-". Ans. — 2a~^b^x~^. 9. Find the value of (16 a-''"+^ a-^")^ . Ans. ± 4 a;"'+^ a-'. 10. Kequired the value of (Sa'^^'^d^y. 1 m 2 Ans. ^ra'h^d". 160 ALGEBEA. 1 ' *i 11. Required the with root of —„(j>: — y)». Ans. a "' (x — y)\ 12. Required the square root of (a + xf V^ x /. Ans. ± (« -j- iB) i ar* y^. SQUARE ROOT OP POLYNOMIALS. 224. The manner of forming the square of a polynomial must, by reversing the process, lead to the discovery of its root. If we take any binomial, as a -\-i, we have (a-f J)« = a'' + 2fflJ + J«; and the last two terms of this fixpressioa factored give (2 a -f- 5) h. 1. Let us now reverse the involution, and discover how the root a -\-h may be derived from the square d'-{.2ah-\-W. OPERATION. The square root of the ^2 is the first term of the r ' required root. Subtract- T^ "^ ing the square of a from " ' the given polybomial, we have 2 a J 4" *^ ^^ (2 a + J) b, for a remainder or dividend. Dividing the first term of the dividend, 2ab, by 2 a, which is double the first term of the root, we obtain b, the other term of the root, which, connected to 2 a, completes the divisor, 2a-\-b. Multiplying this divisor by the last term of the root, b, and subtracting the product, 2ab -\-b% from the dividend, we have no remainder. By a like process, a root consisting of more than two EVOLUTION. 161 terms may be found from its square, since all such roots can be expressed in a binomial form. Thus, a -\-b -\- c = {a -\- h) -\- c, and its square, a2_^2ffl& + 6'' + 2ac-^-2Jc4-'»' — ^^'"* — ^^■ Ans. m^ — 2w — 4. 168 ALGEBRA. 3. Extract the square root of a* — 2 o^ a; + 3 a' a;" — 2aa^ -\-x*. Ans. a^-^ax -{- o^. 4. Required the fifth root Qf 32 as? — 80 oJ* + 80 a? — 40a^4- lOo:— 1. Ans. 2a;— 1. 229i When the index of the required root is a multiple of two or more numbers, we may obtain the root by siteces- sive extractions of the simpler roots. For, Since (Art. 201) reversing the involution, we have 5^"^ =; ^"fS; or. The ranth root of a quantity is equal to the, m/~b. 5. Find the sum of 3 Vl and 2 V^. Ans. IVT^ 6. Find the sum of (¥yY and (hff. Ans. Q,^y)(hy)^. 7. Add 8^1, — i-v/~12, 4V''27, and — 2 x/"^. Ans. 14^ VT. 8. Add ^/48aJ^ a/ 75a', and \/3a(a — 9J)». Ans. (6 a — 5 5) x/^- 9. Eequired the sum of iy ", and — 1/^6" Ans: (3«-l)^"||. 10. Eequired the sum of 3 (242 a" 5«)^ and 11 (2c^V)^. Ans. 11 {Za'W-^-ab) {2ab)^. 11. Eequired the value of v'^ + 2 \/72 + a */bo?. Ans. 2^^ + 12\/"2 + ax\/b. 12. Eequired the value of 4^ 125 b* — 250 a'i' + 4/ 64 a' 6— 128 a*. Ans. (56 + 4a) ^ 6 — 2a. SUBTEACTION OF EADICALS. 242. When two radicals are similar, the common radi- cal part, with the difference of their coefScients, will con- stitute their difference. But, while they are dissimilar, their difference can only be indicated. RADICALS. 177 Hence the following RULE. Reduce each radical, if necessary, to its simplest form. If, then, the radicals are similar, find the difference' of the coefpr- dents, and to the result annex the common radical; but if they are dissimilar, indicate the subtraction by the proper sign. Examples. 1. Find the difference between >s/lQiaa? atxA t^4Aaa?. OPERATION. ^/ la^aa? = VSCa:^ X 3a = 6a: \/ 3 a V i&aa? = V' 16a;^ X 3a = 4 cc y/Ya Difference = 2 a; V 3 a Find the difference between 2^320 and 3-^40. Ans. 2^T. 3. From n,/ 4caoi? take Zic y/ 2 a. Ana. — 1 x is/ a. 4. Find the difference between^! and ■•J^^^. Ans. -ft^f^- hi Find the difference between ^'4^a^b and f/^a^J. 6. Simplify 4 ^/ 1 + a;" — ii^^~^l. ' Ans. 4 (V a;' + 1 — 4^x^+1). 7 . Simplify 8 6 (2 a" ff")* _ Y (2 b= J")^ + 8 a (2 a^ ¥)^. Ans. 4o5(2a?iS)^. 8. Eequired the difference between m i/l-\-l — \ and „ jT+(fy . Ans. (m* - J) ^mi + nK 178 ALGEBKA. MULTIPLICATION OF EADICALS. 243i The multiplication of radicals depends upon the principle, that The 'product of like roots of two quantities is equal to the same root of their product (Art. 212). That is, let a and b be any two quantities ; then we have _ i 1 i, iiyab = i^aX'iyb, or (a 5)" = a" X *", and, conversely, _ j_ j_ 1^ X^aX^b = i^ab, or a" X *" = (« *)"• Therefore, if c ill/ a and dis/b be two radicals with coefBcients and a common index ; then (Art. 63), c ^~a X diliTb = cX dX <^^ X /l/b = cdii/ab. Hence the RULE. Reduce the radical parts, if necessary, to those having a common index ; then multiply the coefficients together for the coefficient of the product, and the parts under the radicals for the radical part. Examples. 1. Multiply 3 y/~2a by 2 ^Ta. OPERATION. .8/ 3^/2a X 2^ 3a = i^^^^c? X 2i^^''a\ = 3X2 4/2'a'X S^a'' = 6aJ/T2^ 2. Multiply Xy^ by i^yl. Ans. ^"rf^. EADICALS. 179 3. Multiply 5 a* by 3 a*. Ans. It >^~a^. 4. Multiply 4V^ by 3 \/^- Ans. 24 V^- 5. Multiply i4/1 by f ^f Ans. ^ ^"l5. 6. Multiply (a + i)* by (a 4-5)*. Ans. ^ (a + bf\ 1. Multiply h-^i^lir^ by a^b^c. C 12/ Ans. T3 V a 4^- 8. Multiply 4«y/|| by ^a^± Ans. |f. 244i When either or both of the radicals are con- nected with other quantities by the sign -|- or — , each term, of the multiplicand must he multiplied by each term of the mvMipUer (Art. 68). 9. Multiply a + 2 a^I by a — V^- FIRST OPERATION. SECOND OPERATION. a -\- 2»/b a +2 J* a — \/& a — Ifi c?-\-1ais/b a=+2aJ^ — as/b — 2i>/W — a^ — 1li a^-l- a\/6 — 25 a^— ab^ — %b 10. Multiply x — s/xy-{-y by y/x-\-»>/y. Ans. X nj X -\- y i>/ y. 11. Multiply n/a + c 4/b by \/« — ci^l>. Ans. a — / 6 be two radicals with coefficients and a common index ; then erf V ah dy/1 Hence the cd „/ab „,- RULE. Seduce the radical parts, if necessary, to those having a common index ; then divide the coefficient of the dividend by the coefficient of the divisor, and the radical part of the divi- dend by that of the divisor, and prefix the first quotient to the last, written under the common index. Examples. 1. Divide \&a^i^~9^ by ^a»/b. OPERATION. 16 o' ^ 'V _ 16 g' ^"6« _ 16 a= e/T* _ g^ RADICALS. 181 2. Divide 6 ^"96 by 3 ^"8. Ans. 4 y/~B. 3. Divide 4.4/12 by 2 ^"Ts. Ans. 2 ^"I. 4. Divide \/ 96 a« by 4 a. Ans. \/ 6. 5. Divide (ar'/)* by (a;V)^- Ans. {3?f)~^. 6. Divide 2 ^Tc by i »^~ac. Ans. ~4^1^¥?. Sac ^ LI. L 7. Divide bc{c^x^)" by c (ax)". Ans. J (a'^a;)". 8. Divide \/ aa^ hy \/ bx. ■ ^ # — t 10. Divide \/~20 -\- \/l2 by V^ + \/¥. Ans. 2. 11. Divide a-\-b — c -\- 2/^/ ab by \/ a-]-*/ b — a^c. Ans. \/ a -|- a/ J -1- 'V' "• 12. Divide 4^ cf I/' -{- aH* by ^' 'a? — V »b Ans. 2«J^^. POWERS OF RADICALS. 246i Involution of radicals depends upon the same general principles as involution of rational quantities. That is, let a represent any quantity ; then, we have the wth power of \^a, or (S/d)" = Xy'a X i>/a X \^'a to w factors =-^a", or, ( ^"j = a" X a"" X o"* to n factors = a"". 16 182 ALGEBRA. Therefore, it b Jit/ a be a radical with a coeflacient, then (Art. 200), (i Xya)" = 5" X v^'a" = J" v'"""- Hence the RULE. liaise the coefficient, if any, to the required power, and write the radical pari, raised to the same power, under the given If a quantity have fractional exponents, the involution - may be performed by multiplying them by the exponent of the required power (Art. 203). If the radical has quantities connected by -|~ or — > perforija the involution by multiplication of the several terms. Examples. 1. Eaise fv^S to the third power. OPERATION. (I V'sy = (I)' X V~s'' = ^ V"2T = jVV9 X 3 = f\/3 Ans. 2. Eaise 9 x /t^ 3 abf to the second power. Ans. 81bx^4^9aH. 3. Eaise li/ aW to the second power. Ans. h s/ a. 4. Eaise ' ^ \/ 3 to the third power. Ans. \ s/Z. 5. Eaise — is/ a? to the fourth power. Ans. a^/^/a^. 6. Find the square of 2 + V sT Ans. 7+4 v' 3. 7. Find the cube of a — hx^. Ans. a^ — Zd'hx^-^ZaVx — Vx^. 8 Find the cube of 2 y/ a — h. Ans. 8 (a — b) i>/ a — h. RADICALS. 183 ROOTS OF RADICALS. 247i Evolution of radicals depends upon the same general principles as evolution of rational quantities. That is, let a jepresent any quantity ; then, by Art. 229, we have for the »»th root of the wth root of a, \l nt ffwi/ ml — tiy a =. Ill/ a, or 4 / a» = amn. Let it now be required to find a root of a radical with a coeflScient, as the cube root of 8 \/ a', or the square root of 20 v' 5 a ; then, by Art. 212, we have Also, by Art. 236, V'20V5a = V4X-5v'5ai; and \/4: X v'SV'Sa =: 2v'\/125a = 2 4'125a. Hence the RULE. Eoctract the required root of the coefficient, if any, when a complete power of the required degree, otherwise introduce it under the radical sign. Extract the required root of the quantity under the radioed sign, when a complete power of the required degree, otherwise multiply the index of the radical by the index of the required root. When only a factor of the coefficient is a complete power of the same degree as the root required, take the root of that factor, and introduce the other factor under the sign. When a quantity is affected by a fractional exponent, its evolution may be performed by dividing this exponent by the index of the required root (Art. 223). 184 ALGEBRA. Examples. 1. Find the cube root of the square root of a. Ans. ^ a. 2. Find the cube root of 64 ^T. Ans. 4 ^T. 3. Find the square root of c^ i^ 5. Ans. a i^ b. 4. Find the square root of 24 v' 3 a. Aps. 2 4/ 108 a. 5. Find the cube root of {x •\- y) s/ x -\-y. Ans. >^ X -^-y. 6. Find the cube root of t i / 5- -^i^s. \s/ ba. 7. Find the fourth root of 2 /^ a^ — x. Ans. /^ 4 (o» — a:). 8. Find the mth root of 5 (a + 5)*. 1 j_ Ans. 6'" (rt + 5)2™. 9. Find the square root of 9a + 36 (3aa;)^ + 108a;. Ans. 3a^ + 6 (3a;)t EATIONALIZATION. 248. Eationalization is the process of removing the radical sign from a qu^intity by multiplication. It is often useful to transform a fraction having an irrational denominator, into one whose denominator ' is rational. This is especially advantageous when the nu' merical value of the quantities is required. 249. -A factor may he found which will rationalize any monomial. EADICALS. 185 The quantity \/«, or a*, is made rational by multiplying it by, \/a, or o*. For, ^ a X v'« = «» or a^ X o^ ^ «• Also, i^a, or a* is made rational by multiplying it by (fi. For, a* X a^ = «• Likewise, a" is made rational by being multiplied by a^. For, a* X «^ = «• That is, y^e required factor is the same as the given quantity, but with such a fractional exponent as, when added to the fractional exponent of that quantity, shall he equal to an integer. 250> -4 factor may be found which will rationalize any binomi>/ a -\- i^ b, or a^ -\- 5 . Here, since w ^ 6, an even number, we have from (1), x" w" (fi JS Sl 41 32 23 14 5 ^-— ^ = Vr I = a* — a^S^ + a^ i^ — a^ J^r i „i 57 _ jt ^ + 2/ a* + 6* = fll _ a2 5^ + a* i* — a J + a* F — i*, the rationalizing factor ; and (a* + b^) {J — aHi-\- J b^ — ai + a^b^ — 5^) = a* _ jf = a^ _ J2, the rationalized result. 251. A trinomial in the form of \/a-{-\/b-{-\/c may be transformed into a binomial form by multiplying it by RADICALS. 187 itself with the sign of one of the terms changed, and then rationalized. Thus, (,\/a-|-\/6-j- V'c) Qs/a-\-/^b — s/c) = a-\-h — c-\-2\/ab, (a-|-J—c+2\/a6) (a4-6—c—2\/«5) = («+&— c)^—4a5, or a''-\-¥-\-e' — 2ab — 2bc — 2ac, a rational result. Examples. 1. Find a factor which will rationalize 3 \/ a". Ans. /s/ a. 2. Find a factor which will rationalize a -\- \/ b. Ans. a — \/ b. 3. Eationalize ^ 5 -j- ,^ 3. Ans. 5 — 3, or 2. 4. Find a factor which will rationalize f^ (a — 6)". Ans. ^ (a — b)\ 5. Find a factor which will rationalize a^ — ■ 6*. Ans. ai- + a=5* + a*J^4-aJ=' + a^6^ + 6^. 6. Eeduce -^ to a fraction whose denominator shall be y/ a rational. 4 X ^ = *-^, Ans. Y. Eeduce j= to a fraction whose denominator shall be y/ a rational. j^^^ ^ V "' ^ a 8. Eeduce ^= tc a fraction whose denominator shall be y' a rational. ^g, ^ V' °"~' . 188 ALGEBRA. 3 9. Keduce -^= = to its simplest form. y/S — v/2 Ans. v'5+\/2. 10. What fector will simplify -^=z =e ? V'i + V'c Ans. ^J.=^l ^b — ^c 11. Rationalize \/8 — \/l — v'S. Ans. 4. 12. Reduce ^ ' to a fraction 'whose de- nominator shall be rational. . a — \/ a^ — ^ X 13. Find the approximate value of -r= to three V'S + l places of decimals. 1 y/l — 1 /3 —^ t/1 — 1 ^ : — v/3 + 1 " v's — 1 ^"^ ^ = hlllpl = .566, Ans. 14. Find the approximate value of — - — = to four places of decimals. Ans. .8918. IMAGINARY QUANTITIES. 252i An Imaginary Quantity is an indicated even root of a negative quantity. Thus, \/ — 4 and v — "^ are imaginary quantities. In contradistinction, all other quantities, , rational and irrational, are called real quantities. RADICALS. 189 253a When an expression consists of a real part joined with an imaginary part, the whole is spoken of as imagi- nary, on account of the presence of the -latter. Thus, a -\- b \/ — 1, considered as a whole, is imaginary. 234. Let a denote any real quantity ; then the square roots of — a^ will be expressed by ± V — V. Now, — a" may be considered as the product of o° and — 1 ; so if we suppose that the square roots of this product can be formed, in the same manner as if both factors were positive, by multiplying together the square roots of the factors, we have ± /\/ — a? = ± \/a^ X V— 1 = ± a V— 1, in which V' -^ 1 is the imaginari/ factor, and a its coeffi- cient. Thus, we may regard v' — 1 as a universal factor of every imaginary quantity, and use it in our investiga- tions as the only symbol of such a quantity. 255t Imaginary quantities may be added, subtracted, and divided the same as other radicals, but with regard to multiplication, the common rule requires some modifi- cation. If we take the product of two imaginary quantities in which the imaginary parts are equal, it is evident that the sign of the product is changed by removing the radical. Thus, b\^ — a X C'V'' — *" = Jc(^-a) == — ahe. But if we take two' unequal imaginary quantities, \/^-« and V' — i>, by the common rule (Art. 243) we have (\/^-a) (yy/ — b) =3 \/ab. Now, since the quantity whose root is to be extracted was not produced by that root, but from two unequal facr 190 ALGEBRA. tors, it does not immediately appear whether the result obtained is to be taken positively or negatively. We may, however, resolve the imaginary quantities into two factors, of which one is \/ — 1 (Art. 254). Then we have (+^/-a) (+ V-6) = (+Va X V-l) (+A/i X V-l) = -ir\Uh X {sf—^y = J^^'abX (—1) = — ^ ah. Hence it appears that the result is properly negative. In like manner, it may be shown, that ( — \/ — a) ( — \/ — *) = — * = x^-\-2xik/y-\-y. ■ (1) By Art. 262, a = s?-[-y, (2) and \/* = 2a;\/y. (3) Subtracting (3) from (2), we have a — \/l ^=^ a? — 2xs/y -\- y, (4) whence, ^ a — t^b =. x — ^ y. (5) In like manner, it may be shownj if Sj a-\-y/b = is/x-\-i»/y, then \J a — /^l = i>/x — s/y. 264. In the solution of certain equations, it frequently becomes necessary to extract the square root of binomial surds in the form of a ± \/&. 265. To obtain a rule for the operation, when possible, assume Va + V6 = Va: + Vy, (1) where one or both terms of the second member are irra- tional. But by Art. 263, \la — is/b = s^x — i>/y. (2) Multiplying (1) and (2) together, we have 1^ a^ — b = x — y. (3) Squaring equation (1), we obtain « + V* = x-\-2 f/l^^y; whence, by Art. 262, a =^ x-\-y. (4) Adding and subtracting (3) and (4), and factoring, EADICALS. 195 y = "^^J^^. (6) Let, now, a* — 6 be a perfect square, whose root is c ; and substituting in (5) and (6), we have a-\-c , a — c X = —^, and y = -^-. Substituting these values of x and y in (1) and (2), we obtain Hence, the square root of a binomial of the form of a ± is/h may be found, when a^ — h is a perfect square, and Cl) and (8) are formulas which apply to any par- ticular example, by substituting the particular values for a, b, and e. Examples. 1. Eequired the square root of 3-|-2v'2. Here a = 3, -%/! = 2 >/2 = \/"8, or b = 8, and = ^/9 — 8 = 1- Hence, ^3 + 2V2 = ^i±i + y/i=i=V2+l, Ans. 2. Eequired the square root of 11 -|- V"72. Ans. 3 4- V 2". 3. Required the Square root of 6 + v' 20. Ans. 1 rf- ■%/ 5. 4. Eequired the square root of 1 + 4 v' — 3. Ans. 2 + ^ — 3. 196 ALGEBBA. 5. Eequired the square root of 43 — 15 \/ 8. _ Ans. 5 — 3v'2. 6. Required the square root of 12 + 2 \/ 35. Ans. tj Ans. \/5 + 'V^'f- t. Eequired the square root of 2)n + 2 \/ »»* — »*. Ans. /^ m -\- n -\- /»/ m, — n. 8. Required the square root of — 2 \/ — 1. Here a is equal to 0, and the required root is 1 — »/ — 1. 9. Required the square root of r + s V «* — *'• Ans. - + -^isf a^ — l?. RADICAL EQUATIONS. 266i Radical Equations are those containing radical quantities. 367< The solution of a radical equation consists in rationalizing the terms containing the unknown quantity, and in determining its value. The following examples containing radical quantities re- duce to simpk equations. 1. Given t,/x-\-l=y/ x — 3 + 2, to find the value of x. OPERATION. s/x + 1 = v'a; — 3 + 2 Transposing and uniting, \^ x = /^^ x — 3-f-l Squaring, x = x — Z -\-2k/ x — 3 + 1 Or, X = x — 2-\-2 V« — 3 Transposing, 2 = 2 n/ x — 3 Dividing by 2, 1 = \/ « — 3 Squaring, 1 = a: — 3 Whence, a: = 4 RADICALS. 197 2. Given ^^=^ — = i=r-i^ — , to find the value of x. OPERATION. Clearingoffractions, a;+42v'«+152 = x-{-3it>= (a;^ _|_ lOa: + 1)^^ to find x. Ans. a: = ^. 11. Given \/ x — 32 = 16 — a/'x, to find a;. Ans. X = 81. 12. Given i/-^- + i/-^ = d 4^, to find x Ans. X = a (|±i). 13. Given a-\-x = ^ a" -\- x v' S" + a;", to find a:. Ans. a; = a. 4a 14. Given (a;^ ^Ba')i—(J — 3 a^)^ = ?-^ , to find X. . 9(i'V Ans. a; = 4 (6 — ay QUADRATIC EQUATIONS. i, A Quadratic Equation is an equation of the second degree (Art. 146), or one in which the square is the highest power of the unknown quantity ; as, aa? = h, and ai'^^ 8a; = 20. Quadratic equations are divided into two classes, Pure and Affected. 200 ' ALGEBKA. PUEE QUADEATIC EQUATIONS. 269t A Pure Qtjadbatic Equation is one which contains only the square of the unknown quantity ; as, ex^ = ab, and x^ = 400. Equations of this kind are sometimes called incomplete equations of the second degree. 270i A pure quadratic equation can always, by trans- position and reduction, be exhibited under the general form x^ = a, in which a may represent any quantity, real or imaginary, positive or negative, integral or fractional. Thus, the equation _ 50=^-4 = — + 2-, by clearing of fractions and transposing, reduces to x^ = 16. Equations of this kind, therefore, have sometimes been denominated binomial, or those of two terms. 271. A binomial quadratic equation in the form x' = a (1) is readily solved, the value of x being evidently found by taking the square root of both members. Thus, X = ± Va, (2) where the double sign is used because the square root of a quantity may be either positive or negative (Art. 213). The sign of x may also be negative, but still x will* be either equal to -\-\^a, or — ^a. Either of these roots substituted in equation (1) will satisfy it. Hence QUADRATIC EQUATIONS. 201 A pure quadratic elation has two roots, equal in numerical value, but with opposite signs. 1. Given Sx" -{- 1 z= ^-^ -\- 35, to find x. OPEKATION. Clearing of fractions, 12 a:" + 28 = 5 a:' + 140 Transposing and uniting, *! x' = 112 Or, x^ = 16 Extracting the square root, x :=■ ±4. From the foregoing illustrations is deduced the following RULE. Seduce the given equation to the form x^ = a, and then extract the square root of both members. EXAMFLKS. 2. Given 4 a!* — 1 = 29, to find the values of x. Ans. a; = ± 3. 3. Given Sa;^ + 5 = 3a;* + 55, to find the values of x. Ans. a: ^ ± 5. 4. Given 1x^ — 6 = Sa;* -|- 11, to find the values of x. Ans. a: = ± 2. 5. Given ■^—, 1- r — ^ = ^i to find the values of x. 4 -J- W 4 — X 3 Ans. ar = ± 1. 6. Given a? — ab ■=■ d, to find the values of x. Ans. ar= ±. isf d-\- ab. If Given ax^-{-n = m — c, to find the values of x. Ans. X Im — c . — n 202 ALGEBRA. 8. Given J - 3 + ^ = ^ - -^ + -|- to find the values of x. Ans. a; = ± 3. 9. Given 13 — \/ 3 a;^ + 16 = 5, to find the values of X. Ans. a; = ± 4. 2a 10. Given x 4- is/ x^ -\- a = , to find the values of X. Ans. X = ± i\^ 3 a. 11. Given ^^=r == = ^, to find the 1— \/l— K* l + y/l— 2^ ar" values of a;. Ans. x = ± ^. 12. Given ^- — ^ \- - — — — = i/ 7, to find the values of a;. Ans. x= ±v'24a — 144. 272i Simultaneous Equations (Art. 163) sometimes pro- duce pure equations after elimination. The methods' of elimination are the same as in simple equations. IB. Given x^ — 9y^ = 63,* and | = — 2y, to find the values of x and y. OPERATION. 3^—9f= 63 (1) I = -2y (2) Clearing (2) of fractions, x ^ — 4y (3) Squaring (3), 7? = \Qf (4) Substituting the value of a;= in (1), 16^—9^' = 63 (5) Or, If = 63 Whence, / = 9 Extracting the square root, y = ± 3 Substituting the value of y in (3), a; = =F 12. 14. Given 3x-{-Sy = 5x, and (a; -|- y) a; = 960, to find the values of x and y. Ans. a; = ± 24 ; y = ± 16. QUADRATIC EQUATIONS. 203 15. Given Bx" -{- ii/ = 1Q, and 3/ — 11 a;' = 4, to find the values of x and y. Ans. a;=±2;y=±4. 16. Given a^ — xif = 54, and xy — y^, = 18, to find the values of x and y. Ans. x = ± 9 ; y = ±3. 17. Given x^ -{-xy = 60, and xy-\-y^ = 84, to find the values of x and y. Ans. x = ± 5 ; y ^ ±1. 18. Given x^ — f = 2268, and ^ = 28 y, to find the values of x and y. Ans. a: = ± T2 ; y ^ ±54. PKOBLEMS LEADING TO PURE EQUATIONS. 1. Bought a lot of flour for $ 115, and the number of dollars per barrel was to the number of barrels as 4 to T. How many barrels were purchased, and what was the price of each ? SOUITION. Let X = the no. of dollars per barrel, 7 X then -r = the number of barrels. 4 Therefore, ^ = 1'75 Clearing of the fraction, 7 a;' = 700 Or, a? = 100 Extracting the square root, x = 10, no. of dollars per barrel, 7 J. whence, — =: 17 J, the number of barrels. 2. I have three square house-lots, of equal size. If I were to add 193 square rods to their contents, they would be equivalent to a square lot whose sides would each measure 25 rods, Required tl^e length of each side of the three lots. Ans. 12 rods. 204 ALGEBEA. 3. A and B lay out money on speculation ; the amount of A's stock and gain is $ 27, and he gains as much per cent, on his stock as B lays out. B's gain is $ 32 ; and it appears that A gains twice as much per cent, as B. Ee- quired the capital of each. Ans. A's capital, $ 15 ; B's, $ 80. 4. There are two square fields, the larger of which con- tains 25,600 square rods more than the other, and the ratio of their sides is as 5 to 3. Eequired the contents of each. Ans. The larger, 40,000 square rods ; the smaller, 14,400 square rods. 5. Bought sugar at such a rate, that the price of a pound was to the number of pounds as 4 to 5. If the cost of the whole had been 45 cents more, the number of pounds would have been to the price of a pound as 4 to 5. How many pounds were bought, and what was the price per pound ? SOLUTION. Let X = the no. of pounds, and 7/ = the price per pound. Then a;y = the whole cost. Therefore, y = — - (1) And X = —^ (2) Or, Sar" == 4a;y+ 180 (3) 16 ar* Substituting value of y in (3), Sa;'' = — 1- 180 Or, 25a^ == 16a:=+900 Transposing and uniting, 9 a:'' = 900 Or, ^ — 100 Extracting the square root, x = 10, the no. of pounds. Substituting the value of a; in (1), y = 8, the price per pound. 6. A and B engage in a speculation. A disposes of his share for $ 11, and gains as many per cent, as B QUADRATIC EQUATIONS. 205 invested dollars. B's gain was $36, and the gain upon A's investment was 4 times as many per cent, as upon B's. How much did each invest ? Ans. A invested $ 5 ; B, $ 120. 7. What two numbers are those whose difference mul- tiplied by the less is 42, or by their sum is 133 ? Ans. ± 18 and ± 6. 8. Two workmen, A and B, engage for a certain num- ber of weeks, at different rates. At the end of the time, A, who had been absent 4 weeks, received $ T5 ; but B, who had been absent t weeks, received only $48. Now, if B had been absent only 4 weeks, and A 7 weeks, they would have received exactly alike. How many weeks were they engaged for, how many did each work, and what had each per week ? Ans. Engaged for 19 weeks ; A worked 15 weeks, B 12 weeks ; A received $5, B $4, per week. Here, the conditions lead to the equation 25 (x — *IY = 16 (a; — 4)^, which, by taking the square root of each member, gives 5 (a; — 7) = 4 (a; — 4). AFFECTED QUADEATIO EQUATIONS. 273i An Affected Quadratic Equation is one which con- tains both the second and first powers of the unknown quantity ; as, a? -\- ax = h, and aa? -\-bx — e z= hx^ — ax-{-d. Equations of this kind, containing every power of the unknown quantity from the first to the highest given, are sometimes called compUle equations. 274. Any affected quadratic equation can always, by reduction, be exhibited under the general form a?-\-px = q, 18 206 ALGEBRA. where p and q are understood to represent any quantities, positive or negative, integral or fractional. For we can reduce it to this form, by bringing all the terms containing the unknown quantity to the first ^em- ber of the equation, and the known quantities to the other, and then dividing by the coefficient of a?. Equations of this kind, therefore, have sometimes been denomipiated trinomial, or those of three terms. 275i To solve an affected quadratic equation, it is obviously required to take the square root of its members. But the member containing the unknown quantities alone cannot be a perfect square (Art. 224), nor can it often be so rendered by transposition. It is therefore necessary to have recourse to methods of completing its square. FIRST METHOD OF COMPLETING THE SQUARE. 276. Suppose we have given any quadratic equation, a? -\- px = q, in which it is required to find the values of x. Adding l-\ to both members (Art. 90), we have where the first member is a complete square. Extracting the square root, we obtain ^+l = ±\/»+?. Transposing, x = —^ ± U q -\-^. Here, since px must be twice the product of the two terms of the root of the completed square (Art. 90), QUADRATIC EQUATIONS. 207 and X being one of those terms, -^ must be the other. We therefore made x''-\~px a perfect square by adding to it (f) I or f> which had also to be added to the second member to preserve the equality. The double sign is used in the result, since the square of both +y/? + |^ and -y^^+f is +(? + ?): consequently, Emery quadratic equation may have two roots. 1. Given -2 a? + 12 a; -f 36 = 356, to find the values of X. OPERATION. 2a? +12* + 36 = 356 -By transposition, 2 a;^ + 12 a; = 320 Dividing by/ 2, a? + 6 a; = 160 Completing the square, a;'' -|- 6 a; -|- 9 = 169 Extracting the square root, a; -|- 3 ^ ±13 Transposing, a; = -^ 3 ± 13 Whence, x = 10, or — 16. Here, taking 13, the positive root of 169, we find a; = 10 ; but taking — 13, the negative root, we find ar = — 16. 2. Given x'^ — 12 a; -|- 30 = 3, to find the values of x. OPERATION. a?