CORNELL UNIVERSITY LIBRARY MATHEMATICS f OSIjEI-L. UNIVERSITY LIBRARY 3 1924 063 723 393 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/cu31924063723393 o o jy: i^ L E T E A L GE B E A SCHOOLS AND COLLEGES, A* SCHUYLEE, M. A., Prnfesaor of Mathematies and Logic in Baldwin University; AjitUor of Higher Arithmetic and Princvples of Logic. CINCINNATI: WILSON, HINKLE & CO. NEW YORK: CLARK & MAYNARD. Entered according to Act of Congress, in the year 1870, by WILSON, IIINKLE & CO., In the Clerk's Office of the District Court of the United States for the Southern District of Ohio. STF.UEOTYPF.P AT THE FUANKLIN Tll'K FnUXPIIV, CINCINNATI. PREFACE In preparing an Algebra, complete in one book, it has been the aim of the author to render the work suiEciently elementary for beginnere who have a practical knowledge of Arithmetic, and suf- ficiently advanced for those .who intend to pursue the Higher Mathematics. The work has been called out by a recognized want: One book embracing the essential principles of the science, clearly explained and illustrated by appropriate examples. In this way, it is hoped that the multiplication of classes may be avoided. Indeed, there seems to be no reason why two books should be required for Algebra, except perhaps the immaturity of some who make haste to begin the study. Ill the preparation of the work, care has been taken to render the demonstrations clear, and to illustrate the principles by their application to a variety of carefully graded examples. Theorems and Factoring, subjects intimately related, are treated in immediate connection. In Equations, each numerical problem is immediately followed by a general one of like nature, the solution of which gives a formula from which the answer to the numerical problem may be deduced. The student is thus early trained to generalize and to apply formulas. In the solutions^ the equations have been numbered and the operations concisely indicated, thus aflFording models to guide (ill) IV PREFACE. the student in giving to his own solutions the form suitable for publication. The Problem of the Lights has been generalized, and all the equally illuminated points have been found, which will add greatly to the interest of the discussion. By the method of proof known as Mathematical Induction, the Binomial Theorem is first demonstrated when the exponent is a positive integer; afterwards, by the principle of Indeter- minate Coefficients, the theorem is established for any expo- nent. Ratio, Proportion, Variation, the Progressions, Logarithms, and the Theory of Equations, have all received careful atten- tion. The proposition. Every equation has a root, which is usually assumed, has been rigorously demonstrated. The beauty of the type and the arrangement of the matter will add much to the pleasure of the study. A. SCHUYLER. Baldwin Univeusity, Berea, O., Feb. 7, 1870. INDEX Definitions, .... Summary of Algebraic Symbols, Definitions Continued, . Addition, Subtraction, , Multiplication, Multiplication by Detached Coefficients Division, ..... Division by Detached Coefficients, Theorems and Factoring, The Greatest Common Divisor, The Least Common Multiple, Fractions, .... Vanishing Fractions, Equations of the First Degree, Elimination, .... Symmetrical Equations, Indeterminate Equations, Indeterminate Problems, Incompatible Equations, Problem of the Couriers, Involution, Evolution, Radicals, Imaginary Quantities, Inequations, . Equations of the Second Degree, (V) PAOE Recurring Equations, 204 Binomial Equations, .•.....■•• 208 General Discussion of Quadratics, .,,.... 212 Problem of tlio Lights, .... .... 218 Quadratics Involving two or more Unknown Quantities, . . 224 Ratio, . . 233 Proportion, ....... .... 236 Variation, ..... ... . 247 Harmonical Proportion, 254 Arithmetical Progression, 265 Geometrical Progression, . 262 Harmonical Progression, 270 Permutations, 273 Combinations, . . ........ 275 Indeterminate Coefl&cients, . ....... 279 Binomial Theorem, . 284 Differential Method of Series, ... .... 288 Logarithms, . ....... . . 295 Theory of Equations, ......... 312 Derived Functions, 319 Permanences and Variations, 340 Transformation of Equations, ........ 342 Limits of the Roots, 348 Sturm's Theorem, .... ... . . 352 Horner's Method of Approximation, 358 Cubic Equations, • 363 Biquadratic Equations, 366 ALGEBEA. 1. Definitions. 1. Algebea is that branch of Mathematics which treats of the general relations of quantities by means of symbols. 2. The symbols employed in Algebra are the figures, letters, and signs used in expressing quantities, operations, and relations. 3. The symbols of quantity are figures and letters. 1st. Known quantities are usually denoted by figures, 1, 2, 3, . . . , or by the first letters of the alphabet, a, b, c, . . . 2d. Unknown quantities are usually denoted by the final letters, . . . x, y, z. 3d. The symbol, 0, called zero, denotes the absence of quantity, or a quantity less than any assignable quantity. 4th. The symbol, co, called infinity, denotes a quantity without limit, or a quantity greater than any assignable quantity. 5th. The symbols, '■ "■ '"■ . . . , called accents, and the symbols, j, 2, 3, . . . „, called subscripts, are attached to letters to denote quantities symmetrically arranged, or un- known quantities whose values have been determined. 8 yUjfjrajOIl^l. The expressions, a', a", a'", . . . , are respectively read, a prime, a second, a third, ...» and ai, aj, as, ■ ■ ■ «nj are re- spectively read, a one, a two, a three, . . . , a n. 4. The symbols of operation are the signs of addition, subtraction, multiplication, division, involution, and evo- lution. 1st. The sign of addition is the vertical cross, +, called plus. Thus, a-\-h, read a plus b, indicates that h is to be added to a. 2d. The sign of subtraction is a short horizontal line, — , called minus. Thus, a — b, read a minus b, indicates that b is to be subtracted from a. 3d. The sign of multiplication is the oblique cross, X , or the dot, [.]. Thus a y. b, or a . b, read a multiplied by b, or a times b, or a into b, denotes the product of a and b. If either factor is denoted by a letter, the sign of multi- plication is usually omitted. Thus, 5 a, denotes the same as 5 X a, or 5 . a, and a b denotes the same as a X 6, oi a . b. 4th. The signs of division are, -;-, :, — , ), | . Thus, each of the expressions, a n- 6, a:b, j-, b)a, a\b, denotes that a is to be divided by b. 5th. The sign of involution is a small figure or letter, called an exponent, placed at the right and a little above a symbol of quantity. Thus, ^- ^' ^ ", are signs of involu- tion, and the expression a', a^, a^, . . . a", denote, respect- ively, the first, second, third, .... and nth powers of a. The exponent, ', is usually omitted; thus a' denotes the same quantity as a. DEFINITIONS. 9 6th. The sign of evolution is the symbol, ^, called the radical sign, or the fractional exponent, -'»'••■«. The degree of the root to be extracted is indicated by the index of the radical, a small figure or letter placed in the angle of the sign, or by the denominator of the fractional exponent. Thus, f^, ^a, . . . i/^ or a\ a*, . . . a", denote, respect- ively, the square root of a, the cube root of a, ... . the n"' root of a. The index, ^, is usually omitted. Thus, ^^denotes the same as y'^. 5. The symbols of relation are the signs of equality, inequality, ratio, proportion, variation, aggregation, contin- uation, and deduction. 1st. The sign of equality is two equal lines, parallel and horizontal, =. Thus, a; = a is read x is equal to a. 2d. The sign of inequality is an angle placed thus > or thus < - Thus, .r > a is read x is greater than a, and a; < i is read X is less than b. 3d. The sign of ratio is the colon, [:]. Thus, a:b denotes the ratio of a to b, and is equivalent to a -=- 6, or to j. 4th. The sign of proportion is the colon, double colon, and colon, : : : : . Thus, a : b : : e : d, read a is to 6 as c is to d, is a proportion, and denotes that the ratio of a to 6 is equal to the ratio of c to d, and is thus equivalent to the equation, a -^b ^^ c -^ d, 01 to J = -^. 5th. The sign of variation is the symbol, oc . Thus, axb, denotes that a varies as b. 10 ALiijtjmiiA. 6th. The signs of aggregation are the bar, | , the vincu- lum, , the parenthesis, ( ), the bracket, [ ], and the brace, j \. Thus, each of the expressions, , A, « + 6, (a -\- b), [a -|- ^3> W -\- b\, denotes that a -j- 6 is to be treated as one quantity. If but one sign of aggregation Ife required, () is commonly used; if two, first (), then []; if three, first (), then [], then { }; if more, they are repeated in the same order. Thus, (a + 6)c, [(a + 6)c+d]e, } [(« + 6) c + d] e +/1 gr. 7th. The signs of eontinuation are dots, . . . , or dashes, , or etc. Thus, a, a', a', . . . , or a, a", a^, , or a, a^, a\ etc., read a, a', a', and so on. 8th. The sign of deduction is three dots placed thus, . •. , and read therefore, hence, or consequently. 9th. The sign of reason is three dots placed thus, ■.", and read because or since. 2. Summary of Algebraic Symbols. 1. Symbols of Quantity. 1st. Of known quantities, \ ^ig^^^s, 1, 2, 3, . . . (. First letters, a, b, c, . . . 2(1. Of unknown quantities — Final letters, . . . x, y, z. r Positive limits j 3d. Of limits of quantity, -< in' (. Negative limits \ I 00. 4th. Of symmetrical or determined f a', a", a'", . . . quantities, la,, a.^, a^, . . . DEFINITIONS. 11 2, Symbols of Operation. 1st. The sign of addition, +. 2d. The sign of subtraction, — . 3d. The signs of multiplication, X, • • 4th. The signs of division, -^, :, -, ), (_. 5th. The signs of involution, '■ ^' s.-". n., . ^ w f v. -r, -r, ■ • ■ i"/. The signs oi evolution, \l t \ i 6th 3. Symbols of Relation. 1st. The sign of equality, ^. 2d. The signs of inequality, > , < . 3d. The sign of ratio, : . 4th. The sign of proportion, : : : : . 5th. The sign of variation, oc. r The bar, The vinculum, . The parenthesis, ( ). The bracket, [ ]. The brace, ! }. 6th. The signs of aggregation. 7th. The signs of continuation, -etc. 8th. The sign of deduction, . ■ . . 9th. The sign of reason, • . ■ . 3. Definitions Continued. 1. A co-efficient of a quantity is a factor showing how many times that quantity is taken. Thus, in the expression, 12 ALiUJiUliA. 5x, is the co-efficient of a;; also, in the expression, ax, a is the co-efficient of a;, or 1 understood, the co-efficient of ax. The co-efficient equals the number of additions plus 1. Thus, 3a; = a; + a; + a;, and aa;:=a; + a; + a;-j-...to a terms. 2. An exponent of a quantity is a small figure or letter placed at the right, and a little above the symbol of thu quantity, denoting how many times that quantity is to be taken as a factor. Thus, a', (x -\- y)", denote, respectively, the cube of a, the »"' power ot x -\- y. The exponent equals the number of multiplications plus 1. Thus, a* = a X ffl X a, and a'' = a X a X a • • • to n factors. 3. A power of a quantity is the product of factors each equal to the given quantity. Thus, a' ^ a X » X a, is the third power or cube of a, and a" is the n"' power of a. 4. A root of a quantity is one of the equal factors of that quantity. Thus, a is the cube root of «*, and x is the n"' root of a;". 5. The reciprocal of a quantity is unity divided by that quantity. Thus, - is the reciprocal of a. 6. A term is an expression not connected with any other by the sign + or — , or it is any one of the parts of an expression which are so connected. Thus, 6xy is a term, and 3a^, — Gab and + 3c are the terms of the expression, 3a2 — 6a6 -f 3c. 1st. A positive term is one which has the sign -{-, cx- pressed or understood. Thus, + 3aa; and lab are positive terms. 2d. A negative term is one which has the sign — . Thus, — 7xy is a negative term. DEFINITIONS. 13 3d. A simple term is a single expression, not having parts connected by the signs + or — . Thus, 5ax is a simple term. 4th. A compound term is a collection of terms represented as one quantity by one or more of the signs of aggregation. Thus, [(m -\- n) p -{- r\ s \a s. compound term. 5th. A numerical term is a nvimber. Thus, 4, $15 are numerical terms. 6th. A literal term is one containing one or more letters. Thus, ah, Ix^, — %a^x are literal terms. 7th. Similar terms are abstract numbers, similar denom- inate numbers, or literal terms containing the same letters, each affected with the same exponent. Thus, 4, 5; 17, — $5; 5a'62c^ _7a3j2c. 8th. The degree of a term is the number of its literal fac- tors, and is determined by finding the sum of the exponents. Thus, 5a* 6^ c is of the sixtli degree, since 3 -)- 2 -|- 1 = 6. 9th. Hmnogeneous terms are those of the same degree. Thus, 3a^6, and — 606^ are homogeneous terms. 10th. A monomial is a single term, simple or compound. Thus, 7a^ b, and 8 (a + ^) *re monomials. 11th. A polynomial is an expression of two or more terms not represented as one quantity by a sign of aggregation. Thus, 7a + 5c — Qd is a polynomial. 12th. A binomial is a polynomial of two terms. Thus, a + & is a binomial. 13th. A trinomial is a polj'uomial of three terms. Thus, a -\-b — c is a trinomial. The terms of a polynomial may be written in any order, if the sign of each term be retained. Thus, a — 6 -(- e = — b-j-e-\-a =, etc. 14 ALGEBRA. 7. A proposition is somethiug proposed or stated. Propositions are definitions, axioms, postulates, theorems, problems, lemmas, corollaries, or scholiums. 1st. A definition is such a description of an object as will distinguish it from all other objects. 2d. An axiom is a self-evident truth. 3d. An absurdity is a self-evident falsity. 4th. A postulate is a self-evident possibility. 5th. A theorem is a truth which requires demonstration. 6th. A -problem is something proposed for solution. 7th. A lemma is an auxiliary theorem or problem. 8th. A corollary is an obvious consequence. 9th. A scholium is a note or remark. 10th. An hypothesis is something assumed. 11th. A formula is a theorem expressed in algebraic lan- guage. 8. A demonstration is the proof of a proposition. Demonstration is direct or indirect. 1st. A direct demonstration is one which commences with known truths and combines them in a logical process till it establishes the proposition to be proved. 2d. An indirect demonstration, called also the reductio ad absurdum, proves a proposition true by showing that the supposition that it is false involves an absurdity. 4. Axioms. 1. If equals be added to equals, the sums will be equal. 2. If equals be subtracted from equals, the remainders will be equal. 3. If equals be multiplied by equals, the products will be equal. EXERCISES. 15 4. If equals be divided by equals, the quotients will be equal. 5. The same powers of equals are equal. 6. The same roots of equals are equal. 7. The whole is greater than a part. 8. The whole is equal to the sum of all the parts. 9. Things equal to the same thing are equal to each other. 5. Postulates. 1. Quantities of the same kind can be added together. 2. One quantity can be subtracted from another of the same kind. 3. A quantity can be taken any number of times. 4. The number of times that one quantity is contained in another of the same kind can be found. 5; A quantity can be divided into any number of equal parts. 6. The ratio of two quantities of the same kind can be found. 7. Any power of a quantity can be found. 8. Any root of a quantity can be found. 6. Exercises in Algebraic Notation. 1. Write the sum of x and y. 2. Write the sum of x and y multiplied by their differ- ence. 3. Write the sum of x and y divided by their difference. 4. Write a-{-h divided by c. 5. Write a, -\-h divided by c. 6. Write the square root of x-\- y. 7. Write the square root of x, -\- y. 8. Write a, -\-h multiplied by a, ■ — h. 9. Write a-\-h multiplied by the square root of a — h. 16 ALGEBRA. 10. Write a, -\- b multiplied by the square root of a, — b. 11. Write a homogeneous binomial. 