— 12a; + 30 = 3' By transposition, a? — 12 a; = — 2T Completing the square, a? — 12 a; + 36 = 9 Extracting the square root, x — 6 = ±3 Transposing, a; = 6 ± 3 Whence, a; = 9, or 3 ; where both values of x are positive. 208 ALGEBRA. 3. Given — Sar*— Tx = Jj"-, to find the values of x. OPERATION. — 3a:''— 7a: = Af Dividing by — 3, of -\-\x ^ — ^ Completing the square, a:^ + Ja; -|- || =7^5- Extracting the square root, x-\-\ =. ± f Transposing, a; = — J ± f Whence, a; = — §, or — | ; where both values of x are negative. The results obtained in each of these operations may be readily verified. Hence the following General Eule. Reduce the given equation to the form x' -f" P ^ = I- Complete the square by adding the square of half the coefficient of ^ to both members. Extract the square root of both members, and solve the simple equation thus produced. Sometimes the first member of the equation reduces to a perfect square, and the second to 0. Then the root may be found directly ; but x will really have but one value. Thus, a;''— 6a: = — 9 by transposition becomes a:^— 6a:-|-9 = 0, where the first member is the square of a: — 3, whence a: = 3 ± = 3. It is, however, convenient in this case to consider the quadratic equation as having two equal roots. QUADEATIC EQUATION& 209 Examples. 4. Given o^ — 6 a; + 12 = 4, to find the values of x. Ans. a: := 4, or 2. 5. Given o:^ — 8x + 50 = 98, to find the values of x. Ans. X = 12, or — 4. 6. Given Sai" — 3 a; + 6 = 5^, to find the values of x. Ans. ar = §, or ^. t. Given — Sar' + 36 a; — 105 = 0, to find the values of X. Ans. a; = t, or 5. 8. Given 4 a; ^ 46, to find the values of x. X Ans. x= 12, or — f. 9. Given 6 a; — 30 = 3 a;", to find the values of x. Ans. a; = 1 ± 3 y/ —\. _1_ Qni - 10. Given ^ _ f -I- 20J = 42§, to find the values of x. Ans. x=:1, or — 6J. 11. Given a^ -\- ax :^ b, to find the values of x. •Ans. a; = — l + ^J + J, or — | — ^S + J. 277t Since the equation x' -{-px = q is the general expression of any affected quadratic equation reduced to that form (Art. 276), we may use its roots, as the general formulas for the roots of any affected quad ratio equation, and write out its roots, by substituting the particular values of p and q in these formulas. Hence, Write for the roots of an equation in the form of x^ -\- p x = q, half the coefficient of x with a contrary sign, plus or minus the square root of the sum of the second member and the square of said half coefficient. 18* 210 ALGEBBCA. 12. Given x^ -\- i,x = 140, to find the values of x. Here ^ = 2, and q = 140, whence, by substitution, X = —2 ± V 140 + 4 = 10, or — 14. 13. Given 2 a;'^ + 8 a; — 20 = tO, to find the values of x. Ans. X ^ 5, or — 9. 14. Given 2a^ — 2x — f = 0, to find the values of x. Ans. »;:=§, or — ^. 15. Given a;^ -f- 6 a; -j- 2 = — 6, to find the values of x. Ans. a; = — 2, or — 4. 16. Given 5x — = 2 x A r — , to find the values of x. Ans. a; = 4, or — 1. SECOND METHOD OF COMPLETING THE SQUARE. 278i Any affected quadratic equation may be solved by the first method, since its rule is general ; but it is some- times more convenient to employ another mode of com- pleting the square, knovm as the " Hindoo method." An affected quadratic reduced to three terms, cleared of all fractions, and divided by the greatest common divisor of its terms, may thus be reduced to the form a a;^ -|- J a; = c, where a, h, and c represent any whole numbers whatever which have no common divisor greater than unity. To make the first term of this equation a square, and the second term divisible by 2, we multiply both members by 4 a, and have ^c?3? -\- 4:abx = 4a c. Now, 2 a a;, the square root of ia^s^, must be the first term of the root, and — - — , or 2aJa;, must be the prod- QUADRATIC EQUATIONS. 211 2 txh X uct of its two terms ; hence, -;; , or b, mast • be the ' 202; second terA of the root ; therefore i' must be added to complete the square, giving 4:0.^0^ -\-4.abx-\-l^ = 4.ac-\-W, •where the first member is a complete square, equal to (2 a a: + Vf- Whence, taking the square root, we have 2ax-\-h =z ± \/ 4=ac -{-b\ or, by transposmg, x = 1- ■ — . It will be observed that the quantity added in complet- ing the square is the square of the coefficient of x in the original equation ; and that the introduction of fractions has been avoided in completing the square. As a, b, and c in the equation may have any value whatever, the process is general. Hence the RTILE. Reduce the equation to the form a x^ -f- '^ ^ = ^> '"'* iJohich a, b, and c are prime to each other. MuUi'ply both members of the eqvtation by foue times the coefficient of is?, and add to each the square of the coefficient of X. Extract the square root of both members, and solve the simple equation thus produced. Any quadratic equation may also be solved by using the coefficient of a;^ as a multiplier, instead of four times that coefficient, and adding the square of one half the coefficient of x, instead of the square of that coefficient. If the coefficient of x is an even number, this method will avoid fractions in completing the square. 212 ALGEBRA. The iormula x = — ^± V^^°<' + ^ may be used for the solution of an equation in the form aa? -\-hx-=c, in the same manner as in Article 277. same manner as the formula x = — ■^ ± t / ? + j" '^ used Examples. 1. Given a? — 5a;-|-4^0, to find the values of x. OPKBATION. a;2_5a: + 4 = By transposition, a? — bx = — 4 Multiplying by 4 times the coeflScient of 7?, we have 4 a^ — 20 a: = — 16 Adding to each member the square of the coeflScient of a;, 4 a^ — 20 a; + 25 = 9 Extracting the square root, 2 a; — 5 = ± 3 Or, 2a; = 5±3 = 8, or2 Whence, x = 4, or 1. 2. Given 3 a;'' = 42 — bx, to find the values of x. Ans. a; =3, or — 4|. 4. Given — r-r -\ — = 3, to find the values of x. - ' 1 I a; . 3. Given -«/ (4 + a:) (6 — a;) = 2 a; — 10, to find the values of x. Ans. x=b, or 3^. 6 , 2 Ans. a; = 2, or — ^. 6. Giyen 2a;^ — 'ra;-|-3 = 0, to find the values of x. Ans. a; = 3, or ^. 6. Given ac3? — icx-\- adx = bd, to find the values of a;. Ans. x = -, or . a c 7. Given llOaj*— 21a; + 1 = 0, to find the values of X. Ans. a; = ^, or y\. QUADRATIC EQUATIONS. 213 8. Given 125 a? — T a: = Hi, to find the values of x. Ans. x = %, or — ^. • 9. Given ax^ — 2 6 a; = c, to find the values of x. Ans. x = - ± - kl ae -\-W. a ay ' THIRD METHOD OF COMPLETING THE SQUARE. 279f An affected quadratic may be exhibited under the form a^x^ -\-hx = c, (1) when the coeflScient of the highest power of the unknown quantity is a perfect square. Adding some quantity, as m^, to the first member to complete its square, we have d'x^-{-hx-\-n? = c-\-m\ (2) Now, from the nature of the square of a binomial (Art. 90), twice the product of the roots of the extreme terms must equal the second term. Therefore, 2 {ax ")(, m) ■= bx, whence m=z~-, and m^=(-— ). 2a' \2 a/ Substituting the value of m in (2), we obtain where the square is complete. Hence the EXILE. Reduce the ffiven eqtMtion to the form a'x' + bx = c, either hy multiplication or division. Divide then the second 214 ALGEBRA. term hy twice the square root of the first, and add the square of the quotient to both members of the equation. Extract the square root of both members, and solve the simple equation thus produced. Examples. 1. Given 4^0? -\-\&x = 33, to find the values of x. Ans. a; = f , or — Jgi. \/ 4 a:^ = 2 a; ; — — = 4 ; and 4^ ^ 16. Hence, the addi- tion of 16 will complete the square. 2. Given Saj^ + 2 a: + 6 = 11, to find the values of x. Ans. a; = 1, or — ^. 3. Given 16a:^-j" 1^^ = 34, to find the values of x. Ans. x = 1^, or — 2. 4. Given 9 x~^ — 12 x~^ = — 3, to find the values of x. Ans. a; ^ 1, or 3. 5. Given ^^ ) K ^ ) __ g ^q g^^ ^jjg yalues of x. Ans. a; = ^, or ^. 6. Given —^ -| — g— = — 4, to find the values of x. 16 Ans. a; = — 1 J, or — 5J-. Given — — -j- -— r j = 0, to find the values of x. Ans. a; = '7, or — 11 S' 280. The following equations may be solved by either of the preceding methods, . preference being given to the one best adapted to the example considered. Special methods and devices may also be employed whenever any advantage can thereby be gained. QUADRATIC EQUATIONS. 215 EXAUFLSS. 1. Given a;'' — a; + 3 == 45, to find the values of x, Ans. a; = 1, or — 6. 2. Given (x — 12) (a; -j- 2) = 0, to find the values of x. Ans. X = 12, or — 2. 3. Given 2a? — x=z2l, to find the values of x. Ans. a; = J, or — 3. 4. Given 4 a: p— =: 14, to find the values of x. Ans. a; = 4, or — J. 5. Given "^ -| 7— = 9, to find the values of x. OPERATION. ipc-.iy x-i _ 2 "r~r~ — ^ Multiplying by 2, (aj— 1)^ -}- ^ (a;— 1) = 18 Considering {x — 1)^ a simple quantity, ' and completing the square, {x-\y + i{x-\)^^= 18 + tV = ^«/- Extracting the square root, (a; — 1) -)- 1- = ± -'/- Or, a: — f = ± V- Whence, x ^ ^ ±, ^- Or, a; = 5, or — J. 6. Given -^ — = — , to find the values of x. X XT y Ans. a; = 3, or f|. T. Given a; + \^ 5 a; -|- 10 = 8, to find the values of x. Ans. X = 18, or 3. 8. Given ar + >/ 10 a: + 6 = 9, to find the values of x. Ans. X := 25, or 3. 9. Given 48 a;"^ + 82a;-i = 11, to find the values of x. Ans.- a; = 4, or — .l^-j-' 216 ALGEBRA. 10. Given 81 a^ — 36 a; -j- 4 = 49, to find the values of X. Ans. x^l, or — |. 11. Given V 3 a: — 5 = ^^-^^i^, to find the val- ues of X. Ans. a; = 6, or — 2. 12. Given - -H 20 a; = 3 ar* — 80, to find the values of X. Ans. a;= 10, or — 2f 13. Given |^ + ^ = 2 (^-±|), to find the val- ues of X. Ans. , a; == 5, or 0. X ^ 13 14. Given --[-- = --]--, to find the values of x, . 1 ± \frs ' Ans. X = — r^ — . 21 X 7 Ans. x = — 2, or — 16. 15. Given ^ — - = 3f , to find the values of x. 16. Given a? — 2aa; + a* — J^=0, to find the val- ues of X. Ans. X = a ±, h. 17. Given 6 a; — -33 = Sa^, to find the values of x. Ans. a; = 1 ± \/— 10. 18. Given — t-it-i — = - + r + -» to find the values a-\-b-\-x a'6'a;' of X. Ans. X ■= — a, or — h. 19. Given a? — 2ax •\-W = 0, to find the values of x. Ans. X =■ a ±, ti/ a^ — S". 20. Given „ ^° J'^ = ^, to find the values of x. 3a — ix 4 . • 3o o Ans. a: = — , or -. 21. Given 6 a; -j 44 = 0, to find the values of X. Ans. a; = T, or |. QUADEATIC EQUATIONS. 217 22. Given 3 a; = 10 + -, to find the values of x. ' 4 Ans. a: = 6 ± 2 4/— 1. 23. Given | (a;^ — 3) = ^^, to find the values of x. Ans. X = il, or — 1^. 24. Given k^ — 4aa: + a2 + J2 = 0, to find the values of X. Ans. a; = 2 a ± -v/Sa" — 6^ 25. Given 3 a; — r + 3 = 14, to find the values of x. 4 Ans. a: = 6 ± \/— 8. 26. Given ar'-f-Ja; + ca; = a^ — 016 + '="' — ^''^ to find the values of x. Ans. x z=. a — h, or — a — c. 27. Given aJa;*-! = — ^I— , :, to find the values of a;. • 2a — 6 30 + 26 Ans. X = , or r* . ac he 28. Given a? — 8 a; = 14, to find the values of x. . Ans. X = 9.4112+, or — 1.4772+. 29. Given (7—4 V 3) a;' + (2 — is/1) x = 2, to find the values of x. Ans. a; = 2 + sT^, or — 2 (2 + ^/S)- 30. Given — i^r — I -. — = 1, to find the values X — 2a ' X -\- a of X. Ans. a; = |(l ± \/ — 27). PKOBLEMS LEADING TO AFFECTED QUADRATIC EQUATIONS. 1. There is a certain number whose square increased by 30 is equal to 11 times the number itself. Kequired the number. 19 218 ALGEBKA. SOLUTION. Let X = the number. Then, a;= + 30 = 11a: Or, ar"— 11a; = — 30 Completing the square, a? — 11 a; -|- if i = J Whence, x y^ = ± i And X ^ 6, or 5. The number is either 6 or 5, for each of these values satisfies the conditions of the problem. 2. Find two numbers such that their sum shall be 15, and the sum of their squares 117. Ans. 6 and 9. 3. A person cut and piled two ranges of wood, whose united contents were 26 cords, for 356 dimes ; and the labor on each of them cost as many dimes per cord as there were cords in its range. Kequired the number of cords in each range. Ans. 10 cords and 16 cords. 4. Bought a watch, which I sold for $ 56, and thereby gained as much per cent, as the watch cost me. Eequired the amount paid for it. SOLUTION. Let X ^ the amount paid, in dols. Then x = the gain per cent., and — — X a; = — = the whole gain, in dollars. Therefore, — - = 56 — x Or, x^-\- 100 a; = 56aO Completing the square, a;^ + 100 a; + 2500 = 8100 Whence, a; + 50 = ±90 And X = 40, or — 140. Only the positive value of x is here admissible, as the negative result does not answer to the conditions of the problem. The cost, therefore, was $40. QUADEATIC EQUATIONS. 219 5. A grazier bought a certain number of oxen for $240, and having lost 3, he sold the remainder at $ 8 .a head more than they cost him, thus gaining $ 59 by his bargain. What number did he buy ? Ans. 16. 6. The plate of a rectangular looking-glass is 18 inches by 12, and is to be framed with a frame all parts of which are of equal width, and whose area is to be equal to that of the glass. Kequired the width of the frame. Ans. 3 inches. T. A merchant sold a quantity of flour for $39, and gained as much per cent, as the flour cost him. What was the cost of the flour ? Ans. $ 80. 8. There are two numbers, whose difierence is 9, and whose sum multiplied by the greater is 266. What are those numbers ? Ans. 14 and 5. 9. A and B gained by trade $ 1800. A's money was in the firm 12 months, and he received, for his principal and gain, $ 2600. B's money, which was $ 3000, was in the firm 16 months. What money did A put into the firm? Ans. $2000. 10. A merchant bought a quantity of flour for $ '12, and found that if he had bought 6 barrels more for the same money,, he would have paid $1 less for each barrel. How many barrels did he buy, and what was the price of each ? Ans. 18 barrels, at $4 per barrel. 11. A square court-yard has a gravel-walk around it. The side of the court wants 2 yards of being 6 times the breadth of the gravel-walk, and the number of square yards in the walk exceeds the number of yards in the perimeter of the court by 164 yards. Eequired the area of the court. Ans. 256 square yards. 12. My gross income is $ 1000. After deducting a per- centage for income tax, and then a percentage less by one than that of the income tax from the remainder, the income 220 ALGEBRA. is reduced to $ 912. Eequired the rate per cent, at which the income tax is charged. . Ans. 5 per cent. 13. I have a rectangular field of corn, which ccnsists of 6250 hills, but the number of hills in the length exceeds the number in the breadth by 75. Of how many hills does the length and breadth consist ? Ans. Length, 125 hills ; breadth, 50 hills. 14. A certain company agreed to build a vessel for $ 6300 ; but, two of their number having died, those that survived had each to advance $ 200 more than they other- wise would have done. Of how many persons did the company at first consist? Ans. 9 persons. 15. A detachment from an army was marching in regu- lar column, with 6 men more in depth than in front ; but when the enemy came in sight, the front was increased by 870 men, and the whole was thus drawn up in 4 lines. Eequired the number of men. Ans. 3712 men. 16. A has two square gardens, and the side of the one exceeds that of the other by 4 rods, while the contents of both are 208 square rods. How many square rods does the larger garden contain more than the smaller ? Ans. 80 square rods. 17. A certain farm is a rectangle, whose length is twice its breadth ; but should it be enlarged 20 rods in- length and 24 rods in breadth, its contents would be doubled. Of how many acres does the form consist .'' Ans. 20 acres. 18. A square court-yard has a rectangular gravel-walk around it. The side of the court wants one yard of being six times the breadth of the gravel-walk, and the number of square yards in the walk exceeds the numter of yards in the perimeter of the court by 340. What is the area of the court and width of the walk ? Ans. Area of the court, 529 square yards ; width of the walk, 4 yards. QUADRATIC EQUATIONS. 221 19. A merchant bought 54 bushels of wheat, and a cer- tain quantity of barley. For the former he gave half as many dimes per bushel as there were bushels of barley, and for the latter 4 dimes per bushel less. He sold the mixture at $ 1 per bushel, and lost $ 51.60 by his bargain. Eequired the quantity of the barley, and its price. Ans. 86 bushels, at $ 1.40 per bushel. 20. A lady wishes to purchase a carpet for each of her square parlors ; the side of one of them is 1 yard longer than the other, and it will require 85 square yards for both rooms. What will it cost the lady to carpet each of the rooms with carpeting 40 inches wide, at $1.15 per yard? Ans. Larger room. $11.11^; smaller, $56.10. • EQUATIONS IN THE QUADKATIC FOKM. 281 • An equation has the quadratic form when it is. expressed in three terms, two of which contain the un- known quantity ; and of these two, one has an exponent twice as great as the other. The quadratic form, then, is ai^" -\- p af = q, or a x^" -\- baf' = c, where n may have any value whatever, and x may also represent either the unknown quantity, or some expres- sion containing it. 282i The rules already given for the solution of qua- dratics, will apply to equations having the same form. For, in the equation, x^" -\- px" ==. q, assume y = a:", then Whence, by rule. Art. 211, we have 19* 222 ALGEBRA. Extracting the nth root of both members, - 4. Given 1 10^ + y=3ajy> Ans. ■[ ^ = 2, or -i. (. 3^ — a;=2 ) ly=i,OT^. 5. Given j»^ + ^^ = 84) Ans.-[^=±^- \xy + f = Q(iS \y=±b. 3 y = 3, or — 5 6. Given W^^ + y' = lO) Ans.|^=±8. 7. Given K^+^,^= ^n. Ans.-[^ = 3,or2. UV + *Y = 468J _ ly = 2, or3. 8. Given |»=^+/ = 25)_ Ans. j^ = ± ^' <"• ± 3. • ( 2a;y=:24i (y=±3, or±4. ]- + ' = - I -(x ' y 2 ^ (a^w + a;/ = 162) 9. Given Ans. i''= ^' ^'^ 3' o'" f (- 3 ± \/n). ty=3, or 6, or ^(—3 tVIT). 10. Given -ja; — y ~ .r + ?/ ~ 3 [- . Ans. -^*~*°' ( 3^^ + 3,^—45) iy=±3. QUAPKATIO EQUATIONS. 237 11. Given |^ + ^'=^^y|. Ans.j'^^^'""- (.a; — y=^xy) ty = 2, or — 13 p,. (3a:= — 2a;« = 15) 13. Given -^ ' > . (2a: +3y = 12) a / 2 6 — a" 2' = 2^ 2— ■ Ans , j a; = 3, or — 1^. " \y = 2, or m. 14. Given 1^^ t^t^'"'!^'^]- Ans J a: = ± 2, or ± ^ ^/n. \y = ±3, or T 5\\/31. 1, xa = y^ 15. Given -^ (a; -(- y) (» — x — jr) = 3 a; = ± 1, or ±9. Ans. }y ^ ±2, or T 6. U = ± 4. PROBLEMS LEADING TO QUADEATIC EQUATIONS. 1. A says to B, "The sum of our money is $ 18 " ; B replies, " But if twice the number of your dollars were multiplied by mine, the product would be $ 154." How many dollars had each ? Ans. A, $7 ; B, $11, 238 ALGEEEA. 2. The difference of two numbers is 5, and the sum of their squares is 193. What are those numbers ? Ans. 12 and t. 3. A and B have each a small field, in the shape of an exact square, and it requires 200 rods of fence to enclose both. The contents of these fields are 1300 square rods. What is the value of each, at $2.25 per square, rod ? Ans. One, 1 900 ; other, $ 2025. 4. There are two numbers whose product is 11, and the difference of whose squares is to the square of their difference as 9 to 2. Required the numbers. Ans. 11 and 1. 5. Two gentlemen, A and B, were speaking of their ages. A said that the product of their ages was ^50 ; B replied, that if his age were increased 1 years, and A's were less- ened 2 years, their product would be 851. Required their ages. Ans. A's, 25 years ; B's„ 30 years. 6. John Smith's garden is a rectangle, and contains 15,000 square yards, exclusive of a walk, 1 yards wide, which surrounds it, and contains 3696 square yards. Re- quired the length and breadth of the garden. Ans. Length, 150 yards ; breadth, 100 yards. 1. What two numbers are those whose difference mul- tiplied by the less produces 42, and by their sum 183 ? Ans. 13 and 6. 8. A man bought 10 ducks and 12 turkeys for $22.50. He bought 4 more ducks for $ 6 than turkeys for $ 5. What was the price of each ? Ans. Of a duck, $0.75 ; of a turkey, $1.25. 9. A man purchased a farm in the form of a rectangle, whose length was four times its breadth. It cost ^ as many dollars per acre as the field was rods in length, and the number of dollars paid for the farm was four times the QUADRATIC EQOATIONS. 2S9 number of rods round it. Required the price of the farm, and its length and breadth. Ans. Price, $ 1600 ; length 160 rods, breadth 40 rods. 10. I have two cubic blocks 'of marble, whose united lengths are 20 inches, and contents 2240 cubic inches. Required the surface of each. Ans. 864 square inches ; 384 square inches. 11. A's and B's shares in a speculation altogether amount to $ 500 ; they sell out at par, A at the end of 2 years, B of 8, and each receives in capital and profits $ 297. How much did each embark ? Ans. A, $275; B, $225. 12. A farmer bought as many sheep as cost him $'300; after reserving 15 out of the number, he sold the re- mainder for $270, and gained $0.50 per head on them. How many sheep did he buy ? Ans. 75. 13. A person has $ 1300, which he divides into two portions, and loans at different rates of interest, so that the two portions produce equal returns. If the first por- tion had been loaned at the second rate of interest, it would have produced $36, and if the second portion had been loaned at the first rate of interest, it would have produced $49. Required the rates of interest. Ans. 7 and 6 per cent. 14. Two men, A and B, bought a farm of 200 acres, for which they paid $200 each. On dividing the land, A says to B, " If you will let me have my part in the sit- uation which I shall choose, you shall have so much more land than I that mine shall cost 75. cents per acre more than yours." B accepted the proposal. How much land did each have, and what was the price of each per acre ? Ans. A, 81.867 acres, at $2,443; B, 118.133 acres, at $ 1.693. 240 ALGEBRA. 15. A and B- start at the same time from two distant towns. At the end of 'J days, A is nearer to the half-way house than B is, by 5 miles more than A's day's journey. At the end of 10 days they have passed the half-way house, and are distant from each other 100 miles. Now it will take B three days longer to perform the whole jour- ney than it will A. Required the distance of the towns, and the rate of walking of A and B. Ans. Distance of the towns, 450 miles ; A's rate, 30 miles, and B's rate, 25 miles^ per day. THEOEY OP QUADRATIC EQUATIONS. 292i Every complete quadratic equation, it has been shown (Art. 2 74), may be reduced to the form x^ -\- px = q, where p and q represent known quantities, positive or negative, integral or fractional. This form may also rep- resent a pure quadratic, by considering p = 0. The values of x in this equation are and, since no other values necessarily result from the general equation, we infer that Boery equation of the second degree has two roots, and only two. 293. If we add the two roots of the general equation (Art. 292), their sum is — />, and if we multiply them together, their product is QUADRATIC EQUATIONS. 241 That is, 1. The algebraic sum of the two roots of a quadratic equa- tion is equal to the coefficient of the second term, with its sign changed. 2. The PRODUCT of the two roots is equcd to the second mem- ber, with its sign changed. 294t Eepresenting by r and r' the two roots of the quadratic equation x^ -\- px — q = 0, We have x :^ r, or x ^ r', that is, X -»— r = 0, or ar — r' z= 0. Multiplying the last two expressions together, we have {x — r) (x — r') = 0, a;° — (r -J- r') as -(- r r' =: 0. But, by Article 293, r -\- r' =: — p, and r r' = — q ; whence, x' -\- px — q = (x — r) (x — r') :=: 0. That is, ff all the terms of a quadratic equation be transposed to the first member, it may be resolved into the two binomial factors formed by subtracting each of the two roots of the equation from the unknown quantity. 295f A Quadratic Expression is a trinomial which con- tains the first and second powers of some letter or quan- tity. The principle established in the last article furnishes a method of resolving any quadratic expression into simpld binomial factors. 21 242 ALGEBRA. Examples. 1. Eesolve 7? — 4 a: — 60 into binomial factors. OPERATION. Assume \a;^ — 4 a: — 60 = Whence, a;N= 10, or a; = — 6 Hence, by Art. 2M:, a;y— 4a; — 60 = (a; — 10) (a: + 6). 2. Eesolve Z^ — 10a; — 25 into binomial factors. OPERATION. 3a;2 — 10a; — 25 Separating the factor 3, we have 3(^_H_'_f) Assuming the quantity within the parenthesis equal to 0, and transposing, we obtain „ 10 a; 25 ^ §- = T Whence, a; =: 5, a; = — \ And, by Art. 294, Sa;!* — 10a; — 25 = 3 (a; — 5) (a; + f). 3. Eesolve 0^ — 6 a; -|- 8 into binomial factors. Ans. (a; — 4) (a; — 2). 4. Eesolve x^ -\-1%x -\-\% ~~ a ' , c — a; ± /^ whence, -■ = — r^. From this equation we obtain as the values jaf a; : — \^a-\-\/br _ cy/g _ ^ / y/g J \ X = ^= V'g Since both a and b are positive, the two values of x are both real. Hence, There are two points of egwal illumination on the line of the lights. Since there are two lights, c must always be greater than 0, consequently neither a, h, nor c can be 0. The problem, then, admits properly of only these three dif- ferent suppositions : — 1. a >S; 2. a < i; and, 3. a=S. 250 ALGEBEA. We shall now discuss the values of x, under each of these suppositions. 1. a > S. In this case, the first value of x is less than c ; because -z=y"-^, being a proper fraction, is less than 1. This value of X is also greater than ^c; because the denomi- nator being less than twice the numerator, the fraction is greater than ^. Hence, the first point of equal illumina- tion is at C, between the two lights, but nearer the lesser one. The second value of x is greater than e ; because" _ " _ , being an improper fractidn,' is, g^eat^r than 1. y/ a — y/fi '•-•■ " "J .'- ■",*"-'■- Hence, the second point is at C, in the prolongation of the line AB, beyond the lesser light. These results agree with the supposition. For, if a is greater than b, ■ then B is evidently the lesser light. Hence, both points, of equal illumination will be nearer the second light, and since the two lights emit rays in all directions, one of the points must be in the prolongar tion of A B beyond both lights. 2. a < b. In this case, the first value of x is positive. It is also less than J-c; because -i=H- — =;, having the denomi- nator greater than twice the numerator, is less than ^. Hence, the first point of illumination is between the lights, but nearer A, the lesser light. The second value of x is negative, because the denomi- nator \/fl! — *i/b is negative ; which must be interpreted as measuring distance from A towards the left (Art. 302).- EATIO AND PROPORTION. 251 Hence, the second point is at C", in the prolongation of the line at the left of the lesser light. These results correspond with the supposition, the case being the same as the preceding one, except that A is now the lesser light. 3. a = J. In this case, the first value of x is positive, and equal to c ^x, and both reduce to - . Hence, the first point of equal illumination is midway between the two lights. The second value of x is not finite ; because yJa — ^V having h/ a = \/^> reduces to — ■=.ix> , which indicates that no finite value can be assigned to x. Hence, there is no second point of equal illumination in the line AB, or its prolongation, at a finite distance from it. These results agree with the supposition. For, since the lights are of equal intensity, a point of equal illumi- nation will obviously be midway between them ; and it is evident that there can be no other like point in their line. The preceding discussion illustrates the precision with which algebraic processes will conform to every allow- able interpretation of the enunciation of a problem. EATIO AND PROPORTION. 305, The Katio of one quantity to another of the same kind is the quotient arising from dividing the first quan- tity by the second (Art. 158). Thus, the ratio of a to 6 is r . or « = *• 252 ALGEBRA. "06. The Terms of a ratio are the two quantities re- quired to form it. Of these the first is called the antece- dent, and the second the consequent. Thus, in a : b, a and b are the terms, of which a is the antecedent, and b the consequent. 807. A Pkoportion is an equality of ratios. Thus, if the ratios a : b and c : d are equal, they form a proportion, which may be written a : b ■=. c : d, or a :b : : c : d. S®8. The Terms of a proportion are the four terms of its two ratios. The first and third of the terms are called the antecedents; the second and fourth, consequents; the first and last, extremes ; the second and third, means ; and the terms of each ratio constitute a couplet. Thus, ivi a : b : : c ■. d, a and c are antecedents ; b and d, consequents ; a and d, extremes ; b and c, means ; a and b, the first couplet ; and c and d, the second couplet. 309. A Proportional is any one of the terms of a pro- portion, and a Mean Proportional between two quantities is either of the two means, when they are the same quantity. Thus, in the proportion a : b : : bid, the fourth term, d, is the fourth proportional to a, b, and c, taken in their order ; and in the proportion a : b : : b : d, b is a mean proportional between the quantities a and d. 310. A Continued Proportion is one in which each consequent is the same as the next antecedent ; as, the proportion a;i :: b : c : : c : d : : d : e. RATIO AND PBOPOETION. 253 PROPERTIES OF PROPORTIONS. 31 !• When four quantities are proportional, the product of the extremes is equal to the product of the means. Let a, h, c, d be the four quantities ; then, since they are proportional (Art. 301), a c _ whence, ad = be. Hence, if three quantities be in continued proportion, the product of the two extremes is equal to the square of the mean. Thus, if a ih : : h : c, then ac = JJ = &*. Likewise, if three terms of a proportion be given, the fourth may be found. Thus, if a : b : : c : x Then ax -^ be be Whence, x = — . 312. -5^ the product of two quantities be equal to the prodwt of two others, two of them may be made the extremes and the other two the means of a proportion. Let adz=bc; dividing by bd, we have - z=z -; whence, a : b : : e : d. 31 3« ^ four quantities be in proportion, they wiU be in pro- portion by ALTEKNATioN ; that is, the antecedents will have the same ratio to each other that the consequents have. 22 254 ALGEBRA. If a : b : : e : d, then a : c : : b : d. For, - =1 - ; multiplying by - , we have - = ^ ; whence, a : c : : h : d. 314. If four quantities he in proportion, they will he in pro- portion hy INVERSION ; that is, the second term will have the same ratio to the first that the fourth has to the third. If a : h : : c : d, then h : a : : d : c. a c _ 5 ~ d' For, r = -, ; dividing 1 by each member, we have 6 _ d. a c ' whence, h : a : : d : c. 313t ^ four quantities he in proportion, they will he in pro- portion hy COMPOSITION ; that is, the sum of the first and second terms will have to the second the same ratio that the sum of the third and fourth has to the fourth. If a : h : : c : d, then a -\- h : b :: c -{- d : d. For, jT ^ ^; adding 1 to both members, we have a , , c,, + 6 c -i-d whence, a -\- b . b :: c -\- d : d. 316. If four quantities he in proportion, they will he in pro- portion hy DIVISION ; that is, the difference of the first and second terms wiU have to the second the same ratio that the difference of the third and fourth has to the fourth. If a : h : : c : d, then a — b ; h :: c — d : d. For, r = ;j ; subtracting 1 from each member, Tve have RATIO AND PKOPOETION. 255 a , c , a — 6 c — d 6-1 = 5-1' ""^ -^ = -1-' whence, a — J:J :: c — d : d. 317. If four quantities he in proportion, the sum of the first and second is to their difference as the sum of the third and fourth is to their difference. If a : b : : c : d, then a -\-b ; a — h :: c -\-d : c — d. By Art. 315, 4-' = ^^, and, by Art. 316, — r — = — j — ; therefore, a -\- b a — 6 c -\- d _ c — d a -\-b c -|- d_ ~1 '~ ^F~ "~ ~~d ' d~' °^ a^^^ — c^^ ' whence, a-\- b : a — b :: c -\- d : c — d. 3I8t Quantities which are proportional to the same quan- tities are proportional to each other. If a : b : : e :f and c : d : : e : f, then a -.h : : c : d. For, I = J, and I = 1; therefore \ = \; whence, a yh : : c : d. 319. If any number of quantities are proportional, any antecedent is to its consequent as the sum of all the Antecedents is to the sum of dU the consequents. If a : 5 : : c : d :: e : f, then a :b ■.: a-\-c-\-e : b-\-d-\-f. For, by Art. 311, ad = be, and af =^ be ; also, ab = ba. Adding, ah -\- ad -\- af = bar-\-bc -\- be, or, a{b-\- d -\-f) = S (a + c + e) ; whence, by Art. 312, a : b :: a-|-c-|-e : b -\- d -\-f. 256 ALGEBRA. 320i When four quantities are in proportion, if the first and second he multiplied or divided hy any quantity, as also the third and fourth, the resulting quantities will be proportional. If a : b : : c : d, then ma : mb :: nc : nd. For, r = -, ; therefore — r = ^ ; ' b d mb nd' whence, ma : mb :: nc : nd. T ,., abed In like manner, — : — ::-: -. m m n n Either m or n may "be made equal to unity ; that is, either couplet may be multiplied or divided without mul- tiplying or dividing the other. 321 • When four quantities are in proportion, if the first and third be mvUiplied or divided by any quantity, as also the second and fourth, the resuMng quantities will be proportional. If a : b : : c : d, then ma : nb :: mc : nd. ■„ a c ,, J. ma mc .ma mc lor, ^ = - ; therefore ^- = — =-, and — r =: — -.; b d b d ' nb nd whence, ma : nb : : mc : nd. T Ti abed In like manner, — : - : : — : - . m n m n Either m or m may be made equal to unity. 322. If^ there be two sets of proportional quantities, the products of the corresponding terms will be proportional. If a : b i : c : d and e : / : : g : h, then ae : 'bf : : eg : dh. For, 7 = J and -7=: f ; therefore ?4. = ^ ; b d /A bf dh' whence, ae i bf :: eg : dh. RATIO AND PROPORTION. 257 323t If four quantities be in proportion, like powers or roots of these quantities will he proportional. If a:b::e:d, then a" : J» : : c" : d". _i 2 For, ^ = 1 ; therefore ?_ = f- and ^ = IL , where b d J" £?» i i 6" rf» w may be whole or fractional ; whence, 11 1^ 1 a" : If : : c" : d", and a" : fi" : : c" : A Progression is a series, or a succession of terms increasing or decreasing according to some uniform law. There may be distinguished three kinds of Progression : Arithmetical, Geometrical, and Harmonical. AEITHMETICAL PEOGEESSION. 334< An Arithmetical Progression is a series that in- creases or decreases from term to term by a common difference. Its formation may be considered to be by the continual addition of the common difference ; therefore, when the series is increasing, the common difference will be positive, and when decreasing, it will be negative. Thus, 1, 3, 5, 7, 9, 11 is an increasing arithmetical progression, in which the common difference is -1-2 ; and 19, IT, 15, 13, 11, 9, is a decreasing arithmetical progression, in which the common difference is — 2. 335i In an Arithmetical Progression, the last term is equal to the first term plus the product of the common difference hy the number of terms less 1. Let a denote the first term, d the common difference, and / the last term ; then the progression will be a, (a-{-d), (a-\-2d), (a -|- 3 rf), . . . . That is, the wth term of the series will be a -\- (n — 1) d. PEOGEESSIONS. 263 Also, by putting I for the rath term, we have I = a -j- (« — !)'«?, where d is either positive or negative, according as the series is an increasing or decreasing one. 336i In an Arithmetical Progression, the sum of the terms is equal to the product of one half the number of terms hy the sum of the extremes. Let S denote the sum of the terms; then ,S' = a+(a + «Q4-(a + 2(^) + + Z, or, writing the terms in the reverse order, .S- = l-\-(l — d) J^(l—2d)-\- + a. Therefore, by adding these equations, term by term, 2^= (a + + (« + + (« + + + (« + 0- Here, (a -|- 1) is taken as many times as there are terms, or n times ; whence, 2 5 = M (a + 0' or S z= ^n {a-{-J). In the course of the demonstration, it may, also, be readily seen, that The sum of any two terms equidistant from the extremes, is equal to the sum of the extremes. 337. To insert a given, number of arithmetical means between two given terms. Let a and I be the two given terms, and m the number of terms to be inserted. Then we are to find m + 2 terms in arithmetical progression, a being the first term, and I the last. Let, now, d denote the common differ- l — a ence ; then ? = a -|- {m-\-\) d ; whence d = ^ ■ ^ . This determines d, and the m required means will be a-\-d, a-\-2d, a -\-3d, a-\-md. 264 ALGEBRA. That is, The given first term plus the common difference will give the first of the required means ; the first mean phis the common difference will give the second mean ; and so on. 338. The formulas established in Arts. 335, 336, 1 = a-\-{n — l)d, (1) S=in(a + t), (2) are fundamental; and, since they constitute two indepen- dent equations, together containing all the five elements of an arithmetical progression, when any three of these are given, the other two may be readily determined. Thus, from these two equations we deduce other formu- las ; and there are in all twenty, as exhibited in the fol- lowing TABLE. No. 10 Giveu. a, d, n /, df n d,\S a,n,l a^UfS I, n, S a,d,l a,l,S a,d,S l,d,S Re- quired. l,S a,S a, I d,S d,l d, a n,S n, d n, I n, a fa' r Formulaa. l = a+(n — \)d; S = ^n[2a+ (n — \)d\. a = l — (n — l)d; 5 = Jn [2i— (n — 1) d]. ^ _ 25 — «(n — l)rf . ^_ 25-f H(ra — 1) d 2n d = I — a ^_ 2(g-a«) . n(n— 1) ' n(n—l) ' „- 2-5 . 2n a^^A-l. ^_ JlJ^a)(l-a + d) 2d ^^ (l + a)(l — a) 2S—(l+a) ' ^^d-^a±^Ad-2al±SdS , = „+(„_„i 2d > -ry. , d+2l±^(d+2lf-%dS . „ , , ,,^ ~ , a = «-(n — ija. PKOGBESSIONS. 265 Hence, for the solution of questions in Arithmetical Progression, the following RULE. Substitute in the fundamental formulas, or such as may he deduced from, them, the given quantities and reduce the result. Examples. 1. If the first term of an arithmetical progression is 5, and the common difference 3, what is the 7th term ? Here, 0=5, rf = 3, and w =: 7, to find I. Whence ^ = 5 + (7 — 1) 3 = 23, Ans. 2. Eequired the 4th term of an arithmetical progres- sion, of which the first term is 10, and the common differ- ence — 2. Ans. 4. 3. If the first term is ^, the common difference — ^, and the sum of the series — 1^, what is the number of terms ? Ans. 12. 4. The first term is 3, the number of terms 20, and the sum of the terms 440. Eequired the common differ- ence. Ans. 2. 5. The sum of n terms of a series, of which the com- 3 n* mon difference is 3, is — — \-2n. What is the first term ? Ans. J. 6. Insert three arithmetical means between — 9 and 18. •Ans. — 2i, 4^, Hi- 7. Eequired the first and last terms of a series of which the common difference is 5, the number of terms 6, and the sum 321. Ans. 41 and 66. 8. Find the sum of the odd numbers from 1 to 100. Ans. 2500. 23 266 ALGEBRA. 9. A debt can be discharged in a year by paying $ 1 the first week, $ 3 the second, $ 5 the third, and so on : required the last payment, and the amount of the debt. Ans. Last payment, $103; amount, $2f04. 10. A person saves $270 the first year, $210 the second, and so on. In how many years will a person who saves every year $ 180 have saved as much as he ? Ans. 4. 11. Two persons start together. The one travels ten leagues a day, the other eight leagues the first day, which he augments daily by half a league. After how many days, and at what distance from the point of de- parture, will they come together ? Ans. After 9 days, at a distance of 90 leagifes. GEOMETRICAL PROGRESSION. 339. A Geometrical Peogression is a series, each terra of which is equal to the preceding one, multiplied by a constant factor. The constant factor is called the ratio of the progres- sion. Since the successive terms of the progression may be considered as derived from the first by continually multi- plying it by the ratio, therefore, if the first term is posi- tive, the series is increasing when the ratio is greater than 1, but the series is decreasing when the ratio is less than 1. Thus, 3, 6, 1-2, 24, 48, is an increasing geometrical progression, in which the first term is 3, and the ratio 2 ; and 9; 3, 1, ■^, ^, is a decreasing geometrical progression, in which the ratio is ^. PROGRESSIONS. 267 340i In a Geometrical Progression, the last term is equal to the product of the first term by the ratio raised to a power whose exponent is one less than the number of terms. Let a denote the first term, r the ratio, and I the last term ; then the progression will be a, ar, ar^, ar^, That is, the nth term of the series will be Also, by putting I for the wth term, we have I = ar^-^. 341 1 To find the sum of a given number of terms in a Geo- metrical Progression. Let To insert a given number of geometrical means between two given terms. Let a and I be the two given 4erms, and m the number of terms to be inserted. Then we are to find m -\-'2 terms in geometrical progression, a being the first term, and I the last. Let, now, r denoie the ratio ; then (I Xm+l - I This determines r, and the m required means will be ar, ar^, ar', ar" ; that is, The given first term multiplied by the ratio will give the first of the required means ; the first mean muUiplied by the ratio will give the second mean ; and so on. PROGRESSIONS. 269 344. The formulas established in Arts. S40i 341, I = at^-\ (1) ■ S = ^, (2) are fundammUd ; and' since thiey contain the five elements, if any three of these are given, formulas may be ded^uced" for finding the other two, as in the following TABLE. Re- quired. 10 l,r,n r,n,S a,n,l a,n, S l,n,S a,r,l a,l.S a,r, S l,r.S l,S a,S a, I r,S r,l r,a n,S n, I n,l ^ I . " r»-i' 5 = or" — o r — I ■ lt» — l a = (r — 1)5. r^ — 1 -li' r" — 1 aT« — rS=a — S; i(S— i)»-i = o(S— a)"-!. Zr — a log r „ = log ^ — log" +1 . r= ?~^ log (S—a) — \oe{S—iy ' S — l log[a+(r— 1)5]— log a. , _ a+{r—l)S n = — 2-t ; ,. * . log r . r n = log '-log [ ^' — ('>-l ) -Sj+i ; a _ Zr-(r-l)S. log r The formulas for n, exhibited in the table for conven- ience of reference, require for their solution methods not yet given, but which will be presented in their proper place ; and those containing the nth degree of the unknown quantity may present diflSculty, unless n is less than 3. 23* 270 ALGEBRA. Hence, for the solution of questions in Geometrical Pro- gression, we have the following RULE. Substitute in the appropriate formula the given quantities, and reduce the result. Examples. 1. If the first term of a geometrical progression is t, the ratio 3, and the number of terms 5, what is the last term ? Here, a =i t, r = 3, and w = 5, to find I. Whence I = T (3)»-i =7X3*= 56T, Ans. 2. If the first term of a geometrical progression is 7, the ratio 3, and the number of terms 5, what is the sum of the series ? Ans. 84t. 3. If the ratio is 2, the number of terms 6, and the last term 128, what is the first term ? Ans. 4. 4. If the first term is 2, the last term 43H, and the number of terms 8, "what is the ratio ? Ans. 8. 6. Insert three geometrical means between J and 128. Ans. 2, 8, 32. 6. If the first term is 2, the ratio 4, and the number of terms 12, what are the last term, and the sum of the series ? Ans. Last term, 8388608 ; sum of the series, 11184810. 1. Eequired the sum of the series ' 1, ^, ^, to infinity. Ans. 1^. 8. Find the sum of f, 1, f, to six terms. Ans. 4j-Vj. PEOGEESSIONS. 271 9. Find the value of the decimal .3S3 to infinity. Here, .333 = ^ _|_ ^g.^ 4_ ^^3^^ Whence, ^ = ^^^ -=- (1 — ^) = | = ^, Ans. 10. Find the value of .1212 to infinity. Ans. ^*5. 11. Required the sum of 1 — i -{- i — i -\- to infinity. Ans. §. 12. A person who saved every year half as much again as he saved the previous year, had in seven years saved $ 2059. How much did he save the fiz-st year ? Ans. $ 64. 13. A gentleman, by agreement, boarded 9 days, pay- ing 3 cents for the first day, 9 cents for the second day, 27 cents for the third day, and so on, in this ratio. Re- quired the cost. Ans. $295.23. HARMONICAL PROGRESSION. 345i Three or more quantities are said to be in Har- MONicAL Progression, when their reciprocals form an arith- metical progression. Thus, 1, 1, i, ^, are in harmonical progression, because their reciprocals, 1, 3, 5, T, are in arithmetical progression. 34Si If three consecutive terms of an Harmonical Progres- sion he taken, the first has the same ratio to the third that the difference of the first and second has to the difference of the second and third. 272 ALGEBRA. For, if a, h, c are in harmonical progression, their recip- cali fore, rocals, -, T, -, will be in arithmetical progression; there 1 _l _ 1 _l c b h a Clearing of fractions, we have ah — ac = ac — he Or, a {b — c) ^ c (a — h) Dividing by c (5 — c), " = |^^ Whence, a : c :: a — J:5 — c. 347t From the preceding it follows that every series of quantities in harmonical progression may be easily con- verted into an arithmetical progression, and then the rules of the latter may be applied. There will be found, how- ever, no general expression for the sum of an harmonical series. Thus, 1. Given the first two terms of an harmonical progres- sion, to find the wth term. Let a and h be the first two terms, and I the nth term ; then - and =- are the first two terms of an arithmetical a progression, and -j is its wth term ; therefore, , 1 1 a — b 6 a ah Whence, 1 li/ -,\ a — b (n — l)a — (n — 2) 6 I a ^ ^ 'ah ah Therefore, I = („_ i) „!_(„_ ^p • (1) PKOGKESSIOTtS. 273 2. To insert m harmonical means between a and I. Here, if d be the common difference of the reciprocals of the terms, we have r a — / a — I fc)\ °'"' • '^ — {n — l)al — (m + 1) ai ' ^^> whence the arithmetical progression is found ; and by in- verting its terms, the harmonical means will be ascer- tained. Examples. 1. The first term of an harmonical progression is 1, and- the second term ^ ; find the fifth term. Ans. |. 2. Insert two harmonical means between ^ and ^ ; and also between 4 and 5. Ans. ^, ^ ; 4f , 4^-^^. 3. Continue the series 3, f, |, for two terms. Ans. /x. A- PEOBLEMS kequiking the application of the principles of the progressions. 1. It is required to find four numbers in arithmetical progression, such that the product of the extremes shall be 45, and the product of the means 1%. SOLUTION. Let X be the first term, and y the common difiference ; then the numbers will be X, x-{-y, x'-^2y, x-\-^y; and by the conditions, 274 ALGEBRA. x^J^Sxy =45 (1) (2) 2/ = 32 y^ ±4: Subtracting (1) from (2), Whence, Substituting the value of y in (1), we have x" ± 12 a; = 45 Whence, ^ = ±3, or ±15. Hence, the four numbers are 3, 1, 11, and 15. 2. It is required to find three numbers in geometrical progression, such that their sum shall be 14, and the sum of their squares 84. SOLUTION. Let X and y denote the two extremes ; then \/xy is the mean, and by the conditions. X -\ \-\/xy-]ry = 14 (1) x-" + ccy + / = 84 (2) Dividing (2) by (1), x - - A/xy 4- y = 6 (3) Adding (1) and (3), x-\-y = 10 (4) Subtracting (3) from (4), /1,/xy = i (5) Or, xy= 16 (6) Therefore, (x -\- yY — 'txy = = 100 — 64 = 36 0) Whence, x—y= ±6 (8) From (4) and (8), x = 8, or 2 ; y = 2, or 8. Hence, the three numbers are 2, 4, and 8. 3. Find four numbers in arithmetical progression, such that the sum of the squares of the first two terms shall be 34, and the sum of the squares of the last two 130. Ans. 3, 5, 1, and 9. 4. The sum of three numbers in harmonica! proportion is 13, and the product of their extremes is 18. What are the numbers ? Ans. 3, 4, and 6. Let X and y denote the extremes: then -f ^ is the mean. x-\-y PKOGKESSIONS. 275 5. A gentleman set out from Boston for New York. He traveled 25 miles the first day, 20 miles the second day, each day traveling 5 miles less than the preceding. How- far was he from Boston at the end of the eleventh day ? Ans. 6. Suppose a number of stones were laid a rod distant from each other for twenty miles, and the first stone a rod from a basket. What length of ground will that man travel over who gathers them up singly, returning with them, one by one, to the basket? Ans. 128,060 miles 2 rods. T. If a number of workmen be 2 days in raising the tenth foot of a tower which is to be 100 feet high, how long will they be in building the tower, the time of raising any foot being in proportion to its height ? Ans. 1010 days. 8. Five persons undertake to reap a field of ST acres. The five terms of an arithmetical progression, whose sum is 20, will express the times in which they can severally reap an acre, and they all together can finish the job in 60 days. In how many days can each, separately, reap an acre ? ' Ans. 2, 3, 4, 5, 6, days. 9. Suppose the elastic power of a ball that falls from a height of 100 feet, to be such as to cause it to rise 0.9375 of the height from which it fell, and to continue in this way diminishing the height , to which it will rise, in geometrical progression, till it comes to rest. How far will it have moved ? Ans. 3100 feet. 10. If a man travel 20 miles the first day, 15 miles the second, and so continue to travel 6 miles less each day, how far will he have advanced on his journey at the end of the. 8th day? Ans. 20 miles. 11. The sum of the squares of the extremes of four numbers in arithmetical progression is 200, and the sura 27^ ALGEBRA. of the squares of the means is 136. What are the num- bers ? Ans. 2, 6, 10, and 14. 12. After A had traveled for 2f hours, at the rate of 4 miles an hour, B set out to overtake him, and went 4|- miles the first hour, 4| the second, 5 the third, and so on, increasing his speed a quarter of a mile every hour. In how many hours would be overtake A ? Ans. 8 hours. 13. The sum of the first and second of four numbers in geometrical progression is 15, and the sum of the third and fourth is 60. Eequired the numbers. Ans. 5, 10, 20, and 40. 14. The greatest of three numbers in harmonical pro- gression is the product of the other two ; but, if one be added to each, the greatest beco,mes the sum of the other two. What are the numbers ? Ans, 2, 3, and 6. PERMUTATIONS AND COMBINATIONS. 348. The different orders in which quantities can be ar- ranged are called their Permutations. Thus, the permutations of the letters a, h, c, taken two at a time, are ab, ha ; ac, ea ; be, cb ; and taken three at a tiipe, are abc, acb; bap, Ifca,- cab, cba. 349. The CoMSfNATioNS of quantities are the different collections that can be formed out of thein, withqi^t re- garding t)ie order in which they are placed. PERMUTATIONS AND COMBINATIONS. 277 Thus, the combinations of the letters a, h, c, taken two at a time, are ah, ac, he, ah and ha, though different permutations, forming the same combination. To find the number of permutations of n quantities taken r together. Let a, h, c, d, h he n quantities. The number of permutations of these n quantities, taken singly, is evi- dently n. The number of permutations of n quantities, taken two together, is n (w. — 1). For, since there are n quantities, if we put one of these, as a, before the others, we obtain ah, ac, ad, ak, I or n — 1 permutations in which a stands first. Proceed- ing in the same manner with h, we have n — 1 permuta- tions in which h stands first. Similarly, there will be n — 1 permutations in which c stands first, and so on. Hence, taking two together, the whole number of per- mutations will be n (w — 1). The number of permutations of n quantities, taken three together, is w (ra — 1) (« — 2). For, it has just been shown that out of « quantities we can form n (n — 1) permutations, each of two quantities. Hence, out of w — 1 quantities we can form (n — 1.) (n — 2) permutations, each of two quantities ; put a before each of these and we have (n — 1) (n — 2) permutations, each of three quantities, in which a stands first. Similarly, there are (n — 1) (n — 2) permutations, each of three quantities, in which b stands first ; as many in which e 24 278 ALGEnRA. stands first ; and so on. Hence, taking three together, the whole number of permutations will be n (n — 1) (n — 2). In like manner, we can prove that the number of per- mutations of n quantities, taken four together, will be n (n — 1) (n — 2) (n— 3). In examining the results obtained, we perceive such a relation existing between each class of permutations and the numerical part of its corresponding expression, that the negative number in the last factor is I less than the number of quantities taken together. Hence, we may con- clude that the number of permutations of n quantities, taken r together, is n (n — l) (n — 2) ..... (n — r-\-l). (1) 851 1 If we suppose all the quantities to be taken to- gether, or r = n, formula (1) becomes n (n — 1) (n — 2) (n — 3) 1, (2) or, by inverting the order of the factors, 1X2X3 (« — 1) «. (3) That is. The number of permutations of n Utters, taken n together, is equal to the product of the natural numbers from 1 up to n. For the sake of brevity, this result is often denoted by \n_ ; thus \n_ denotes the product of the natural numbers from 1 to n inclusive. S52. To find the number of combinations of n quantities, taken r together. The number of combinations of n quantities, taken sin- gly, is evidently n. PERMUTATIONS AND COMBINATIONS. 