7. Exercises in Finding Numerical Values. Find the numerical value of the following expressions, if a = 2, 6 := 3, c = 4, d^ 5, a; = 8, y=Q. 1. a-\- b — e. 2- (x-\-y) (x — y). 3. (x + y)x — y. 4. x^xy — y. 5. [(a + 6) c + £«](» — 2/). 6. \\_ (a + h) c- d-\x + y \y. 7. i/a^ -\- d X i/a;2 +2/'- 8. \'(a + 6)i/(a; + 2/)(a;-2/)-6. 9. a + 6 V(a + f^) i/a;!/ — c + d. Am. 1. Am. 28. Am. 106. Alls. . 50. Am. 50. Am. 756. Am. 30. Am. 5. Am. 23. ADDITION. 8. Definitions. 1. The sum of two or more quantities is their simplest ex- pression. 2. Addition is the process of finding the sum of two or more quantities. 9. Illustrations. 1. Find the sum of 3a ^ 6 and 5a^b. OPERATION. 3a^ b 3 times any quantity + 5 times the same Srt^fe quantity := 8 times that quantity. 8a^b ADDITION. 17 2. Find the sum of Safc^ and — bab"^. OPEEATION. 8a6^ 8 times any quantity — 5 times the same — 5a&^ quantity = 3 times that quantity. 3a62 3. Find the sum of — 9mji and 4m?i. OPERATION. — 9mn- — 9 times any quantity + 4 times the same 4mw quantity = — 5 times that quantity. — bmn 4. Find the sum of ax and hx. OPERATION. ax a times x -\- h times x = (a -\- h) times a;, a hx_ and 6 are regarded as co-efficients of x. (a -f ■ 6) a; 5. Find the sum of 6a, 8a, — 5a, and — 6a. OPERATION. 6a 8a 6a + 8a = 14a. — 5a — 6a =; — 11a, and — 5a 14a _ lu = 3a. — 6a 3a 6. Find the sum of a, — 6, and c. OPERATION. a — 6 -f- c Connect the quantities by their signs. 7. Find the sum of 2ab -\- 3xy, 4ab -\-2xy, and ab — OPERATION. 2ab + 3xy 4a6 -)- 2xy ab — 8xy + 3p lah — Bxy + 3^ C. A. 2 18 ALGEBRA. 10. Rule. 1. Write the quantities to be added so that similar terms, if any, shall stand in the same cohimn. 2. If all the terms of a column have like signs, find the sum of tlieir co-efficients, prefix the common sign, and annex the com- mon literal part. 3. If aM the terms of a column have not like signs, find tlie sum of the co-efficients of tJie positive terms, also the sum of the co-efficients of Ike negative terms, take their numerical difference, prefix the sign of the greater, and annex the common literal part. 4. Bring down, with their proper signs, the terms, if any, which are not similar to any of the others. 11. Examples. 1. Add 3a6 + 5a2 6, M> — Sa^ h, Bab — a^ b, and Sab + 2a2 b. Am. 20ab — 2a^ b. 2. Add 5xy + 4x^z — Sxz"^, Ixy + 9x''z — xz", and — lOxy + 20a;2 z — 15xz^. Ans. 2xy + 33a;2 z — IQxz"^. 3. Add lOx^y—lOxy^, 15x'^y—5xy', x^y + xy^, and — '2x'y -{- 5xy^. Ans. 2ix^y — dxy^. 4. Add 3a + 6 — c, 9a— 66 — 10c+/, 8a + 66 — 8c -|- 3/, 4a — 6 -(-c + g, and 6a — 36 — 8c. Ans. 30a — 36 — 26c + 4/+ ^. 5. Add a + 6, c — d, e -)-/ — g. Ans. a-f-6-fc — d-|-e+/ — g. 6. Add 5(a + 6) — 6(x — y), 10 (a + 6) — 8 (a; — y), and 20 (a + 6) — 10 {x — y). Ans. 35 (a -+■ 6) — 24 (a; — y). 7. Add 8a'6 + 6a6S 10a3 6 — 18a58, 4a8 6 — 20a68, 9a36 — 8a63, and — Ba^ 6 + 15a53 +c-\-h. Ans. 25a3 b — 25a63 -f c + A. SUBTRACTION. 19 8. Add ax, hx, ex, and dx. Ans. (a -\- h -\- e -\- d) x. 9. Add a(x-{-y) and h (x -[- y). Ana. {a -{- b) (x -\- y). 10. Add ax -\- by, ex -\- dy, ex -\-fy, and gx + hy. Ans. (a H- c + e + gr) a; + (6 + d +/+ h) y. 11. Add 5i/a + 6, i&ya^b, ly'a-[-b, —lOy'a-{-b, and 18 i/a+7- Am. 26 1/0+^. 12. Add a y/x + y, b j/a; + y, c 1 ''iJT+y, and rf i/i~+l/7 J.?is. (a + 6 + c -|- d) i/a; -)- y. 13. Add am + bn and 6m -f- cm. Am. (a -\- b) m -\- (a -\- h) n. 14. Eeduce (a -\- b) m -\- (a -^ b) n, by regarding m and n as co-efRcients of (a -j- 6). J.ns. (a -f 6) (m -|- m). 1 5. Add ax -\-by -\- ez, bx -\- ey -\- at, and ex -)- ay + 62. ^m. (a + 6 -)- c) (a; + 2/ + z). 16. Add abx"" + oc)/" + 602", act;" + icy" + a6z", 6ca;" + oJy" + aez'', labx'- + 2acy* + 2ba£\ 2ac.v" + 2662/" + 2ahz", 2bc.r:" + 2al)y" + 2aez''- Ans. 3 (06 + ac + 6c) (.-c" + y" + z"). SUBTRACTION. 13. Definitions. 1. Subtraction is the process of taking one quantity from another. 2. The quantities considered are the subtrahend, the min- uend, and the difference. 1st. 7'he subtrahend is the quantity to be subtracted. 20 ALGEBRA. 2d. The minuend is the quantity from which the subtra- hend is to be subtracted. 3d. Tlie differenee is the quantity which added to the sub- trahend will give the minuend. 13. Illustrations. 1. From a subtract b. OPERATION. a Since the terms are dissimilar, the subtraction b can only be indicated. a — b PROOF. b Adding the difference, a — b, to the subtra- g — b hand, b, the sum is the minuend, a. a 2. From a — b subtract c — d. OPERATION. a — b Subtracting c from a — b, we have — d a — b — c ; but we have subtracted too * — b — e -\- d much hj d; . " . the remainder is too small by d; . • . a — b — e -\- d is the true re- mainder. The signs of the minuend are retained, but those of the subtrahend are changed. PROOF. c — d Adding the difference, a^b — c -\- d, to a — b — c -\-d the subtrahend, c — d, the sum is the miu- a — b uend, a — b. Hence, for all values of a, b, c, d we have the formula (1) a — b — (c — d)^a~-b~c + d. If {^^2} (1) becomes (2) a-i—d)=a + d. SUBTRACTION. 21 3. From 5a — 66 subtract 3a — 26 — e. OPERATION. 5a — 66 Conceiving the signs of the subtrahend 3a — 26 — c changed, we have 5a — 3a = 2a, — 66 -(- 26 2a — 46 + c = — 46, and + c. 14. Rule. 1. Write the subtrahend under tlie minuend so tJiat similar terms, if any, shall stand in the same column. 2. Conceive the signs of all the terms of the subtrahend changed, and proceed as in addition. 15. Examples. 1. From Sax — 46?/ -|- 3cz take ax + 3by — 5gz. Ans. lax — Iby -|- %cz. 2. From lOa^ 6 + 156^ c — Sc^ d take 5a2 6— 106^ c + c^d. Ans. 5a'^ 6 + 256^ c — 9c^ d. 3. From bx'^y — 6xy^ + Sy^z — 5ye^ subtract — Ax^y -\- 5xy^ -j- 5y^s — Gye'' — m -j^ n. Ans. 9x^ y — llxy'^ — 2^/^ z -{- yz^ -\- m — n. 4. From a^ + 2a6 + 6^ take a^ — 2a6 + 6^. Ans. 4a6. 5. From 6*^ 63 c — 7a6c3 + 9ac take 4a2 6^ c + Bahe" —■ Sac. Ans. la"^ b^c— lOabc^ + 17ac. 6. From 5a™ — 66™ take a™ + 6'" — c". Ans. 4a™ — 76™ + c". 7. From x"'y'' take x'"if — tf"z". Ans. f^K 8. From ax -\- by take ca; + dy. Ans. (a — o) X -\- (6 — d) y. 9. From a* -|- ay + az take 6a; + 6y + bz. Ans. (a — 6) (a; + 2/ + 2). 22 ALGEBRA. 10. From (a -(- 6) j/k take (a — b) y'x. Aiw. 26|/i. 11. From a ^/i + 6 i/y + c j/z take rf v'^ + e i/j/ +/lA ^m. (a — d) v/» + (6 — e) |/y + (c — /) 1/2. 12. From 3aa; + Say + 3(k take 36* + 2,hy + 862. .4*48. 3 (a — 6) (a; H- 3/ + 2). 16. Principles relating to the Signs of Aggregation. 1. a-\- (h — c) = o + 6 — c. This formula is true, since the quantity within the paren- thesis is to be added to the preceding quantity. Hence, if a quantity within a sign of aggregation is preceded by the sign -{-, the sign of aggregation can he omitted without any change of the signs of the terms within the sign of aggrega- tion. Conversely, any number of the terms of a polynomial can be enclosed by a sign of aggregation preceded by the sign -\-, witJwiit any change of tlie signs of the terms. 2. a — (b — c) = a — b -{- c. This formula is true, since the quantity within the paren- thesis is to be subtracted from the preceding quantity. Hence, if a quantity voithin a sign of aggregation is preceded by the sign — , the sign of aggregation can be omitted, if the sign of every term enclosed by the sign of aggregation be changed. Conversely, any number of terms can be enclosed by a sign of aggregation preceded by the sign — , if the signs of all the terms placed within tlie sign of aggregation be changed. The sign preceding the sign of aggregation is the sign of operation. The signs within the sign of aggregation are the signs of the quantities. The signs after the sign of aggregation has been omitted are the essential signs. MULTIPLICATION. 23 If the sign of operation is -(-> the essential signs are like the signs of the quantities. If the sign of operation is — , the essential signs are un- like the signs of the quantities. 17. Examples. Omit the signs of aggregation in the following examples : 1. a — 6-f (c — d — e+/). Ans. a — 6 + c — d — e -{-f. 2. a-\- ( — 6 — c — d -\- e). Ans. a — h — c — d -\- e. 3. a — ( — 6 — e-\- d — e) . Ans. a -\-h -{- c — d -{- e. 4. a + 6_[c-d_(e_/)]. Ans. a -\- b — c -\- d -\- e ■ — /. 5. a- J6 + [d-e- (/-(,)]!. Ans. a — b — d -\- e -f/ — f-b-\-...toa terms = -j- ab. ■■■ (4) -X- = +. From (1) and (4), and (2) and (3), we have the laws: 1st. Like signs give -\-. 2d. Unlike signs give — . 20. Consequences. 1. Changing the sign of either factor, changes the sign of the ■product. ' 1st. If the signs are alike, the product is -\- ; chang- ing either sign, the signs will be unlike, and the product — . For, ' 2d. If the signs are unlike, the product is — ; chang- ing either sign, the signs will be alike, and the product -|-. For, i MULTIPLICATION OF MONOMIALS. 25 ' 2. Changing the signs of both factors, does not change the sign of the product. ' 1st. If the signs are alike, the product is + ; chang- ing both signs, the signs will still be alike, and the product +. 2d. If the signs are unlike, the product is — ; chang- ing both signs, the signs will stilL be unlike, and the product — . 3. if tihe number of minus factors of a product is odd, the product is — , otherwise the product is -\-. For, the product of the positive factors, if any, is +, and this by one minus factor, — , by two, -f-, by three, ^, by four, +, in general, by an odd number, — , by au even number, -|- Corollary. — An odd power of a negative quantity is — , an even power, +. 21. The €o-efflcients and Exponents. 1. 5a62 X iabc = 5abb X 4abc = 5 X iaabbbe = 20a2 fes g_ 2. 4a™J"eP X Sa'^b'^ed^ = 12ffl™+"6"''2c'''''i d^. Hence, the co-efficients are multiplied, and the exponents of tJie same letters are added. 22. Eule. 1. Multiply Hie co-efficients, prefixing Hie sign — , if the num- ber of minus factors is odd, otherwise the sign -\-. 2. Annex all the letters common to two or more of the factors, giving to each an exponent equal to the sum of its exponents in the factors. 3. Annex the letters not common with their exponents. C. A. 'A. 2G ALGEBRA. 23. Examples. 1. Sa^ft^e and 7a2 6''c2. Ans. ^ba^ h^ c^ . 2. — Ta'i^cS and Saft^c^f. ^m. — 56a8 6*c*d 3. lOa^ he and — 3a6s c. ^m. — SOa" 6" c^. 4. — bx'y^ and — Sx^y^z. Ans. Ibx^y^z. 5. a'^h, ah''',- ac''', —ad, — a^, — 6^. ^jis. — a'^b^c^d. 6. 22;22/> 3a;2/^, xz^, — xz, — x^ z. Ans. Gx^y^z''. 7. a™, 6", — c", — aP, b", —d". Am. — a'"+''i"'-P(f d«. 8. 7 (n + 6)'", — 8 (a + 6)", — (a + 6). ^n.s. 56(a + 6)'"-"'+'. 9. 3 (a 4- Z.) (c + d), — 5 (a + 6)8, 5 (c + d)2. ^m. — 75(a-|-6)*(c + d)3. 10. 5 (a + 6)"' (c + dy\ — 4 (a + 6)3, — (c + d)2. .4m. 20 (a + bj"'^ (e + dy''^. MULTIPLICATION OF POLYNOMIALS. 34. Illustrations. 1. Multiply a — 6 by c. OPERATION. We first multiply a by e, and obtain ae; a — b but since a is too great by b, the product, ac, _c is too great by be. Subtracting be from ae, ac — 6c we have ae — be for the true product. 2. Multiply a — 6 by c — d. OPERATION. We first multiply a — 6 by c, and a — 6 obtain ae — he; but since e is too ^ — d great by d, the product is too great ae — be — ad + hd hy a~h multiplied by d, which is ad — bd. Subtracting ad — bd from ac — 6c, we have ae — he — ad -\- bd for the true product. UULTIPLIGATION OF POLYNOMIALS. 27 These examples can be thus performed: 1. (a — 6) c = ac — ho. 2. {a ~h) {c — d) =. {a ~h) c — {a — h) d = ao — he — (ad — hd) =ae — be — ad -\- bd. Therefore, for all values of a, h, o, and d, we have the formula, (1) (a — b) (c — d) = ae — be — ad -{- hd. Hence, for the multiplication of polynomials, as for that of moncMiaials, we have the laws for the signs : Isi, Like signs give -\-. 2d. UnLike signs give — . The laws for the signs in the multiplication of monomials are involved in those for the multiplication of polynomials ; for, 1st. If { ^ ^ ^ I - (1) becomes (2) + a X + c = -f- ac. 2d. If j * "" ^ I , (1) becomes (3) — 6 X + c = — 6c. 3d. If I ^ ^ ^ I , (1) becomes (4) -^ a X — d = — ad. 4th. If {"^^l' '^^^ becomes (5')~bX—d= + bd. 3. Multiply a^ + 2ab + h^ by a'^ — 2ab + i^. OPERATION. a^ + 2ah + b^ a^—2ab + 6^ a* + 2a^b + a^"^ — 2a^b — 4a262 _ 2ah^ a^b^ + 2ah^ + h* a" — 2a262 _^ ^4 28 AhGKBHA. 35. Rule. Multiply all of tiie terms of the multiplicand by each term of the multiplier, and add tlie results. 26. Examples. 1. Multiply X -{- y hy X + y. Ans. x^ -j- 2xy -\- y^. 2. Multiply X — y by x — y. Ans. a;^ — 2xy + y^- 3. Multiply x -\- y hy X — y. Ans. x^ — y^. 4. Multiply 3a;2 + 2y^ by Bx^ + 2y\ Ans. 9x* -\- 12x2 ?/2 _|_ 4y4 5. Multiply Qx'^y -\- 5xy^ by Gx^y — 5xy^. Ans. SQx* y'^ — 25x'^ y" . 6. Multiply a2 ^ a6 + 62 by a^ _ ^jj ^ 52, Ans. a* + a'^b'' + 6*. 7. Multiply 2a2 — 3ab + 46^ by Sa^ — 6ab — 26^. Ans. 10a* — 27a^b + Sia'^b'^ — ISaft^ — 86". 8. Multiply x^ + 3x''y+ Sxy^ + y^ by a;^ _ Sx^y-\- 3xy" — y^. Ans. x" — 3x* y'^ -{- Sx'^ y'^ — y". 9. Multiply 2x3 ^ 4^2 _|_ gj. _|_ jg by 3.t — 6. Ans. 6x* — 96. 10. Multiply x-\-y, x — y, x^ -)- xy + 7/2, and x^ — ■ xy -f- 1/2 together. Ans. x" — j/" . 11. Multiply 4(a + 6)2— 3(c + d)8 by 4 (a + 6)2 + 3 (c + (0". .4ns. 16 (o + 6)4 — 9 (c + d)". 211 12. Multiply X'" + 2/'' by .x™ — y\ ^l,ks-. x2"' — ?/ 13. Multiply X + a by X + 6. Ans. x2 + (« + 6) a; + ah. 14. Multiply ^x™ + 5x" + r by ^x" + yx" + r. Ans. p^x'^'" + 2j)gx'"+" + 2prx™ + q^ x2" + 25rx" + r^. 15. Prove that (a + 6 + c) (a — 6 -)- c) = a2 _j_ j2 _j_ ^2^ if ac ^. 62. MULTIPLICATION OF POLYNOMIALS. 29 MULTIPLICATION BY DETACHED CO-EFFICIENTS. 37. Illustrations. 1. Multiply a;' -)- 3a;2y + 8xy^ + y^ by x + y. OPERATION. 1 + 3 -(- 3 4- 1 The co-efficients are detached from 1 -|- 1 the literal part and multiplied as in I + 3 -)- 3 + 1 multiplication of polynomials. By + l + 3-|-3H-l simple inspection, we see that the ex- l-(-4-|-6-(-4-|-l ponents of x in the product are 4, 3, 2, 1 ; those of j/, 1, 2, 3, 4. x is not in the last term, nor y in the first term. . ■ . x"^ -\- 4x^ y -\- Gx'^ 2/^ + 4a;i/* -j- y'^ = the product. 2. Multiply x^ -\- xy -\- y^ by k — y. OPERATION. 1 + 1 -|- 1 The co-efficients are detached and mul- i — 1 tiplied, and the exponents determined by 1 + 1 -{- 1 ' simple inspection. The terms Ox'^y and — 1 — 1 — 1 Ox!/^ are omitted since they are 0. 1 -)- -f- — 1 . • . .T^ — y^ "^ the product. 3. Multiply a;' -(- .ti/^ -(- y^ by a;^ + *^2/ ~l" V^- OPERATION. The terms Ox^ y in the 1 -(- + 1 -|- 1 multiplicand, and Oa^^ in 1 + 1 + -(- 1 the multiplier, are to be sup- 1 + -)- 1 -f 1 plied ; then the process is as 1 + + 1 -f- 1 before. 1 + + 1 + 1 ••• .-B^ + ai^y + a;*2/2 + 1+1 + 1 + 3 + 1 + 1+1 3a;3 2/3+.i;2 2/*+x2/5 +^6 the product. 30 ALGEBRA. 28. Rule. 1. Arrange the terms of the factors according to the regular descending or ascending powers of a certain letter, supplying missing terms with the co-efficient 0. 2. Detadi the co-effidents, and multiply as in the multiplica- tion of polynomials. 3. Restore the letters with their exponents determined by the law of the product. 29. Examples. 1. Multiply x'^ -\- xy -{- y^ by a; + y. Ans. x^ + 2x^y -\- 2xy^ -\- y^. 2. Multiply a;8 + Sa;^^/ + ixy'^ + y^ by a;^ + 2xy + y^. Ans. x^ + 5x*2/ + 10a;3 2/2 + lOx^y^ + bxy*^ -\-y^. 3. Multiply a2 _^ a6 + h^ by a'' — ah -\- b\ Ans. a* + aH'' +b^. 4. Multiply a;* + y* by a;* + an/ + y^. Ans. x^ -j- x^ y -\- X* y^ + x^y* -\- xy^ + y'. 5. Multiply x^ -\- Sx'^ y + Sxy^ -\- y^ by a;* — 3x^2/ + Sxy'^ — y^. Ans. x'^ — Sx^y^ ^Sx^y^ — y". 6. Multiply x" + x2y2 _^ yi ^y a;2 _(- a;?/ -)- y2. ^M. x«+a;S2/ + 2x*2/2 +x32,3 4-2x23/* + w/^ +2/". DIVISION. 30. Definitions. 1. Division is the process of finding one of two factors when their product and one of the factors are known. 2. The quantities considered are the dividend, the divi- sor, the quotient, and the remainder. DIVISION. 31 1st. Tke dividend is the given product. 2d. The divisor is the given factor of the dividend. 3d. Tlie quotient is the required factor of the dividend. 4th. The remainder is the dividend minus tlie product of the divisor and integral part of the quotient. It is an ac- cidental quantity occurring only when the dividend is not a multiple of the divisor, in which case the complete quotient is a mixed quantity. DIVISION OF MONOMIALS BY MONOMIALS. 31. The Signs. Since, - -j-aX+i = + o* — a X -i-b — — ab -j- a X — b^ — ab ^ — aX — b = + ab^ (1) +ab (2) -ab (3) — ab-i- -{- a 1(4) +ab ■ + a= + b^ — a= + b — a = — b From (1) and (2), (3) and (4), we have the laws: 1st. Uike signs give -\-. 2d. Unlike signs give — . 32. Consequences. 1. Changing the sign of either divisor or dividend changes the sign of the quotient. ' 1st. If the signs are alike, the quotient is -|- ; chang- ing either sign, the signs will be unlike, and the quotient — . 2d. If the signs are unlike, the quotient is — ; changing either sign, the signs will be alike, and the quotient -j-. 2. Changing the signs of both divisor and dividend does not diange Hie sign of tlie quotient. For, - For, 32 ALGEBRA. ' 1st. If the signs are alike, the quotient is + ; chang- ing both signs, the signs will still be alike, and the quotient +. 2d. If the signs are unlike, the quotient is — ; changing both signs, the signs will still be unlike, and the quotient — . 33. The Co-efflcients and Exponents. 1. 10a'" ~ 5a" = Sa'"-", since 5a" X 2a"'-^ = lOa"'. Hence, 1st. The co-efficient of the quotient is equal to the co-effi- cient of the dividend divided by the co-efficient of the di- visor. '2d. The exponent of a letter in the quotient is equal to its exponent in the dividend minus its exponent in the di- visor. 2. ^ = a»-" = a»; but^=l; .-. a» = 1. Hence, a" a" ' 1st. A quantity with an exponent is equal to 1. 2d. A factor with an exponent 0, can be retained, omit- ted, or introduced, at pleasure. .a"' a" "' n™ 1 ;, by dividing both terms by a'" /7 /I — r — ^ —^ — , by dividing both terms by a". 'i J I a™ _ 1 a" a" a'"-" 1 Hence, 1st. A quantity with any exponent is equal to the recip- rocal of the same quantity with the sign of its exponent changed. Dl VISION. 33 2d. A factor may be transferred from one term of a frac- tion to the other, if the sign of its exponent be changed. 3d. By this transfer, an expression can be freed from negative exponents. Thus, a^ 6-3^-, ^^^=^^. 30a5 62c--5a2 62c2d2 = 6a3 60c-id-2^?^. 34. Rule. 1. Divide the eo-efficient of the dividend by the co-efficient of the divisor, prefixing the sign -j-, if the signs are alike; — , if the signs are unlike. 2. Annav all the letters common to the dividend and divisor, affecting each with an exponetit equal to its exponent in the div- idend minus its exponent in the divisor, omitting any letter whose exponent becovies 0, unless it is desired to retain a trace of that letter. 3. Annex the letters of the dividend not found in tlie divisor with their e.xponents. 4. Annex the letters of the divisor not foimd in the dividend with the signs of their exponents changed. 5. Free the result from negative exponents, if any. 35. Examples. 1. Divide 16a»6*c by bah'^c. Aiis. Sa^fcs. 2. Divide SGaft" c^ by Sa^ 62 erf. 12^2 /. Ans. 12a-H'- cd-' ■- ad 12 3. Divide 84ffi3 63 g by 7(j4j3 (.2^2 j^^s 4. Divide 32(1"' 6" c" by 8a" 6" c''. Ans. 4tt'"-"B«- 34 ALGEBRA. 16c« 5. Divide — lUa^ 6' c" by Qa^ b^ cd^. Am. — ^i^2gr 6. Divide 28a'" 6" c" by — 4a-'" 6"" c-". Ans. — Ta^-'J^" c^". 7. Divide — 14a'"6" by — 7a"6"'c-''. Ans. 20"-%"-'"' Dividing the dividend 1 _ ^ _L °-i_ ^ _j_ . . by the first term of the 62 a a a" b^ b b^ b^ — 6 a a a divisor, we have the quo- J ^ tient 1. Multiplying the divisor by this term of the quotient, and subtracting the product from the dividend, we have the remainder — 6. Dividing this re- -|- — mainder by the first term of the divi- sor, and continuing the process, we find the quotient to be an infinite series whose law is, any term is found by multiplying the preceding term by . 41. Eule 1. Write the quotient as a fraction having the dividend for the numerator and the divisor for the denominator. 43. Rule 2. 1. Divide the dividend by the first term of the divisor for the first term of the quotient. 2. Multiply the divisor by the first term of the quotient and subtract the product from the dividend. 3. Divide the first term of the remainder by the first term of the divisor, and continue the process, at pleasure, or till the law of the quotient is obtained. 43. Examples. 1. Divide a by a — b. a~h ■ ^ a^ a^^ a^ ^ DIVISION OF POLYNOMIALS. 37 2. Divide a hy a -{- b -j- c. 3. Divide 1 by a; — y. 4. Divide 1 hj x -\- y. Ans. — , =;1 — -— -4-— ^----^. . . a; — 1/ a; a,-* a;* , 1 It/, 1;^ ^ws. — - — = - — -^ + -J a; -|- 2/ X x' x'^ 5. Divide a by 1 — a. Ans. :j ^ as + *^ + ''^ + • DIVISION OP POLYNOMIALS BY POLYNOMIALS. 44. Illustration. Divide a^ — Sa^ 6 -f 2,ab^ — ¥ by a — 6. OPEEATIO> «3 — 3a2 6 + 3a62 _63 a — b a^ — a^ 6 rt2 _ 2a6 + 62 — 2an + 3a62 — 2a2 6 + 2a62 a62 — 63 Arranging the polynomials with reference to the descend- ing powers of a, and dividing the first term of the dividend by the first term of the divisor, we find a^ for the first term of the quotient. Multiplying the divisor by this term of the quotient, and subtracting the product from the dividend, and proceeding as before, we find the entire quotient to be a2_2a6 + /)2. 38 ALOEBBA. 45. Rule. 1. Arrange the dividend and divisor with reference to the ascending or descending powers of the same letter. 2. Divide ■ -\- a;™" r y r -\- . DIVISION BY DETACHED CO-EPFICIENTS. 47. Illustration. Divide a* + 2a^ b^ + 6* by a^ _(- 52, 1 + 0+1 OPEEATION. 1+0+2+0+1 1 + 0+1 1 + + 1 0+1 + Supplying the missing terms, we + + have a* + Oa^ b + 2a2 52 _j_ Qofis + 6^ 1 + 0+1 to be divided by a^ + Oab + fe^. 1 + + 1 Detaching the co-efficients and divid- ing as in division of polynomials, we find the co-efficients of the quotient to be 1, 0, 1; the expo- nents are determined by inspection. Then restoring the literal part, omitting the term whose co-efficient is 0, we have the quotient a' + 6^. ni yiiiiUJS BY DETACHED CO-EFFICIENTS. 41 48. Rule. 1. Arrange the dimdend and divisor according to the powers of the same letter, supplying missing terms with the co-efficient, 0, 2. Detach the co-effi^ents and divide as in division of poly- nomials. 3. Restore the letters with their exponents determined by the law of the quotient. 49. Examples. 1. Divide ic* + Ax^y -\- Qx^y^ -\- 4xy^ -\- y^ hy x -{- y. Ana. x^ -\- 3x''y + Sa;?/^ + y^.. 2. Divide x^ — y^ by x — y. Ans. x'' -\- xy -\- y^. 3. Divide x^ -\- x^ y -]- x* y'' -\- 3x^ y^ -\- x'^y* -{- xy^ -\- y^ by x^ + x^y -\- y^. Ans. x^ -)- xy'^ + y^. 4. Divide x^+5x*y+10x^y^ + 10x^y» +5xy* + y^ by a;2 -|- 2xy -j- y^. Ana. x^ -)- 3x^y -\- 3«i/^ -\- y^. 5. Divide a* + «= 6^ + b* by a^ -]- ab + 6^ Ans. a^ — ab -j- b^. 6. Divide x^ + x^ y -\- x* y^ -\- x'' y* -{- xy^ ^ y'^ by x* -)- y*. Ans. x^ -\- xy -\- y^. 7. Divide x'' — 3x* y'^ + Sx^ y* — y^ by x^—3x''y-{- Sxy^ — y^. Ans. x^ -\- 3x^ y -\- Bxy^ -\- y^ . 8. Divide x^ + x^y + 2x'^y^ + x^y^ + 2x^y* + xy^ +yO hy X* -\- x'' y^ -\- y* . Ans. x^ -{- xy -{- y^ . 9. Divide »* — y^ by a; — y. Ans. a;' -f- x^y -j- x^y^ + x'^^y^ -{- x^y* -f- x'^y^ + ni^y^ + 2/'- 10. Multiply x^ —y^ by a;* — y^, and divide the product hyx'^-\-y'^. Ans. x^ — x^y'^ — x'^y^-\-y^. C. A. 4. 42 ALGEBRA. THEOREMS AND FACTORING. 50. Theorem I. The square of the sum of two quantities is equal to the square of the first, plus twice the product of tlie first by the second, phis the square of the second. a -\- b Let a denote the first quantity and b the a -\- b second; then a -\- b will denote their sum. a^-\- ab Now, the square of the sum is the product ah -\- b'^ obtained by multiplying the sum by itself. a^-\-2ab-\- b^ By actual multiplication, we find that the product is a^ + 2ab -\- b^ of which a^ is the square of the first, 2a6, twice the product of the first by the second, and 6^, the square of the second. Hence, the the- orem is proved, and we have the formula, (a + by^a'^ + 2ab + b\ Conversely, the sum of the squares of two quantities, plus twice their product is equal to the square of iheir sum. Reversing the formula, and writing (a -\- b) (a -{- 6) for (a -|- 6)^, we have the formula for factoring : a2 -)- 2ab + 6^ = (« _^ 5) (a -f- b). 51. Examples. 1. 3a; (a + by = 3a; (a^ -\- 2ab + b^) =r Sa^a; + Qabx + 362 a;. 2. 4m^pq -\- 8mnpq -|- 4*1^^3' = 4pq (m^ -f- 2mn -|- n.^) = 4pq (m, -\- n) (m -\- n). 8. Develop (x-\-yy. 4. Square m -\- n. THEOREMS AND PAOTORINO. 43 5. Square '6a^b -{- Aa'^b^. 6. Develop 4a6 (c^ -^ d^)'-- 7. Factor x^ + 2xy + y"^. 8. Factor m^ -\- n^ -f 2mn. 9. Factor 9a*6' + 24a''6* + 16a*6\ 10. Develop [(a + 6) + (c + ci)] [(a + J) + (c +cZ)]. 11. Factor a^ _j_ 2a6 -f 62 -f 2ac + 2ad + 26e + 26d + e- + 2cd + d2. 52. Theorem II. Tlve square of tiie difference of two quantities is equal to the square of the first, minus twice tJie product of the first by the second, plus the square of the second. Let the student demonstrate this theorem, and obtain the formula, (a — J)2=ra2— 2a5 + 52. Corollary 1, — The sum of the squares of two unequal quantities is greater than twice their product. Since the square of any quantity, positive or negative, is positive, (a — 6)^ is positive; .". its equal, a^ — 2a&-|-6^, is positive. .-. a2 + 62->2a6, if a>6 or a<6. Corollary 2. — The sum of the squares of two equal quan- tities is equal to twice their product. lia=b,a—b = (i,.-. (a — by=0, .• . a^ — 2ab + b^ =0. .-. a2-f-62=2a6, if a = 6 Conversely, the sum of the squares of two quantities, minus twice their product, is equal to the square of their difference. Reversing the formula, we have a2 — 2a5 + 6^ ^ (a — 6) (a — b). 44 ALGEBRA. 53. Examples. 1. Develop (x — y)"^. 2. Square 2a — 36. 3. Square 5a^ — 4ab. 4. Develop 5a (a — 6) 2. 5. Develop 6xy (2x — Si/)^. 6. Factor x'^ — 2xy + j/^. 7. Factor ia'' — 12a6 + 962. 8. Factor 25a* — 40a3 6 + IGa^ 62. 9. Factor 5a» — 10a2 6 + 5a62. 10. Factor 24x'' y — 72x^ y^ + bixy'. 11. Develop [(a — 6) — (c — d)] 2. 12. Factor a^ — 2a6 + 6^ — 2ac + 26c + 2ad — 2hd -|- C2 _|_ cJ2 __ 2cci. 54. Theorem III. Ti^e product of the sum and difference qf two quantitisB is equal to the difference of their sgua/reg. By actual multiplication, we find (a + 6) (a — 6)=a2— 62. Conversely, tlie difference of the squares of two quantities is equal to the product of their sum and difference. By reversing the above formula, we have a2— 62== (a + 6) (a— 6). 55. Examples. 1. Develop (x -\-y) (x — y). 2. Develop (2a + 36) (2a — 36). THEOREMS AND FACTORING. 45 3. Develop (m^ -)- w^) (m^ — re^). 4. Develop (j9^ + 9^) (i? + 9) (p — -\- a"'P-"b" ^itp—m ^MP—lm'Ln _l_ ™mp— 3?h /j2» 1st rem. = — a""-"" 6" + 6"" 2d remainder = a™P-2™62n _|_ jfp 3d remainder = — a'"''-^"'b^" + b"" We see that the first term of each remainder of an odd order is negative. . • . If ^ is odd, the p"' remainder = — a"*""'" 6^"+ 6"?= 0. n™P _|_ hnp ■ _ CJ"\ — ZL—i ^mp— m (jWtp— 2mJ'! ] _j_ l)''P-^_ . • . (2) a"* + 6"'= (a" + 6") (a™""™ — a'"''-'i'^b" +...). Scholium. — Let the student state the laws of the quotient. 62. Examples. 1. (a" + 6«) -f- (a3 + 62). 3. (a^ + 6«) -f- (J + &t). 4. (a2> -)-6i4)-=-(a3-f-62). 5. (a;3'"''-)-2/6"P)-;-(a;'"+i/2''). G. (a;' +2/' »)-=-(«'? -I- 2/V). 7. (a;2" + 2/20) ^(a;s + 2/4). 8. Factor a' + 6'. 9. Factor a;'« +y"2. 10. Factor aJ'o + J/^ 11. Factor a« +b^. 12. Factor a^o + b^". 13. Factor x^s +1/2 >. 14. Factor x^'^ +y^«. THEOREMS AND FACTORINO. 49 63. Theorem VII. The 'product of two binomiak whose first terms are equal is equal to the square of the first, plus the sum of the second terms into the first, phis the product of the second terms. (1) {x + a) (x — 6) = x"- -\-{a — h)x — ah. Remarh. — The sum of a and — 6 is a — 6, and plus this sum into a; is -j- (a — 6) a;. The product of a and — 6 is — ah, and plus this product is + ( — ah) or — ah. The second member of formula (1) is of course found from the first by actual multiplication. Conversely, a trinomial whose first term is a square, and the square root of whose first term is a factor of the second term, can be resolved into two binomial factors of which the first term in each is the square root of the first term of the trinomial, and tJie second terms the two quantities whose sum is the co-efficient of ilie second term of the trinomial, and whose product is tlie third term. Reversing formula (1), we have (2) .r2 + (a—b)x—ah = {x-\- a) (x — b). 64. Examples. 1. Develop (x + 5) (a; + 3). 2. Develop (x — 7) (a; — 8). 3. Develop (a; — 9) (a;+ 5). 4. Develop (x + 11) (x + 5). 5. Develop (a;— 7) (a;+ 9). 6. Develop (a;^ — 5) (a;^ — 8). 7. Develop (a:™ - 5) (a;" + 6). C. A. 5. 50 ALGEBRA. 8. Factor x2 _|-2a;— 15. 9. Factor x'^--4cx— 45. 10. Factor x^—lbx-\- 56. 11. Factor x"^ + 13a; + 42. 12. Factor a;4 — 13a;2 +40. 13. Factor x^ — ISx^ + 77. 14. Factor x^ — 15a;* + 54. 65. Miscellaneous Examples. Factor or develop the following: 1. (a2 — 62)*. 2. [(a + 6) (a — 6) + c] [(a + 6) (a — 6) - c]. 3. (a+6-|-c)2— d^. 4. (a + 5 + c) (a + 6 — c). 5. (a + 6)2 — (a — 6)2. 6. 5a (a — 6)2. 7. Ila5 6 — 176a65. 8. 11a (a2 -I- 462) („ _j_ 26) (a — 26). 9. a;>« + a;9. 10. a5 _f- 326S. 11. (a« — 1) (a6 + 1) (a' 2 + l). 12. 256a8 — 65616«. 13. ai6— 6i«. 14. a* + a2 62+6''. 15. x^ -\- x^y -{- xy^ -\- y^. 16. 5aa;2 + 65aa; + 210a. 17. (a + 6)3 — (c + d)3. 18. x" —xy* —x*y -\- y*. THE GREATEST COMMON DIVISOR. 51 THE GREATEST COMMON DIVISOR. 66. Seflnitions. 1. A divisor of a quantity is a quantity which will divide the given quantity without a remainder. 2. A common divisor of two or more quantities is a di- visor of each of those quantities. 3. The greatest common divisor of two or more quanti- ties is the greatest quantity which is a divisor of each of those quantities. 4. A multiple of a quantity is a quantity of which the given quantity is a factor. 5. A composite quantity is a quantity which can be re- solved into integral factors differing from the quantity itself and unity. 6. A prime quantity is a quantity which can not be re- solved into integral factors differing from the quantity itself and unity. 7. Two or more quantities are relatively prime when they have no other common integral factor than unity. 67. Notation. Let c. d. denote a common divisor, and g. c. d. denote the greatest common divisor. 52 ALGEBRA. FIRST METHOD— BY FACTORING. 68. Principles. 1. Any common factor of two or more quantities is a c. d. of those quantities, since a factor of a quantity is a divisor of that quantity. 2. The product of two or more common prime factors is a c. d., since this product is a common factor. 3. The product of all the common prime factors is the g. c. d., since this product is the greatest common factor. 69. Illustrations. .1. Find the g. c. d. oiSOa^ b*c^ and 105a* fe^d. OPEEATION. 30ci'6*e2 =. 2 X 3 X 5a^^o^ ] f 3 X Sa* b^ ^ 15a^ 62 cM f 3 X ba^b^ d } ' ' 1 = g. c. d. 105a*b^d ^3X5X7a*bH } " ' (. = g 2. Find the g. c. d. of 2a'-—2b\ 2am + 2bm, and 2a2 + 4ab ^ 2b^. OPEEATION. 2a2 — 262 = 2 (a + j) (a — 6) ^ 2am + 2bm = 2m (a + b) [.-. 2(a + 6) = g. c. d. 2a2 + 4a6 + 262 =: 2 (a + 6) (a + 6) 3 70. Rule. 1. Resolve the quantities into iJisir prime factors. 2. Take tlie product of all the common prime factors. GREATEST COMMON DIVISOR. 53 71. Examples. 1. Find the g. c. d. ot'iWa^h^c and llbba'^ h^ d"^ . Ans. 105a^6". 2. Find the g. c. d. of 2na''-h^e, lOOla^Jc^d, and 154a2 62c. Ans. lla^be. 3. Find the g. c. d. of a^ — 2ah + 6^ a^d a^ __ 62_ ^ris. a — 6. 4. Find the g. c. d. of a* —6* and a^ + 2a6 + i^, 5. Find the g. c. d. of a;^ + 8a; + 15, x'^—^x — 15, and z"^ +2x — 3. ^rw. a; + 3. 6. Find the g. c. d. of a;2 + 8a;+]5, a;2 4-2a; — 15, a;2 + 4a; — 5, and x'^ — 2x — 35. Ans. x-\- b. 1. Find the g. c. d. of 3x {a-^hy and 2ax (a^ — J^). J.ns. aa; -(- &»;■ 8. Find the g. c. d. of a;« — llx^ + 30, a;« — 13x3 + 42, and x^ -\-x^ — 42. Ans. x^ — 6. 9. Find the g. c. d. of a< + a^ft^ _^ 6" and a^ — 2aH + 2ab^—bK Ans. a^ — ab + h^ . 10. Find the g. c. d. of x^'" + .x™ — 30 and x^m _ gf — 42. . J.ns. .X™ + 6. 11. Find the g. e. d. of 3ax« — Bay« and 3ax*y + Saxy*. Ans. 3ax' + 3ajf*. 54 ALGEBRA. SECOND METHOD— BY DIVISION. 72. Principles. 1. A divisor of a quantity is a divisor of any multiple of that quantity. Thus, d is a divisor of dq and of mdq. 2. A c. d. of two quantities is a divisor of their sum. Thus, d is a c. d. of dq and dq', and a divisor of dq -)- dijf. 3. A c. d. of two quantities is a divisor of their differ- ence. Thus, d is a c. d. of dq and dg', and a divisor of dq — dqi . 