279 ■ The number of combmations of n quantities, taken two . ,, . n (n — 1) together, is \ y, ^ For the number of permutations of n quantities, taken two together, is n (n — 1) ; but each combination of two quantities admits of two permutations ; hence, the number of combinations is the number of permutations divided by 2. The number of combinations of n quantities, taken three together, is " ^"i~ '^ ^^ T ^^ " For, there are n (n — 1) {n — 2) permutations in n things taken three together ; but each combination of three quantities admits of 1X2X3 permutations ; hence, the number of combinations is that of the permuta- tions divided by 1X2X3. In like manner, ft may be shown that the number of combinations of n things taken r together, is n(n—V) (n — i) („ — r + l) ' 1X2X3X . . . . y.r • ^ ' 353< The number of combinations of n quantities taken r together, is the same as the number of combinations of n quan- tities taken n — r together. For, it is evident that for every combination of r quan- tities which we take out of n quantities, we leave one combination of n — r quantities, which contains the re- maining quantities. Examples. 1. How many changes can be rung with 10 bells, taken 7 at a time ? Here, w ^ 10, r = 7, and n — r -\- 1 = 4 ; then, by formula (1), 10X9X8X':X6X5X4 = 604800, Ans. 280 ALGEBRA. 2. In how many different orders may Y persons be seated at table ? Here, by formula (3), 1X2X3X4:X5X6X'7 = 5040, Ans. 3. How many different combinations can be made with 5 letters out of 8 ? Here, n = 8, r = 5, and n — »■ -|- 1 := 4 ; then, by formula (4), 8 X 7 X 6 X 5 X 4 _ . 1X2X3X4X5-^®' ^"'- 4. How often can 4 students change their places in a class of 8, so as not to preserve the same order ? Ans. 1680. 5. From a company of 40 soldiers, how many different days can a different picket of 6 men be taken ? Ans. 3838380. BINOMIAL THEOEEM. 354i The Binomial Theorem is a formula, by means of which any binomial may be raised to any given power, without the use of the ordinary process of involution. We have already seen that (a; -j- a)^ = x^ -\-2xa-{- a^, and that (a: + a)' =^ s? -{- ^ x^ a -\- Z x a^ -\- a^ ; our object now is to find an expression equal to (x -\- ay, where n is any positive integer. 3S5> By actual multiplication we have BINOMIAL THEOREM. 281 (x -\- flj) (x + flTj) ^ a;'' 4- (af + Oj) a; + ai «2, (a; + flj) (a; + fla) (x + ag) = a? -\- (a^ + a^ -{- Wj) a;' + («i«2 + «2«3 + «i«3) a; + aia^ag, (x -\- Oi) {x + aa) (« + «3) (a: + 04) = a;« + («i + «2 + «3 + «4) a:' + («1 «2 + «1 «3 + «! «4 + «2 «3 + a2 «4 + «3 «4) a:" -|- (fill 0.2 03 -|- «! «2 0^4 -f- <^1 «S «4 -f- «2 "s "4) 3! -|- flj (J^ '"3 <''4- Now, in these results we see that the following laws hold : — 1. 77ie number of terms on the right is one more than the number of binomial factors which are multiplied together. 2. The exponent of x in the first term is the same as the number of binomial factors, and in the succeeding terms each exponent is less than that of the preceding term by unity. 3. The coefficient of the first term is unity ; the coefficient of the second term is the sum of the second terms of the binomial factors ; the coefficient of the third term is the sum of the pro- ducts of the second terms of the binomial factors, taken two together ; the coefficient of the fourth term is the sum of the products of the second terms of the binomial factors, taken three together, and so on ; the last term is the product of all the second terms of the binomial factors. It will now be proved that these laws always hold, whatever be the number of binomial factors. Suppose the laws to hold when n — 1 factors are multiplied to- gether ; that is, suppose (x-\-ai) (x-j-a^) ... (x-\-a„_i) = a?-'' -\- p^x^-"" -\- p^:£'-^ + /'3a^-' + --- +/'»-! > where px = the sum of the terms «i, a^, «n-i- P2 = the sum of the products of these terms, taken two together, p, = the sum of the products of these terms, taken three together. 282 ALGEBRA. Pn-i = the product of all these terms. Multiply both members of this identical equation by another factor, x -\- a„. Thus, + (Ps +P'2 «») a:^"' + +Pn-i an ■ Now, Pi + a,i = "i + «2 + + a»-i + «» = the. sum of all the terms Oi, a^, a„ ; Pi-\-pia^ = ^2 + «„ (Ol + «2 + + On-\) = the sum of the products, taken two together, of all the terms «i, a^, a„ ; Pt+Piaa ^ Pi-\-an ("lf'2 + fl'2"3 + «1 «3 + ) = the sura of the products, taken three together, of all the terms aj, a^, . : . . . a^. . Pa-iO,^ = the product of all the terms. Hence, if the laws hold when n — 1 factors are mul- tiplied together, they hold when n factors are multiplied together ; but they have been proved to hold when 4 factors occur, therefore they hold when 5 factors occur, and so on ; thus they hold universally. The result for the multiplication of n factors may be written thus. The number of terms in qi is obviously n ; the number of terms in jj is the same as the number of combinations of the n quantities ffli, a2> "„, taken two together, that is, — - — ~ ; and so on. Now suppose a^, a^, a„ each equal to a ; thus gi becomes n a, q^ becomes "^" ' a\ and so on ; and we obtain (Art. 353), BINOMIAL THEOREM. 283 ix + ay = x''-{-nax'-^ + " ^""^^ .^ .^2 4. ^X?^ — (" — 2) ^b f^ + ... _^«(n— 1)^^,^_|_^^„_.^_|_^,. or, exchanging the letters a and x, ^ ^1.2 ^ 1.2.3 + + '!L^lZZ}la'x^ + naX^-^ + 3f, (A) in which n, So' ®*'°"' ^^® considered the coefficients. This is the Binomial Formula. The series on the right of the sign of equality is called the expansion of (a + a;)". 336> In the preceding formula, suppose a = 1, and we have (l + xy = l+nx+"-^^x^ + ^-^^=^(^a? + 1.2 I 1.2. .'L^_i)af-2 + Ma;»-i + x". (B) §57. In formula (B), suppose x changed into — x, and it becomes (l_.).^l_.. + !L^)..^-Cn-l ) (n-2)^3^..... (cj where, since every odd power of — x is negative (Art. 202), the signs are alternately plus and minus. 358. The expansion of a binomial can always be re- duced to the case in which one of the two terms of the binomial is 1. For, (a+a;)» = a"(l4-?) = a''(l+2,)», if y = ^. Expand now (l-^-y)", and multiply each term by a", and we have the expansion of (a -\- x)". 284 ALGEBRA. J. If in formula (B), which is true for all values of X, we put a:=l, we have 2«=l + n + "-^^ + + »4^) + n + l; that is, when both terms of a binomial are positive, the sum of the coefiScients of the expansion is equal to 2 raised to the same power as the binomial. In like manner it may be shown, by putting a; := 1 in formula (0), that, when one of the terms of the binomial is negative, the sum of the coefBcients of the expansion is equal to zero. From the preceding principles may be readily deduced some important precepts, with relation to the expansion of any binomial, (a -|- a;)". 1. The leading letter enters cdl the terms except the last, and the foUowing letter all except the first. 2. In the successive terms, the exponents of the leading letter, beginning with the exponent of the power, decrease by 1, while those of the following letter, beginning in the second term with 1, increase by 1 ; and the sum of the exponents in any term is equal to the exponent of the given power. Thus, the expo- nents are a", a''-^x, a^'-^x^, a"-^a:«, a^x""-^, a«^~'^, x", where the sum in any term is n, or the exponent of the power. 3. If the coefficient of any term be multiplied by the exponent of the leading letter of the same term, and the product divided by the number of the term from the left, the result will be the coefficient of the term next following. Thus, The coefficient of the first term is 1, of the second — ^— , or n, of the third ^ — -, and so on. 1 * BINOIUAL THEOREM. 285 4. The number of terms is always one more than the ex- ponent of the power. 5. If both terms of the binomial are positive, aU the terms of the power are positive ; but if the second term of the bino- mial is negative, the even terms of the power will be negative. 6. The coefficients of any two terms taken equidistant fvom the beginning and end of the expansion are the same. For, the rth and {n — r -\- l)th terms are equidistant from the first and last terms, and their coeflScients are, respectively, the number of combinations of n quantities taken r — 1, and n — {r — 1), together (Art. 353). 7. The general term of the expansion, or that from which the others can be deduced, is rth term — »("-!) ("-2) (n-r+2) „_,^i ^ rttt term _ 1.2.3 (r—l) " ^ • For, the expansion of (a -\- a;)" exhibits the exponent of a in any term, n diminished by a number which is 1 less than the number of the term ; that is, the exponent of a for the second term is n — 1, for the third n — 2, and so on ; for the rth term it would be n — r -f- 1 ■ Also, the last factor of the coeflScient of any term has for its numerator n diminished by a number which is two less than the number of the term, and for its denominator a number 1 less than the number of the term. Finally, the expo- nent of a; is 1 less than the number of the term. 361> By the sixth precept, the latter half of an expan- sion may ,be written out from the first half; and by means of the seventh precept, any single term of the expansion may be found, independently of the rest. Examples. 1 . Kaise a -|- a; to the fifth power. By 2, Art. 360, the terms without the coefficients are 286 , ALGEBRA. a*, a^x, a'a;^ o?s^, aa^, sd' ; and by 3, Art. 360, the coefiScients are 1X55 X_4 10 X 3 10 X 2 5X1 . 1, —^—, 2 ' 3 ' 4 ' 5 ' or, 1, 5, 10, - 10, 5, 1. whence, by prefixing the coefiScients, we obtain (a-\-xy=a^-\-da*x-\- 10 a^x''-^- 10 a'^x' + 5 a a;* + sc^, Ans. Here, « -f- 1, or the number of terms, is equal to 6 ; therefore, it was only necessary to determine the coefiS- cients for the first 3 terms, and repeat the same numbers in the reverse order for the others. 2. Eaise a — 6 to the fifth power. Ans. c^ — 5aH-\-10a^l>' — 10a^li'-{-5ab* — ¥. 8. Find the 10th term of the expansion (a — a;)^'. Here, n = 18, and r = 10 ; and substituting these in the general term (1, Art. 360), we have 18.17.16.15.14.13.12.11.10 , „ .aoon 9 9 1.2.3.4.5.6.7.8.9 «^ = -48620a»a=». 4. Find the 49th term of (a — x)^. Ans. 1225 a' x"^. 5. Eaise a + a; to the seventh power. Ans. a'-\-1 a«a; + 21 a^x^-{-35 a*x?-\- 35a'a;« + 21 a'xf -\-1aii^-{-x'. 6. Find the 9th term in the development of (a; -|- «)*". Ans. 45 a;^ a*. • Y. Raise 1 -|- a: to the sixth power. Ans. 1 + 6a;+15a;'' + 20a;»+15a^+6ai« + a;«.- BIXOMIAL THEOREM. 287 S62i If either or both terms of a binomial have co- efficients or exponents, the binomial formula may still be applied. 8. Baise e'-^-yz to the fifth power. Here, by inclosing c" and yz, each in a parenthesis, they may be treated as a single literal quantity. Thus, (c^ + yzy= (cy + 5 {cy yz+lQ {c^f {y ^)^ + 10 {c'^Y (y zf -\-bc^(jjzf+{yzy^ c"' + 5c'ya+10c«y2z''+10cV2' + 5c22/V + 2/'a«, Ans. 9. Baise 1 + a;'' to the seventh power. Ans. l + 7a^+21a;*+35a;8+35a^ + 21a;'''+1a:'24-a^*. 10. Required the middle term of {m? -\- n^f. Ans. 10 m^ n^^. 11. Raise 3a'' — 2W to the sixth power. Ans. 129 ai2 — 2916 a^ W + 4860 a' W — 4320 a« W -f 2160 a* IP — bna^l^-\- 64 IP- 12. Required the expansion of (1 — - j . 7 21 £5 I £^_ ?^ _i_L _1 Ans. 1 — -+a^— a^+a;i ar^ + a* a;'' 13. Find the term which involves cFV in the expan- sion of (a* + 3 a &)'. Ans. 18132 a^'i^ / 3 a \2002 14. Write down the 2001st term of {a^ -\-x^^) . . 2002 X 2001 % ~„ Ans. — r^m — «^ a*™. 1 X ^ 15. Raise 1 — ^x to the tenth power. Ans. 1 — 5a; + ^^a:''— 15a:» + -ig5.a4 — -6^a^+-V/a:* 363> The binomial formula may be applied to the ex- pansion of any polynomial whatever, since, by means of the parenthesis, it may be treated as a binomial. 288 ALGEBRA. 16. Kaise 2 a — J + c" to the third power. Put 2 a — b = a, and c' = a; ; then (2ffl ^ J 4- c^y = (a + xy = a' + SaPx -{-3ax'-\-a?. Substituting for a its equal, 2 a — b, and for a; its equal, c', we have (2a_64-c'')' = (2a— 5)'+3 (2a— i)=c= + 3(2a— i)c*+c'. Performing the operations indicated in the second member, (2a_ J + c'')' = 8 a= — 12a=&+ 6a4^ — 6^+ 12a2c''— 12060" -\-3bl'c^ + 6ac* — 3be*-\-c^, Ans. IT. Eequired the term involving (f¥c' in the expan- sion of (a + i + cf. Ans. 90 a'U'cK 18. Expand {a -\- 2 b — c)' by the binomial formula. Ans. a^ -\- Qa'b -\- I2a¥ r\- 8¥ — Sa'c — I2abc — 12 J^c + 3 ac^ _j- 6 Jc" — c«. SERIES. 364i A Series is a succession of terms, so related that each may be derived from one or more of the others, in accordance with some fixed law. The simpler forms of series have already been exhibited in the progressions. 365, A Finite Series is one having a finite number of terms. 366i An Infinite Series is one having an infinite num- ber of terms, as l+a; + a:'' + a;« + SERIES. 289 367. An infinite series is said to be convergent when the sum of the terms cannot be made to exceed numer- ically some finite quantity, and it is said to be (divergent when the sum of the terms can be made numerically greater than any finite quantity. For example, by actual division we obtain 1 l—x = l-\-X-^X^-\-SI?-\- Now, if X is numerically less than 1, by taking a sufiB- ciently large number of terms, the numerical value of the series can be made to approach the numerical value of , to within any assignable limit. Thus, if x = ^, then , = 2 ; and if we take four terms, the series 1 X equals 1.8Y5. If we take five terms, it equals 1.9375 ; and, by taking terms enough, it may be made to differ from 2 by as small a quantity as we please ; or, the series is actually equal to 2, when carried to infinity. But if X is greater than 1, for instance, if it is equal to 2, the fraction equals — 1 ; whereas, if we take four terms, the series is 15 ; if five terms, 31 ; and so on, where, by taking terms enough, the difference between the series and the fraction may be made greater than any finite quantity. In the former case the series is conver- gent, and in the latter divergent. 368. Infinite series may be developed by the common processes of Division and ^fraction of Moots, and by other methods which it will now be our object to elucidate. UNDETERMINED COEFFICIENTS. 369. A method of expanding algebraic fractions and other expressions into a series, simple in its principles, and general in its application, is based upon the following 25 290 ALGEBRA, Theorem., If the series A -f- B x -|- x^ -j- D x* + is always eqwd to zero, whatever may he the value of x, the coefficients A, B, C, D, must each separately be equal to zero. For, Bince the series is to be zero, whatever may be the value of X, we may put a; = ; thus the series reduces to A, which must therefore itsglf be zero. Hence, removing this term, we have Bx + Go? +. Da? + always zero ; divide by x, and B -\-Gx -{- Dof -|- is always zero. Henqe, as before, we infer that B =: 0. Proceeding in this way, the theorem is established. Again, if the series A 4- Bx 4- Ox" -I- Dais 4- , . . . . and A' -\- Bx -^^ C'x^ + B'a? -f are always equal, whatever may be the value of x, then A — A'^ {B — B).x -\-{G—0) a? + ..... is always zero, whatever may be the value of x ; hence we infer that, A — A' = Q, B—B'z=0, C—C'=,0 That is, 77ie coefficients' of like powers of x in the two series are equal. The theorem here given is the Principle of Undetermined Coefficients, which are so called because, being assumed, the coeflScients are at first unknown, or undetermined. 1. Let it be required to expand into an infinite series. SERIES. 291 Assume 1+x = ^ + ^ar -t- Ca? -f Z?af + , (1) CTeariDg of fractions, and arranging the terms according^ to the powers of x, we have +.C af+.- aj + e But, by the principle of undetermined coefficients, the coefficients of like powers of x in the two members of the equation are equal. Therefore A = 1, whence A = I ; A + B=—l, '• B=i^2; C-\-0 — Q, " i?=:— 3;etc. Substituting these values in (1), we obtain 1 —X \-\-x l — 2x-\-2x^ — 2a^-\-. Ans. which may be readily verified by. division. ^. L&% it be required to expand series. Assume c -\-bx Clearing- of fi^actions-, a =^ Ac -{- Be -\-Ah c -\-hx = A-\-Bx-\-Ga?■\■D«?-{- ia.to an infinite x-{- Oc ■\-Bb x'-[-Dc -\- G'b a:^+, Equating the coefficient^, of lilsfe paw.em of x^ ^xe^ oibJiain^ A = a at c- c« Substituting, we haw® a G: aV aV +-^ ; D = —^ •> etc c -\-bx' SB' c +0^ = 0, Substituting, we have x^-^-AB -\-BG ■\-B0 -]-AB 3?-{-AE + BB ■\-BB -l-AB whence it . A = 1 B= 0. 2? = p ^ = — J ; etc. Ans. ^ l-\-x^ = 1 + ia;^ — ia:* + . . . which may be verified by evolution. Similarly may other like expressions be expanded. Hence, the RULE. Assume the given expression equal to a series with unknown coefficients, and clear the equation of fractions, or raise it to the power indicated. Equate the coefficients of like powers of the unknown quan- tity in the two members of the equation, find the value of each, and substitute it in the assumed series. Examples. 4. Expand , , into an infinite series. . a ax , a^ aa? , SEBIES. 293 5. Expand — -r—, into an infinite series. Ans. l + 2a: + 3a;^ + 4a^+ 5a;*4- 6. Expand (o* — a;')^ into an infinite series. . 3? X* 3? -^°^- " "" 2^ ~ 8^» "~ 16^' "~ Here, the operation will be shortened by assuming {a^ — 3?)^ z= A -\- Ba? -\- Cx" -{- Di^ -\- Ea? + . 1 -\- X 1. Expand _ into an infinite series. Ans. \-\-2x-\-23?-\-2a? + 8. Expand ; — . -3 into an infinite series. . 1 6 a; , 24 a" 80 a;' , Ans. -; T- H 5 i — r 9. Expand (a -\- x)~^ into an infinite series. 1 2a; , Sr" in? , Ans. -5 j-4 7 5- + 10. Expand r , into an infinite series. ^"^- -r + 9- + F7 + 8i + Here, taking the usual assumed series, and determining the coefiBcients, we obtain 1=0, which is absurd. That form, therefore, is not applicable in the present case. By division, however, we discover that the first term of the 1 x~^ required development is — , or — . We therefore assume ^^^ = Ax-'-{-B+Ox + I>x^-{-i:a?-\^ 25* 294 AL^BSA. 11. Expanid ■■ ■ j£'^-j_- j, into an infinite senes. Ans. l + 3a;4-4a;24-7a*+lla:* + , 2 12. Expand ^— , into an infinite series. 2^;-" I 4a-i , >8 , 16 a: , 32 a' , Ans. -g- i- -^ -r 27 "T" "sT ~r 243 "1" • DECOMPOSITION -OF tlATlONAL FRACTIONS. 370t The principle of undetermined Cofefllcients may be applied to the Decomposition of Fractions, whose denomi- nators are %spressed by factors, into partial fractions whose sum is equal the given fraction. 2 rt 'g 1. Let it be required to decompose ^ >. Ta'ctorhig a^ — x^ gives (a -\^o:) (a — x). . 2 a — a; A , B Assume now ^ — —5 t^ —r — r ' a' — a? a -\- X ' a — x Ctearing of fractions, and uniting, 2a — x= {A + B)a—(A — B)x Equating the coefficients, A — B= li Ib = ^ Substituting, 2a — X 3 1^ 1 a" — a^ "~ 2"(a^^ ^ S(a — *)' ^' 3 X — ^ 1 2. Decompose ^ , , -3 into its partial fractious. 3a;— 1 ^ , £ , C , D Assume : + ly a; ^^ a= ^ a; 4- 1 a;" (a; + 1)= x ' 3? ' a; + 1 ^^ (a; + 1)» SERIES. 295 Clearing of fractions, 3a:— 1 = Ax {x-\-iy -\- B(x-\-iy + Gx'ix+l) -\- Dx'. Equating the coefficients of like powers of x, A = 5, B= —I, C = —5, D — —4. Substituting, 3a;— 1 5 1 5 4 . Ans. i?(x-{-iy~ X a? a; + 1 (a; + If Hence, the following RULE. Resolve the denominator of the given fraction into ecs inah)/ factors as possible. Take the several factors as denominators of partial fractions, and assume A, B, C, etc., re^icUvehf, ai numerators. Find ike value of these numerators by the prin- ciple of undetermined coefficients, and substitute them for those assumed. ExAMPLiss. 3 -P 5 3. Resolve -5 ^s — j— 77; into its partial fractions. ar — 13 a; -(- 40 "^ . 19 10 Ans. 3 (a; — "8) 8 (a: — 5) * 4. Eesolve ;__^ into its partial fractions. Ans. - + - X — 1? 1,1 1 1 -|-x' 3? 6. Resolve y-^ -^7 v\ ™t° i*^ partial fractions. Qaf — i-) \x — i) A t A Ans. (a: — 2) 2 (a: — 1) ~ 6 (a; + 1) • 3 X -t- 2 6. Resolve ,. T.^, into its partial fractions. X {X -J^ I) A ^ 2 2 ■ 1 ^^^- i~i+T~(a;-|-iy"''(i + l)»- 296 ALGEBRA. BINOMIAL THEOEEM. ANY EXPONENT. 371 • We have seen (Art. 356) that when w is a posi- tive integer, (l+:rr= l+nx + "^'^-'^ x"--\- We shall now show that this formula holds when n has any value, ppsitive or negative, integral or fractional ; that is, we shall prove the Binomial Theorem for any exponent. Let n be negative, or = — n ; then, by actual division, (1 +«)-" = 7--| — r- = —-. 1 = 1 — nx 4- ^ ' '' (l-l-x)" l-\-n-x-\- ' and the exponents increase as before. Let M be a fraction, and be denoted by - ; then, since any power or root of 1 is 1, we assume (1 + a;)« = l-^Ax-\-Bx^ + Ox^-\- whence, (1 + a;)'' = (l-\-Ax + Bx^-\-Ox^-\- )« or, I -\-px-{- = 1+9^3;+ Equating the coefScients (Art. 369), we obtain p =: qA; whence - =: A; and (l+x)li = l^P-x + £x^-{-Gx'-\- Hence, whether the exponent n be positive or negative integral or fractional, the form of the expansion will be (l + x)" = l-}-nx-\-Bx^+ Cx''-^- Dx* + (1) where the coefficient of the second term is the same as the exponent of the power. Thus two terms of the series are determined. SERIES. 297 To find now the other coeflScients, we put x -\- z for x in (1), and, regarding (x-\-z) as one term, we shall have ll + (oc-{-z)Y = l+n(x-{-z)-^B(x+zy+Cix-\-zr + z=l-\-nx-{-Bx^-\-Gi,^ + I>x*-\- -\- (n+2Bx+3Cx'-\-iI>x?+...)z+... (2) which is true for all values of x and z. Kegarding (1 -(- a;) as one term, we shall have [(l+a=)+^]»= (l+a;)« + «(l+a:)«-i« + (3) But (Art. 369), the coefficients of z in each must be. the same, and n (1 +a;)"-i = n'-\- 2 Bx -{- ZOx^ + 4.Da? ■{■ ■ for all values of x. Multiplying by \ -\- x, m (1 + a;)» = » + 2 J5 + n a; + 3C + 2J5 x^ + 4=D + 3C »!» + Equating the coefficients of the same powers of x, we have 2B-\-n = n^, 3G+2B = nB, and 4Z) + 35 = nC. 2B= n(n—l); whence B — " ^" ~ ^^ ; 3C7=5(«-2); whence o- = " C» - i) (» - 2) . 4.D = C7(«-3); whence i? = "^""y.^^rg'V""'^ ' etc. ; whence, (!+,)«= l+„,+!L^,. + !L(!^i)i!^)^ + (A) and- the Binomial Theorem is established in the most general form. 298 ALGEBRA. 372. By substituting - for x, and multiplying both a members by a" (Art. 358), (A) becomes (B) Any power or root of any binomial may be developed by substituting in (A) or (B) thfe given values of n, x, and a. It may sometimes be convenient, however, when n is negative of 'fractional, to make use Of one of the following reduced forms. (l+:.)-|=l-f:.+^^.^-^-^^J^^^.'+.... (E) 373i If X is negative, all odd powers of x will be neg- ative, and therefore the even terms of the above series will change their signs. 374. It has been shown (Art. 360), if w be a positive integer, that the expansion will terminate with the (re+l)th term. If n be either negative or fractional, it is manifest, since no one of the factors of th6 rth term can eVef become 0, the series never terminates ; consequently the develop- ment furnishes an infinite series. It will be seen that each coefficient may be obtained from the preceding one, by multiplying it by certain fac- tors, easily determined. Examples. 1. Expand (1 -\^ x)~^ into an infinite seriesv SERIES. 2&9 Here, h == — 2 ; henbe, by formula (A)-, = 1 — 2x + Bx'' — 4:X^ -]-...'... , Ans. 2. Expand (l=— ar) •' into an infihite Beties. Here, n = — |; hence, hf formula (E) and Art. 373, = 1 + §a: + fa;^ + l^ic' + , Ans. 3. Expand (a -{- x)~'' into an infinite series. Here, the expansiotl biay b§ made direbtly by formula (B), or, after the expression is transformed into -5(1 + -) , (0) may be used. (a + ir)-2 = a^^ — 2a-^x-^Sa-*x^-^iet-^^^ 1 2x , 3a? 4a? , a' a' ' a* a* ' ' 4. Expand (1 -\-2x)^ into an infinite series. Ans. 1 +a; — ia;2 + i«' — !*'+ 5. Expand — — , or (1 — x)~^, into an infinite series. Ans. l-{-x-\-x'-\-si?-\-x*-\- 6. Convert j- into an infinite Series. (1 + a;)* Ans. l — ix + ^\x^ — ^\\x' + 300 ALGEBRA. T. Convert 2 (c -\- x)~^ into an infinite series. 2 4a; , 6a? 8_K» ^ "c^ """ "c* ?^ , 2 4a; , 6a? Sn? , ^ns. -__+ — _ — + 8. Convert —^= into an infinite series. y/l+a? Ans. l-^x^ + fa;^ — A^' + TVffa:* — 9. Convert a^ {a? — y)~* into an infinite series. Ans. a;+X + _|.+_:£ + _^^ + 10. Convert (a" — a:^)"* into an infinite series. Ans. a~^ + f a~i x^ + || a""^*" a;* -f- ^y^ a~ V- jc^ -(-... . 11. Convert - into an infinite series. •^°®" hy 26^ + 86* 166»+1286^~ /• 375. The binomial theorem may be employed to find the "approximate roots of numbers. Let iV be a number whose wth root is required, and suppose N = O" ±, b; then Nn = (a" ± b)n = ah ± -\l. If, now, — be a small fraction, the terms in the expan- sion of (l±^\" converge rapidly, and we obtain an approximate value of (l ± ~\n , and therefore of iV», by taking the sum of a few terms of the development. But if the expansion converges slowly, it will require the sum of many terms to insure a considerable degree of accu- racy. SERIES. 301 12. Find the approximate cube root of 31. Here, ^ 31 = ^ 3= + 4 = 3 (1 + ^Y ; and 3 (1 + ^)* = 3 + sV - ^\%r + ri^\°^ - T^%m^T + The value of the first term is + 3.00000 " second " -j- 0.14815 thii;d " —0.00131 fourth " +0.00060 fifth " —0.00006 The sum of the five terms =3.14138 ■which is the approximate value of .^31 to the fifth deci- mal place, as may be verified by evolution. 13. rind the approximate value of /^9 to five decimal places. Ans. 2.08008. , 14. Find the approximate value of /^ 39 to four decimal places. Ans. 2.080t. 15. Find the approximate fifth root of 260, or (3=+ It), to the fifth decimal place. Ans. 3.04085. 16. Find the approximate value of ^l08, or ^128—20, to the fifth decimal place. Ans. 1.95204. EEVERSION OF SERIES. 376i The Reversion of a Series is the process of find- ing an expression of the value of its unknown quantity in terms of another unknown quantity. Let there be given y^=ax-\-b3?-\-ca?-\- , to revert the series, or to find the value of x in terms of y. In order to express the value of x in terms of y, by the method of undetermined coefiScients, assume x = Ay-\-Bf+0^ + Dy'-^ 26 302 ALGEBRA. Substituting thid valiiS bt x in the equatioii y ■=. ax -\- bx^ + ''^ + <^^* + • ■ • it becomes y = CL-^y -\- aB -\-2hAG + 5J52 -|-3c^''5 4- dA* Equatiilg the like coefficients of y, \-2hAB + cA^ 2/* + tt4= 1 aB-{-lA'' = aC-\-ihAB-\-cA' = aD+ihAC-\-h&-^ZcA^B-\-dA'' = whence A ■ 1 a b a" „ ^ 26' — ac o° „ Tj ^ 56' — 5a6B4-a'rf Hence, 1 6 , , 25^—00 5W — 5abc + a'd !/" + • (A) If the even powers of x be wanting in the series, thus, y = ax-\-b3?-\-eaf-\^. . . » . ^ we assume for x a series containing the odd powers of y, and obtain 1 6 , , 36« — ac , 126'— 8a6c + aV (B) ill like maiiiief, by application of the principle of un- determined coefficients, or by the direct application of (A) and (B) as fbfmulas, other series of similai' forni ttiay be Reverted. ExAltPLBg'. 1. Eevert the series y^x-\-x^-\-a^-\-a!^-\- Ans. x==y-^f-j-f—y*-\. SERIES. 303 2. Eevert the series y = 2x-\-3x^^4:x'-\-5a;^-\- Ans. x = iy — ^f.-{. ^^^y' _ ^^2./ + 3. Revert the series y = l-\-x-{- — '^ — -{- — -\- , or y—l=x + - + ~ + — 4- Ans. a:= (y_l) _ 1 (y_l)2+^(y_l)3_ 1 (y_l)4+ 4. Revert the seriedy = a; — ^a^-j-|«''.^^a;'-j- Ans. a;:=y + ^^ + A2^ + VA/ + 5. Revert the series ax -\- bx^ -\- ca? -\- = my . a . I im^ — a^n „ , cm* — a'mp — 2an(bm' — a'n) ~ , SUMMATION OF INFINITE SERIES. 