4. If the less of two quantities is a divisor of the greater, it is their g. c. d. 6. If two quantities are relatively prime, their g. c. d. is 1. 6. If one of two quantities be either multiplied or di- vided by a quantity which is prime to the other quantity, the g. c. d. will not be changed. Thus, d is the g. c. d. of ad and hd, also of mad and hd, and of d and hd. 7. The g. c. d. of three quantities is the g. c. d. of the g. c. d. of two of them and the third quantity, and, in gen- eral, the g. c. d. of 11 quantities is the g. c. d. of the g. c. d. of (n — 1 ) of them and the remaining quantity. 8. Any common factor of two or more quantities is a fac- tor of their g. c. d., and the product of this common factor by the g. c. d. of the quotients obtained by dividing the quantities by this common factor, is the g. c. d. of the quantities. Thus, m is a common factor of ma'^ — mb'^ and wia^ + 2mab -\- mb^. Dividing by m, the quotients are a^ — 6^ and a^ + 2ab + b^, of which the g. c. d. is a -|- b, and m(a-\-b) is the g. c. d. of ma^ — wb^ and ma^ + 2mab -f mb^. GREATEST COMMON DIVISOU. 5G 73. Proposition. The g. e. d. of Ike divisor and remainder is tlie g.' c. d. of the divisor and dividend. Let us take the less of the two quantities whose g. c. d. is to be found for the divisor, and the greater for the divi- dend, and let d denote the divisor; D, the dividend; q, the quotient; and r, the remainder. Then from the principles of division we have (1) D^dq + r. (2) D — dq^r. The demonstration of the above proposition depends on the following LEMMAS. 1. Any c. d. of d and r is a e. d. of d and D. For, a divisor of d is a, divisor of dq, by Principle 1; .'. a e. d. of d and r is a c. d. of dq and r, and .•. a di- visor of D, by Principle 2, since D ^dq-^-r, and . • . a c. d. of d and D. 2. Any c. d. of d and D is a e. d. of d and r. For, a divisor of d is a divisor of dq, by Principle 1; .-. a c. d. of d and D is a c. d. of dq and D, and .".a divisor of r, by Principle 3, since D — dq = r, and . • . a c. d. of d and r. It has now been proved, 1. That any c. d. of d and r is a e. d. of d and D. 2. That any c. d. of d and D is a e. d. of d and r. Hence, the c. d.'s of d and r are identical with the c. d.'s of d and D; .•. the g. c. d. in the one case is the g. c. d. in the other case. Hence, Tlw g. e. d. of d and r is the g. e. d. of d and D. 56 ALGEBRA. The process is us follows: d)D(q dq r) d iq' rq" r') r (5" r" Let us suppose r" = 0; then by Principle 4, / is the g. c. d. of r and / ; but, by the Proposition, the g. c. d. of r and / is the g. c. d. of r and d; .•. r' is the g. c. d. of r and d ; but, by the Proposition, the g. c. d. of d and r is the g. c. d. of d and D; -•. / is the g. c. d. of d and D. Hence, the first remainder, which is a divisor of the last divisor, is the g. c. d. sought. 74. Application. Find the g. c. d. of the following polynomials : 9x^yz — 30a;* ya + 45iC2/a -|- 24yz and Ibax^ij — SOax^y — 90ax^y + eOccB^y -}- 195axy -H 90ay. Dividing both polynomials by the common factor 3y, which is a factor of the g. c. d., the quotients are Sx^z— lOx^z + 15a» + 82 and 5ax^ — Wax* — SOax^ + 20aa;2 + 65aa; + 30a. Dividing the first of these polynomials by z and the sec- ond by 5a, which will not afiect the g. c. d. by Art. 72, Principle 6, and taking the first result for the dividend and the second for the divisor, we proceed thus: ^s_2x* — 6x« , + 4a:^ + 13a= + 6)^^'-10a;» + 15x + 8(3 3a;° — &x* — 18a;3 + 12a;^ + 39a; + 18 6a;4 + 8x^ — 12a;2 — 24a; — 10 GREATEST COMMON DIVISOR. 57 Since the g. c. d. of d and r is the g. c. d. of d and D, we shall find the g. c. d. of d and r. Dividing r by 2 and multiplying d by 3, and taking the results for d and D, we proceed thus: 3a;* +4x3 -6»2-12a;-5)^'''~ 6a;* - ISx^ + 12a;'2 + 39x + 18 (x 3x^4- 4x* — 6x3 _ 12x2 — 5x 2)— lOx*- - 12x3 _j_ 24x2 + 44x + 18 — 5x*- - 6x3 _)_ 12x2 + 22x + 9 8 — 15x* - — 15x* - - 18x3 _|_ 36a;2 + 66x + 27 (- - 20x3 _^ 30x2 _f. 60x + 25 -5 2x3 _^ 6x2 + 6a. _^ 2 Dividing r by 2 and taking the result for d, and the last d for D, we proceed thus; x3 + 3x2 + 3x + 1 ) 3a;4 _^ 4^3 _ 6x2 — 12x — 5 ( 3x — 5 3x* + 9x3 _^ 9a;2 _^ 3a; — 5x3 — i5a;2 — 15a; — 5 — 5x3 — i5a;2 — 15a; — 5 .•. x3 -|- 3x2 _|_ 3a; _|_ 1 is the g. c. d. of 3x5 2 — . i0a;3 2 _[_ I5a;2 -f 82 and 5ax5 — lOdx* — 30ax3 + 20ax2 + 65ax + 30a. .-. 3?/(x3 -f 3x2 + 3x+l)=3x3 2/ + 9x2y+.9x!/+32/ = g. c. d. of the given polynomials. 75. Rule. 1. Set aside the obvious common factor, if any, a8 a factor of the g. 0. d. 2. Divide the quantities whose g. c. d. ds to be found by the factor set aside, and, if necessary, prepare the quotients for di- vision by rejecting or introducing a factor in eitlier which is prime to Hie other. 58 ALGEBRA. 3. Divide one by ilie. ofJier, the divisor by the remainder, and so on, till there is no remainder. The last divisor will be fJie g. c. d. of the prepared quantities, and this g. c. d. multiplied by the factor set aside will be the g. e. d. of the given quantities. 4. If there are more than two quantities, find the g. c. d. of two of ihimi, tlien of this g. c. d. and Hie third quantity, and so on. 76. Examples. 1. Find the g. c. d. of a* — b* and a* +2a^b^ -j-b*. Ans. a^ + 62. 2. Find the g. c. d. of a;'» +x''y^ + y* and x'^ —y^. Ans. a;2 -)- xy + 2/^- 3. Find the g. c. d. of a* + 3a^ + 4a + 12 and a» + 4a2 + 4(1+3. Ans. a + 3. 4. Find the g. c. d. of x* — a;^ + 2a;2 + a; + 3 and X* + 2x^ —x — 2. Ans. x^ -\-x + l. 5. Find the g. c. d. of x* — 9a;2 + 20 and a;» + 4x'- — 32. Ans. x^ — 4. 6. Find the g. c. d. of x' — ISa;^ + 56, x^ + 4x« ~ 96, and X* — 9a;2 + 20. Ans. x — 2. 7. Find the g. c. d. of 4a2 — 5ab + b'^, 3a3 — 3a^b-{- ab'' —b^, a,nd a* —b*. Ans.a — b. 8. Find the g. c. d. of 6x^ — 4a;'' — 11a; ^ — 3a;2 — 3a; - 1 and 4a;* + 2a;8 — ISa;^ + 3a; — 5. Ans. 2x' — 4a;2 + a; — 1. 9. Find the g. c. d. of a^ ~ a" b — a'^ b^ — ab^ — 2b* and 3a3 — 7a2 6 + 3(iJ2_263. Ans. a — 2b. 10. Find the g. c. d. of a;^ — x''y — xy* + y^, x* — x^y — a;2y2 + xy^, and x^y -\- x^y^ — a^* — y*. Ans. x^ — y^. LEAST COMMON MULTIPLE. 59 THE LEAST COMMON MULTIPLE. 77. Definitions. 1. A multiple of a quantity is a quantity of which tlie given quantity is a factor. 2. A common multiple of two or more quantities is a quantity of which each of the given quantities is a factor. 3. The least common multiple of two or more quantities is the least quantity of which each of the given quantities is a factor. 78. Notation. Let c. m. denote a common multiple, and 1. c. m. denote the least common multiple. FIEST METHOD— BY FACTORING. 79. Principles. 1. The 1. c. m. of two or more quantities contains as fac- tors all the prime factors of these quantities and no other factors. 2. If two quantities are relatively prime, their 1. c. m. is their product. 80. lUnstration. Find the 1. c. m. of a* — b^ and a^ — 2a6 + b^. OPERATION. ai — b-' = (a + b)(a — b) \ ( (a+b)(a—b)(_a—b) + b){a-b) I _ I = (a-6)(a-6)r -1 a2_2a6 + 62=(a — 6)(a— 6)J • ' ( = 1. e, 60 ALGEBRA. 81. Rule. 1. Resolve the qiiantities into their simplest factors. 2. Find the product of the factors, each taken the greatest number of times tliat it occurs as a factor in any 'of — > and -T • be ac ab OPERATION. o a X a a^ be be X a abe b bXb 62 ae ac Xb abc c c X c c2 ab ab X c abc The operations are not usually indicated, thus : abc. o' b^ 0^ be ac ab abc abc abe 68 ALGEBRA. n a c e . adf bcf bde ^- I' j' 7" ^'^- IjF' T17' TT?- d f hdf bdf bdf a — b a + b gg— 2a6 + 6^ a^ + 2ab -\- b" '^- a + b' a — b' **"• 'a^—b^ ' a^—b^ ' ab cd ^ gli ed ah gh ef . a^b'^ef gh c^d^efgh abcdej^ abcdg'^h^ abcdefgh abcdefgh abedefgh ahcdefgh a^ — 6^ a -\- b a — b j„, a fe (" — b) c (a+ 6) a2— 62 a2 — 62 ' ^2 — 62 ' 1 1 1 (a + by(a—by(a + b)(a—b') 4„, (c^-hr (c^ +by a'' - b^ (a2_62)2' (a2_52)2' ((,2_52)2' „ a 6 c . 6a2 362 jQc^ 7. rr- ' T7i — ' ;r-; • -Ans. 56c lOac 3a6 ' 30a6e 30a6c 30a6c .T y X y x-\-y x-\-y X — y x — y Ans. x^ — xy mf — y2 x^-\-m/ xy-\-y^ a;2 — 2/2 x'' — 2/2 x^ — y^ x^ — y'^ 9. _^ 1 a;2 y2 rgS yH A x^ -\- x'^'y -\- xy'^ xy -\- y^ x*-\-x'^y — xy" — y* x*-^x^y — siy^ — y* ADDITION OF PKACTIONS. 101. Definition. Addition of fractions is the process of finding the sum p A — A,««+:« — ADDITION OF FRACTIONS. 69 103. Rule. 1. Reduce the fractions to equivalent fractions having a oommmi denominator. 2. Add their numerators, and write their sum over the common denominator. 103. Examples. ^ a , c ^ I ^ ad -\- be ' h d hd hd hd 2. Add - , - , - and r • Am. — — ■ b a c aba 3. Add — I — ^ and — '- — ^ • Ans. a;+5 X — 5 ' x'^^^b' . .jj a; ^ X ^ 2a;^ — 12a; 4. Add = and ■ Ans. x — 7 ■ a;2 — 12a; + 35 5. Add "—y and '^^^ ■ Ans. ^^ ■ X -{- y X — y x^ — y'' 6. Add :; — ; — -„ and ; r- Ans. 1 + a;2 1 — a;2 1 — a; >4 7. Add — T—, and f • J.m. -;; — r^ a + 6 o — a^ — b^ 2a* + 12a^^ + 2b* « A AA (a— by , (o + by 8. Add ) — -~ and ) — l_^ {a -\- by (a — oj ^ o* — 2a262 + 6 4 9. Add f) :; and f ) rj aiiu ; • a — — c c — a 3a6c — 2ac^ — 2b^c — 2061:; + aad + bH + bc^ ahe — b'^c — ac^^be^ — ahd^b^d^acd — bed 70 ALGEBRA. 10-^dd^._7. + 12'a;^-8a;+15"'^%^- X -9a;+20 ^'^- x^- 3a; - 8a; + 15 SUBTRACTION OF FRACTIONS. 104. Definition. Subtraction of fractions is the process of finding the difference of two fractions. 105. Rule. 1. Reduce the fractions to equivalent fractions having a common denominator. 2. Subtract the numerator of Hie subtrahend from tke numer- ator of the minuend, and write Ike difference over the common denominator. 106. Examples. a c ad 6c_ ad — he a{d — c) — c(b — a) ' h d hd hd hd bd Scholium. — The last form is useful in solving numerical examples. 2. 21 33 21X6 — 33X3 24 39 24 X 39 27 963 3. 65 , , 73 From —a take ^a. 0/ 79 A 244 4. r.om^_^takc^_5. 2x ^^"'' x^ — 12x + 35 5. From^ + ^'take^T^'. x—y x+y Ans. v^^' MULTIPLICATION Or FRACTIONS. 71 6. From ) ^ ( take ) — —ri^ • (a — 0)2 (fls + o) 2 ((4_2a262-(-6* 7. From — —f take — ; — -,• a — c -\- a 2be + 2ad ac — he -\- ad — bd „,-, a -\- b , a — b , , a 4- b a — & 8. From — -^ H —7 take , --: • a — a -f- b a- — b a -{- b Ans. —7 — a -\- b ^'*'- x + 8a;+15' 10. From („-£-J take (^ . 12a°fe + 40a363 + 12a6^ a"— 3a*62 + 3a2J'» — 66' MULTIPLICATION OF _ PEACTIONS. 107. Deflnition. Multiplication of fractions is the process of finding the product of two or more fractions. 108. Illustration. 5^2 2 „ 5 , , 1 „ 5 _ 5 . , 2 5 _ 10 7>^3=3°^7' ^^*3°^7"2T' ' ' 3 °^ 7 " 21 72 ALGEBRA. 109. Rule. Multiply the numerators together for the numerator, and -\- an a -4- n ^ a r„ , ab 4- bn ^ ah -\- an a -\- n ^ a If a > 6, ,„ , , < ,„ , , > or 7— — < r • h^ -\-hn h^ -\-bn + n h 2. If the same quantity be subtracted from both terms of a fraction, the resulting fraction will he equal, to, less, or greater than the given fraction, according as flie numerator is equal to, less, or greater than the denominator. Let 7 be the given fraction, and , 7 the resulting fraction. Reducing to a common denominator, we have a ab — an . a — n db — hn b b^ — bn h — n h^ — hn J ah — bn ah — an a — n a ' b^ — hn b^ — hn b — n b ■rn ^ I <»^ — hn , ah — an a — n ^ a \. A when x = a. /jj4 ^4 2. Find the value of -; > when a; := a. Am. 3a\ ^^ (^:i 3. Find the value of -, --: > when x = a. Ans. 2a 2. (x — a) ^ Ans. (j3 J3 4. Find the value of — ;; — tt > when a = 6. 5. Find the value of -^ ^ { — ; — ; ' when x=--a. x^ — ax'-'~a''x-\-a^ Ans. 0. a," (ji' 6. Find the value of > when x = a. X — a Ans. nvP^'^. /J* 2 QSd 7. Find the value of .^4_2ax3 + 2aB«,_c,4 ' ^hen a; = a. Ans. 00. fi I />» ft T 8. Find the value of X 7 — , — r; ' when a; = a. a — X {a^xy Ans. 1. 9. Find the value ^f "^"'"'~("+p-''" + ^ when x=l. 1 — x^ 4n«. 0. ™^ «j I c^ , ^i^w 10. Find the value of -. ^ - when x = n. (x^—n^y Ans. -:?=• V2n 82 ALGEBRA. EQUATIONS OF THE FIRST DEGREE. 133. Definitions and Classification. 1. An equation is the expression of the equality of two quantities. Thus, a; ^ a is an equation. 2. The members of an equation are the two quantities connected by the sign of equality. 1st. The first member is the part on the left of the sign of equality. 2d. The secoiid member is the part on the right of the sign of equality. 3. An equation of the first degree is an equation which involves only the first power of the unknown quantity and known quantities. 4. A numerical equation is an equation in which all of the known quantities are expressed by numbers. 5. A literal equation is an equation in which the known quantities are expressed, wholly or in part, by letters. 6. An identical equation is an equation in which the members are the same in form or in sense. Thus, x-\- a = x -\- a and (x -\- a)^ ^= x^ -\- 2ax + a^ are identical. 7. Equations may involve one unknown quantity or more than one. TRANSFORMATION OF EQUATIONS. 83 124. Formation of Ec[uatious. 1. Form an equation from x -\- x, if a; = 10. OPEEATION. If a; =10, a; + a; = 10 +10, or a; 4- a; =: 20. 2. Form an equation from x -\- x, if a; ^ 3. 3. Form an equation from x -{- 2x, if a; ^ 8. 4. Form an equation from x -\- 2x -{- 3x, if a; = 5. 5. Form an equation from x -\- 3x -\- 5x, if a; = 12. 6. Form an equation from — ^ — < if a; = 10. „ -c, ^. „ 3a; — 5 , 2a; +10 . „ , _ 7. harm an equation from — 1 r — ' " a; ^15. 8. Form an equation from — |- 7 > if a; : ^ a b ■ah. 9. Form an equation from — --7 + — + > if x=a'^ — 6^. ^ a-\-b a — 6 10. Form an equation from ■ — 7 — > if a;=2a6. teajStsfoemation of equations. 125. Definition. A transformation of an equation is any change in its form which does not affect the equality of its members. S4 ALGEBRA. 126. First Transformation— Clearing of Fractions. The object of this transformation is to make all the terms of the equation entire. ox X Clear the equation -2-+l = n~l"3of fractions. OPEEATION. 3x X -J + 1 = K + 3. Multiply both members by 4, the 1. c. m. q ,, n _i_i9 of the denominators, canceling the de- nominators. The members of the re- sulting equation will be equal. (Ax. 3.) 127. Rule. Multiply both members of the equation by the I. e. m. of the denominators, reducing the fractions to entire quantities. 128. Examples. Clear the following equations of fractions : 3a; + 1 1 ^ 7x + 3 . 1 3 "^6~ 8 "^4" Ans. 24a; + 8 + 4 = 21a; + 9 + 6. 2. 1 + 1 + 1 = 26. Ans. 6a; + 4a; + 3a; = 812. 5a;— 11 11a;— 1 ^ a;— 1 4 12 10 ■ Ans. 75a; — 165 — 65a; + 5 = 6a; — 6. 4. — |- r- + - =:d. Ans. box + aca; + abx = abed, a c a; + g x — a a + 6 a — b Ans. ox + a^ — bx — ab = ax — a^ + 6a; - TRANSFORMATION OF EQUATIONS. 85 6. 3.'(; + 4a;H-5a; = 7ra; — 4x ^+7^" 2 4 2 Am. 12x + 16a; + 20a; = 2a; — 16a; — 3a; + 2a. m w _ p ■ a;— 2 ^a;— 3 a;^— 5a;+6" A71S. mx — 3m -\- nx — 2n=^p. ^ a . h e X -{- a X -\- h x'^ -\- {a -{- V) X -\- ah Atis. ax -\- ah -{- hx -\- ah ^:= 0. x — a ^ + b ^ P + q a+ h'^ a — b~ a2 _ 52 • Ans. ax — a^ — hx -\- ah -\- ax -\- ah -\- hx -\- b'^=p -j- q. Ans. x^ — |)^ — 2"^+ g'^=jp^a;2 — 52^2 — ^4 _|_p2g,2_ 129. Second Transtbrmation— Transposition. The object of this transformation is to bring all of the unknown quantities into the iirst member of the equation, and all of the known quantities into the second. Take the equation (1) mx — q^=^p -\-nx. If q be added to both members, and nx be subtracted from both members, the equality will not be destroyed. (Ax. 1, 2.) Performing these operations, we have (2) mx — nx = p -\- q. 86 ALGEBRA. Comparing (1) and (2), we find that — q in the first member of (1) becomes + g in the second member of (2), and that -|- m; in the second member of (1) becomes — nx in the first member of (2). Hence, Any term may be transposed from one member of an eyttation to the other, if its sign be changed. 130. Rule. Transpose all of the known quantities which are in t/tc first m,emher into ike second, and all of the unknown quantities which are in the second member into Ae first, changing tlie signs of all the terms transposed. 131. Examples. 1. ax — b ^ a -\- bx. Ans. ax-^bx = a -\-b. 2. 