377t The Summation of a Series is the process of find- ing a finite expression equivalent to the series. Different series require different methods of summing, according to the nature of the series, or the law of its for- mation. Methods of summing arithmetical and geometrical Series have already been given (Arts. "336, S4l). Methods applicable to otteir series will now be treated. RECURRING SERIES. 378. A Recurring Series is one in wliich a certain number of coiisecutive terms, taken in any part of the series, bear a uriiform relation tO the term immediately following. Thus, l + 2a;-f 3a:2-|-4jt^-f is a recurring series, in which each term after the fifs't two is equal to the product of the first preceding tdfni 304 AL3EBKA. by 2x, plus the product of the second preceding term by — x'. These constant multipliers, 2x — x^, are called the scali of relation of the terms, and their coefiBcients, 2 — 1, con- stitute the scale of relation of the coefficients of the 379i A recurring series is said to be of the Jirst order when each term, commencing with the second, depends upon the one immediately preceding ; of the second order, when each term, commencing with the third, depends upon the two preceding ; and so on. If the series is of the first order, the scale of relation will consist of one term ; if of the second order, it will consist of two terms ; and, in general, the order and the number of terms in the scale will correspond. S80« To determine the scale of relation of a recurring series. 1. If the series is of the first order, it is a simple geometrical progression, and the scale of relatioo of the terms is the ratio of any two consecutive terms. 2. If the series is of the second order, let a, b, c, d, represent the consecutive coefiScients of the series, and p -\- q their scale of relation. Then we have c = pb-\-qal ^^^ d = pc -\- q b) to determine' p and q ; and reducing, we obtain ad — be , c' — bd p = 7T, and q = rs- ^ ac — ¥ ^ ac — V 3. If the series is of the third order, let a, b, c, d, e, /, represent the consecutive coefficients of the s'e- ries, and p -{- q -\- r their scale of relation. Then we have SERIES. 305 d = p c -\- qh -\- ra\ e=pd-]-qc-\-rby (B) f ^= pe -\- qd -\- r c J from which we can find p, q, and r. To ascertain the order of the series, we may first make trial of a scale of two terms, and if the result does not correspond with the series, we may try three terms, four terms, and so on, till the true scale is found. If we assume the series to be of too high an order, one or more terms of the scale will be found equal to 0, and the~ others will be the true scale. 381. To find the sum of a recurring series whose scale of relation is known. Let a + Ja: -j- ca? -f- d-i? -)- ex* -\- be a recurring series of the second order. Denoting the sum of n terms of the series by «, we have S=a-lrhx-\-ca?-^d3?-\- + Za:»-i p Sx ^^pax-\-phii?-\-pca?-{- -\-ph3?~^-\-plaf q Sx^ = qaa?-\-qh3?-\- -\-^j ^~^ + qh3^-\-qlx^\ Subtracting the last two equations from the first, we have S^-p Sx — q So? = a-\-hx — pax — pla? — qka^' — qlaf*^; for all the other terms of the second member disappear by means of the relation expressed in equations (A). Therefore we have ^ _ a+ (6— pa)a;— (p? + (y^)2:" — g^a:''+' ,^ 1 — px — qn? ' ^ ' the formula for finding the sum of n terms of a recurring series of the second order. But if n becomes infinite, and the series is converging, then the terms which involve as" and a;"+^ must become 0, and we have 26* 306 AL=&EBKA. « _ n + (b — pa)x /j)x the formula for finding the sura of an infinite recurring series of the second order. If g = 0, then 6 = /) a, and we have 1 — px ^ ' the formula for finding the sum of an infinite recurring series of the first order (Art. 342). in like manner we obtain .tj _ g+ Q)—pa)x-\-(c—ph—'qa)^ ,™ '^ ~ l—px — qx' — rx' ' ^^> the formula for the summation of an infinite recurring series of the third order. A recurring series, like other infinite series, originates from an irreducible fraction, called the generating fraction. The summation of a recurring series, therefore, repro- duces the fraction. Prom formulas (D), (E), and (F)j. we infer that the number of terms in the numefatof of that fraction is the same as the order of the series, and in the denominator it is one greater. Examples. 1. Find the sum of 1 -|-2a: + 8a:2 + 28a^-fl00x* + . Here, a = 1, J = 2, c = 8, rf = 28, c = 100, etc. Substituting these values in formulas (A), we obtain jo = 3, and q z^ 2 ; whence, by formula (D), we find _ 1 + (2 - 3) :r _ \.— -± ^ ~ 1 — 3x— 2x2,~~ i — 3x—2a?' Series. SOT 2. Find the sum of 1 + 2a; -|- 3a;'' + 6a!' + 8a;* + ■kna. /^ , . 1 — X ■ — ir 3. Find the sum of t- r; a; + ^s- a;^ n-a^ + Ans. s—, • 4. Find the sum of4 + 9a; + 21a;« + 51a;' + . 4 — 11 a; ■^"^- l_5a;+"6:K»- 5. Find the sum of 1 + 3a; + Sa;^ + Ya;* + . Ans. ,, w - (1 — xf 6. Find the expression from which is derivable 2 — a + 2a2— 5a» + 10a* — l^c^ + 2 + 5a + 5a' -^IIS. (i_|_„)3 •■ 1. Find the expression from which is derivable 3 + 5a; + 7a;^+13a;» + 23a;<+45a;» + . 3 — a: — ear* '^n*' ■l — 2x — x' + ^3?- DIFFERENTIAL METHOD. 382. The Differential Method is the process of finding any term, or the sum of any number of terms, of a reg- ular series, l)y means of the successive diflerenc6s of its terms. 383. If, in any series, we take the first tertn frotn the second, the second from the third, the third from the fourth, and so on, the remainders will form a new series called the Jirst order of differences. If the difiiBrelices be taken in this new series in like manner, we obtain the second o^der vf differences, aiid so on. 308 ALGEBRA. Thus, if the given series is 1, 8, 21, 64, 125, , the successive order of differences will be as follows : — 1st order, 1, 19, 31, 61, 2d order, 12, 18, 24, 3d order, 6, 6, 4th order; Hence, in this case, there will be only three orders of differences. 384 • To find any term of a series. Let the series be a, b, c, d, e, then, the first order of differences is h — a, e — h, d — c, e — d, the second order of differences is c — 2h-\-a, rf— 2c + 5, e — 2d-\-c, the third order of differences is rf— 3c + 3i — a, e — 3cZ+3c — 5, . ... . . the fourth order of differences is e — 4cf+6c — 4i + a, where each difference, although a compound quantity, is called a term. Let, now, d^, d^, d^, di, represent the first terms of the several orders of differences. Then/%; di^b — a; whence b=. a-\- d^; d2 = c — 2b + a; " c = a + 2rfi+l ds = d—3c-\-3b — a; " d= a-\-3di-\-'m^-\-ds; di = e_4rf+6c— 46+a ; " e = a + 4^1+^^+4^3-1-^4; etc. . etc. SERIES. 309 where the coefficients of the value of h, the ' second term in the given series, are the coefficients of the Jirst power of a binomial ; the coefficients of the value of c, the third term, are the coefficients of the second power of a binomial ; and so on. Hence, we infer that the coefficients of the nth term of the series are the coefficients of the (ra — l)th power of a binomial ; whence, representing the nth term of the series by T„, we shall have When the differences at last become 0, the value of the nth term can be obtained exactly ; but, in other cases, the result will be merely an approximate value. 385i To find the sum of any number of terms of a series. Let the series be a, b, c, d, e, (1) Assume the series .0, + a, a-\-b, a-\-b-^c, a-^b-\-c-}-d, (2) where the (re -|- l)th term is obviously equal to the sum of n terms' in the proposed series, and the first order of differences is equal to series (1). Hence, the (re -j- l)th order of differences in series (2) is the nth order in se- ries. (1). If, then, we substitute in formula (A), for a, re -j- 1 for n, a for d^, d^ for d^, etc., and denote the sum of n terms of the proposed series by S^, we shall obtain £, I nCiir-\y J , n(n — 1) (n — 2) , , ,„. Sn = na-\ 'Y:^^-\ 1.2.3 ^ "^ + ^^^ EXAUPLES. 1. Find the 12th term of the series 2, 6, 12, 20, 30, SIO ALGEBEA. ' OPERATION. 2, 6, 12, 20, 30, 4, 6, 8, 10, 2, 2, 2, 0, 0, Hence a = 2, <^ = 4, rfj = 2, and w ^ 12, Then, by substituting in formula (A), we have for the 12th term, 2+(12-l)4+^^^=^ii^^=?^2 = 2+44+110 = 156, Ans. 2. Find the sum of 8 terms of the series 2, 5, 10, 11...... Here, we find a = 2, d^ = 3, 4 =- 2, and w = 8. Then, by substituting in formula (B), we have >^s = 8X2 + Mi^3+ii?^piIi|=a2 = 16 + 84+112 = 212", Ans. 3. Find the first term of the fifth order of differences of the series 6, 9, 11, 3S, 63, 99, Ans. 3. 4. Find the first term of the sixth order of differenafis. of the series 3, 6, 11, 11, 24, 36, 50, 12,, Ans. ^^14., 5. Find the seventh term of the series 3, 5, 8, 12, IT, Ans. 30., 6. Sum the series 1, 4, 10, 20, 35, to the 12th term. Ans. 1365. 1, Sum the, series 1,. 2, 3^ 4, 5, to the 100th term. Ans. 5050. 8. Find the 15th term of the series 1^ 2^ 3^ 4^ Ans. 225. 9. Sum l' + 2'+3' + 4' + 5'+ to the wth term. Apfi.. !:;(«* + 2 »i« + «0- SEEIES., ail 10. Sma. 1 4- ^4 4, 3* 4- 4^ -f. ^^ ^. 65 + to the mth term. * n' . n* , re' re 11. If shot be piled in the shape of a pyramid, with a triangular base, each gide. of whieh exhibits 9 shot, find the number contained in the pile. Ans." 165. 12. If shot be piled in the shape of a pyramid, with a square base, each side of which exhibits 25 shot, find the number contained in the pile. Ans. 5525. II^TEEPOLATIQN. 386i Interpolation is the process of introduoi'iig' between terms of a series other terms conforming to the law of the. series, Ite usual] application is iYi finding iiitermediate numbers between those given in Mathematical Tables, which may- be regarded as a series of equidistant terms. 387. The interpolation of any intermediate term in a series, is essentially finding the, nth term of the s^jcjep,, by the differential method (Art. 384). Thus, Let t represent the term to be interpolated in a series of equidistant terms, and p the distance the term t is removed from, the first term:, «j tb^t is,,p==it — 1. Then, by substituting p for n — 1 in formula. (A), Art. 384, we have the- formula of interpolation, t = a+pd,+?il=fid,+P^l[Pp^d,+ EXAUFLES. 1. Id tjie serigs xV- A', tVv A* iV. .■•••, fin* th^ mj44Je tepQ betwfiqn ^V *9<1 ^- 312 ALGEBRA. Here, the first differences of the denominators are 1, 1, 1, 1, The second differences are 0, 0, 0, whence, c^ ^ 1, and ^2 = 0. Since each interval between the terms is to be reck- oned as unity, the distance from the first term to the required middle term is 2^ intervals, or p = 2^. Make a = 13, the denominator of the first term ; then, by the preceding formula, 13 + 21 X 1 = 15i, and 1 -t- 15^ = /x. or the proposed middle term. 2. Given the square root of 94 = 9.69536, of 95 = g.Uete, and of 96 = 9.'79'796, to find the square root of 94i. Here, the first differences are .05143, .05117 ; and the second difference is ■^.00026; whence, rfj = .05143, and d^ = — .00026. Since the distance is |- of the first interval, p = ■^. Hence, \/"94| = 9.69586 + ^ (.05143) — ^'j (—.00026) = 9.69586-1- .012864-.00002 = 9.'70824, Ans. 3. Given the cube root of 64 = 4, of 65 = 4.0207, of 66=4.0412, and of 67=4.0615, to find the cube root of 66.5. Ans. 4.0514. 4. Required the number of miles in a degree of longi- tude in latitude 42° 30', the length of a degree of longitude LOGARITHMS. 313 in latitude 41° being 45.28 miles, in latitude 42° being 44.59 miles, in latitude 43°, 43.88 miles, and in latitude 44°, 43.16 miles. Ans. 44.24 miles. 5. Given the cube root of 45 = 3.556893, of 47 = 8.608826, of 49 = 3.659306, and of 51 = 3.708430, to find the cube root of 48. Ans. 3.634241. 6. If the amount of $1 at 7 per cent compound interest for 2 years is $1,145, for 3 years is f 1.225, for 4 years $1,311, and for 6 years $1,403, what is the amount for 4 years and 6 months ? Ans. $1,357. LOGARITHMS. i. The Logarithm of a number is the exponent of the power to which some fixed number, called the base, must be raised, to produce the given number. Thus, sup- pose a' = n, then X is the logarithm of n to the base a, and may be written X = ,Iog„ w, where log^ is read logarithm to base a. The subscript is usually omitted when the base is readily understood. 389i If, in the preceding equation, a remaining fixed, n be supposed to assume in succession all positive values, the corresponding values of x, taken together, will con- stitute a system of logarithms. But, since a may be made any positive number greater than unity, there may be different systems of logarithms. 27 314 ALGEBRA. THE COMMON SYSTEM. 390. The base, in the common system of logarithms, is 10. Hence, since 10" = 1, is the logarithm of 1 ; 10^ = 10, 1 " " 10 ; ■ 10^ = 100, 2 " " 100 ; 10' = 1000, 3 " " 1000 ; etc. etc. It thus appears that, in the common system, the loga- rithm of every number between 1 and 10 is some number between and 1 ; that is, a proper fraction. The loga- rithm of every number between 10 and 100 is some num- ber between 1 and 2 ; that is, 1 plus a fraction. The log- aritlim of every number between 100 and 1000 is some ixumber between 2 and 3 ; that is, 2 plus a fraction ; and so on. 391 1 By means of negative exponents the application of logarithms may be extended to numbers less than 1. Thus, since IQ-^ = 0.1, — 1 is the logarithm of 0.1 ; 10-2 = 0.01, —2 " " 0.01; 10-' = 0.001, —3 " " 0.001 ; etc. etc. From this it appears that the logarithm of every num- ber between 1 and 0.1 is some number between and — 1 ; that is, — 1 plus a fraction. The logarithm of every number between 0.1 and 0.01 is some number be- tween — 1 and — 2 ; that is, — 2 plus a fraction. The logarithm of every number between 0.01 and O.OOl is some number between — 2 and — 3 ; that is, — 3 plus a fraction ; and so on. 392. The logarithms of all numbers which are not exact powers of 10, being incommensurable with those LOGARITHMS. 315 numbers, can have their values only approximately ob- tained, and therefore are usually expressed by means of a decimal fraction. 393. The Chakacteristic of a logarithm is its integral part. The decimal part is sometimes called the mantissa. 394f 7%e charaeteristic of the logarithm of any integer, or MIXED NUMBER, is ^ositive, and equal to the number of integral places less one. This is evident from Art. 390. 395t 27ie characteristic of the logarithm of any decimal FRACTION is negative, and numerically one more than the number of ciphers between the decimal point and the first significant figure. This is evident from Art. 391. PROPEKTIES OF LOGARITHMS. The logarithm of 1 in any system is equal to 0. For, in the equation a^ = m, where x is the logarithm of m, whatever be the value of the base a, make a; = 0, and we have a"- = m = 1, or log 1 = 0. 397. The logarithm of the base itself, in any system, is 1. For, make ar = 1 in the equation a'' = m, and we have a^ :^ m = a, or log a = 1. 398. A negative quantity has no real logarithm. For, since the base a is positive, all its powers must be positive (Art. 202). 399. ^e logarithm of a product is equal to the sum of the logarithms of its factors. 316 ALGEBEA. For, let m = flf*, and « = a" ; then, multiplying these equations, member by member, we have wra = a'^a" = «*+". Therefore, log (mn) = a; -|- y = log m -f- log n. 400. The logarithm of a quotient is equal to the logarithm of the dividend diminished hy that of the divisor. For, let m = a", and n = a" ; then, dividing the first equation by the second, member by member, we have m^ a^ ,_y n oM Therefore, log I — j=a; — y = log m — log n. 401 • The logarithm of any power of a number is equal to the logarithm of the number multiplied by the exponent of the power. For, let m = a* ; then, raising both members to the rth power, we have m^ = (a'Y = a". Therefore, log (m"") := xr = r log m. 402i Hhe logarithm of any root of a number is equal to the logarithm of the number divided by the index of the root. For, let m = a' ; then, extracting the rth root of both members, we have X li/m = if/ a" = o^ Therefore, log (^"^) =- = ^l^^. LOGARITHMS. 317 403 • The arithmetical mean of the logarithms of two num- bers is the logarithm of the geometrical mean of those numbers. For, let a* = m, and a" = n; then 0*+" =: WW, and a ^ ■=. sf mn, or, I°g"' + t°g» 3^ 1 ^^^. (^rtg. 33^^ 343.) 404i From the foregoing, it is evident that multipli- cation, division, involution, and evolution may be per- formed by means of logarithms ; hence they are often of great practical utility in shortening numerical calcula- tions. In computations by logarithms, negative quantities are used as if they were positive (Art. 398), the sign of the result being determined irrespective of the logarithmic work. COMPUTATION OF LOGAEITHMS. 405t The properties just demonstrated are true of every system of logarithms ; but their application to numerical calculations supposes the construction of a logarithmic table. A table of logarithms may be computed by means of Art. 403 ; but the following is a more convenient method. Let us resume the equation cP = n, (1) where x is the logarithm of n, to the base a. Assume a = \-\-m, and w = 1 +^, then (l+»i)» = l+;j. (2) 27* 318 ALGEBEA. Eaising both members of this equation to the nth power, we have for every value of n. Expanding both members of the last equation by the Binomial Formula, we obtain = i+«;>+"-^/>'+ "^"7.yr'^ ^' + Omitting 1 from both members, dividing by n, and fac- toring the first member, we find - ("^ + ^^ '"^ + ^"' ~l .^^"3' ~ '^ "^ + ) Make « = 0, and the last equation reduces to ^("^-Y + T- ) =P-2H- (2) But, by equation (2), X = log (1 -\-p). Hence, if we make the reciprocal of the other factor of the same member equal to M, equation (3) becomes logil+p) = log n = M(p-P~+^- ), (A) which is an expression for the logarithm of any number n. This expression is composed of two factors ; the one within the parenthesis, dependent upon the number, and the other, M, dependent upon the base of the system.. The factor dependent upon the base is called the modulus of the system of logarithms. LOGARITHMS. 319 THE NAPIERIAN SYSTEM. 406. The Napierian system, or that invented by Baron Napier, has the modulus arbitrarily assumed as unity, or M= 1. Denoting the logarithms in the Napierian system by lege, we have, from formula (A), l0ge(l+;') =J»-|-'+f'- (B) This series may be applied to the calculation of Na- pierian logarithms, if jo is a proper fraction, but unless p be very small, it converges so slowly that a large num- ber of terms are required to obtain the accuracy desirable ; and if p be greater than 1, the series becomes divergent, and consequently altogether unsuitable. It can, however, be converted into another more con- venient series. Thus, l0ge(l+i»)=p-f'+f'- (1) Substituting — p for p in this equation, \og,(l-p)=-p-^-Pl- (2) Subtracting equation (2) from equation (1), a,nd observ- ing that log, (1 +p) - log, (I -p) = log, (J-+^), we obtain iog,(ii|) = 2(j+f+|:+ ) (3) Let, now, , """^ = 1 4- - , or p = - — r-- ; then 1 ~p '2 ^ 2z-|- 1 ' log, (1 +j) = loge ('-±-') = loge (^ + 1) - log, Z. 320 ALGEBRA. Substituting these values in equation (3), we have log.(.+l)-log..=2(^ + 3-^3 + 5(24:i)a+--)(C) 407. This series, (0), may be employed in computing the Napierian logarithm of any number, when the like logarithm of the preceding number is known. But, by Art. 396, the logarithm of 1 is ; therefore, making z -\- 1 successively equal to the prime numbers 2, 3, 5, 1, etc., and observing that the logarithm of any composite number is equal to the sum of the logarithms of its prime factors (Art. 399), we obtain the following Napierian Logarithms. '°^^2 = 2(l+3i3, + ^,+ A, + ) =0.69314t log. 8 = log,2 + 2(i + 3-^3+54a+^+.-) - 1.098612 log, 4 = 2 loge 2 — 1.386294 log, 5 = ]og,4+2(-;+3-^+^+^,+-) = 1-609438 log, 6 = log, 2 + log, 3 = l.MlTSg log,Y=log,6+2(i+^ + ^,+ ....)-1.945910 log, 8 == 3 log, 2 == 2.079442 log, 9 = 2 log, 3 — 2.197225 log, 10= log, 2 + log, 5 =2.302585 etc. etc. etc. Napierian logarithms are sometimes called hyperbolic log- arithms, from having been originally derived from the hyperbola. They are also sometimes called natural log- arithms, from being those which occur first in the inves- tigation of a method of calculating logarithms. LOGARITHMS. 321 COMMON LOGARITHMS. 4©8i To compute common logarithms, it is necessary to find the value of M, the modulus of the system, when the base is 10. Let 10" = n, (Ijr where x is the common logarithm of n. Then, taking the Napierian logarithm of each member, by Art. 401 we have X loge 10 = logs «, whence x = g^ = log, n X j^, (2) where = — — is the factor depending upon the base (Art. 405), and, therefore, the modulus requiued. Hence, M log. 10- But, by Art. 407, loge 10 = 2.302585, whence, M= ^^^^^^.^ = ASmi5. (3) 409i The common logarithm of a number may he derived from the Napierian logarithm of that number, by multiplying it by .4342945. For, by equation (2), Art. 408, the common logarithm of a number is equal to the Napierian logarithm of that number multiplied by the modulus of the common system. Thus, representing a common logarithm by log, log 2 = 0.693147 X .4342945 = 0.301030 logs == 1.098612 X .4342945 = 0.477121 etc. etc. etc. 322 ALGEBRA. 410. A table of common logarithms, exact to any num- ber of decimal places, may also be constructed directly from the series already employed. For, multiplying both members of (0) by the modulus of the common system, we have log (z + 1) — log z = •868589(2-^^+3-^^3+5(^+7(2^^ (D) Make « + 1 successively equal to 2, 3, etc., and we find log 2 = .868589 (1 + ^^ + ^ + ) = 0.301030 logs = log2+.868589g + 3i^ + g-^,+...) = 0.mi21 log 4 = 2 log 2 = 0.602060 log 10= log 2 + log 5 =1.000000 etc. etc. etc. 411a In the common si/stem of logarithms, if the logarithm of any number be known, we can immediaiely determine the logarithm of the product or quotient of that number by any power o/" 10. For, log (m X 10") ;= log m + log 10" = log m + n, and log (t^) = log m — log 10" = log m — n. That is, if the logarithm of any number be known, we can determine the logarithm of any number which has the same succession of figures, but differs merely by the The logarithm of 15 is 1.176091 1.5 tt 0.176091 .15 it T.176091 .015 it 2.176091 LOGAEITHMS. ' 323 Here, the negative sign placed over the characteristic indicates that it alone is negative, the decimal part being always positive. Common logarithms are often called Briggs's logarithms, from the name of their inventor. Base of the Napierian System. 412. Let ^ z= 10, where e is the Napierian base, and x is the Napierian log- arithm of 10. Then, taking the common logarithm of each member, we have a; log e = 1, or log e = j^^ = .4342945 (Art. 408.) That is, the modulus of the common system is the com- mon logarithm of the Napierian base. From Arts. 409, 410, it is evident that the base of the Napierian system is between 2 and 3. Calculated to seven places of deci- mals it is found that e = 2.7182818. TABLES OF COMMON LOGARITHMS. 413. A Table of Logarithms usually contains all the whole numbers between 1 and a given number, with their logarithms. The characteristics of the first 100 numbers are inserted, but for all other numbers the decimal part only of the logarithms is given, while the characteristic is left to be supplied by inspection (Arts. 394, 395). The numbers are in the column headed N, and their logarithms, or the decimal parts of their lonfarithms, are opposite on the same line. Sometimes there is a column headed D, in which are the mean or average differences of the ten logarithms against which they are placed. 324 ALGEBRA. TABLE I. N. Log. N. Log. N. Log. N. Log. N. Log. 1 0.000000 21 1.322219 41 1.612784 61 1.785330 81 1.908485 2 0.301030 22 1.342423 42 1.623249 62 1.792392 82 1.913814 3 0.477121 2S 1.361728 43 1.633468 63 1,790341 83 1.919078 4 0.602060 24 1.380211 1.397940 44 1.6-13-153 64 1.806180 84 1.924279 5 0.698970 25 45 1.653213 65 1.812913- 85 1.929419 6 0.778151 26 1.41497S 46 1.662758 66 1.819544 86 1.934498 7 0.843098 27 1.431364 47 1.672098 67 1.826075 87 1.939519 8 0.903090 28 1.447158 48 1.681241 68 1.832509 88 1.944483 9 0.954243 29 1.462398 49 1.690196 69 1.838849 89 1.949390 10 1.000000 30 1.477121 50 1.698970 70 1.845098 90 1.954243 11 1.041393 31 1.491362 51 1.707570 71 1.851258 91 l.k901I 12 1.079181 32 1.505150 52 1.716003 72 1.857332 92 1.963788 13 1.113943 33 1.518514 53 1.724276 73 1.863323 93 1.969483 14 J.146128 34 1.531479 54 1.732394 74 1.869232 94 1.973128 15 IS 1.176091 35 1.544068 55 1.740303 76 1.875061 95 1.977724 1.204120 36 1.656S03 66 1.74S18S 76 1.880814 96 1.982271 n 1.230449 37 1.568202 67 1.755875 77 1.886491 97 1.986772 IB 1.255273 38 1.579784 68 1.763428 78 1.S92095 98 1.991226 19 1.278754 39 1.591065 69 60 1.770852 79 1.897627 99 1.993633 20 1.301030 40 1.602060 1.778161 80 1.909090 100 2.000000 TABLE IL N. 1 i 2 3 4 j 5 6 7 8 9 D. 100 000000 000434 000808 001301 001734 002166 002598 003029 003461 003891 432 1 4321 4751 5131 5609 6038 6466 6894 7321 7743 8174 428 2 8600 9026 9451 9876 010300 010724 011147 011570 011993 012415 424 1 ^ 012837 013259 013680 014100 4521 4940 6360 6779 6197 6616 420 4 7033 7JS1 j 7863 8284 8700 9116 9532 9S47 020361 020776 410 5 021189 021603 022016 022423 022841 023252 023664 024075 4486 4896 412 6 5306 6715 ; 6125 6333 6942 7350 7757 8161 8571 8978 409 7 9384 9789 030195 030600 031004 031108 031812 032216 032619 033021 404 8 033424 03J826 j 4227 4628 6029 6430 6830 6230 6629 7028 400 9 110 7426 7825 1 8223 85)0 9017 9111 9811 040207 010602 040998 397 393 041393 041787 1042182 042576 04:969 043362 013756 044148 044610 044932 1 5323 5714 6105 6493 1 6885 7275 7664 8053 8442 8830 390 2 9218 9606 9993 050380 ' 050766 031153 051633 031924 052309 052694 386 3 063078 053463 053846 4230 4613 4990 6378 5760 6142 6524 333 4 6905 7286 ; 7666 8046 , 8426 8805 9185 9563 9942 060320 379 5 060SS8 061075 ,061452 061829 002206 062682 062953 063333 063709 40S3 376 6 4468 4331 ' 5;0S 5580 5953 6326 6699 7071 7443 7815 373 7 8186 8557 ! 8928 9298 9663 070038 070407 070776 071145 071514 370 8 071882 072260 '■ 072617 072986 073352 3718 4085 4451 4816 5182 366 9 5347 6912 1 6276 6010 7004 7368 7731 8094 8457 6819 363 120 079181 079543 j 0799P4 080266 1 080626 030987 031347 031707 082067 082428 360 LOGARITHMS. 325 The logarithms in the preceding tables are taken in the common system, and are sufficient to illustrate the use of a larger table, such as that appended to the author's Trigonometry. In Table II. the first two figures of sev- eral successive numbers, and of the logarithms, when the same, are left to be supplied. 4I4i Numbers may be formed in Table II. by annex- ing each figure at the top of the table to the figures in the column at the left, and the decimal of the logarithm of the number formed will be in a line with the figures taken at the left, and in the column with the figure taken at the top. Thus, The logarithm of 1129 is 3.052694 10.95 " 1.039414 .1209 " T.082426 415. When a number is not in the table, if its factors are there, its logarithm may be found by taking out from the table the logarithms of the factors, and adding them together (Art. 399). Thus, since 1248 = 1208 X 6, log 1248 = log 1208 + log 6 = 3.082061 4- 0.118151 = 3.860218. 416. When a number is intermediate between those in the table, its logarithm may be found by interpolation. For, the tabulated logarithms are of the nature of a reg- ular series, hence they may be interpolated by the for- mula in Art. 381. Thus, Let it be required to find the logarithm of 11.881. Here, 11.881 is a number between 11.88 and 11.89 ; then log 11.89 — log 11.88 = 1.015182 — 1.014816 = .000366. Hence, put d^ = .000366, and p = -1, and we have 28 326 ALGEBRA. a = log 11.88 = 1.074816 pdi = .000366 X -t = .000256 a-\-pd^ = logll.SSt = 1.0750t2 This process is based upon the supposition that the differences of logarithms are proportional to the differ- ences of their corresponding numbers, which is not strictly correct, yet sufficiently exact for practical pur- poses. The difference d^ might have been taken directly from the table, being found in a line with the logarithm of 11.88, and in column D. 417. When a given logarithm is intermediate between those found in the table, its corresponding number may be found. For, this is obviously the converse of the last case, and, therefore depends upon a process based upon a like sup- position. Thus, Let it be required to find the number corresponding to the logarithm 1.075072. The decimal of the given log is .075072 " the log next less " .074816, corresp'ng no. 11.88 Difference, 256.0 _ Diff. betw. log 11.88 and log 11.89, 366 Whence, the required number =: 11.887 Examples. 1. Find the logarithm of 1095. Ans. 3.039414. 2. Find the logarithm of .0735. Ans. 2.866287. Here, .0735 = 1^^ (^"^t^- ^^^' ^1^). 3. Find the logarithm of 1728. Ans. 3.287543. LOGARITHMS. 327 4. Find the logarithm of .00016. Ans. 4.204120. 5. Find the logarithm of 2.34. Ans. 0.369216. 6. Find the logarithm of |J. Ans. IMdOIL Here, by Art. 400, log f J = log 31 — log 94. 1. Find the logarithm of ^^j. Ans. 2.44TT37. 8. Find the logarithm of 11^. Ans. 1.245'727. Here, IT^I = W- ; log V;^ = log 81 + log 5 — log 28. 9. Find the logarithm of 1.03t5. Ans. 0.015988. 10. Find the logarithm of 104.857. Ans. 2.020597. 11. Find the number whose logarithm is 0.025306. Ans. 1.06. 12. Find the number whose logarithm is 3.010724. 13. Find the number whose logarithm is 0.009663. Ans. 1.0225. 14. Find the number whose logarithm is 3.009264. Ans. 1021.56. EXPONENTIAL EQUATIONS. 418t An Exponential Quantity is one that has an un- known quantity for an exponent ; as, 1 a", a", af, etc. 419i An Exponential Equation is one containing an exponential quantity, as a* = b, a;* = a, etc. Equations of this kind may be readily solved by log- arithms. 328 ALGEBRA. 420. To solve an exponential equation in the form of a" = b, take the logarithm of each member, and we have X log a = log b ; , log 6 whence, x = j^^. Examples. 1. Find the value of a; in the equation 2* = 1024. Here, a: log 2 = log 1024 ; 1 log 1024 3.010300 Whence, x = -"^p- = ^^^^^ = 10, which may be verified by involution. 2. Find the value of x in the equation 3* = 15. Ans. 2.465. 3. Find the value of x in the equation 5" = 100. Ans. 2.861. 4. Find the value of x in the equation ab''=:c. Ans. '°g^-;°g". log 6 4 5. Find the value of x in the equation 3* = 9. Here, . raisi-ng both members to the a;th power, we have 3* = 9", or 81 = 9" ; and log 81 = a: log 9 ; whence, a: = 2, Ans. 6. Find the value of x in the equation 6" =: ^. Ans. 3.911. T. Find the value of x in the equation 9'" = 1000, there being given the logarithm of 3.14 = 0.496930. LOGARITHMS. 329 • Here, x is the exponent of the exponent 3 ; hence, taking the logarithm of each member, we have 3* log 9 = log 1000 ; 3' = -f~y^ = 3.14 ; X = — f ' = 1.04, Ans. logs 8. Find the value of x in the equation - — -^^ = n. Ans ^°g('^» — g )— '°g <' log 6 9. Prom the formulas given in Arts. 340, 341, obtain the value of n in terms of a, r, and I. Ans. n = '°g'-'°g° +l. log r ' 10. Given a, r, and iS', to find n. Ans. n = '°g ['' + ('•— 1) -S] — log a log r 11. Given a, I, and S, to find n. Ans log^-logg ■ ^ ^°^- log(S-a)-Iog(S-0^ 12. Given r, I, and S, to find n log r ' 421 • To solve an exponential equation iu the form of yf = a. Taking the logarithms of both members, we have X log X = log a, where the value of x may be obtained by trial, as follows : Find by trial two numbers nearly equal to x, substitute them for x, and note the results. Then, As the difference of the results is to the difference of the two assumed numbers, so is either error to the correction required in the corresponding assumed number. 28* 330 ALGEBRA. Examples. 1. Required the approximate value of x in the equation a;" = 100. Here, ^ x log x = log 100 ^ 2 ; and upon trial x is found to be between 3 and 4, but greater than 3.5, and less than 3.6. Assuming, then, 3.5 and 3.6 for the two numbers, jire have, riEST SUPPOSITION. SECOND SUPPOSITION. log X = log 3.5 = 0.544068 log x = log 3.6 = 0.556303 Multiplying by 3.5 Multiplying by 3.6 a; log a; = 1.904238 a; log a; = 2.002691 Diff. results : Diff. assumed nos. : : Error 2d result : Its correction. .098453 : 0.1 :: .002691 : .00273 Whence, a; = 3.6 — .00213 = 3.59727. 2. Required the approximate value of x in the equation x" z= 10. Ans. 2.506. 3. Find the value of x in the equation x'^ = 256. Ans. 4. 4. Required the approximate value of x in the equation x" = 5. Ans. 2.129. COMPOUND INTEREST AND ANNUITIES. 422t Logarithms can be used very advantageously in many of the operations relating to Compound Interest, Annuities, and Increase of Population, as well as Geo- metrical Progression (Art. 404) ; but when the unknown quantity occurs as an exponent, they become essential to a direct solution (Art. 420). LOGARITHMS. 331 423i To find the amount of a given sum for any time, at compound interest. Let P = any principal, in dollars, r = the interest of one dollar for one year, and n = the number of years P draws interest. If -4 = the amount of P for n years, at the rate r, we have yl = P (1 + r)» ; (A) whence, log A = log P -\-n log (1 -\- r) (1) log P = log ^ — » log (1 + r) (2) log(l+>-)= ^°^^7^°^^ (3) log A — log P log (1 + r) (4) 424. Suppose the interest is due more frequently than onee a year, for example, quarterly ; then 4 n will be the number of payments and - the rate of interest, and the amount of P in « years becomes i^+d' Similarly, if interest be due q times a year, at com- pound interest, we have i^+d" (B) 425i To find the amount of an annuity, left unpaid for any number of years, at compound interest. Let a represent the annuity, and A its amount at the end of n years, at the rate r. At the end of the first year, a is due ; at the end of the second year, a-\-a{l-\-r) is due; and so on. Hence, the amount due at the end of n years is • 332 ALGEBRA. a + a (1 + r; + a (1 + r)= + + « (1 + »•)"-', a geometrical series, whose sum gives (Art. 341) ^ ^ "J(i + 0"-i]. (C) 426i ^o _^mrf whence P = ° ."T , Z" . r (1 -\- r)" 427t Formula (D) may be thus expressed, p __ f a (D) 7-(l+r)"- . Suppose, now, the annuity to be perpetual, or « = oo , and we have ^="- (E) 428t Problems relating to the natural increase of popu- lation in a country, being analogous to those in compound interest, are solved by the same formulas. Problems. 1. Find the value of n in (0). Ans. „=:J°g(°+;^)-I°If. log (1 + r) 2. Find the value of n in (D). Ans. „ = 1^^- '°g>-^). log (1 + r) LOGARITHMS. S33 3. In how many years will any sum of money double itself, at 6 per cent., compound interest ? . log 2 — log 1 , , - ^°«- log 1.06 = 11-9' y^*'"^- 4. Required the amount of $ 16 for 30 years, at 5 per cent., compound interest. Ans. $69.15. 5. What is the amount of $ 500 for 9 years, at 6 per cent, per annum, interest semiannually ? Given 2.930036 = log 851.21. Ans. $851.21. 6. In what time will $2233.57, at 6 per cent., com- pound interest, amount to $ 4000 ? Given log 2233.5'? = 3.349000. Ans. 10 years. 1. What principal, at 6 per cent., compound interest, will amount to $4000 in 10 years? Ans. $2233.5'?. 8. At what rate per cent, will $2233.5Y amount to $ 4000, in 10 years, at compound interest ? Ans. 6 per cent. 9. If the population of a certain country increases at the rate of 20 per cent, every 10 years, in what time will it double itself ? Ans. 38 years, nearly. 10. What is the present value of an annuity of $ 500* a year, to be continued 10 years, reckoning at 6 per cent., compound interest ? Ans. $ 3680.04. Given 0.253060 = log 1.79085 ; log .79085 = T.898095 ; 3.565854 = log 3680:04. 11. Required the present value of a perpetual lease, yielding $ 1200 a year, payable semiannually, interest being at the rate of 5 per cent. Ans. $ 24000. 12. The present value of an annuity, to be continued 10 years, at 6 per cent., compound interest, payable annually, is $ 3680.04 ; required the annuity. Ans. $ 500. 334 ALGEBRA. 13. A gentleman dying, left for the benefit of his widow $8000 in an annuity ofiSce, paying 5 per cent., compound interest ; in how many years will this sum be exhausted, if she draws out, annually, $ 850 ? Ans. 13 years 12 days. 14. Suppose a salary of $ 600 per year to remain unpaid 5^ years ; to what sum will it amount, at 6 per cent, per annum, compound interest, if the salary and interest are payable quarterly?. Ans. $8875.60, nearly. Given 0.142252 = log 1.38156. GENERAL THEORY OF EQUATIONS. 429. The General Form of a complete equation of the Bth degree is a;"4-;?a^-i + goif—'' + -]- t x^ -{- u x -{- v = 0, in which n is a positive integer, and the number of terms is n -f- 1- The known quantities p, q, t, u, v are either positive or negative, integral or fractional ; but the coefficient of a:" is understood to be -(- 1, unless otherwise indicated. The absolute term, v, is independent of x, but it may be regarded as the coefficient of aP (Art. 71). 430. In reducing an equation to the general form, all the terms must be transposed to the first member, and arranged according to the powers of x. If a;" has a coeffi- cient, it may be removed by dividing the equation by that coefficient ; and if any power of x is wanting, it may be supplied, with the coefficient 0. Thus, wh4n reduced to the general form, the equation 4 a;^ — 6 = 3 a;* becomes a;* ± Ox^ — ^x^ ± 0a: + 2 == 0. GENERAL THEORY OF EQUATIONS. 335 431 • A EooT of an equation is any real or imaginary expression, which, being substituted for its unknown quan- tity, satisfies the equation, or makes the first member equal to (Art. 148). Every equation must have at least one root, for, from the nature of an equation, in order to be true, there must be some quantity that will satisfy it, when substituted for its unknown quantity. 432i A Function of a quantity is any expression which contains that quantity, and is, therefore, dependent upon it. Thus, of, X -\- a, and 10a;' — ax^-\-2i, are functions of x. 433i A Permanence of sign occurs when two successive terms of a series have the same sign ; and A Variation of sign occurs when two successive terms of a series have different signs. GENEKAL PEOPERTIES. DIVISIBILITY OF EQUATIONS. ,434t If z. ii a root of an equation in the form afl_|_ joa:»-i -|- yaf-^ -j- -\- to? -{- ux -\- v = Q, then the first member is divisible by x — a. It is evident that the division of the first member by x — a may be carried on until x disappears from the re- mainder. Let Q represent the quotient, and B the re- mainder, which must be independent of x ; then the given equation may be made to take the form (x — a)Q-\-Ii = 0. 336 ALGEBRA. But if x = a, then (x — a) Q = 0, and, consequently, B — Q; that is, a; — a is a factor of the first member of the given equation. 435 1 If the first member of an equation in the form af" -j- pa^~^ + qai"'^ ■\- -\- tx^ -];- ux -\- v ^ is divisible bff x — a, then a is a root of the equation. For, if the first member of the given equation is divisible by X — a, then the equation may be made to take the form (x- a) Q=Q. ' But the substitution of a for x causes the first member of this form to become ; therefore it will also satisfy the given equation in its original form (Art. 431). Examples. 1. Prove that 3 is a root of the equation 3» — %x'^-\- 11a; — 6 = 0. Here, dividing a;' — %x''-\-\lx — 6 by x — 3, we ob- tain x^ — Bx -\- 2, and no remainder. 2. Prove that — 1 is a root of the equation a;* + 1 = 0. Here, (a^ + 1) -^ (x -\- 1) = a:' — a; + 1. 3. Prove by division that 1 is a root of the equation a;S_^a;=_l'7a:+ 15 = 0. 4. If — 2 is a root of the equation a^ — 3 a!" -f- 4 a; + 4 = 0, by what factor must its first member be divisible ? 5. What number is a root of the equation (a: + 3) (a:'' — 2a: + 5) = 0? GENERAL THEORY OF EQUATIONS. 337 NUMBER OF ROOTS. 436> M)ery equaiion of the nth degree, containing but one unknown quantity, has a roots, and no more. Let a be a root of the equation a^+;>a:"-i + yx»-2 + ^ tu^ -\- ux -\- v = 0; then, by Art. 434, the first member is divisible by x — a, and the equation may take the form (x — a) (a!»-i +/)ia^-= -f -{- u^x -\- v^) =0. This equation may be satisfied by making either factor of the first member equal to ; hence, a?'-»4-^ia:»-2 + ^ u^x -\- v^ = 0. Bat this equation must have some root, as b, and may be placed under the form (a: — J) («"-« + ^aa:»-= + + «j,a; + «j) = : and be satisfied by making the second factor equal to ; and so on. Since each of the factors, x — a, x — J, etc., contains only the first power of x, it is evident that the original equation can be separated into as many such binomial factors as there are units in the exponent of the highest power of the unknown quantity, and no more ; that is, into n factors, or (a; — a) (x — b) (x — c) {x — l) = 0. Hence, by Art. 435, the given equation has the n roots a, -b, e, I. Moreover, if the equation had another root, as r, then it must contain another factor, x — r ; which is impos- sible. 29 338 ALGEBEA. 437. It should be observed, that, the n binomial factors of which the general equation of the nth degree is com- posed are not necessarily unequal; hence^ two or more of the roots of an equation may be equal. Thus, the equation x^—&x'-{- 12a: — 8 = may be factored so as to take the form (x — 2) (x — 2) (x — 2) — 0, or (x — 2)' = ; and, hence, the three roots are 2, 2, and 2. 438. An equation is satisfied by making anw owe of the factors of its first member equal to ; but two or more factors cannot each be made equal to at the same time, and for the same value of x, unless there are equal roots. 4S9i It will be seen that any equation, one of whose roots is known, may be readily depressed to another of tfie.jiext,, lower degree,, which shall contain the remaining roots. Hence, if all the roots of an equation are knowi;i except two, those may be obtained from the depressed equation, by the rules for quadratics. Examples. , 1. One root of the equation a;' -\-2sc^ — ; 23 a: — 60 = is — 3 ; required the depressed equation, and the remain- ing roots. _ * Here, dividing a?-\-2x^ — 23 a; — 60 by x-{-Z, we obtain a? — X — 20 ; hence x^ — x — 20 :;== is the depressed equa- tion. Solving this equation by the rules for quadratics, we obtain 6 and — 4 as the remaining roots. Note. If we were to divide the original equation b^ a; -}- 3 == 0, we should obtain a? — a;— r 20= ^; that is, when a; -(- 3 = 0, the original equation is satisfied, whatever the value of the. other faoton, af — a; — ,20. , "yj^^en a;-|-3 = 0, x^^x^^ 20, or ^, must be equal to — ^, fo^ the same value of x. But the original equation is satis- fied by making either of its factors equal to 0'; hence, we may take a:'' — a: — 20 = 0, instejld of a; -^- 3 = (Art. 438). GENERAL THEO&lf 6f' EQUATIONS. 3S9 ^. One' foot of the eqtiatioii a;*— 19a; + 30 =± (J is 2 ; required the depressed equation, and the remaining roots. Ans. Equation; a;''-|- 2a; — 15 = 0; Boots, 3 and — &i 3. Eequired the three roots of the equation x^ = o', or a? — a* = 0. , _j° Ans. a and - (-^1 ± V — 3). 4. One root of the equation a;" + a;" — 16 a; -]- 20 = is — 5 ; required ihe ifeihainihg roots. AnS. 2 &nd 2. 5. Two roots of the equation a;* — Sa;^ — 14 a;' + 48 a: — 32 = are i and 2 ; required the remaining roots. Ans. 4 and — 4. 6. One root of the e'c[uati6ii x* — 1 a? -{-3x-^S = is 1 ; what (equation contaiiis the remaining roots ? FORMATION OF EQUATIONS. 440« -An eqvMian having any given roots may he formed by subtracting each root from the unknown quantity, and placing- the product of these binomial factors equal to 0. Fbr, it is Evident, from principle's already festablishedr that an equation having the n roots a, b, c, , . . ^ . I may- take the form (x — a) (x — b) (x — c) (as — = 0- After performing the multiplication indicated, the equa- tion will assume the form a;"+;)x»-^-f-^af-^4- + = 0, that is, if the absolute term of an equation be wanting, at. least one root must be 0. 443> Every rational root of an equation is a divisor of the last term. 444> When all the roots of an equation but two are known, the coeflScient of the second term of the depressed equation (Art. 439) can be found by subtracting the sum of the known roots, with their signs changed, from the coefficient of the second term of the original equation. The absolute term can be found by dividing the absolute term of the original equation by the product of the known roots, with their signs changed. Examples. 1. What is the sum of the roots of the equation ^ — *lx-\-% — 0? 2. Are 2, 3, and 4 the roots of the equation as* . — 8 a:^ + 26a; — 24 = 0? (Art.. 441.) If not, how must the equation be changed in order that they shall be its roots ? 3. What are the second and last terms of the equation whose roots are 3, 5, — 2, and f ? . Ans. — ^x' and —20. 4. What are the third and last terms of the equation whose roots are — 2, — 3, and 4 ? Ans. — 14 a; and — 24. 29* ^43 ALGBBSA- 5. Write the equg-tjop wiose roats are 1, 2, and 3. Ans. a^ — 6s:?'+lla: — 6 = 0. 6. Wihat are the roots of -the equaftion x* -\- 8 1^ -{- 10 x' = ? Ans. 0, 0, a,nd —4 ± \/6. T. Two roots of the equation a;* — 3 a? — lix^ -\- 4i8x — 32 = are 1 and 2 ; obtain the depressed equation by .ArJ. M^. FRACOIONAL ROOTS. 445> An equation whose coefficients are all integers, that of the first term 'being unity, cannot have a root equal to ,« ra- tional fraetion. 'If possible, let j- , a rational fraction in .its lowest terins, be a rpot of the .g^u^ljion of ,-4-,;??f-V+ qa?--^ -!-...;.-+ ta?-i{-ux -\-v .= ^0, in which p, q, . . . . . t, u,v axe integers. Then.^;) +^l^i^ +,^j + ^.^^ + «^y+.=0. Multiplying the equation by 5"""^, and transposing, Now, sin.ce ,a and h haive no common factor, ,«" an(i 5 can have none ; hence, we have a fraction in its ,lowest terms equal to an entire quantity, which is impossible. Therefore, no .root of the equation can be a rational fraction. 446. An equation •vy'hose coefficients are whole num- jbers, although its .raiiionaZ roots are likewise whole num- bers, may have real roots whose value may be expressed in the form of 'irrational fractions, or approximately ex- ,pregsed by-nie^ns of decimal .fractions. A real root which cannot; be exactly expressed 'in num- beys, is said -to jbe incommensurable . GENERAL THEORY OF EQUATIONS. 343 IMAGINARY ROOTS. 447. If the coefficients of an equation be real quantities, imaginary roots enter it hy pairs. .Suppose a-\-h ^/ — 1 to be a root of Ijhe equation a^^^jpaf-^ 4- ^a*-2 + .'..;. + t'^ + "«; + w = 0. If we substitute this root in place of x, and develop each expression by the binomial theorem, we find that all the odd terms of each series either contain the even powers of by/ — 1, or simply the powers of a, and are therefore real, while all the even terms contain the odd powers bf b />/ — I, and are therefore imaginary. Eepresenting the sum of all the real quantities by P, and that of the imagi- nary quantities by Q \/ — 1, we have -P+ (?\/— 1 = 0. This equation can be true only w;hen both P = 0, and''0 = b. If we now substitute a — 6 \/ — I for x, we find that the series differ from the former only in having their even terms negative. Hence, we obtain P— In any complete equation, the sum of the num- ber of variations and of permanences is equal to the number of_terins less pne, or to the degree, of ;the equa- tion (Art. 429). Hence, when the roots are all real, the number of positive roots is equal to the number of variations, a,nd the number of negative, roots is equal to the.nuinber of permanences (Art. 436). 45^> A complete equation whose terms ,, are , all pos- itive 'can have no . positive root ; and one whose terms are alternately positive and negative can have no negative root. Examples. 1. What is the limit to the .number of ppsitive and of negative ropts in the equation a? — 11 a;"-!- 43 a;: — 65 = 0? Ans. It can have no negative root, and not more than 3 positive ones. "2.* The roots of the equation s^ ± Ox'' — Bx. — 2 = are all real; what signs must they take? Ans. Qne positive ; two negative. GENERAL THEORY OF EQUATIONS. 347 3: The roots of the equation a:' — lOx + S = are all real'; what signs must they take? f" Ans. Two positive ; one negative. 4. The roots of the equation sc^ — 1 a:'' + 36 = are all real ; what signs must they take ? ' 5. What are the signs of the roots of the equation "f — !i^ — 4 = 0? Ans. One positive ; two imaginary. 6. If the roots of the equation a^ — II x* -}- 11 3^ -{-11 x^ — lla;-|-l:^0 are all real, what signs must they take? Ans. , Four positive ; one negative. T. What are the signs of the roots of the equation a?+16a; + 55 = 0? ' ~ TEANSFOEMATION OP EQUATIONS. 456t The Transformation of an equation, as the term is here used (Art. 149), consists in ch'anging the' equation into another of the same degree, whose roots shall bear a specified relation to those of the given equation. 457. To transform an equation into another which shall have the same roots with their signs changed Let the given equation be sf-^paf-'^ + 5-2^-2 4- -^ta?-\-ux-\-v := 0, Put X = — y ; then, whatever value x may have, y will have the same, with its sign changed. Substituting this value of X, we have i-y)" +pi—yY-^+q (— y)"-' + -\-tf—uy-irv=Q, in which the terms having an odd exponent are negative, and the others positive. If n is even, the equation is in Its proper form ; but if n is' odd,' all the signs should be changed ; in bither case, then, we have 348 ALGEBRA. y —pt~^ + ?y-' — ± (+ *3'' — "J' + ") = *'• Hence, to effect the required transformation, we may simply Change ike signs of the alteniate terms. Examples. 1. The roots of the equation je* — lra; + 6 = arel, 2, and — 3; find the equation whose roots are — 1, — 2, and 3. ■ Here, as the complete equation is a?±Oa^ — 7a;-|-6 = 0, we must change the signs of se' and 6 ; hence the required equation is a^ — 1 x — 6 = 0. 2. Two of the roots of the equation a;^ — 2 3? -\- x — 182 = are 4 and — 3 ; find the corresponding equa- tion whose roots are 3 and — 4. 3. If the roots of the equation a^ — 6a^ + 3a;+10 = are 5, 2, and — 1, what are the roots of the equation a? + 6a^+3a;— 10 = 0? 458. To transform an equation into another whose roots shall be some multiple of those of the given equation. Let the given equation be of -\- p x"--^ -\- qx"-^ -\- -\-tx^-{-ux-{-v = 0. Put y^mx, or a; = -^, and we have (1)"+^ (ir ■+ » (r + +'© + •(£)+'=»■ Multiplying by m", y" ■\-pmy^^ -{-qm^jf-^ -{- -(- In a similar manner we may transform an equa- tion into another whose roots shall be equal to those of the given equation divided by some quantity. Examples. 1. Transform the equation a^ — 2a!^ — a;+2 = into another whose roots shall be twice as great. Here, if we substitute | for x, we obtain ^ — |- — f 2 o 2 A + 2 = 0, which, multiplied by 8, becomes ^ — 4y^ — 4y + 16 = 0. The coefficients of the transformed equation may also be obtained by multiplying the coefficient of x^ by 2, that of x by 2^, and the absolute term by 2'. 2. Transform the equation a:* — *l x -\- Q = Q into an- other whose roots shall be 3 times as great. Ans. 2/" — 21y + 54 = 0. 3. Transform the equation a? — 'I x — 6 = into an- other whose roots shall be 4: times as great. Ans. f — n2y — 384 = 0. 4. Transform the equation a? — fa; — 27 = into another whose roots shall be one third as great. Ans. / — ^y — 3 = 0. 460< To transform an equation containing fractional coeffi- cients into another whose coefficients shad he entire, that of the first term being unity. If we assume m (Art. 458) equal to the least common multiple of the denominators found in the equation, it will always remove them ; but often a less number will produce the same result. Hence we . Assume such a factor that its proper power (Art. 458) will cancel the denominator found in any term. Examples. Transform the following equations into others whose coefficients shall be entire. 30 350 ALGEBEA. I- 1. a^ — icf—^x + .ji^ = 0. Here, the denominators, when factored, are 3, 3^ . 2', 3' . 2^ ; hence the multiplier 8.2, or 6, will remove them. Multiplying the four terms, respectively, by 1, 6, 6^, and 6^ we obtain ^ — 2^ — y-f"2 ^ 0, whose roots are 6 times as great as those of the given equation. 2. x^ + fa: — i = 0. Ans. / + 3y— 14=0. ' '>' ''• t' . ■ ' 3. a:* — 5a^+^j5.a;2_2^.a; + § = 0. Ans. yt — 10 / +, 35 / — 50 3^ -t- 24 = 0. 4. jc^ ar + r = 0. Ans. V — bmy 4- abn'=:0. 5. a?-\-^^x' — ix — ^-^ = 0. Ans. /-f^/'— 180^ — 900 = 0. 461 1 Tb transform an equation . into another whose roots shall be the reciprocals of those of the given equation. Let the given equation be a^-{-paf'-r^-\-q3r-'-\- + y-' + !?y-' + + t'f + u'y + v- = 0, (3) in which p' = nr -\- p ~2 q' = n '^— r'-\-p(n—l)r-\-q t'. = n ^^^.r^^J^pin-l) "^r^^q («-2) '^^-^4- + « u' = nr^^-\-p(jir—l)r^^-^q{n—2)r^^-\- + 2tr-\-u v' =r"-\-pr^^-{- qr^^-\- .-[■tT^-{-ur-\-v. Thq roots. of (3) are evidently less by r tha^ ,tbpse of (1). If —r be, substituted, for, r, the roots ,of .the tran^form^jd equation will ^e^.gr^atfir byr than .those of (1). 352 ALGEBRA. When n is quite small, the substitution may be effected as above ; but for equations of a higher degree, a less tedious method is desirable. 