5a; — 3 + a; = 2a; -f- 8. J.?is. 5a; -|- a; — 2x = 8 + 3. 3. 7a; + 4 — a = & — 3*. Ans. 7x + 3a;= a + 6 — 4. 4. (a-\-h)x — e = a-\-h — (a — b)x. Ans. (a -f- 6) X + (a — 6) a; = a -f- 6 + "• 5. mx — nx — p = q-{- rx. Ans. mx — nx — rx=p -\- q. 132. Third Transformation— Reduction. The object of this transformation is to collect the terms as much as possible. Thus, take the equation 5.t + 3a; = 30— 6. By reducing, we have 8a; = 24. TRANSFORM A TION OF EQUA TIONS. 8 7 Also, take the equation mx — nx =p -\- q. By factoring, we have (m — n) a;=p -|- q. 133. Rule. Reduce both members as much a« 'possible by adding and factoring. 134. Examples. 1. 9a;— 5a; = 9 + 7. Ans. 4x=U. 2. 8a; + 7a; = 25 + 5. Ans. 15a; = 30. 3. ax -\-bx — a^b. Ans. (a-\-b) x = a -(-6. 4. px — p^qx-\-bp. Ans. (p — q)x = 6p. - GH) rr + a; = — ^t___ j_ j^^ 2ax = 2b. a — 135. Fourth Transformation— Dmding by the Co-efflcient of oe. The object of this transformation is to find, from the reduced equation, the value of the unknown quantity. Take the equation 8a; = 24. Dividing both members by 8, we have a; = 3. [Ax. 4. J 88 ALGEBRA. Also, take the equation (m — 11) x=p -\- q. Dividing both members by m — n, we have m — n Take the equation — 3a; = — 12. Dividing both members by — 3, we have x = 4. Again, take the equation — x = — a. Dividing both members by — 1, we have X ^ a. 136. Rule. Divide both members by tJie co-efficient of the unknown quantity. 137. Examples. 1. (m — n)x=^p — a. Ans. x = ? • m — n 2. 6a; — 5 = 3x + 10. Ans. x = 5. 3. y— 8 = | — a; + 2. Ans. x ^ 9. ax bx , . ab (e-i-d) 5. '^-(a-b)x-'- = l. Ans. x = ~^-(^+^ a ^ ' d d(l—a^-\-ab-) MXAMrLMiS. 89 138. Yeriflcation. The value of the unknown quantity is verified, if, when substituted for the unknown quantity, it satisfies the equa- tion ; that is, causes the two members to be equal. Take the equation (in — n) X =p -\- q. X m — n This value of x is thus verified : m — n or, p-}. q^p + q. (m — n) " " = « -f o. m — n 139. Rule. Substitute Ike value of the unknown quantity for the unknown quantity in the given equation; and if it causes the two members to be equal, Hie value of the unknown quantity will be verified. 140. Examples. 1. ax — bx = e — d. Aiu. x = ; • a — 2. 7 — 5 = ^- Ans. a; = 60. 4 6 3. |— :^ — 3 = 5. Ans. x = 820. 8 10 X , X 4. ±-^± = c. Ans. X = abc a b ' ' a+b 5.-x — \x — 7 = 8. Ans.x=iiiO. 3 4 C. A. 8. 90 AHiUBRA. SOLUTION OF EQUATIONS. 141. Definition. The solution of an equation is the process of finding the value of the unknown quantity. 142. Rule. Clear the equation of fractions, transpose, reduce, and divide by the co-efficient of the unknown quantity. 143. Examples. 1-1-7-2 = 1-1. Ans.x = 20. 2 4 5 ^- 2+3=4+2- Am.x = -. 3. 5^-3^ = 5x-10. Ans.x-=B. . 5a; — 1 9a; — 7 5 — 9a; , 4. — = = — = — — Ans. a; = 3. ( 11 ^ 2a; — 6 3a; x — 4 „ . „ ^■—5 r3"~9~=^- Ans.x = 13. . a; 2a; — 9 5a; + 8 6. J g = — g— • Ans. x = 1|. _ 5a; — 7 2a; + 7 „ , , , 7. — 2 o — = 3a; — 14. Ans. a; = 7. a 26 — a; a; — 11 o- a; — = — Ans. x = S. q ^ + 1 5 — a; _ . x + 2 . ._ »• 2 ^ — ^^ 3 Ans. X— 13. 3a; -11 28 -9a; .59 . . 10- 4 g = « — ^ ■ ^ns. a; = 4. SOLUTION OF EQUATIONS. 91 11 2a; — 1 3a; — 2 _ 5a; — 4 _ 7x+6 3 4 6 ~T2~' J.JIS. a; = 4. 12. (a; + a) (a; + 6) = (a; + c) (a; + d). . ed — ab a ■ + b—c~d 13. TO n Ans. bm — an a; — a x — h TO — n 14. a;-l a;— 2 .T — 2 a;— 5 x — 6 x—7 Ans. a; = 4^. a; — 3 a;— 6 15. a a a a x — 8 Ans. x = 5. x — 2 x — 4 a; — 6 16. 2a; — 9 27 x — S 4 , X 25 "^18" — 3a; 3 Ans. a; = 9. 17. X X m n ,X X p 1 = r. A ViA V mnpgr npq — mpq -\- mnq — mnp 18. (a; + a) (a; — a) = (a; — a) (a; — 6) + aJ + 62_ Ans. x = a -\- b. 19. 2+3 = .^e_ _L _2_ . Ans. ^^ff ^?— 2p?r + y>g^ _ a; — r a;— j? a; — q p^ — pr — qr-\-q'^ ^31 53 20. 7^ =a;^ — aa;+ a^. Ans. x = a4-b. X -\- b 01 5a; — 1 11a; — 3 13a; — 15 _ 24 — 61a; „„ ^^- "TlT ^ 3 ~ 35 ~''^- Jh«. a; = 9. 22. 2^_i = £_r?. ^ns. .. = ^. 92 ALGEBRA. 23. £5_r = ^^-a. a^,, ^ ^ idle, -^ p + q p + q 11. A can do a certain work in 10 days, B in 12 days, and C in 15 days. In what time can they together do the work? ^918. 4 days. 12. A can do a certain work in a days, B in 6 days, and C in c days. In what time can they together do the work? . abc , Ans. 1 — ; — r days. bc-\- ac-\- ah '' 13. A, who has 3 hours to spare, wishes to enjoy as long as possible the company of B, who leaves immediately in a coach which travels 8 miles per hour. How far can A ride so as to walk back in time, at the rate of 4 miles per hour ? Ans. 8 miles. 14. A rode a certain distance at the rate of m miles per hour, and walked back at the rate of n miles per hour, and performed the whole journey in t hours. How far did he ride? . mnt ., Ans. ; — miles. m -\- n 15. The interest on ■§■ of a certain capital, at 5 % + the interest on f of it at 6 % is equal to $840. Required the capital. Ans. $15000. PROBLEMS. 97 16. The interest on - of a certain capital, at r % -|- the interest on the remaining part, at r' ^ = i. Required the capital. lOOdi Ans. nr -f- (d — n) r' 17. A man rows a boat with the tide 7 miles in 40 minutes, and returns against a tide j as strong in 1 hour. What is the rate of the strongest tide ? Ans. 2|- miles. 18. A man rows a boat with the tide m miles in t hours, and returns against a tide - as strong in f hours. Required the rate of the strongest tide. . mn (tf — t) ., 19. The diiference of the squares of two consecutive numbers is 21. What are those numbers? Ans. 10 and 11. 20. The diiference of the squares of two consecutive numbers is d. What are those numbers? d—1 , d+1 Ans. — ^ — and — x 21. The dift'erence of two numbers is 5, and the differ- ence of their squares is 45. What are those numbers ? Ans. 2 and 7. 22. The difference of two numbers is d, and the differ- ence of their squares is d'. What are those numbers? , d'—d'' J d'+d^ Ans. -^^—and^^. 23. Divide 45 into three such parts that ^ of the first, ^ of the second, and ^ of the third shall be equal to each other. -4ns. 10, 15, 20. C. A. 9. ^^ ALGEBHA. 24. Divide a into three such parts that — of the first, 1 1 ^ m ' — of the second, and — of the third shall be equal to each n p ^ other. , am an av Am. — , — . — ■ , — > -—'— m-t n -\-p m -{- n -\- p m + n-j-y 25. Divide 72 into four such parts that if the first be increased by 5, the second diminished by 5, the third mul- tiplied by 5, and the fourth divided by 5, the results shall be equal. Ans. 5, 15, 2, 50. 26. Divide a into four such parts that if the first be increased by n, the second diminished by n, the third mul- tiplied by n. and the fourth divided by n, the results shall be equal . an an , a an^ ■^'^- (n -t 1)2 "' (n+iy ^'*' (w + 1)2 ' l^ir+iy ' 27. The greater of two numbers is three times the less; but if each be increased by 15," the greater will be twice the less. Required the numbers. Ans. 15, 45. 28. The greater of two numbers is m times the less; but if each be increased by p, the greater will be n times the less. Required the numbers. 29. A's money + 2 times the sum of B's and C's = I B's money + 3 times the sum of A's and C's = $10500 ; C's money + 4 times the sum of A's and B's = $12000. They together have $4500. What has each ? Ans. A has $1000 ; B, $1500 ; C, $2000. 30. A's money -f I times the sum of B's and C's = js ; B's money + m times the sum of A's and C's = q; PROBLEMS. 99 C's money -(- n times the sum of A's and B's = r. They together have s. What has each ? . . , p — Is. -r, q — rm . „ r — ns Ans. A has f r > B, ^ > C, :; 1 — I 1 — m 1 — w 31. If 12 oxen eat the grass of 3^ acres in 4 weeks, and 21 oxen eat the grass of 10 acres in 9 weeks, how many oxen will eat the grass of 24 acres in 18 weeks, the grass being at first equal on every acre, and growing uniformly? Let X ;=: the growth on 1 J., in 1 wlc., the grass at first on 1 A. being regarded as the unit. 40 Then q- a; = the growth on 3 J A. in 4 wk. — ^t ^: the grass on 3^ A. + the growth in 4 wk. o 10 + 40a; , ,,-, .• . , = what 12 oxen eat m 4 wk. 3 10 + 40a; what 1 ox eats in 1 wk. 144 90a; — the growth on 10 A. in 9 ivk. 10 + 90a; = the grass on 10 A. + the growth in 9 wk. 10 + 90.'«; = what 21 oxen eat in 9 wk. 10 + 90a; , , , .-17 — -J— — = what 1 ox eats in 1 wk. lo9 10 + 90x ^ 10 + 40a; 189 144 _ J^ ■■• ^'"^12' 10 -I- 40a; 5 , , , .-17, — -L— — =-—:=: wliat 1 ox eats m 1 wk. 144 54 5 5 —7 X 18 = 77 = what 1 ox eats in 18 wk. 54 o 100 ALGEBRA. 1 24 18 7^ X Y" X ^ = 36 = the growth on 24 A. in 18 wk. 24 + 36 = 60 =: the grass on 24 A. + the growth in 18 wk. 5 60 -=- ^ = 36 = the number of oxen that eat the grass on o 24 A. -\- the growth in 18 wh. 32. If a oxen eat m acres of grass in q weeks, and 6 oxen eat n acres in r weeks, how many oxen will eat p acres in 8 weeks, the grass being at first equal on every acre, and growing uniformly ? , anpqr — bmpqr -\- bmprs — anpqs mnrs — mnqs ELIMINATION. 147. Definition. Elimination is the process of combining equations involv- ing two or more unknown quantities so as to cause the disappearance of one or more of the unknown quantities. ELIMINATION BY ADDITION OR SUBTEAGTION. 148. Illustrations. f(l) x + y = s.\ 1. Given < y Required x and y. (.(2) x-y=d.) SOLUTION. (l) + (2) = (3) 2x = s + d. .-. x = is + id. (l)-(2) = (4) 2y = »~d. .-. y = is-id. ELIMINATION. 101 (1) 2^ + 72/ = 29.) 2. Given I )■ Required x and y. \(T) Zx+by = 21.) SOLUTION. (1) X 3 = (3) 6a; + 21y = 87. (2) X 2 = (4) 6x +- 10?/ = 54. (3) - (4) = (5) lly = 83. [Ax. 2.] ••• 2/ = 3. Substituting this value of y for y in (1), we have 2a; -)- 21 = 29. Transposing and reducing, we have 2x= 8. . • . a; ^ 4. r (1) 5a; + 6y = 49. ) 3. Given I \ Required x and y. ((2) 7a;-42/=19.j SOLUTION. (1) X 2 = (3) 10a; + 12y = 98. (2) X 3 = (4) 21a; — 12y = 57. (3) -f- (4) = (5) 31a; = 155. .-. a; = 5. This value of x substituted in (1) or (2) will give 2/ = 4. 102 ALGEBRA. r(i) a-B + 51/ =: 61. ) 4. Given I \ Required x and ii. ((2) zx + ^^m.) SOLUTION. (2) X 2 = (3) 6x + 82/ = 76. (1) 6a; + 52/ = 61. (3)-(l) = (4) 32/ = 15. •■• 2/ = 5. Substituting in (1) or (2), we find a; = 6. r (1) 8a; + 62/ - 64. 1 5. Given < y Required x and 2/. ( (2) 5a; — 32/ = 13. J SOLUTION. (1) -r- 2 = (3) 4a; + 32/ = 32. (2) 5a; — 32/ = 13. (3) + (2) = (4) 9a; = 45. . • . a; ^ 5. Substituting in (1), (2), or (3), we find 2/ = 4. 149. Bule. 1. Prepare the equations, if neceasary, either by mvUvplmttion or division, so that the co-effieients of the quantity to be dimi- nated shall be numerically equal. ELIMINATION. 103 2. Then, if these Go-efficients have unlike signs, add the equa- tions; hut if tlwy have like signs, subtract one equation from the oilier. 3. Find the value of the unknown quantity in the resulting equation. 4. Substitute the value of the unknown quantity found in one of the equations involving two unknown quantities, and find the value of the other unknown quantity. 150. Examples. ( 3x + % = 19. ) 1. Given \ [ Eequired x and y. i4x-2y^S. ) Ans. a; = 3, y ^ 2. Cbx +ll2/=69. ) 2. Given < [ Required x and y. (7a; — 52/= 15.) Ans. X ^ 5, y = i- CQx + Qy = 39. ) 3. Given ] [ Required x and y. ldx + 3y = 21.) Ans. X ^ S, 2/ = 2. C4x + Qy^ 58. ) 4 Given \ [ Eequired x and y. (ix — 3y = 34. ) Ans. X = 7, y =~ 5. (¥ + iy = Q- } ^ . , ^ 5 Given ->' >■ Required x and y. lix + iy = 5. \ Ans. X =■- 8, 2/ = 6. 6. Given ] [ Required x and y. (.> + ^ = 8- ) Ans. a; = 21, y = 15. 104 ALOEBBA. 7. Given 8. Given 9. Given -f- % = ?n. I + dy = n.) Required x and y. . dm — bn em — an Ans. X --^ — ; ; — . y - ad — he he — ad mx — tiy - px —qy = s. '■I Required x and y. , qr — ns pr — ms Ans. X = -' ) y =^ -^ mq — np inq — np X ■[■ y =-- s. I V Required x and y. Ans. ^^ «(«-¥ , y^H^l^s) ^ a — b a — b 10. Given < n , 1 - + - = a. X y ^x y Required x and y. Ans. x^ a+ 6 y-- a — 6 ELIMINATION BY SUBSTITUTION. 151. Illustration. r (1) 2a; + Sy = 19. ) Given \ \ Required x and y. ((2) 3a; + 22,= 21. ^ SOLUTION. By transposing 1x in (1), we have 32/ = 19 — 2a!. ELIMINATION. 105 19 — 22; .-. (3) .- 3 Substituting tlais value of y in (2), we have 3. + 2 Q^-^) = 21. Clearing of fractions, and developing, 9a; + 38 — 4x = 63. Transposing and reducing, 5a; = 25. . . X :^ 5. Substituting in (3), we have 19 — 10 „ y = ^ = 3. 152. Rule. 1. Find from one equation the value of one unknown quantity in terms of the known and other unknown quantities in that equation. 2. Substitute this value in the other equation for the unknown quantity which it represents. 3. Find the value of ike unknown quantity in the resulting equation. 4. Substitute the value of the unknown quantity thus found in the expression for the other unknown qimntity, and reduce. 106 ALGEBRA. 1. Given 2. Given 3. Given 153. Examples. x-\- 2y = 13. 2a; + 2/ = 14. 4a; + 3y==20. 2a; + 62/ = 25. 7a; + % = 130. 9a; + 72/ = 126. Required x and y. Ans. a; = 5, 2/ = 4. Required .i; and y. Ans. a; := 2^, 1/ = 3^. Required x and y. Ans. X = 7, y = 9. f 11a; — 13j^ = 0. ^ 4. Given ] /'^ Required x and 2/- (l3a: — ll2/ = 48. ) 5. Given 6. Given Am. a; = 13, y = 11. Required x and y. X + y = s. X — y ^ d. Ans. x = ls + ^d, y = ^s — ^d. mx -\- ny = a. px+ qy = b. Required x and y. . aq — bn bm ■ Ans. X = —^ , y = mq — np mq — np 7. Given ( ax -{- by = c. (^ mx = ny. Ans. x- Required x and y. en an-\- bm y^ an -\- bm ELIMINATION. 10 ' X . y \-^ = a. m n 8. Given - - Required x and y. - p q > J _ __ amnp — bm/pq bnpq ■ — amnq .a-7hSt X ) U np — mq np — mq 9. Given ■ '7 + 1 = "- 1 a b ^ = 1. <■ m n ' Required x and y. Am. x = oftcm abm an -{- bm ' an -\- bm 10. Given a , b 1 — ^m. X y c , d - + -=n. ^ X y - Required x and y. Ans. X ad — 6c ad — be = T 7~ > V^ ■ dm — bn an — cm ELIMINATION BY COMPAKISON. 154. Illustration, ((1) 4x + % = 27.-) Given I r Required x and y. [(2) 3x+5y = 34.) SOLUTION. 27 — 4x From (1) we find (3) y = „ , ..s 34 — 3a; From Q2) we find (4) y = — g — ■ ■ 108 ALGEBRA. .-.by Ax. 9, 34^^2L=J,. Clearing of fractions, we have 102 — 9x = 135 — 20a;. Transposing and reducing, we have 11a; = 33. .-. a; = 3. Substituting in (3) or (4), we find y = 5. 155. Rule. 1. Find from each equation ihe value of the same unknown quantity, in terms of the known and other unknown quantities in that eqiudion. 2. Write these values equal to each other. 3. Find the value of ike unknown quantity in the resulting equation. 4. Substitute the value of the unknown quantity tlvus found in either expression for the other unknown quantity, and reduce. 156. Examples. r Ix -^by = 74. \ 1. Given ■; y Required x and y. ( bx + 7y = 70. 3 Ans. X --= 7, y = 5. 2. Given 3. Given 4. Given 5. Given 6. Given 7. Given 8. Given ELIMINATION. 13x + Vly = 304. 17a; + 19y = 362. 109 Required x and y. 9x— lly^O. 12a; — 10y = 42. ClOOx— 502/ = 500. 1 150a; — lOOy = 500. f-+72/ = 99. - + 7a; = 51. 3 ,2 ... ma; + % =^ .P- ) nx + cy =^ q. ) Ans. a; = 9, j/ = 11. Required a; and y. Ans. a; = 11, y = 9. Required x and y. s. a; := 10, y = 101 - Required x and y. A71S. .a; =: 7, J/ = 14. - Required x and y. Ans. X = 12, y := 9. Required x and y. . cp — bq mo — np Ans. X = —^ f- , y = -— ' /- . mm hr, ■' gjjj, Jjj inx := ny. y z^ ax -{- b Ans. X cm — bn Required x and y. bn bm m — an m — an no ALOEJSiiA: 9. Given i -A't Reaui ^■1/ ax -\-uy :=q. Required li and y. a(b — d)^ b — d Ans. X = — "r^ — '-tt > y 10. Given — + — = a. » 2/ ^ =6. X y Required a; and y. . m^ — n^ n"^ — m^ Aim. X = -■ ;— . y ■■ am — bn an — bm ELIMINATION BY INDETEBMINATE MULTIPLIERS. 157. Illustration. f(l) 5x+3y^M.) Given \ [ Required x and y. (_ (2) Sx+ 5y = 30. ) SOLUTION. (1) X TO = (3) 5ma; + 3my = 34m. (2) Sx+ 5y = 30. (3) _ (2) = (4) (5m — 3) a; + (3m — 5) 1/ = 34m — 30. 1st. Assume 3m — 5 = 0. 5 and (4) becomes 16 80 a; = 5. ELIMINATION. 2d. Assume 5m — 3 = 0. ... .=|. and (4) becomes 16 48 Ill .-. 2/^3. 158. Rule. 1. Multiply either equation by m, and from the resulting equation subtract the other equation. 2. Assume the eo-effieient of one unknown quantity in the resulting equation equal to 0, from which deduce the value of m whidi substitute in the preceding equation, and reduce. 1. Griven 2. Given 3. Given 159. Examples. 4a; + 5y = 41. ) V Required x aud y. 5.r + 4y = 40. ) Ans. a; = 4, y = b. 5a; + 72/ = 50. \ 6a; + 5!/ = 43. 3 8a; + 9?/ = 43. 11a; + 13?/ = 61. r Required x and y. Ans. a; ^ 3, y = b. Required x and y. Ans. X ^ 2, y = 3. 112 ALGEBRA. ( 12x + 7y = 88. ) 4. Given J. [ Required x and y. (. 7x — 5y = 15.) Ans. X = 5, y = 4c. C ax + by==p. ] 5. Given J. y Required x and y. {_ ex + dy = q. ) Am.x = ^^S, y = ^^. ad — be ad — oc ELIMINATION BY THE GBEATEST COMMON DIVISOE. 160. Illustration. r (1) 4a; + 52/ — 41 = 0. ) Given | > Required a; and y. ( (2) 5* + 42/ — 40 = 0. 3 If the value of y were found and substituted in these two equations, we should have two equations involving x and giving the same value for x. Since the iirst members of these equations are divisible by x minus the value of x (Art. 145, Prop. 2), they must have a common divisor involving x. Let us now proceed as in finding the g. c. d. , and put the final remainder involving y equal to 0, which must be the case since the equations have a common divisor. Multiplying (1) by 5, and dividing the result by (2), and writing the remainder equal to 0, we have 20a;+ 25^ — 205 20a; +16?/— 160 92/ — 45 = • ■ • 2/ = 5. 5a; + 42/ — 40 ELIMINATION. 113 161. Bnle. Transpose the second members, proceed with the resulting first memhers as in finding ilie greatest common divisor, and place the remainder involving one unhnown quantity equal to 0, which will give the value of that unknown quantity. 163. Examples. r 3a; + 42/ = 39. 1 1. Given < \ Required x and y. ( bx-\-Qy = 61. ) Ans. a; = 5, y = %. r 39; + 73/ = 42. ) 2. Given \ \ Required x and y. ( lx-\-2,y = 58. ) Ans. X = 7, y ^ 3 5x + 7y = 184. ) 3. Given ■{ > Required x and y. .lla;+ %=296. 3 Ans. a; = 13, y = 17. 3a;-2y = 0. ) 4. Given < > Required x and y. 5a; — 43/=— 20. ) a; = 20, 2/ = 30. f mx -\- ny = a. 5. Given < [- Required x and y. (_ px+ qy = b. . aq — hn hm — aip Ans. X = --* ) y = ^ • mq^np mq — np C. A. 10. 114 ALGEBRA. - Kequired x, y, and 2. EQUATIONS INVOLVING THREE OR MORE UNKNOWN QUANTITIES. 163. Illustration. (1) Sx + 2y + 5z= 32. Given ■ Required x, y, z. ^2x— y-\-2z = 12. Am. a; = 5, y = 4, 2 = 3. ' cue -^ by = c. 4. Given ^ bx -\- ck^= e. V Required x, y, z. SYMMETEICAL EQUATIONS. 168. Definition. Symmetrical equations are those equations which resolve themselves into each other when certain permutations are made of the letters entering them. Symmetrical equations admit of elegant solutions. Let them be carefullv studied. SYMMETRICAL EQUATIONS. 119 169. Illustrations. (1) x + y=:a.^ 1. Given <{ (2) y + z = b. . (3) z +x^c. , - Required x, y, s. If X be changed to y, ?/ to a, a to x, a to b, b to c, and c to a, (1) will resolve itself into (2), (2) into (3), and (3) into (1). SOLUTION. (1) + (3) — (2) = (4) 2x = a + e — b. a -\- c — b (1) + (2) - (3) = (5) 2y=a + b-c. a -j- b — c (2) + (8) — (1) = (6) 28 =: 6 + c — a. b 4- c — a 8= — ■ ■ 2 We can obtain the value of y from that of x, and z from that of y, by permuting as above. ' (1) i + i = «. X y 2. Given <| (2) ~ + - =^b. y z X z ' Required x, y, z. 120 ALGEBRA. SOLUTION. [(l) + (3)-(2)]-^2 = (4)^ = "+°-^ Taking the reciprocals, we find 2 Iso , we find a-\- c — 2 b' c ^~6+a- 2 e+ 6 — a ' a; + 2/ = 7. ^ 3. Given < 2, + g = 6. ^ a; + z = 6. ■ Required x, y, z. J.WS. a; = 4, 2/ = 3, a = 2 '11 8 ~ a;"^i/ 15" 4. Given ■{ 11 12 y'^ z 35 ■ 1 1_10 ^ a; z 21 '' ' Required x, y, z. J.M. a;=3, 2/ = 5, z = 7 r 1 + 1 + 1=13. 1 u^ x^y 12 5. Given < 1 + 1+1=47_ a; ^2/^2 60 1 + 1 + 1-19. y^z^U 20 ^ Required m, x, y, z. 1+1+1=? Iz ^ u^x 3 1 0' Ans. M = 2, a! = 3, y = i, 2 = 5. 6. Given < PROBLEMS. u-\-v-{-w-^x-\-y-- 15. V -\- w -\- X -\- y -\- z ^^ 20. w -j- X -{- y -{- z +M=19. X -{- y -\- z +M + w=18. y -\- z -(- It -|- -u -J- ™ = 17. z -|-M+'y + w-|-a;=16. Arts. M == 1, ^; = 2, M = 3, a; = 4, 2/ 121 Required u, V, w, X, y, z. 5,z = 6. 170. Problems. 1. What two numbers are those whose sura is 40 and difference 10? Ans. 25 and 15. 2. What two numbers are those whose sum is s and difference d? Ans. ^s -\- ^d and ^ s Id. 3. A person has two horses, and a carriage which is worth $150. The first horse and carriage are worth twice the second horse, and the second horse and carriage are worth three times the first horse. What is the value of each horse? 4ns. |90, |120. 4. A person has two horses, and a carriage which is worth e dollars. The first horse and the carriage are worth wi times the second horse, and the second horse and the car- riage are worth n times the first horse. What is the value of each horse? . (m -\- 1) o (n-\-l)c J —a £ , inn — 1 mn — 1 Ans. 5. A and B can do a certain work in 12 days ; B can do it alone in 30 days. How long will it take A alone to do it? Ans. 20 days. C. A. 11. 122 ALGEBBA. 6. A and B can do a certain work in m days ; B can do it alone in n days. How long will it take A alone to do it? , mw , Ans. days. n — m •' 7. A and B can do a certain work in 24 days, A and C in 30 days, B and C in 40 days. How long would it take each to do it? Ans. A, 40 days ; B, 60 days ; C, 120 days. 8. A and B can do a certain work in m days, A and C in n days, B and G in p days. How long would it take each to do it? jy^s. A, — '^^^^ ; B 2»M£__. ^ 2mn 'np-^-invp — mn 'np-{-mn — inp 'mp-{-mn — np' 9. A has two kinds of money; 10 pieces of the first or 20 pieces of the second make a dollar. How many pieces of each kind must he take in order that 15 pieces may make a dollar? Ans. 5 of the first, 10 of the second. 10. A has two kinds of money ; m pieces of the first or n pieces of the second make a dollar. How many pieces of each kind must he take in order that p pieces may make a dollar? Ans. — ^ of the first, — ^^ '-^ of the second. 11. A's money + 2 times the sum of B's and C's = I B's + 3 times the sum of A's and C s = $10500, and C's + 4 times the sum of A's and B's = $12000. What has Am. A has $1000, B $1500, C $2000. INDETERMINATE EQUATIONS. 123 12. A's money + I times the sum of B's and C's ^ p dollars ; B's + w times the sum of A's and C's = q dollars ; C's + n times the sum of A's and B's = r dollars. How many dollars has each? Ans. A's = B's ^ mnp -j- Ir -\- Iq — Imr — p — nlq Im -\-ln-\- mn — 2lmn — 1 mnp -\- Imr -\- q — mr — mp — Inq 2mln -{-1 — Im — In — mn p, np -\- Imn -\-nq — r — mnp — Inq Im -\- mn -\- In — 2lmn — 1 INDETEEMINATE EQUATIONS. 171. Definition. An indeterminate equation is an equation in which the unknown quantities have an infinite number of values. 172. Propositions. 1. One equation containing two or more unknown quantities is indeterminate. Thus, take the, equation (1) 2a; — Sy = 15. (2) ,_ 15 + 3y _ Let Then 2/ = 1, 2, 3, 4, 5 .... x = 9, 10^, 12, 13^, 15 Any two corresponding values of x and y will satisfy equa- tion (1). 124 ALGEBRA. the equation (1) 3x + 4y- -5z^ 20, ■ (2) 20- -Jy + 5z 3 Now, we can assign values at pleasure to y and z, and deduce the corresponding values of x. Then, the corre- sponding values of x, y, and z will satisfy equation (1). 2. Equations are indeterminate when tlie number of unknown quantities involved exceeds the number of equations. For, by eliminating, we can reduce the equations to one equation involving two or more unknown quantities, which is indeterminate, as before shown. It will not do to supply the lack of equations by deducing other equations from those already given ; for the resulting equations would not be independent. Thus, if we have (1) 2x+3y = 6. and multiply by 2, we have (2) 4x-{-Qy= 12. Now, in attempting to eliminate, we find that both x and y disappear at the same time. 3. An equation of Hie first degree involving but one unknown quantity may be indeterminate in eo^sequenae of certain rela- tions existing between the hnovm quantities. Thus, take the equation (1) mx -\- p = nx -\- q. ... (2) x^^^^. m — n INDETERMINATE EQUATIONS. 125 I^ow, i£ q^p and n = m, we shall have (3) ^ = 1 This value of x is indeterminate ; but the above relations reduce (1) to (4) mx + p = mx -\- p. But (4) is an identieal equatimi, and may be satisfied for any value of x ; that is, (4) is an indeterminate equation. An identical equation affirms no new fact, but merely that a quantity, is equal to itself. In this light it is not to be regarded as an equation, in the ordinary sense, express- ing a relation between two different objects of thought. We then have no equation and one unknown quantity; that is, the number of unknown quantities exceeds the number of equa- tions. 4. Two equations involving two unknown quantities may he. indeterminate in consequence of certain relations existing between ilie known quantities. Thus, take the equations (1) mx -\- ny = r. (2) px + qy = s. .-. (3) X-. qr- — ns mq- — np ms ■ — pr and (4) y ■■ mq — np Now, if we have (5) qr = ns, and (6) np = mq. 126 ALGEBRA. then, by multiplying (5) and (6) together, and reducing, we have (7) pr = ms. These relations reduce (3) and (4) to a; = ^, and 2/ = ^. But from (5) and (6), we find »is , mq ma a = — > and p = ^= — r ^ n r These values of p and q reduce (2) to (1), and we then have one equation involving two unknown quantities, which is indeterminate, as has been shown. All the cases of indetermination, with respect to equa- tions, therefore resolve themselves into this : Indetermination arises ivhen the number of unhnoum quanti- ties exceeds the number of equations. INDETERMINATE PROBLEMS. 173. Definition and Remarks. An indeterminate problem is a problem which admits of an infinite number of solutions. When an indeterminate problem is stated, it will be found that the number of unknown quantities exceeds the number of equations. We may often limit the number of solutions by imposing the condition that the values of the unknown quantities shall be integral numbers. INDETERMINATE PROBLEMS. 127 174. Illustrations. 1. How many sheep at |4 per head, and cattle at $25 per head, can be bought for $1000? SOLUTION. Let X = the number of sheep, and y the number of cattle. Then (1) 4x + 252/ == 1000, 25 .-. (2) x = 250 — -^y. Since x and y are integral and positive, y must be a mul- 25 tiple of 4, and "ry <. 250. Let y= 4, 8, 12, 16, 20, 24, 28, 32, 36. Then X = 225, 200, 175, 150, 125, 100, 75, 50, 25. 2. Find two whole numbers such that 12 times the one minus 13 times the other equals 9. SOLUTION. (1) 12a;— 13?/=: 9. . .2^ 13y + 9 „ y+9. • • ^^^ ^— 12 ~^+ 12 V + 9 Since x and y are whole numbers, ..„ must be a whole number. Then let V+9 12 y = 12n — 9. (2/ = 3, 15,27, 39.... Letm = l. 2, 3, 4.... Then ] (a; = 4, 17, 30, 43.... ^28 ALGEBRA. 3. There are three whole numbers whose sum is 90, and 2 times the first + 3 times the second + 4 times the third is 200. What are those numbers? SOLUTION. (1) x+ y+ 2 = 90. (2) 2x + 3y + 4z = 200. (2) — (1)X2 = (3) y + 22 = 20. .-. z = lQ-iy. Let y= 2, 4, 6, 8, 10, 12, 14, 16, 18. Then 2=9, 8, 7, G, 5, 4, 8, 2, 1, and X = 79, 78, 77, 76, 75, 74, 73, 72, 71. 175. Examples. 1. The sum of three whole numbers is 11 ; and if the first be multiplied by 3, the second by 5, and the third by 7, the sum of the products will be 57. What are the numbers? T a; =: 4, 3, 2, 1. Ans. J 2/ = 2, 4, 6, 8. [ 2 = 5, 4, 3, 2. 2. How many calves at $3^, sheep at $1^, and lambs at $^ per head, can be bought for $100, the number bought being 100? ( Calves, 1, 2, 3. ^"8- } Sheep, 47, 44, 41. (. Lambs, 52, 54, 56. 3. What number between 12 and 24, when divided by 2, 3, and 4, will give 1, 2, 3 for the respective remainders? Ans. 23. INCOMPATIBLE KqUATlUJSH. 129, 4. Bought three kinds of tea at 6, 9, and 12 shillings per pound. How many pounds of each kind did I buy, provided the entire amount was 40 pounds, and the entire cost £17 8s? 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10,- 9, 8, 7, 6, 5, at 6s. 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, at 9s. 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, at 12s. 5. A number is expressed by 3 digits whose sum is 20, and if 16 be subtracted from the number, and the remainder divided by 2, the digits will be inverted. What is the number? An&. 974. INCOMPATIBLE EQUATIONS. 176. Definition. Incompatible eq[uations are those equations which can not be satisfied for the same values of the unknown quan- tities. 177. Propositions. 1. If the number of independent equations exceeds the number of unhnown quantities, these equations vjiU, in general, he in- compatible. Thus, -^ (2) x — y = 2. ((3) ^ = 2. From (1) and (2) we find a; ^ 5 and y = 3, which will not satisfy (3) ; from (2) and (3) we find a; = 4 and y^2, which will not satisfy (1); from (1) and (3) we find a; = 5^ and f^ 2|, which will not satisfy (2). 130 ALGEBRA. 2. ]ff the number of independent equations exceeds Hie num- ber of unknown quantities, suck relations between Vie knowi quantities can be found as will make tiie equations compatible. f (1) x + y= s. Thus, -^ (2) x — y^d. 1(3) = = ,. From (1) and (2) we find ic = i s + i d. y = is — id. Substituting these values of x and y in (3), we find s+ d '^-T^d- This relation of q, s, and d will make the equations com- patible. Thus, take the equations a) x + y = 9, (2) x-y = S, (3) -y = 2, in which this relation exists, since P 9 + 3 Now, from any two of (1), (2), and (3) we find a; = 6 and y = B, which satisfy the other, or the equations are com- patible. PROBLEM OF THE C0VB,1ER& 131 178. The ProMem of the Couriers. Two couriers were traveling the same road, the one at tlie rate of a miles per hour, the other at the rate of h mUes per hour. At a certain time the distance between them was d miles. When were they together? C A B C Suppose the courier at A travels a miles per hour, and that the courier at B travels h miles per hour. Let the distance AB be denoted by d, and suppose it was noon when the couriers were at A and B respectively. Let t denote the interval of time from noon to the time when they were together. Suppose they traveled toward the right and met at C. Now, since the rate multiplied by the time equals the distance, we shall have at = AC and U = BC. But AC — BC = AB. . ■ . at — U ^ d, or (a — h) t = d. Let us now discuss the value of t on the following sup- positions : (1) a > 6 and d > 0. The supposition a > 6 means that the courier at A travels faster than the courier at B. 132 ALGEBRA. The supposition d > means that the couriers are not together, and 'that the distance from A to B is estimated to the right. In this case, both terms of the fraction r > which a — expresses the value of t, are positive ; hence, ( is positive. The positive value of t indicates that they were together after noon, which corresponds to the assumption that they met at C, at the right of A and B ; and this ought to be the case, since at noon the more rapid traveler was following the other, and must overtake him after noon, (2) a < 6 and d > 0. Under this supposition t becomes negative, which indi- cates that they were together before noon. This result corresponds to the facts of the case ; for since 6 > a the courier at B travels faster than the courier at A, and the interval which separates them constantly increases; hence, they can not be together after noon. At noon, the distance between them is d; a little before noon, it must have been less than d ; and at a certain time before noon, it must have been 0. (3) a = & and d > 0. Under this supposition t = r = tt = «>. a — b This indicates that they will never be together, which result evidently corresponds to the facts of the case; for, since they are separated by a given interval d, and travel at the same rate, they will continually be separated by the interval d, or will never meet. POWERS OF MONOMIALS. 