463. If we restore x, by substituting x — r for y in (3), we obtain (a; — r)» + ^' {x — rf-'^ + + m' {x — r) + »' = 0, (4) which is evidently identical with (1), and must reduce to the same form when developed ; hence, whatever is proved of (4), is also true of (1). If we divide (4) by x — r, we obtain {x — r)"~^ -\- p' (x — r)""^ -\- + «', and a re- mainder of v' ; if we divide the last quotient by x — r, we obtain a remainder of u' ; and so on, until we obtain all the coefiBcients of (3) as remainders. Hence, to effect the required transformation. Divide the first memher of the given equation by its unknown quantify minus the proposed difference, if the roots of the trans- formed equation are to he less than those of the given equation, {or by its unknown quantity plus the difference if they are to be greater), and the remainder will be the absolute term of the transformed equatioti. Divide the quotient just found by the same divisor, and the remainder will be the coefficient of the first power of the new unknown quantity, and so on, until all the coefficients are thus obtained, in an inverse order. These successive divisions may be written out in full, but the work can be greatly abbreviated by the use of SYNTHETIC DIVISION. 464i This method of ■ dividing, which was first sug- gested by Horner, is an abridgment of the method by detached coeflScients (Art. 81). Although synthetic di- GENERAL THEORY OF EQUATIONS. 353 vision may be employed when the divisor has more than two terms, yet we shall have occasion to apply it only in cases where the divisor is a binomial, the coeflScient of whose first term is unity. Let it be required to divide a^ — 19 a! -|- 30 by x — 2 OPERATION BY DETACHED COEFFICIENTS. 1 ± — 19 -|- 30 1 — 2 4-2- -f2- -19 - 4 ■ 15 4- 30 15-f 30 1-f 2 — 15 x^ -\-2x — 15 ^ Quotient. It will be observed that the first term of each partial product, and the last term of each partial dividend, may be omitted ; for they are merely repetitions of the figures standing above them. Also, the remaining term of each partial product may be added to the corresponding" term of the dividend, provided we change the sign of the second term of the divisor before multiplying. The work now stands thus : 1 ± — 19 -1- 30 + 2 1+2 1-f 2 — 15 — 15 30 As the first term of the divisor is 1, it is usually omit- ted, and the first terms of the dividends constitute the 30* 354 ALGEBRA. quotient. Raising, now, the oblique columns, we have the following concise form for the OPERATION BY SYNTHETIC DIVISION. Dividend, 1 ± — 19 -f 30 i + 2 Partial Products, + 2 + 4 — 30 Quotient, 1 + 2 — 15, -|- Remainder. Here, we use only the second term of the divisor, with its sign changed ; each term of the quotient is the sum of the vertical column under which it stands, and each term of the second line is obtained by multiplying the next preceding term of the quoljient by the divisor, as written. * Synthetic division is especially convenient when there are several successive operations in which each quotient becomes in turn a dividend. 465. The transformation of equations into others whose roots shall differ from those of the given equation by a given quantity, may now be effected by the process of synthetic division, or otherwise. Examples. 1. Transform the equation a? -\- x^ — 10 a; -|- 8 = into another whose roots shall be less by 4. FIEST OPEKATION. a? = iy 4- 4)'" = = y+. .12/ + 482, + 64 x" = (y + 4)^ = _ f f + %y + 16 — 10 a; = -10(y + 4) = = — lOy — 40 8 = ■ 8 2^ + 13/ + 462, + 48 = GENERAL THf OgY 0? EQUATIONS. 355 SECOND OPKKATION. a^+ a^_10a: + 8 x — 4 a?—'L7? «2 + 5a; + 10 a; — 4 ba? — lQx 8 40 jc? — 4a: ; F+9 f-f^ b3? — 2Qx 9a; + lQ 9*— 36 1 10a; + ■ 13, 3d rem. 10 a; — 46, 2d remainder. 48, 1st remainder. THIRD OPERATMN, I +1—10 + .8 I +4 _+ 4 +20 -It ,40 - 1st quotient, 1 +5 +10, +48, 1st remainder, or v'. + 4= +36 2d quotient, 1 + ^' + ^^> 2d remaipder, or «'. + ^ 3d quotieii^t, 1, tJ- 13, 3^ rfimfiind)?r, or/. Hence, y* + 13 y'' + ^^ y + ^^ = is the required equq-tion. 2. Transform .the equation a*^ x^ — 2 a; — 1 = into another whose roots shall be greater by 2. OPEEATION. 1 ±0 — 2 — 2 — 2 — 4 — 2 — 1 — 2 — 1 + 4 — 6 +16 + 3 • — 8 4- 8 —22 + 15, 1st remainder, or »'. + 11 + 12 — 30, 2d remainder, or «'. — 6 + 23, J 3^( remainder, or t'. — 2 — 8, 4th remainder, or «'. 356 ALGEBRA. Therefore, y* — S^/^ -f 23^'' — SOjf -f- 15 = is the re- quired equation. In the foregoing solution the coefficient 1 is not repeated, and the columns of figures are separated ; in other respects it is like the preceding one. This method may also be applied when the coefficient of the first term of the given equation is not unity. lla^4-6a; 3. Increasi — 1 = by 6 3 the re 0.5. — 11 — 3 )ots of the equation Qx' — 1] OPERATION. -}- 6 —1 —0.5 -1- 1 —6.5 6 — 14 — 3 -{-13 , — ^.5 = w' -1- 8.5 6 — It , — 3 , -f-21.5 = u' 6 , —20 = t'. Therefore, Gy* — 20y'' -|- 21.5y — t.5 = is the re- quired equation. 4. Transform the equation a^ ■ — 9x^ -\-26x — 24 = into another whose roots shall be less by 1. Ans. 3^ — 62^-1-11^-6 = 0. 5. Diminish the roots of the equation a^ — 13a:''-|- 56a; — 80 = by 3. Ans. 2^' — 4/ + 5y — 2 = 0. 6. Increase the roots of x* — 5a;^ -|- 4 =: by 2. Ans. 2^— 82/^-1- 19/— 122^ = 0. 466i To transform a complete equation into another whose second term shall be wanting. To make y = 0, it is only necessary to make nr^pz=Q (A'rt. 462), or r = — ". We must, then, substitute y — ^ for X ; that is, GENERAL THEORY OF EQUATIONS. 357 Substitute for the unknown quantity a binomial consisting of a new unknown quantity, and the coefficient of the second term, taken with a contrary sign, and divided by the number denoting the degree of the equation. This substitution may be effected in either of the ways mentioned in Arts. 462-465. It is evident that the sum of the negative roots of the transformed equation must be equal to the sum of the positive' roots (Art. 442). Examples. Transform the following equations into others whose second terms shall be wanting. 1. a? — 6x'-\-9x — 6 = 0. Assume x == y -\- ^ ^ y -\-2. FIRST OPERATION. a? =, y^J^Qyi^Uy-^- 8 — 6a:='= — 6/ — 24y — 24 -\-9x = 9y + 18 — 6 = — 6 f ± Oy— 3y— 4 = SECOND OPERATION. 1 —6 + 9 —6 +2 + 2 — 8 +2 — 4 + 1 , — 4, 1st remainder. + 2 — 4 — 2 , — 3, 2d remainder. + 2 0, 3d remainder. Hence, the transformed equation is ^ — Sy — 4 = 0, whose roots are less by 2. 3'58 ALGEBRA. 2. J—px-{-q = 0. Ans. y^ — iP' -{^ q = 0. 3. a^4-6a;= — 3a;-}-4 = 0. Ans. /— 152^ + 26 = 0. 4. x*—8a?-j-193^—l2x = 0. Ans. 2^— 5/-|-4=0. 5. a:* — 8a^ + 23a!'' — 30a:+15 = 0. Ans. y — / — 2y— 1 = 0. 6. a^ + 0^ + 4 = 0. Ans. / — ^y + ^'^"^ 0. DEEIVED POLTliTOMIALS. 4o7i A I)ekived P'olynomial is one that may be ob- tained from any given polynomial, containing only a single unknown quantity, as x, by ihnltii)lying' each tterm by the exponent of x in that term, and then diminishing the ex- ponent by 1. The second derived polynomial is the derived polynomial of the first derived polynomial ; the third is derived in like manner from the second, and so on, each being one degree lower than the preceding. Derived polynomials are also called derivatives. 468t A Derived Equation is one whose first member is a derivative of the first member of another equation. The first derived equation is also called the limiting or separating equdtion, when used to separate, ot determine the limits of the roots of an equation. Examples'. 1. Obtain the successive derived polynomials of je'4"^*' -f 3a;-f 9. r First, Zx'-\-lQx-\-Z. Ans \ ^6^o?i-^ -\-p (m— 1) In— 2) r^" + ..... -f 2« ^3 = w (m— 1) (m— 2) r^-\-p (n— 1) («— 2) (J^— 3) ?*-*+.. . . . It will be seen that R, the absolute term, or coefficient of jf, takes the same form as the original equation, (I) (Art. 462), r being substi,tuteji for x ; Ri takes the form of the first derivative of (1), R.^ that of its second deriva- tive, and so on. Hence the coefficients of the transformed equations of Arts. 462-466 may be obtained By mfeans of the law of derived polynomials. EXAUFLGS. 1. Traiisiform the equation s? — 1 x — 6 = into an- othet whose roots shall be greater by 3. Here, r = — 3. The successive derivatives of the first member are 3 a? — 7, 6ar, and 6. Hence, R = j^ — T r — 6 = — 2T -f 21 — 6 = — 12 ^1 = 3r2 — 7 = 27 — 7 = ^20 ifj = 6 r = — 18 ; -R3 = 6. Substituting these values in (5), we obtain — i2+'202?— ijV+iy = 0, or f —9^ + 2i)y— 12 — 'ii. 360 ALGEBRA. 2. Transform the equation a:* — 8 a? + 24 a;" — 32 a; — 65 = into another whose roots shall be less by 4. Ans. ^-\-8f-\-24:y^-\-32y — 6b = 0. 3. Transform the equation a? — 1 x^ -\- 36 = into another whose second term shall be wanting. Ans. 3/» — ^1/ + W- = 0. EQUAL ROOTS. 470. If we consider that the roots of the general equa- tion a^4-j9a:»-i + ya;"-2-j- -\- tx' -\- ux -\- v = Q (1) are a, h, c, d, , we have (Art. 440) the identical equation af+j9af-i+g'a;»-2+ = (x—a){x—h){x—c)(,x—d) Substituting x-\-y for x, we have (a;-|-3,)«_|_p(a;4-y)»-i-f- = (x—a^)ix—h-{^)(x—c-\-y) Developing each member of the equation, the coefScients of y must be equal, or, »a;»-i+ (ra — I) j9 af-^ + (w — 2)g'a*-' + = (x — a) (x — b) (x — c) . . . . -\- (x — a) (x — b) (x — d) -\- {x — a) (x — c) {x — d) y to » — 1 factors. -|- (a; — J) (x — c) {x — d) . . .. + Now, if b^^a, it is evident that every term of the second member of this equation is divisibl^i by x — a, and consequently its first member is divisible by the same factor ; that is, if equation (1) has two roots equal to a, its first derived equation has one root equal to a. In the same manner, if cz=b = a, that is, if (1) has GENERAL THEORY OF EQUATIONS. 361 three equal roots, Its first derived equation has two of them. In general, If an equation has n roots equal to a, m roots equal to d, etc., then its first derived equation must have n — 1 roots equal to a, m — 1 roots equal to d, etc. 471, To determine whether an equation has equal roots, and if so, to find them. From the principle demonstrated in the preceding Ar- ticle, it is evident that we must Seek the greatest common divisor of the first member of the given equation and its first derivative: If there is no common divisor, there can he no equal roots. If there is a common divisor, put that divisor equal to 0, and the solution of this equhtion will give the required roots. The number of times that each of the eqital roots occurs in the original equation is one greater than in the equation formed of the common divisor. When the greatest common divisor is of a high de- gree, it should be compared with its own derivative, in order, if possible, to obtain a common divisor of a lower degree. When the equal roots are known, it is evident that the given equation may be depressed to one containing only the remaining unequal roots (Art. 439), by dividing by the product of all the binomial factors corresponding to its equal roots If the first member of the given equation be divided by the greatest common divisor, the resulting equation will contain all the different roots of the original equation, but without the repetition of any. Examples. Determine whether the following equations have equal roots, and obtain all their roots. 31 362 ALGEBRA. 1, ai^— Ua?-\- 61 x' — S4:X-}- 36 = 0. Here, the first derived polynomial is 4:3? — 42 ar' -j- 122 X — 84, and the greatest common divisor, obtained according to Art. 108, is a:" — 1 x -\- 6. The equation ar" — 7a;-|-6^0 gives, according to the rules for quad- ratics, or by factoring, a; = 1, or 6. Therefore, 1, 1, 6, and 6 are the roots of the given equation. ■ 2. a;* — 8a^ + 13ar — 6 = 0. Ans. 1, 1, and 6. 3. a? — 1x^-\-16x — 12 = 0. Ans. 2, 2, and 3. 4. a;*— a^— 18x''+52ar— 40=0. Ans. 2, 2, 2, and —5. 5. a^— 13 a;* + 61 a:'— 17 la;'' + 216a;— 108 = 0. Ans. 2, 2, 3, 3, and 3. 472i If an equation has two roots equal in magnitude, but opposite in sign, the signs of its alternate terras may be changed (Art. 457), and the common divisor of the two equations will contain those roots. 6. a;*+3a;»— 13a;''— 27a; + 36 = 0. Ans. 3, — 3, 1, and — 4. SIGNS OP POLYNOMIAL FUNCTIONS. 473t A polynomial of the form ai"+_pa:"-' + ja?—" + ..,.. + and write the signs of the results in a row. If any other number be substituted in like manner, The difference between the number of variations in the first row of signs and that in the second, is equal to the number of real roots of the given equation comprised between the quantities substituted for x. This is Sturm's Theorem, the demonstration of which depends upon the following principles. 1. Two consecutive functions of the series X, Xi, Xj, ^^'Cannot both become for the same value of x. If Xi = and .^ = 0, then, by equation (2), Art. 47t, Xs = ; and if ^ = and Xg = 0, then, by (3), X4 = ; and so on, till X„=0. But it is impossible that the last remainder shall become 0, for any value of X, because it is independent of x. 2. When either function of the series, except X and X^, 31* 366 ALGEBEA. becomes for a particular value of x, the two functions adjacent to it must have different signs. If ^2 := when x = h, then neither Xi nor X^ can be for that value of x, and, by equation (2), Xj = — X^; that is, Xi and X^ must have different signs. It is evident that X^ and Xg will also continue to have different signs for values of a; a little greater or less than h ; for no change in the signs of Xj or X3 can take place until after the value of x is so much changed that one of them becomes (Art. 473). 3. When either function of the series, except X and X„, changes its sign for different values of x, the numbek of varia- tions is not affected by the change. If c and d, when substituted for x, give different signs for X2, conceive all possible quantities between c and d to be substituted for x, in regular order, and Xj changes its sign in passing through (Art. 473). Hence, when Xj changes its sign, as well as just before and just after the change, Xj and X3 must have different signs, and the sign of Xi, whether -|- or — , must, therefore, in both cases, be like one of those adjacent to it, and unlike the other, giving, with Xj and X^, one permanence and one variation. The only effect of changing the sign is that the permanence and variation exchange places. The last function, X^, is independent of x, and there- fore can never change its sign, whatever value may be given to x. Hence a change in the number of variations can be caused only by a change in the sign of the origi- nal function, X. 4. When the original function, X, changes its sign for suc- cessive increasing values of x, the number of variations is diminished by one. Let a be a root of the original equation, or a value of X which causes X to become 0, and let ^i represent the GENERAL THEORY OF EQUATIONS, 367 value of Xi when a; =: «. If we assume x := a -\- tf and x = a — y, y -being too small for X^ to become 0, then the sign of X^ for each of these values of x will be the same as for x^ a, that is, the same as the sign of A- But if a-\-y be substituted for x in X, we obtain (Art. 469) ^ + Ay+^f+~f + +y; or, since A is merely a substituted for x in X, and there- fore equal to 0, y(^. + t^+,V+ )• In the same manner, when x=^a — y, X becomes y{-A + ^y-^,f+ ). As these series are finite, y may be made so small that the sign of each shall depend upon that of Ay (Art. 475). Hence, X has the same sign as A-^ when a; is a little larger than a, and a different sign when it is a little smaller. But ^"1 has the same sign as A^. Hence, if x is increasing, X and X^ have a different sign before X becomes 0, and the same sign after. A variation is there- fore changed to a permanence. The truth of the theorem is evident from the foregoing principles. If x take all possible values from p to q, p being less than q, one variation is changed to a per- manence each time that X passes through and changes its sign, and only then, for no change of sign in any of the other functions, X^, X^, etc., can affect the number of variations. Hence the number of variations lost in passing from ^ to 5' is exactly equal to the number of real roots of the equation X=0 comprised between p and q. 3o8 ALGEBRA. 479. When — oo and -|- oo are substituted for p and q, the whole number of real roots of the equation X = becomes known (Art. 413). The substitution of — oo and will give the whole number of negative roots, while and -j- oo will give the whole number of positive roots. When the roots are all real, Descartes' Rule (Art. 452) will answer the same purpose. The substitution of various numbers for p and q will show between what numbers the roots lie, or fix the limits of the roots. 480. It is evident that X and X^ must change signs alternately, as they are always unlike just before X changes. Hence, when the roots of X:^0 are all real, each root of Xj = must be intermediate in value be- tween two roots of X= 0. For this reason the first derived equation is often called the limiting or separating equation (Art. 468). 481. In the process of finding X^, Xg, etc., any posi- tive numerical factors can be introduced or rejected at pleasure, since the sign of the result is not affected thereby. In this way fractions may be avoided. In substituting — oo and -j-oo, the first term of each function determines the sign (Art. 415). Examples. 1. Determine the number and situation of the real roots of the equation a;' — 4a:^ — a:-(-4=:0. Here, the first derived polynomial of the first member (Art. 467) is 3a;''— 8a;— 1. Multiplying a^ — 4:X^ — x -|- 4 by 3, to make its first term divisible by 3 x^, GENERAL THEORY OF EQUATIONS. 369 3a:2_8a; — 1) Sa;' — 12a;'=— 3a;+12 (a; — 2 So?— Sx" — X — 4*2— 2a; + 12 3 — Qx^— 3a:-f 18 — eai^+iea;-}- 2 — 19a:+16 .-. ^2= 19a;— 16 19a; — 16) 3a:"— 8a;— 1 (3a;— 104 19 57 a:* — 57 ar' — 152a; — 48a; 19 — 104 a; — 19 19 — 1976 X — 1976 a; + 361 1664 — 2025 .-. X3 = +2025. X = a;' — 4a;« — a; + 4; iC,= 19a;— 16; :Si r= Sar" — 8a;— 1 ; X3 = +2025. The last step of the division may be omitted, for we can easily determine what the sign of X^ must be, when — 104 X — 19 is obtained. We first substitute — 00 for x in each function, and obtain three variations of sign ;. in like manner -(- 00 gives no variation ; hence the three roots are all real. By sub- stituting 0, and comparing with the former results, or by applying Descartes' Rule, we learn that one root is negative and two are positive. We therefore substitute various numbers to determine the limits of the roots. The following table presents all of these results in a connected form. 370 ALGEBRA. X X, X, Xs X =z — <»> — + — +. 3 variations. X =z 2, — + — +. 3 variations a; = — 1, + — + a; = 0, + — — +- 2 variations x=l, — + + a; = 2, — — + +, 1 variation. x = Z, — + + +, 1 variation. a; = 4, + + + x = b, + + + +. no variation a: = oo, + + + +, no variation If only — 2 and 6 had been substituted, it would at once be seen that there must be three real roots between those numbers, for the former produces three more va- riations than the latter. But we also find that — 1, 1, and 4 each reduce X to 0, and are, therefore, the three roots required. Note. To obtain all the relations of signs, we might substitute — 0.5 and 0.9 for x, the former giving -| — | 1-, 2 variations, and the latter -\ 1 — |-, 2 variations. These will serve to illus- trate the demonstration of the third and fourth principles, Art. 478, although they are of no service in obtaining the roots of the given equation. 2. Determine the number and situation of the real roots of the equation a;* — 3a:' + 3a;^ — 3a;-|-| = 0. Here, X =a;*— 3a?+3a:'' — 3a:-|- J ; Xi.= ^a? — 9x^ + &x — 3; X3 = — 92a;+129; .3^ = 3a;2+18a; — 31; X4 = — 1163. X Xi X^ Xg X^ a; = — 00, -|-_-|--)-_ a; = 0, + 1- - a; = 1, H h ■ x = 2, + + -!-_. « = +«>, + + + — — 3 variations. 3 variations. 3 variations. I variation. 1 variation. GENERAL THEOEY OF EQUATIONS. 371 Substituting — oo and -)- oo, we obtain a difiference of two variations ; hence there are two real and two imagi- nary roots. Substituting 0, we find the real roots to be positive. The substitution of 1 and 2 shows that both real roots lie between 1 and 2. Note. Xj, X^, and Xg all appear to change signs at once, and affect the number of variations ; but the substitution of numbers between 1 and 2 will show that X changes sign twice between those numbers, and the other functions conform to the principles already stated. Articles 473 and 474 give no evidence that two roots lie between 1 and 2, if the signs of X alone are considered, but Sturm's Theorem detects them at once. 3. Determine the number and situation of the real roots of the equation sc^ — a^ — 2a;-|-1^0. Ans. Three ; one between and 1, one between 1 and 2, and one between — 1 and — 2. 4. Determine the number and situation of the real roots of the equation a^ — Ta;+7 = 0. Ans. Three ; two between 1 and 2, and one between — 3 and —4. 5. Determine the number and situation of the real roots of the equation a;* — 23^* — 5a^ + 10a: — 3 = 0. Ans. Four ; between and 1, 1 and 2, 2 and 3, — 2 and —3. 6. Determine the number and situation of the real roots of the equation si^ — 2a; — 5^0. Ans. One, between 2 and 3. Y . How many real roots has the equation 2 a;* — 3 a;' + 17 a?' — 3a;+ 15 = 0? Ans. None. 8. Determine the number and situation of the real roots of the equation a:* — 4a;^ — 3 a; + 27 :;= 0. Ans. Two ; one between 8 and 3, and one between 3 and 4. 372 ALGEBRA. SOLUTION OP HIGHER NUMERICAL EQUATIONS. 482. The real roots of the higher numerical equations must be obtained by tentative methods, or by methods which involve approximation. Equations of the third and fourth degrees may be considered as included in the class of higher equations ; for their general solutions are complicated, and only of limited application (Art. 608- 514). No general solution of an equation higher than the fourth degree has yet been obtained. If the proposed equation contains equal roots, they may be removed (Art. 4'71) ; and the student should also make use of any other principles found on the preceding pages, whenever the process can thereby be abbreviated. COMMENSURABLE EOOTS. 483. Any equation containing fractional coeflBcients may be transformed into another whose coefficients shall be entire, that of the first term being unity (Art. 460), and such an equation cannot have a root equal to a rational fraction (Art. 445) ; hence, to find all rational or commensurable roots, we have only to find all integral roots. 484. Every rational root of an equation is a divisor of the last term (Art. 443) ; hence, to find the commensurable loots of an equation. Ascertain hy trial what integral divisors of the absolute term are roots of the equation. The trial may be made by substituting each divisor, both with the positive and the negative sign, in the equation (Art. 431), or by dividing the first member of SOLUTION OF HIGHER NUMEEICAL EQUATIONS. .373 the equation by the unknown quantity minus the sup- posed root (Art. 435). In substituting very small num- bers, such as ± 1, the former method may be most convenient ; but, when a true root has been used, the latter method will give at once the depressed equation, which may be used in obtaining the other roots (Art. 439). 485i When the number of divisors of the last term is large, this process of successive trials becomes tedious, and another method may be adopted. If a is a root of the equation si^-\-px''-\-qx^-\-tx-\-ii = 0, then a*-\-pa^-\-qa^-\-ta-\-u := 0. Transposing and dividing by a, - = — t — qa — pa" — or, and - is a whole number. a In the same way, if we represent - -f- < by t', ^ = —q—pa — a', and - is a whole number. a Proceeding in this way, we find that when a is a root of the equation, it must satisfy the following conditions, l + t = t'. ' '~+p=p', where each expression represents an entire quantity. Hence the following 32 374 ALGEBEA. RULE. Divide the absolute term of the equation hy one of its integral divisors, and add to the quotient the coefficient of the first power of tlie unknown quantity. Divide this sum by the same divisor, and, if the qiiotiewl is a whole number, add to it the coefficient of the second power of the unknown quantity. Proceed in the same manner with each coefficient, in regular order, and, if the divisor is a root of the equation, each quo- tient will be entire, and the last quotient, added to the coeffi- cient of the highest power of the unknown quantity, will be 0. When the coefficient of the highest power is 1, the last quotient must be — 1 ; and when any power is wanting, it must be supplied with the coefficient 0. Examples. 1. Given a^ — 6 a:'' + 2t :c — 38 = 0, to find the val- ues of X. By Descartes' Rule (Art. 452) it is evident that the equation can have no negative root : and as the sum of the coefficients is not 0, it is also evident that 1 cannot be a root of the equation. The only remaining divisors of the last term are 2, 19, and 38. Dividing the first member by a; — 2 (Art. 464), 1—6 + 27 — 38 I + 2 -f 2 — 8 + 38 1 _ 4-1- 19 , we obtain x" — 4 a; + 19, and no remainder: hence 2 is a root of the original equation, and x^ — 4 a; +19 = is the depressed equation. Solving this by quadratics, we obtain a; = 2±v' — 15 for the other roots, and 2 is therefore the only rational root. SOLUTION OF HIGHER NUMERICAL EQUATIONS. 375 2. Given 8 a!* — 4a? — 14a;= + a: + 3 = 0, to find the values of x. Putting « = I , we obtain (Art. 460) 3^ — /-'?y'' + y + 6 = 0. (1) The divisors of 6 are ±1, ±2, ±3, and ± 6. By substituting -\- 1 and — 1 for y, it is readily seen that both are roots of the equation, and the other two roots can be found from the depressed equation. But all of the rational roots may be obtained from (1) by the method just explained. Divisors, +6, +3, +2, +1, —1, —2, —3, —6 1st Quotients, Quo. +1, + 1, + 2, + 2, + 3, + 3, + 6, —6, —3, + 7, -5, -2, -2, -1 -1, 2d Quotients, Quo. — r. + 1, — 6, + 2, -5, + 7, +5, +1, 0, —2, —6. — 7 3d Quotients, Quo.— 1, 4th Quotients, Quo.+l, -2, — 3, -1, 0, 0, +2, +3 -1, +1, +2 — 1, —1, —1 0, 0, As 6, 2, — 3, and — 6 give fractional quotients, they cannot be roots of the equation, and are rejected. 3, 1, — 1, and — 2 give entire quotients, and the fourth quo- tient for each is — 1 ; hence they are the four values of y, and f, ^, — J, and — 1 must be the four values of x. Find all the commensurable roots of the following equations, and also all of the roots, when possible, by methods already given. 3. a?-)-6a:''+lla;+6 = 0. Ans. — 1, — 2, and — 3. 4. a^ + Sa:? — 4x— 12 = 0. Ans. 2, —2, and —3. 5. a:* — 9a:'+23a;''— 20a;+15 = 0. Ans. 3. 376 ALGEBRA. 6. !ii? — 3x'-\-x-\-2 = 0. Ans. 2, 1.618, and —0.618. 1. a;* — 4«= — 8a:+ 32 = 0. Ans. 2, 4, and — 1 ± \/ — 3. 8. a:' — 7 a:" + 36 = 0. Ans. 3, 6, and —2. 9. a^— 6a;2+10a; — 8 = 0. Ans. 4andl±v'— 1. 10. a;< + a;=' — 29a:2— 9a; + 180 = 0. Ans. 3, 4, — 3, and — 5. 11. aj' — 6a;''+ 11a; — 6 = 0. 12. 4a;= — 16a:'' — 9a: + 36 = 0. Ans. §, 4, and — f. 13. 2a;« — 3a;''+ 16a:,— 24 = 0. Ans. f , and ± 2 V — 2. 14. «» — 2a;»— 16 = 0. Ans. 2. INCOMMENSURABLE ROOTS. 486i If a higher numerical equation is found to con- tain no commensurable root, or if, after removing the com- mensurable roots, the depressed . equation is still of a higher degree, the irrational or incommensurable roots must next be sought (Art. 446). The integral parts of these roots may be found by Sturm's Theorem, or by Art. 4'(4, and the decimal parts by Arts. 515, 516, or by HORNER'S METHOD. 487i Suppose that a root of the equation af-\-psif-^-\-qsi^-^-\- , + tx^-\-ux-\-v = (1) is found to lie between a apd a -{- 1. Transform the equation into another whose roots shall be less by a (Art. 462), and we shall have one of the form y _|_ p'y"-^ -{- y'y-" -f ■^t'y'-\-u'x-\-v' = 0, (2) SOLUTION OF HIGHER NUMERICAL EQUATIONS 377 one of whose roots is less than 1. If that root is found to be between the decimal fraction a' and a' -\- .1, trans- form equation (2) into another whose roots shall be less by a', and we shall have one of the form a" +/>"2?'-^ + 9"a"-2 + -j- t"x^ + u"z + v" = 0, (3) one of whose roots is less than .1. If that root is found to be a little greater than a", proceed with (3) in the same manner as before, and we obtain a; = a -|- «' + o" + • ' ■ • • to any required degree of accuracy. As z, in equation (3), represents a small fraction, its high- er powers will be so small that they may be neglected, and its value is nearly ;; ; therefore a' , a", may be found in this way, with more and more accuracy the smaller they become. Hence, a positive incommensurable root of an equation may be obtained approximately by the following RULE. Find hy Sturm's Theorem, or hy trial, the initial part of the root, and transform the given equation into another whose roots shall be less hy that initial part. Divide the absolute term of the transformed equation by the coefficient of the first power of its unknown quantity, for the next figure of the root. Transform this last equation into another whose roots shaU be less by the figure of the root last found, divide as before for the next figure of the root, and so on. 488. A negative root may be found by changing the signs of the alternate terms of the equation; and then treating the root as a positive one (Art. 457). 378 ALGEBRA. 489. The first division will not always give the correct figure. If the wrong figure is used, the mistake will become evident in obtaining the next figure, and the cor- rect one may be obtained by trial, or by Sturm's theorem. When the equation has two roots whose initial parts are the same, it is best to determine the next figures by ■ Sturm's theorem, or by Art. 414. 490t The last two term.s of the transformed equation should have opposite signs, in order that the quotient should be the proper figure to add to the part of the root already found, for it has been shown that a" = r, (Art. 487). Examples. 1. Solve the equation a? — x'^ — 2x-{-l = 0.. By Sturm's theorem we learn that the roots of this equation are all real, two being positive and one negative, situated between 1 and 2, and 1, and — 1 and — 2. To find the first root, we substitute y -\- \ for x. This transformation may be efiected by either of the methods given in Arts. 462-465, but the method by synthetic division is to be preferred. + 1 — 1 — 2 + 1 +1 — 2 — 2 — 1 +1 + 1 + 1 + 1 + 2 — 1 The transformed equation, whose root is less than 1, is •then / + 22^2 — y — 1 = 0. The second paragraph of the rule will not apply to this SOLUTION OF HIGHER NUMERICAL EQUATIONS. 379 equation (Arts. 489, 490). We try .9, and find it too large, and then take .8, as follows, 1 ■2 — 1 2.24 1.24 2.88 — 1 .8 ,8 .992 2.8 .8 — .008 3.6 .8 4.12 4.4 The transformed equation, whose root is less than .1, is then s? + 4.4z2 + 4.12 2 — .008 = 0. Dividing .008 by 4.12 gives us in this case .002, nearly, the next two figures of the root. Hence, x= 1.802. The other positive root is itself less than 1. Its first approximate value, .5, is found by trial to be too large, and we therefore take .4 as the first figure. Carrying this out according to the rule, we obtain .445, nearly. Changing the signs of the alternate terms of the original equation (Art. 488), its root is found to be 1.24'7. Hence the three roots of the original equation are 1.802, .445, and — 1.24t. 2. Required the real roots of the equation a;*_4a^ — 3a; + 2Y = 0. The successive transformations are usually written in connection, as in the following form, where the different coefficients are indicated by the *. The work may also be contracted by dropping such decimal figures from the right of each column as are not needed for the requii:ed degree of accuracy. By Sturm's theorem, or by Arts. 453, 4T4, the equation is found to have two real and two imaginary roots, the former being situated between 2 and 3, and 3 and 4. 380 — 4 ±0 — 4 — 4 — 4 *0.00 .84 ALGEBRA. — 3 — 8 + 27 2.26745 2 — 22 — 2 2 — 11 — 8 * 5.0000 — 3.7664 2 * — 19.000 .168 * 1.2336 — 1.0987 2 2 — 18.832 .344 * .1349 — .1267 *4.0 .2 .84 .88 * — 18.488 .176 * .0082 — .0072 4.2 .2 1.72 .92 — 18.312 .194 * .0010 — .0009 4.4 .2 *2.64 .29 * — 18.118 .025 .0001 4.6 .2 2.93 .30 — 18.093 .025 *4.80 .06 3.23 .30 * — 18.07 4.86 .06 *3.5 5. In like manner the other root is found to be 3.6796. Obtain the real roots of the four following equations. 3. a^ — 2a; — 5 = 0. Ans. 2.09455. 4. a:' + a:^ — 500 = 0. Ans. 7.61728. 5. a?—1x-{-1 = 0. Ans. 1.3569, 1.6920, and —3.0489. 6. a;' — 17a;= + 54a;— 350 = 0. Ans. 14.95407. 7. Obtain one root of the equation x* — 3a;^-|~'^^* — 10000 = 0. Ans. 9.88600. APPENDIX. RECURRING OR RECIPROCAL EQUATIONS. 491. A Recurring Equatioj^ is one in which the coefficienis of any two terms equally distant from the extremes of the first mem- ber are equal. The equal coefficients may have the same sign, or opposite signs; but a part cannot have the same and a part opposite signs, in the same equation. Also, if the degree be even, and the equal coefficients have opposite signs, the middle term must be wanting. 492 • If any quantity is a root of a recurring equation, the recip- rocal of that quantity is also a root of the same equation. Let x''+px''-^ + qx^-' + ±( ^qi? -{-px+l) =0 (1) be the equation. If it be transformed into another whose roots are the reciprocals of those of the given equation (Art. 461), the trans- formed equation will coincide with the given equation, after changing signs throughout, if necessary. The two equations must therefore have the same roots. If a is a root of the given equation, - is a a root of the transformed equation, and therefore - is a root of the a given equation. In like manner, if 6 is a root of any recun-ing equa- tion, - is a root of the same equation, and so on. h 493i On account 'of the property just demonstrated, recurring equations are also called reciprocal equations, the former term relat- ing to their coefficients, and the latter to their roots. 494« One root of a recurring equation of an odd degree is — 1 when the equal coefficients have the same sign, and -j- 1 when they have opposite signs. 382 ALGEBRA. A recurring equation of an odd degree, as aJ!'»+i_j-^a:2'»_[-5a;2'»-i_^ ±( qx>+px+l) = 0, (2) has an even nuniber of terms, and may be written in one of the following forms, (j^m + i^l-j _^jt,(^m^a.) _^ 5(a;2">-i_|_a^)^ = • (^m + i_ ij j^p (x^'^ — x) + jCr""-! — K^) + =0. If — 1 be substituted for x in the first form, or -}- 1 in the second, the first member will become ; hence — 1 is a root of the first and -(-la root of the second. If equation (2) be divided by a; ± 1, both forms will reduce to the following form, ^m_^p^m-i_^^jam-i_^ -\. qld' + px -\- 1 = 0, (3) a recurring equation of an even degree, in which the equal coeffi- cients have the same sign. Hence, a recurring equation of an odd degree may always be depressed to one of an even degree. 495. ■ Two roots of a recurring equation of an even degree are -\- 1 and r— 1 when the equal coeffideiUs have opposite signs. As the middle term must be wanting (Art. 491), such an equa- tion may be written in the form {x"^—l)-\.px(x""-'—l) + qx'(3?'^-*—l)+ = 0, (4) which is evidently divisible by both x — 1 and ai-f-l, or by x? — 1 (Art. 99). Hence, -|-1 and — 1 are both roots of the equation. If equation (4) be divided by x' — 1, it will be depressed two degrees, and become a recurring equation of an even degree, in which the equal coefficients have the same sign. Hence, every re- curring equation may be depressed to the form of equation (3). 496. Every recurring equation of an even degree, whose equal coef- ficients have the same sign, may he reduced to an equation of half that degree. Divide the equation by k"", and it may be written thus, APPENDIX. 383 the middle term, if present, bgcoxuing a known quantity. Put X -\-l = y x' + L = f — -2 3? ^^-^ = r-^r-^^ Substituting these values in equation (5), we have an equation of the form After this equation is solved, we can immediately find x from the equation x -\- - = y. X 497. It thus appears that any recurring equation of the (2?n-l-l)th degree, one of the (im-\- 2)th degree whose equal coefficients have opposite signs, and one of the 2mth degree whose equal coefficients have the same sign, may each be reduced to an equation of the mth degree. Examples. 1. Given ar* — 5a' + 6a;= — 5a; + 1 = 0, to find x. Dividing by ai», a;2_ Sa; + 6 — 5 + 1 = Or. (..+ ^)-5(. + i) + 6 = Substituting y for a; + _, and ^ — 2 for 3?-\- —, X ar f,-5y + i = Whence, y = 4, or 1 !C 4- - = 4, gives X = 2 ± v' 3 X 384 ALGEBRA. ! + 1 = 1, gives a; = i (1 ± v'^^)- X- That 2 — y/ 3 and - ^ - are reciprocals of 2 + v' 3 and 1 + ^-3 may easily be shown by rationalizing the denominators 2 of L^ and ^ (Art. 250). 2-{-\/3 1 -|-^/ — 3 2. Find the roots of the equation x^ — Ha;'+17a:'-|-17a:' -11^+1 = 0. ^„3. _i, 9±t/^ ^ ^^^ 3±j^ 3. Given a:5 + 2a^ — 8z» — 3a? + 2a;+l = 0, to find its roofs. Ans. — 1, 1, 1, and '^lAAll. 2 4. Given a? — ar' + a* — i? -\- x — 1 = 0, to find its roots. Ans. —1, 1, ±\/^^, and ^±>/ — ^, 6. Given a:* -|- pa? +2'^ 4" 1 = 0, to find its roots. Ans. — 1 and ^ (1 —p ± ^p'—2p — 3). 6. Given 61* -^5^ — 38 a? -{- 5 x -\- 6 = 0, to find its roots. Ans. 2, J, — 3, and — j-. 7. Given 5 a:* — 51 a;* + 160 a:= — 160 ar^ + 51 a; — 5 = 0, to find its roots. Ans. 1, 5, -J, and 2 ± »/ 3. 8. Given a;*-}-5a^-|-5a;-)-l = 0, to find its roots. Ans. X = — f ± i y/'ss'i ^ (42 zp 10 \/"33)*. BINOMIAL EQUATIONS. 498i A Binomial Equation is one which can be reduced to the form a;" ± w = (1) Some of the roots of binomial equations have already been ob- tained ; for pure equalions are binomial (Art. 269). We propose to state only a few general properties, and obtain the remaining roots of binomial equations of the lower degrees, since binomial equations APPENDIX. 385 in general are most readily .solved by means of trigonometrical for- mulas. 499< The roots of a binomial equation are all unequal. For, s" ± B and re a;*-', its derived function, evidently have no common divisor (Art. 471). 500i Every binomial equation may be transformed to a recurring equation. Put x^v^y, or x^ = vy", and (1) becomes, after dividing by v, y"±l = 0, (2) which is a recurring equation whose middle terms are wanting. It is evident that if the roots of (2) are found, those of (1) become known ; hence, (2) is used in place of (1). 501 • If 2m-|-l and 2 m represent n when odd and even, re- spectively, then (Art. 453) j("" + '-|- 1=0 has only one real root, — 1; y"" + ^ — 1 = has only one real root, -\-i. ^''' -|- 1 = has no real root. ^jm — 1=0 has two real roots, -^-1 and — 1. 502i If a be one of the imaginary roots of the equation x" — 1 = 0, then any power of a, is also a root. For, since a" = 1, a"" = 1, a'"=l, etc.; also, o-"=l, „_an ^ i^ g^ Therefore, a-", a"', 1, a, a', a', are all roots of the equation. 503 • If a. be one of the imaginary roots of the equation x" + 1 = 0, then any odd power of a, is also a root. For, since 0" = —!, 0'"=— 1, 0'" = — 1, etc.; also, a-" = — 1, are a-'" = — 1, etc. Therefore, a-\ a-\ a, a^ a', . all roots of the equation. 504. Although the roots of a;" ± 1 = may take an infinite variety of forms, as shown in the last two Articles, yet all these dif- ferent forms must be capable of reduction to n values essentially different (Art. 436). 505i When n is composed of two or more integral factors, as in the equation a^"' ± 1 = 0, let xP ^ y, and we have f±\ = 0. 33 886 ALGEBRA. Haying found the q values of y, put x^ equal to each of them, find the p roots of each equation, and we have the p q values of x. When n contains no other prime factors than 2, 3, and 5, the roots of a binomial equation may. always be obtained by means of ■ quadratics. 506. Since the n roots of equation (1) are expressed by i/ ^ v, it. follows that any quantity has as many roots of a given degree as there are units in the index of that degree. No more than two of these roots can be real, however. Examples. 1. What, are the roots of the equation a^ -(- 1 = ? Dividing by x", 3^ -\- — = 9. x' Substituting y for x-\--, ^ — 2 = X Whence, , y =5j ± ^ 2 Replacing x ■\- -, z':fsy'2 = — 1 X Therefore, a: = ± \f^^\ ^ ±\yf^±\ /^^ 2. What are the five fifth roots of 32 ? The equation a:*= 32 may be transformed into if — 1 = 0, by substituting ly for x. One root of this equation is evidently 1. Di- viding ■^ — 1 by 3/ — 1, we obtain t + t-Vf + y^'i^ = 0- This equation may be solved in the same manner as the la^, by substituting z for y -\- -, or as follows, y sr' + iz^ + i = iyiv's y = i(— l±v/5± is/— 10 T 2 V 6) a; = 2, or ^ (— 1 -f- y/ 5 -f ^/ — 10 — 2 y^l). or or or 1 (— 1 + V^"5 — \/— 10— 2 y/l), i (— 1 — V''5 + V— 10 + 2 fh-), \{c-\ — (A) and (B) are different forms of Cardan's formula for the solu- tion of cubic equations. In the former, both double signs are alike, that is, either both -f- or both — . (B) is derived from (A), by rationalizing the denominator (Art. 250), that is, multiplying both numerator and denominator of the fraction by I ? q: , /T.4-P.\ ' \ 2^Y4~27/ The double signs have also been dropped. After y is known, x is easily found. From the above operation it will be seen that any cubic equation may be made to take a quadratic form, by the following process. RULE. Transform the given equation into another in which the second power of the unknown quantity shall be wanting (Art. 466). Substitute for the unknown quantity a new unknown quantity, together with its reciprocal multiplied by one third of the coefficient of the second term of the equation with its sign changed. APPENDIX. 389 Irreducible Case. 509. When p is negative, and ^ > ^, Cardan's formula contains 27 4 imaginary, expressions, and yet it is easily shown that the three roots df the equation are then real and unequal. The result of Cardan's solution contains imaginary expressions, unless two of the roots are imaginary or equal. The three roots of the equation y'-\-py-\-q=0 may be repre- sented by a-\-\l3b, a — y/ 3 6, and — 2 a, where o and 6 may be either positive or negative, and 6 may be (Arts. 442, 447-450). Then, by Art. 441, p = o» — 3 6 — 20"— 2a/36 — 2a''4-2av'"36 = — 3a' — 36 q = 2a (a-\-/3b) (a — /Jb) = 2a (o»— 36) ?y== — (a«+6)' = _a» — 3o*6 — 3a=6« — 6» ll = o^Co' — 3 5)" = a» — 6a'6 + 9a'5« i-{-^= a'(a' — 3 6)'— (o« + 6)' =. —Qa*h -\- GaW — V n/: p 'Lj^iL = (So' — 6) J — h. Therefore, when the roots are real and unequal, 6 is positive, and Cardan's formula must contain imaginary expressions ; but When two of the roots are imaginary, 6 is negative, and the radical of the second degree in Cardan's formula becomes real. When two of the roots are equal, 6 ^^ 0, and the radicals of the second degree disappear from both the roots and the formula. 5I0i Although the formula -is irreducible by the common rules when the three roots are real and unequal, yet there are means by which it may be reduced, and shown to be real. When the cube root of the binomial surds is extracted, by developing info a series or otherwise, the imaginary expressions *ill have different sighs, and cancel each other. Thus, the equation s^ — 6a;-f-4 = gives i = (— 2 + 2 /--I) + ( — 2 — 2 y/ — 1)* which is proved by trial to be the same as l-f'V' — 1 + ^ — V' — 1 = 2. Also, ar"- 30a; — 86 :s=0 -gives a; = (18 + 26 y/ -- 1)' + (18 — 26 y/— 1)» == Z -\- ^ — l~i~3 — V' — ^ = ^- ^^ 'hAYS no genial ruiie for 33* 390 ALGEBRA. the extraction of the cube root of a binomial snrd; but it may always be developed into a series by the binomial theorem. The table of sines and tangents may also be employed in solving both the reducible and irreducible cases. 511. As any quantity has three cube roots (Art. 506), z evi- dently has six values (Art. 436) ; but the use of the double sign in (B) -would not increase the number of values of y, since one sign must be the opposite of the other, y has only three values, one de- rived from the common cube root of each expression, and two derived from the other cube roots. After the real root is found by means of the common cube root, the other two are readily found by solving the quadratic obtained by depressing the equation (Art. 444). 513t It will be seen that this formula may always be used when p is positive (Art. 453). It can also be employed when p is nega- tive, provided that 27 g* is numerically greater than 4p'. 513i The following examples can be solved by the method pointed out in the rule (Art. 608), or by a direct substitution in formulas (A) or (B), after removing the term containing a;^. Examples. . 1. Obtain the roots of the equation a;* — 3ar'-[-4 = 0. Substituting ^ -|- 1 for a;, ■tf — 3 .y -f- 2 = Substituting z -\- - ioi y, 2?-}-- + 2 = ■ z z" Or, 2?+ 2z»+l = Whence, z" -}- 1 = z = — 1, or liV'- 2 2/ = z + l- = -2, orl±i^Zz£ + LiJ^IZ_3= 2,orl,orl. z 2 ' 2 X = y -{-X = — 1, or 2, or 2. The value of y may also be obtained by substituting — 3 for ^, and 2 for q, in either (A) or (B), when ^j 'La.K. becomes 0, and y 4 ' 27 disappears. After a; = — 1 is obtained, the other values of x might be found by dividing a? — 3a;'4-4 by x-\-\, and solving the equation a;' — 4 a: + 4 = 0. APPENDIX. 391 2. Solve the equation a;* — 9 a; -j- 28 = 0. Substituting z + £ for a;, z" -f- 28z' -^f- 27 = z Solving by quadratics, s? = — 1, or — 27 Whence, z = — 1, or — 3 z = z+£ = _l + _L,or— 3 + -L = —4. ' z ' — 1 ' — 3 Dividing i? — 9 a; -j- 28 by x -\- i, and solving the equation a? — 4 a; -|- 7 = 0, the other roots are found to be 2 ± y/ — 3. a; = — 4 may also be obtained by substituting directly in either (A) or (B). 3. Obtain one root of the equation a;°-j-6a; — 2 :=: 0. Ans. ^1 — ^"2. Obtain the roots of the following equations by Cardan's method. 4. a;' — 6a: + 9 = 0. Ans. —3, or ^^ V~^ 2 5. a^ — 6a:2 + 57a;— 196 = 0. Ans. 4, or 1 ± 4 y/" .6. a:» — 43^^ — 3a:+ 18 = 0. Ans. 3, 3, or — 2. 7. a:*— 24a;2— 24a; — 25 = 0. Ans. 25, or ~ -^ ^ ^ ~ ^, 2 8. a;»+ 9a?— 21a;+ 11 = 0." Ans. 1, 1, or — 11 9. a;'— 2a;='+ 2a; — 1 = 0. Ans. 1, or ^ ^ ^ 2 10. a;» -{- 6 a; — 20 = 0. Ans. 2, or — 1 ± 3 y^ — 1 Note. J£ l.-^ P is positive, but not a perfect square, it does not always follow that tie real root of the equation is irrational. The square and cube roots may be obtained approximately by means of decimals, and the decimals in the two parts of the formula may destroy each other, as imaginary quantities always do in the irreducible case. Sometimes • the cube root of the binomial surd may be ex- tracted. Thus, (10 +. 6 v''3)^ 4- (10 — 6 v'l)* =1+1/1 + 1-/3 = 2. 392 ALGEBRA. BIQUADKAHC EQUATIONS. 514> General soluiions of biquadratic equations have been obtained by Descartes, Simpson, Euler, and others. Some of them require the second term of the equation to be removed, while others do not. All of them depend upon the solution of a cubic equation by Cardan's rule, and will of course fail when that fails (Art. 509). In general, if an equation has no equal roots, these methods cannot be applied without the inconvenience of the irreducible case, unless two of the roots are real and two imaginary. Hence they are practically of little value (Art. 482), especially as numerical equations of all degrees can readily be solved by Horner's method (Art. 487), or by one of the following methods of approximation. APPROXIMATION BY DOUBLE POSITION. 515f Find two numbers, a and 6, the one greater and the other less than a root of the equation (Art. 474), and 'suppose a to be nearer the root than 6. Substitute them separately for x in the given equation, and let A and B represent the values of the first member thus obtained. If a and 6 were true roots, A and B would each be ; hence the latter may be considered the errors which result from substituting a and b for x. Although not strictly correct, yet, for the purpose of approximation, we may assume that A : B :: x — a : x — b, whence (Art 316), A — B : A :: b — a : x — a, or (Art. 313), A—B:b — a :: A :x — a, and X — a = — j^ =-^, A — B ' ~ -,-l(ft-°) ' A~B That is, As the difference of the errors is to the difference of the two assumed numbers, so is either error to the correction of its assumed number. Adding this correction when its assumed number is too small, or subtracting when too large, we obtain a nearer approximation to the APPENDIX. 393 true root. This result and another assumed number may now be used as new values of a and b, for obtaining a still nearer approximation, and so on. It is best to employ two assumed quantities that shall differ from each other only by unity in the last figure on the right. It is also best to use the smaller error. This method of approximation has the advantage of being applica- ble to equations in any form. It may, therefore, be appUed to radical and exponential equations (Art. 421), and others not reduced to the general form (Art. 429). Examples. 1. Find one root of the equation 3? -\- s^ -\- x — 100 = 0. When 4 and 5 are substituted for x in the equation, the results are — 16 and -|- 55, respectively ; hence a = 4, 6^5, .4 = — 16, and B = 55. According to the above formula, the first approxima^ tion gives a; = 4-j- = 4.2 -f- As the true root is greater than 4.2, we now assume 4.2 and 4.3. Substituting these values for x in the original equation, we obtain — 4.072 and -|- 2.297; therefore, 4.3 is nearer the true root than 4.2. 99Q7 Hence, a; = 4.3 — — ^ = 4.3 — .036 = 4.264. 6.369 Substituting 4.264 and 4.265 for x, and stating the result in the form of a proportion, we have .0276 -f- .0365 : .001 : : .0276 : .00043. Hence, x = 4.264 -\- .00043 = 4.26443. Find one root of each of the following equations. 2. a;*— 2a; — 50 = 0. Ans. 3.864854. 3. 3?-{-10x'-\-5x — 260 = 0. Ans. 4.11799. 4. a^' + Sa^ + Sa;— 75.9 = 0. Ans. 2.4257. 5. a;* -j- ij X — I = 0. Ans. .66437. 6. K*— 3 a;'— 75a: — 10000 = 0. Ans. 10.2609. 7. a;5+ 2a:*+ 3a;»-f-4ar'+ 5 a; — 54321 = 0. Ans. .8.414455. 394: ALGEBKA. NEWTON'S METHOD OF APPROXIMATION. 516> Find two numbers, one greater and the other less than a root of the equation (Art. 474). Let a be one of those numbers, the nearest to the z-oot, if it can be ascertained. Substitute a-\- y ior x in the given equation ; then y is small, and by omitting yS j*, , as in Horner's method (Ai't. 487), a value of y is obtained, which, added to a, gives h, a closer approximation to the value of x. Now sub3titute S -^- 3 for a; in the given equation, and a second approxi- mation may be obtained by the same process as before. By proceed- ing in this way, the value of the root may be obtained to any required degree of accuracy. The assumed value of x should be nearer to one root than to any other, in order to secure accuracy in the approximation. Examples. 1. Find the real root of the equation af" — 2x — 5 = 0. When 2 and 3 are substituted for x in the equation, the results are — 1 and -\- 16, respectively ; hence a root lies between 2 and 3, and near to 2. Substitute 2 -j- ^ for x, and there results l — \0y—6y^—f= 0, whence, approximately, y = .1. Now substitute 2.1 -|- z for x, and there results .061 -f 11.23 z + = 0, Ofi 1 * whence, approximately, ^ = — ' = — .0054, and 1 \,Zo T = 2.1 — .0054 = 2.0946, nearly. Find one root of each of the following equations. 2. 31? — 3a;-f-l = 0. Ans. 1.53209. 3. a;»— 15a^ + 63a; — 50 = 0, Ans. 1.02804. THE END. "Advance In methcjs of instruction makts new— Improved text-books a necesslf;," ScHOOi. EoAiiDS and Teachers are respectfully Invited to examine the Improved edition, just published, of NEW PRACTICAL ARITHMETIC, in entirely new work, containing many important improvements, calca- ated to inaugurate a much-needed reform, both in the teaching of the science and of the art of numbers. R vias the first book of the kind to give a full and reliable presentation of the Nettie System of Weights and Measures, as legalized by Congress. Also, the first to make written arithmetic intellectual in all its processes, and >ra(^ical in all its applications; And justly claims pre-eminence for enforcing educational results by orderly 'irrangement of topics, and by systematic review questions and exercises. GREENLEAPS NEW PRACTICAL ARITHMETIC 4. complete in itself, and sufficient for Common Schools and Seminaries.. It is the long desired practical course, time saving' and labor saving. It avoids the usual multiplicity of rules, in Multiplication, Division, (eduction, etc., which have been found unnecessarily to tax the memory, -r confuse the mind. It contains no useless lumber of obsolete weights and measures, "pounds, lillings, and pence," " cross multiplication," etc., to discourage and retard .rogress. The estraordinaiy success of the first editions of this new work is oundant evidence of its meeting present educational wants, and of .:;3 being THE RIGHT BOOK AT THE BIGHT TIME. CREENLEAF'S New Pfimafj, iew Intellectual and Hew Practical .a.i2.iti3:is/i:etios, AND Greenleaf's New Elementary Algebra, were unanimously adopted by the Vermont State Boakd of BcccATroN, In February, 1867, to be used in all the Public Schools of that Slite, for a term of five years, as required by law. GHa E E N JL, E A T^ » S mmw w>mA©wmA^ AmssMMWw^ has, within a few months, been adopted for schools in more than 200 CITIES AND TOWNS, in all parts of the country, including adoptions for several entire couNTiEa in the Middle and Western States. GREENLEAF'S ARITHMETICS AND ALGEBRAS, are approved text-books for NORMAL SCHOOLS AND COMMERCIAL COLLEGES in all parts of the country. ©EEEKLEAE'S SYSTEM Is now used, with great acceptance, in the public schools of upwards of 1000 CITIES AND TOWNS IN NEW ENGLAND, and in the public schools of yew York City, Philadelphia, and New Orleans, and is generally introduced in THS MIDDLE STATES, and has a growing popularity in ALL THE A^TESTERlSr STATES. Greenleaf's is the only Mathematical SBiuEg for which the dmnand steadily iHcreases in spite of UNPARALLELED COMPETITION. 3