133 (4) a ^ b or a <^ b and d = 0. Under this supposition t = r = r = 0. which ^'^ a — a — b indicates that they are together at noon, and were not together again either before or after noon ; and this is evidently a correct result ; for since d = 0, they were together at noon, and since a > 6 or a <^b they travel at different rates, and could not have been together again either before or after noon. (6) a = 6 and d = 0. Under this supposition t = j = ^r, which is indeter- minate, and indicates that they are together one time as well as another, that they are always together. This result corresponds to the facts of the case ; for c? ^ indicates that they were together at noon, a ^b indicates that they travel at the same rate ; hence, they were together both before and after noon, and will continue together so long as they travel according to that law. INVOLUTION. POWERS OF MONOMIALS. 179. Beflnitions. 1. A power of a quantity is the product obtained by taking the quantity a certain number of times as a factor. 2. Involution is the process of finding the power of a quantity. 134 ALGEBRA. 3. The first power of a quantity is the quantity itself. 4. The second power, or square, of a quantity is the product obtained by taking the quantity twice as a factor. 5. The third power, or cube, of a quantity is the product obtained by taking the quantity three times as a factor. 6. The n*'^ power of a quantity is the product obtained by taking the quantity n times as a factor. 180. The Exponent of the Power. 1. The square of the cube or the cube of the square is the 6"' power. Thus, (a3)2 _ ^3 X „3 ^ „3+3 ^ a^x2 ^ ^1 , and 2. The m"' power of the n"' power is the win"' power. Thus, (a")™ = a" X a" X a' X ... = a"*"'"'-"-""" '"•°" = a"™. Conversely, a"" — (a")""'. 181. The Co-efflcient of the Power. The co-efficient of the n"' power is the n"' power of the co-efficient. Thus, (2a) 8 = 2a X 2a X 2a = 23a3 z= Sa'. In general, (Sa")" == 5"a^". POWERS OF MONOMIALS. 135 183. The Sign of the Power. 1. Any power of a positi'ue quantity is positive. This is evident from the fact that when all of the factors are positive, the product is positive. 2. Any even power of a negative quantity is positive. Thus, ( — a) ^" will represent any even power of any negative quantity ; but (—a ) 2» = [(— a) 2]" = (a^)" = a^". 3. Any odd power of a negative quantity is negative. Thus, ( — a)^"''"! will represent any odd power of any negative quantity; but 183. Rule. Raise the eo-effteient to the required power, multiply each exponent by the exponent of the power, and give to the result tlie sign plus, if the monomial is positive, or if the monomial is neg- ative and the_power even ; but give to the result the negative sign if the moiwmial is negative and the power odd. 184. Examples. 1. Square Qa^b'^. Ans. Z6a<>b*. 2. Cube — 5a4&7. .4^. —125a^%^K 3. Raise — ■Sa'^b^ to the 4* power. Ans. Sla^J'^. 4. Raise 8a*b'^ to the w* .powei-. Am. 8"a*''b^''. 136 ALGEBRA. 5. Eaise — a^i"' to the 2w"' power. Am. a*''b^'"". 6. Raise — a^b^ to the 10"' power. Am. a^ojzo, 7. Raise a^b" to the mn"' power. J.ns. aSmnjmnp^ 8. Raise 2a™6" to the n"' power. ^ws. 2''a~6"'. POWEKS OF BINOMIALS. 185. Particular Cases. By actual multiplication, we find that ((1 + 6)1 =a + b. (a + 6)2=a2 + 2a6 + 62. (a 4- 6)3 -^ fflS _|_ 3(j2j _j_ 3^62 _,_ 53_ (a + 6)* = a* + 4ci36 + 6a^■^ + 4a63 + 6'*. (ffl -f 6)5 = fflS + 5a46 + 10a362 + lOa^b^ + dab* + b\ 186. The Exponents of the Power. 1. The exponent of the leading letter in the first term is equal to the exponent of the power, and diminishes by unity in each succeeding term to the right, till, in the last term, it is 0, or that letter disappears. 2. The exponent of the other letter is in the first term, or that letter does not appear in the first term ; and in each succeeding term to the right, the exponent of this letter increases by unity, till, in the last term, it is equal to the exponent of the power. POWERS OF BINOMIALS. 137 187. The Co-efficients of the Power. The co-efficient of the first term is unity, and the co-effi- cient of any succeeding term is found by multiplying the co-efficient of the term preceding the term required, by the exponent of the leading letter of that term, and dividing the product by the exponent of the other letter plus 1. 188. Demonstration. We shall prove that if the above laws hold for the n'* power, they will hold for the (n -\- 1)"' power. If these laws hold for the n"' power, we shall have J. ■ ^ J. ■ ^ ■ o We shall find the (n -\- 1)"' power by multiplying the ?i* power by a + 6, thus : a + 6 X . Ji 1 . Z . O C. A. 12. 138 ALGEBRA. The (r + D* co-efficient = ^(^-1).. (n-r + 1) w(n— 1)..(»— r+2) ^ (?t+l)»(n— l)..(w— r+2) "•" 1.2 . . r— 1 1.2.3 . . r " Hence, if the laws hold for the »i"' power, they hold for the (n -\- 1)"' power; that is, if the laws hold for any power, they hold for the next higher power; but, by actual trial, they are found to hold for all the powers up to the 5"' ; and if they hold for the 5"", then they hold for the 6"', and hence for the 7*, and so on to any extent. 189. The Signs of the Power. 1. If both terms of the binomial are positive, the signs of all the terms of the power are positive. 2. If both terms are negative, the signs of all the terms of any even power are positive, and of any odd power negative. 3. If one term is positive and the other negative, the signs of all the terms containing the odd powers of the negative term will be negative, the signs of the other terms will be positive. 190. The Binomial Formula. (a + by = a" + no,"--' b + ??iI??L=il ^"-afea (a — 6)" = a" — na"-^b + '^^^ ~^^ a"-2fe2 _ n^n-JK '^-^) (,n-3i8^. . .+ 6... SQUABE MOOT OF NUMBERS. 139 191. Application Illustrated. 1. (x + yy =x^ + 6xhj + 15a;*y2 ^ 20x^y^ + 15xhj* + 6«y* + y^- 2. (2a2 + 3by = (2«2)3 - 3 (2a^y (36) + 3 (2*2) (36)2 + (36)3 = 8(i6 ^ 36^4^, _|_ 54^252 _j_ 2763 3. (a + 6 + c)3= [(a + 6) + c]3:= (a -|- 6)3+ 3 (a + b^c + 3 (a + 6) c2 + c3 =a3+ Sa^b + 3ab^ + b^+ 8a'-c + 6abc + 362c + 3ac2 + 36c2 + c^. 1. (a + 6)*. 2. (a — 6)5. 3. (X + yy- 4. (a + 6)8. 5. (a;— y)io. 6. (a;+2/)'^ 192. Examples. 7. (3a + 262)3. 8. (5a — 26)4. 9. (2a6 — 3ac)''. 10. (a™ + 6")°- 11. (a + 6 - c)*. 12. (a — 6 + e)*. EVOLUTION. SQUARE EOOT OF NUMBERS. 193. Definition. The square root of a number is a number which multi- plied by itself will produce the given number. 140 ALGEBRA. 194. Pointing into Periods. The square of the units of a number will produce one or two figures, and hence will fall in the first period of two figures at the left of the decimal point. The square of the tens will fall in the second period, the square of the hundreds in the third period, etc. The square of the tenths will fall in the first period at the right of the point, the square of the hundredths in the second period, etc. Hence, in preparing a number for the purpose of extract- ing its root, we point it into periods of two places each, commencing at units, by placing a dot over the right-hand figure of each period. Thus, 3476.8979 and 784.6370. 195. Formulas for Extracting tlie Scjuare Root. The formula (a + by = a'' + 2ab + b'^ may take the form (1) (a + 6)2=a2 + (2a + 6)&. The square of the sum of two quantities is equal to the square of the first, plus twice the first plus the second into the second. (^a + b + cy = l(a+b) + cy = (a+by+l2(a+b) + c-]c. Omitting [(a -|- 6) -|- c]^, and substituting the value of (a -\-b)^ as found in (1), we have (2) (a + 6 + c)2 = a2 + (2a+ 6) b -f [2(a + 6) + c] c. By a similar process, we find (3) {a+b + c-\-d)^=a^+(2a + b)b + [2(a + b)+c']c + [2(a + b + c)+d']d. SQUARE BOOT OF NUMBERS. 141 III general, the square of the sum of any number of terms is equal to the square of the first term, plus twice tlie first plus tlw second into the second, plus twice the sum of tlie first and second plus the third into the third, plus twice the sum of the first, second, and third plus the fourth into the fourth, plus, etc. 196. Application. 1 . Let us find the square root of 4096. OPEEATION. a^ H- (2a, + 6) 6 = 4096 | 64 a2=36_ 2a+b = 124) 496 == (2a +b)b 496 We find a ^ 6 tens, the first term of the root. Subtract- ing a^ from the first period and bringing down the next period, we have 496 = (2a -\- 6) b, the first dividend. The trial divisor 2a = 120. Dividing 496 by 120, we obtain 4^6, and a remainder, as there ought to be, since the trial divisor is too small. The complete divisor 2a -|- 6 = 124, which is contained in 496 just 4 times. Hence, 64 is the square root of 4096. 2. Let us find the square root of 119025. OPERATION. as-i- (2a + 6)&+ [2(a + 6) + c]c = 119025 1345 a^= 9 2a + 6 == 64) 290 = (2a + 6) 6 + 256 =(2a + 6)6 2 (a + &) -I- c = 685) 3425 = [2 (a+b')+c']G 3425 142 ALGEBRA. 197. Rule. 1. Point off the expression into periods of two places each, commencing with units. 2. Find the root of the greatest square in the left-hand period, place this root at Hie right, subtract its square from tlie left period, and to the remainder bring down the next period. 3. Take twice the root already found reduced to the denomi- nation of the next term of the root for a trial divisor, and find the next term of the root; the trial divisor plus the second term of the root will be tJie true divisor; then divide and proceed as before. Sdi. 1. If the number contains decimal places, there will be as many decimal places in the root as there are decimal periods. Sch. 2. If the number is not a perfect square, there mil be a remainder, to which annex a period of ciphers, and proceed as before. Sch. 3. To extract the square root of a common fraction, extract the square root of both terms, or first reduce it to a decimal. Sch. 4. After the process is thoroughly understood, the letters can be omitted. 198. Examples. Extract the square root of the following : 1. 625. Ans. 25. 2. 15625. . Ans. 125. 3. 85264. Alls. 292. SQUARE MOOT OF MONOMIALS. 143 4. 547.56. Am. 23.4. 5. 5499025. Ans. 2345. 6. j\. ^ Ana. |. 7. 61.1524. Am. 7.82. 8. f. Am. .866+. 9. 342694144. Am. 18512. 10. 1143881586576. Am. 1069524. SQUAKE ROOT OP MONOMIALS. 199. Illustration. Since the square of a monomial is obtained by squaring its co-efficientj and multiplying each exponent by 2, and since the square is plus, whether the monomial is plus or minus, it follows that the square root of a monomial is obtained by extracting the square root of the co-efficient, dividing each exponent by 2, and giving to the result the double sign (±), plus or minus. Thus, 200. Eule. Extract Ike square root of the co-efficient, divide each exponent by 2, retain all the letters, and give to the result the double sign ±. 201. Examples. 1. Find the square root of IQa^b'^c^. 2. Find the square root of 25a^b^c' ". 8. Find the square root of 64a''b^c^. 4. Find the square root of 81a2'»64.i 144 ALOEBMA. SQUAEE ROOT OF POLYNOMIALS. 202. Illustration. (a + 6 + c)2 = a2 + (2a + 6) 6 + [2 (a + 6) + c] c. Let us now find the square root of the second member developed. OPERATION. a2 + 2ab + b^ + 2(xc -f- 26c + c^ (a + b -\- c. 2a + b)2ab + b^ 2ah + 62 2a -h 26 + c ) 2ac + 26c + c^ 2ac + 26c + c^ 203. Rule. 1. Arratige the polynomial with, reference to a certain letter. 2. Extract the square root of its first term for the first term of the root. 3. Subtract the square of ihe first term of the root from the given polynomial, and bring down two terms of the polynomial for tlie first dividend. 4. Take twice ihe first term of the root for a trial divisor, by which divide the first term of ihe dividend, and the quotient will be the second term of the root. 5. To ihe trial divisor add ihe second term of the root, and tlie sum wiU be the complete divisor. 6. Multiply the complete divisor by the second term of the root, subtract the product from the dividend, bring down addi- tional terms of the dividend, and proceed as before. CUBE BOOT OF NVMBEBS. 145 204. Examples. Find the square root of the following : 1. 4x* — 12a;-'' + bx'' + 6a; + 1. Ans. 2a;2 — 3x — 1. 2. a;6— 6aa;°+15a2a;''— 20a3a;3-f-15a%2_6aSa._|_(j6_ Ans. x^ — 3ax^-{-3a^x — a^. 3. a" + 4a= +- 10a* + 20a^ + 25a^ + 24a + 16. Ans. a^ + 2a2 + 3a + 4. 4. a* + 4a^6 + Qa^b'^ + 4ab^ + fc*. Jm. a2 + 2a6 + b^. 5. 4a'' + 12a36 + 13a262 + 6a63 + 6* Ans. 2a2 + 3a6 + 6^. CUBE ROOT OF NUMBEKS. 205. Pointing into Periods. Since the cube of units falls in the first period of three figures at the left of- the decimal point, the cube of tens in the second, etc., the number is pointed into periods of three places each, commencing at the decimal point. Thus, 276340789 and 78674.675832. 206. Formulas for Extracting the Cube Boot. (a + 6)3 =a^ + 3aH + 3ab^ + b^; or, (1) (a + 6) 3 = a3 _|. (3^2 + 3^6 _|_ 52) j_ Hence, The cube of the sum of two terms is equal to the cube of the first, plus the sum of three times the square of the first, three timss the first by the second, and the square of the second by the second. C. A. 13. 146 ALGEBRA. Applying this principle, we have [(a + 6) + cy = (a + hy+ [3 {a + hy + 3 (a + 6) c + c^Jc. Omitting the parenthesis in the first member, and substi- tuting the value of (a + 6)^ in the second member as found in (1), we have (2) (a + 6 +c) ' == a-' + (3a2 + 3a6 + 62) 6 + [3 (a + &)^ + 3 (a + 6) c + c'^y By a similar process, we find (3) (a + 6 + c + d)3==a3^(3a2_j-3a6 + 62)6 + [3 (a+ 6)2 + 3 (a + 6) c + e^Jc + [3((i + 6 + c)24-3(a + 6H-c)d+d2]d. The formula expressed in words gives the law. The cube of the sum of any number of terms is equal to the cube of the first, plus the sum of three times the square of the first, three times the first by the second and the square of the second by the second; plus the sum of three times the square of the sum of the first and second, three times Ae mm of ihe first and second by the third and ike square of the third by the third; plus the sum of three times the square of the sum of the first, second, and third, three times the sum of the first, second, and third by tlie fourth and the square of the fourth by ihe fourth, plus, etc. 207. Application. 1. Let us find the cube root of 91125. OPERATION. a" + (3a2 + 3a6 + 6^) 6 = 9il25 (45 a3 = 64 3a2 = 3 X 402 = 4800 3a6 = 3 X 40 X 5 = 600 62= 52= 25 27125 27125 3a2 + 3a6 + 62 = 5425 CUBE ROOT OF NUMBMBK 147 2. Let us find the cube root of 94818816. OPERATION. 94818816(456 a»= 64 3a2 = 3 X 402 = 4800 3a6 =: 3 X 40 X 5 = 600 62 = 52 = 25 5425 3 (a + 6) 2 = 3 X 4502 = 607500 3 (a + 6) c = 3 X 450 X 6 = 8100 c2= 62= 86 30818 27125 3698816 3693816 3 (a + 6)2 + 3 (a + 6) c + e2 = 615636 208. Rule. 1. Point off the expression into periods of three places each, eoiinting from the dewnwd point. 2. Find the root of the greatest cube in the left period, which place at the right, subtract its cube from the left period, and to the remainder bring down the next period, 3. Reduce Ike term of the root already found to the denomi- nation of the next term of the root, and take three times its square for a trial divisor, from which find the second term of the root. 4. To the trial divisor add three times tlie first term of the root by the second, also the square of the second, and the sum will be the true divisor. Divide, and to the remmnder bring down the next period, and continue this process till all of the periods have been brought down. Sch. 1. If the number is wholly or partly decimal, there will be as many decimal places in the root as there are decimal periods used. 148 ALGEBRA. Sch. 2. If the number is not a perfect cube, the process can be continued by bringing down decimal periods of ciphers. Sch. 3. Since the trial divisor is too small, it may give a quotient too large ; if this is the case, it will appear when the true divisor is found. Sch. 4. The cube root of a common fraction can be found by extracting the root of both terms, or by first reducing it to a decimal, and then extracting the root. Sch. 5. If the divisor is not contained in the dividend, place a cipher in the root and bring down the next period ; and, in finding the next trial divisor, reduce the root already found to the denomination of the next term of the root. SrM. 6. We shall now give a contracted method of finding the trial divisor. Take the general formula (a + 6+c + d + ...)3=a3 4- (3a2 _)_ 3^6 -|- 62) ^ + [3(a + 6)2 + 3(a + 6)c + c^]c, + [8 (a + 6 + c)2 + 3 (a + 6 + c) d + d'^'\d+ . . . We find that 3 (a + 6)2 = 3a2 4- Qah + Sfe^, 3 (a + 6 + c)2= 3 (a + 6)2 + 6 (a + 6)c + Sc^, 3(a + 6 + c + d)2=3(a + 6-f c)2+6(aH-6 + e)d + 3d2. That is. Each trial divisor = the sum of the first term of the next preceding true divisor, twice its second term, and three times its ihird term. Two ciphers are annexed to reduce it to the required order. CUBE ROOT OF MONOMIALS. 149' 209. Examples. 1. "What is the cube root of 13824? Am. 24. 2. What is the cube root of 54872 ? Ans. 38. 3. What is the cube root of 117649 ? Ans. 49. 4. What is the cube root of 941192? Ans. 98. 5. What is the cube root of 3048625? Ans. 145. 6. What is the cube root of 102503282 ? Ans. 468. 7. What is the cube root of 341532099 ? Ans. 699. 8. What is the cube root of 491169069? Ans. 789. 9. What is- the cube root of 158252.632929? Ans. 54.09. 10. What is the cube root of 491916472984? Ans. 7894. THE CUBE ROOT OF MONOMIALS. 210. Kule. Extract the cube root of the co-efficient, divide eack exponent by 3, retain all the letters, and give to the result the same sign as iliaJt of the given monomial. 211, Examples. 1. Find the cube root of 8a^6^c'. Ans. lab'^e^. 2. Find the cube root of — ila^b^c^"^. Ans. — 3a6e*. 3. Find the cube root of Qia^b^'^c^^. Ans. ia^b*c^. 4. Find the cube root of — 125a3™66''. ■ Ans. — 5a™62n 5. Find the cube root of ^l^a^^b^^c^"^'. Ans. 6a2P63«'o*^ 150 ALOEBBA. THE CUBE KOOT OF POLYNOMIALS. 213. Illustration. We have already found that (a + 6 + c)s = a3 + (3a2 -)- 3a6 + fe^) 6 + [3 (a + 6)2 + 3 (a + 6) c + c^] c. Let us now find the cube root of the second member. OPKRATION. o8 + C3a'+3a6+52)6+r3(a+&)'+3(«+6)c+c']e |a+6+e «3 3a'+3ffi&+62 I (3a2+3a6+62)6 (3a'+3a6+fe')fe 3Ca +6)''+3(a+6)c+c2 | [3(a+6)2+3(a+6)c+c2]c [3(a+6)^+3(a+ 6)c+c']e 213. Rule. 1. Arrange the polynomial wiik reference to a certain letter, extract the cube root of its first term for the first term of the root. 2. Subtract its cube from the polynomial, bring down three terms of the polynomial for the first dividend, and divide the first term of Hie dividend by the trial divisor, which is three times Hie square of the first term of the root, aiid the quotient win be the second term of the root. 3. To the trial divisor add three times the first term of the root by the second, also the square of tJie second, which will give the true divisor. 4. Multiply the true divisor by the second term of Hie root, mbtraet the product from the first dividend, and bring down additional terms of the polynomial, and the result wiU be the second dividend. HIGHER ROOTS OF NUMBERS. 151 5. Divide tlie first term of the second dividend by the trial divisor, which is three times the square of ihe first term of the root, and the quotient will be the third term of the root. 6. Then find the true divisor by taking three times the square of the sum of i/ie first and second terms of ihe root, three times tJie sum of the first and second terms by the third, also tlie square of the third. 7. MvMply the true divisor by the third term of the root, subtract ihe product from the dividend, and proceed in like maimer till the required root is obtained. 214. Examples. 1. a^ + 3a^b + Saft^ +63. ^,is. a + b. 2. a;^ + 6a;5 — 40a;3 -f 96a; — 64. Ans. x^ + 2x — 4. 3. a;« — 6a;5 + 15a;* — 20a;3 + 15a;2 — 6a; + 1. Ans. x^ — 2x-\- 1. 4. a« + 6a^b + 15a*b^ + 20a^b^ + Ida'^b* + 6ab^ + 6«. Ans. a2 + 2ab + b^. 5. 8a;6 — 66aa;5 + 66aV— &3a^x^-\- 33a%2 — 9a5a;+ a«. Ans. 2x^—3ax + a'. THE HIGHER BOOTS OF NUMBEKS. 215. When the Index of the Eoot is Composite. Let mn denote the index ; then we can apply the princi- ple: The mn"' root of a quantity is equal to the m* root of the m'^ root of that quantity. That is, ■iim/ 7ft I '♦ / — ya= \ya. 152 ALGEBRA. For, let (1) \ya=p. (1)'" = (2) V/^=i3'" (2)» = (3) a = p^". V(3) = (4) mil/ — • Thus, {/a = \i/a, i/a=\y'^, f/a = \V^, etc. 316. Examples. 1. Find the value of 1^20735. Ans. 12. 2. Find the value of i/"1679616. Ans. 6. 6 , 3. Find the value of 1/262U4. Ans. 8. 4. Find the value of ^''387420489. Ans. 9. 5. Find the value of 'v/64339296875X 52521876. Ans. 35. 217. To extract Roots when the Index is Prime- Let n denote the index of the root; then, by reversing the Binomial Formula, Art. 190, we have the rule: 218. Rule. 1. Point off the expresmm into periods of n places each, counting from the decimal point. HIGHER HOOTS OF MONOMIALS. 153 2. Find the greatest n*^ power in the first period, place its n"^ root at the right for the first term of (he root, subtract tlie n* power of this term of the root from the first period, and to the remainder bring down the next period for the first dividend. 3. Reduce the first term of the root to the next lower denomi- nation, and take n times its (n — 1)* power for a trial divisor, from which find the second term of the root. 4. Find the n"' power of the root already found, subtract tlie result from the first two periods, to the remainder bring down the next period for a second dividend, and proceed as before. 319. Examples. 1. Find the value of ^33554432. Ans. 32. 2. Find the value of -1/6103515625. Ans. 25. 3. Find the value of f^936242722357. Ans. 517. 4. Find the value of v/64339296875. J?is. 35. 5. Find the value of '^•36028797018963968. Ans. 32. THE HIGHEE EOOTS OF MONOMIALS. 230. Principles. 1. Every even power of a quantity is positive. 2. Fkery odd power of a quantity has the same sign as the quantity itself. Hence, 1. An even root of a positive quantity is positive or negative. 2. An even root of a negative quantity is impossible. Thus, ■\/ — a, y' — a, {/ — a are called imaqinary quantities. 154 ALGEBRA. 3. Ati odd root of a quantity has the same sign as the quantity itself. Let n denote the index of the root; then we have the rule : 221. Rule. Extract the n"" root of tlie co-efficient, divide each exponent by n, rdain all the letters, give to the result the double sign ± if n is even, but the same sign as the monomial if n is odd. 222. Examples. 1. Find the 5* root of 32a">b^. Ans. 2a26. 2. Find the 6"' root of 64a' ^J' ». Ans. ± 2a^b^. 3. Find the 7* root of — 78125a'6'*. Ans. — 5ab^. 4. Find the n"' root of a^''^'"". Ans. ± a%'". 5. Find the 2?!, root of — a*". Ans. |/=^*^'. THE HIGHER EOOTS OF POLYNOMIALS. 223. Rule. 1. Arrange the •polynomial according to the ascending or descending powers of a certain letter. 2. Extract the n"' root of the first term of the polynomial, place tlve residt at the right for the first term of the root, and subtract the n"' power of this term of the root from the given polynomial. 3. Divide the first term of the remainder by n times the (n — 1)"" power of the first term of the root, and §ie quotient will he the second term of the root. 4. Raise the root already found to the n*'' power, subtract tlie re«uU from the given polynomial, and proceed as before. RADICALS. 155 224. Examples. 1. Find the 4"' root of a;* — 4x^y -f Qx'^y'^ — 4xy^ + y*. Ans. X — y. 2. Find the 5"' root of a^ + 5a*b + lOa^b^ + lOaH^ -f 5ab* + b^. Ans. a + b. 3. Find the 4'" root of I6a* ~ 9GaH + 216a^^ — 216ab» + 816*. Ans. 2a— 36. 4.' Find the 6"' root of a" — 6a^ + 15a* — 20a3 + ISa^ — 6a + 1. -djM. a — 1. RADICALS. 225. Definitions. 1. A radical quantity is an indicated root of an imperfect power. Thus, Vo, F7, 8^, etc., are radicals. Eadicals are also called irrational quantities, or surds. 2. A rational quantity is one which can be expressed without the radical sign or fractional exponent. Thus, 5, 1/9, etc. Rational quantities can always be expressed under the radical form. Thus, 6 = l/25, 8 = f^, etc. By the definition of a root, it follows that Va squared = a, f/^ cubed = a, \/a raised to the n* power = a ; also, the n* root of a" = a, the w"" root of (iXa)" = -^/a, etc. Rad- icals are of different degrees, as the 2d, 3cZ, ... m*. 156 ALGEBRA. 3. The index, denoting the degree of the radical, is the figure or letter placed in the angle of the radical sign, or the denominator of the fractional exponent. Thus, V^, f^a, y'a, or a^, a^, a» , denote respectively the square root of a, the cube root of a, the n"' root of a. A radical of the second degree is often called a quadratic surd; one of the third degree, a cubic surd. 4, The co-efficient of a radical is a factor which is to be multiplied into the radical. Thus, 4 Fa; 4 is the co-efR- cieut of Fa, and 1 understood is the co-efficient of Fa,. 6. Similar radicals are those which have a common index and the same quantity under the radical sign. Thus, ayY, and — c \/F. EEDUCTION OF KADICALS. 226. Case I. To reduce radicals to their most simple form. This reduction depends upon the principle : The n"' root of the product of two quantities is equal to the product of the n"' roots of those quantities, and conversely. For, (1) (yd>y = ah. (2) (y^xV^T^ah. .-. (3) (v^)" = C^^Xi/'5)». v"/(S)-=(4) ^=^xVh. Thus, F'l2a3=V^4^xl/3a = 2aF^. Also 2 V^ia» = 2 Tf 27a« X l''2^ =^ Ga^ ^'^a"-. REDUVTWJS Ul<' liAJjivjiijo. 157 227. Rule. Resolve the given radical into two factors, one of which is the greatest factor which is a perfect power of the degree indicated, extract its root, which multiply by the co-efficient, and place Oie result as the co-efficient of the radical factor. 228. Examples. Ans. lahV'lac. Ans. 5a'^b*e\/2. Ans. Sa'^b^c^ VSac. Ans. 2abf^. 1. Simplify VSa^b^c. 2. Simplify V50a*b<^c^. 3. Simplify l/192a5ft6o'. 4. Simplify fma^b*. 5. 6. Simplify #'500a'69c". Simplify T^243a^6«c5. 7. Simplify r625a*64c. 8. Simplify 3r64ffi5&i0ci2. 9. Simplify 5l^486a='"6i»»c. Ans. da'^b^c^ F4ac^. Ans. Sab^cl^c. Ans. 5ab vc. Aiis. Qah''c'^V2c"-. Ans. ISa^fi"!^. 10. Simplify ly^a^^¥^. Ans. laH'^\/c^. 229. Case II. To pass the co-efficient under the radical sign. Since a = ^^, a^b^^xV^^V^^ 230. Rule. Raise the co-efficient to the n"* poiver, and multiply it into €ie quantity under the radical sign. 158 ALGEBRA. 331. Examples. 1. 5l/2. Am. 1/50. 2. af/f. Ans. ^/a^. 3. SaiXc. ^rw. y'^VcT 4. a^ 1/6. ^?M. y/pS. 5. 3 i/3. ^ns. |/^. 6. p v^. ^ns. i/p^. 7. (a— 6)y^a— B. ' ^m. i/(a — 6)a. 9. ^^VB. Am. '^W. 10. -i^iyF. ^ns. >«. 232. Case III. To reduce radicals to equivalent radicals having a common index. (1) ({/ar = a. (2) ('"i/5)""'=«. -•• (3) (v^)» = (l>^)'""- V^) = (4) p/J- (Tar- But r^)™=7/^xT/«x...=T«"*- REDUCTION OF RADICALS. 159 Hence, we may multiply the index of the radical by m, if we raise the quantity under the radical sign to tJie m*** power; and conversely, we can divide the index by m, if we extract the m*"* root of the quantity under the radical sign. This principle enables us to reduce radicals to a common index, for which we have the rule : 233. Rule. Multiply each index by the number which will give for the product the hast common multiple of the indices, and raise the quantity under each radical sign to tlie power denoted by the corresponding multiplier. 234. Examples. 1. Reduce 1/2, F 3, F 5 to a common index. Ans. "-l/M, '|/8T, 'f 125. 2. Reduce \/a, \/F, y^ to a common index. Ans. y'a^, y^, y'e^. 3. Reduce y'a^, \/a^, y^ to a common index. Ans. ■^^, l/^, i/a^. 4. Reduce ]/2, y^, |/5, f/7 to a common index. Ans. '^/^W, ^^729, 't^md, '^|/343. 5. Reduce i/cT, \/'S to a common index. Ans. T^, T/5^- 160 ALGEBRA. ADDITION AND SUBTRACTION OF EADICALS. 235. Rule. Malce ike radicaU similar, if possible ; take tJie sum or differ- ence of tiieir co-efficients, and annex the common radical; other- wise, connect them by the sign plus or minus. 236. Examples. 1. Find the sum of 3 VB and 7 VF. Ans. 10 VJ. 2. Find the sum of 3 l/6 and l/20. Ans. 5 V5. 3. Find the sum of 7 l/lO and 2 t/90. Ans. 13l/l0. 4. Find the sum of 4 1/ 20 and 3 l/45. Ans. nvT. 5. Find the sum of a vb and c l/6. Ans. (a + e) Vb. 6. Find the sum of 3 Va^b and 2c l/ife! ^7is. (3a + 4c) yT 7. Find the sum of l/| and l/^. Ans. \i l/l5. 8. Find the difference of 7 l/l2 and 4 1/3? 4ns. lOl/a 9. Find the difference of 8 l/32 and 7 l/l8. Ans. nv^. 10. Find the difference of l/f and l/|^. Ans. ^'j I/dT MULTli'LWATlUJS JLl\u juiYioiuiy. IGI 11. Find the difference of 4a v^W and f^MoF. Am. 2a-S^ 'da^. 12. Find the difference of f'iOa^b and f'^a^. Ans. a f^da'^b. 13. Find the difference of -^l&a^b^ and y^c^ST Ans. 2 (a — c) ]/6. 14. Find the difference of 3a i^ and '^/^Top". Ans. 2a ^. 15. Find the difference of 6 /^ and 3 pliJ. Ans. 3 |/a. MULTIPLICATION AND DIVISION OF EADICALS. 237. Rule. Reduce the radicals to a common index, take the product or quotient of the co-efficients for the co-efficient, and the product or quotient of the quantities under the radical sign for the radical part. 238. Examples. 1. Multiply i/a by f^. Ans. ^/a^. 2. Multiply v^ by {/b. Ans. "'I/^^'- 3. Multiply 5 |/2a by 6 i/Ta. Ans. 60a. 4. Multiply 10a i/b by 36 f^ Ans. SOab {/^^K 5. Multiply 5 i/18 by 6|/'2. Ans. 180. C. A. 14. 162 ALGEBRA. 6. Multiply v^, yi, yi, f/d Ans. ^^/a^^c^d^ 7. Multiply V2, VI, V\, V\. Am. ^2. 8. Multiply fq by 't^. Ans. 'V^^- 9. Multiply y^r+J by y'^TS"- Am. yCa + 6)™+" 10. Multiply (a + b) (^/S + v^ by i/H — l/6. -Ans. a" — 62 11. Divide 18 -/TB by 2 ■/S'. Ans. 27 12. Divide i/Si by y^. Ans. S^/W. 13. Divide 12 i/S by 3 f^. Iw. 4 f|, 14. Divide a ^A by o /S'. ^m. ^ :^. 15. Divide a |>^ by c i/3. Ans. " "^ 16. Divide Vk'^jV^ + Vl- -^ws. tV 17. Divide -Ji/i X v^5^ by "\/^'T28 X i/B. Am. ^ *|/|: 18. Divide ^/If X y^ by ^X v^. ^n*- 4 '//588. 19. Divide -^^ by y^j-- Am. ^r' 20. Divide («« — 6^) y'iqry by (a + 6) y^^+l. .4»i8. (a — 6) y^ffi + 6. INVOLUTION OF RADICALS. 163 INVOLUTION OF RADICALS. 239. Illustrations. 2. (a ""r^r = («>/l>^)"'= a- y'K 240. Rule. 1. Jff the index of the radical and the exponent of the power are relatively prime, raise the co-efficient to the required power, also the quantity under the radical sign. 2. ]f tlie index of the radical is a multiple of the exponent of the power, raise the co-efficient to Ike required power, and divide the index of the radical by the exponent of the power. 3. If the index of the radical and the exponent of the power have a common factor, raise the co-efficient to the required power, reject the common factor from the index and exponent, take the remaining factor of the index for the index, and raise the quan- tity under the radical sign to the power denoted by the remaining factor of the exponent. 241. Examples. 1. Square 3 ^5. Ans. 9 f25. 2. Cube 5^3. Ans. 125 V^. 3. Raise x^/y to the w* power. Ans. a" %■ f. 164 ALOEBEA. 4. Square 5 Vz. Am. 25 VZ. 5. Cube 3 i/5. Aw. 27 f5. 6. Raise x "\/y to the n'* power. ^m. a;" ]p/^ 7. Square 7 '-(^5. J.n8. 49 iTS. 8. Raise 3 p^ to the 6"' power. Ans. 729 y^. 9. Raise a 'f/Tto the 12* power. Am. a'^f/P. 10. Raise x ^f/y to the 18"' power. Ans. x'^ ' i/p. EVOLUTION OF KADICALS. 243. Illustrations. 2. (ay^'-^a-'v^ .-. Va'v^^^v'^ 3. iayW)'"-=aF'^/W .-. ^a"'' v/5^= a ^. 243. Rule. 1. If the quantity under the radical sign is a perfeet power of tJie degree of the root required, extract the root of the co- effieient, also of the quantity under the radical sign. 2. If the index of the required root and ike exponent of tiie quantity under the radical sign are relatively prime, extract the root of the co-efficient, and multiply the index of the given radical by the index of the required root. EVOLUTION OF RADICALS. 165 '6. If the index of the required root and the exponent of the quantity under the radical sign have a common factor, extract the required root of the co-efficient, reject the common factor from the index and exponent, multiply the index of the given radical by the remaining factor of the index of the required root, and give to Hie quantity under the radical sign an exponent equal to the remaining factor of its exponent. 244. Examples. 1. Extract the square root of 9 f^. Ans. BVQ. 2. Extract the cube root of 125 V'S. Ans. 5 V2 3. Extract the n"' root of a" y^a". Ans. a y^ 4. Extract the square root of 25 i^. Ans. 5 |/6. 5. Extract the 4* root of 7 f^. Ans. f/l 6. Extract the n"' root of a y^B. Ans. \/a "y^ 7. Extract the 6"' root of a^ ^ ^{/c^ Ans. a^ ^^c^ 8. Extract the 4* root of ais ^^/p^_ j^^s. a" ^/'b, 9. Extract the mn"^ loot of -{/c^. Ans. f/a^ 10. Extract the 12'* root of x-f/y^. Ans. ^i/x^ 11. Extract the 15* root of f{a + by^. Ans. |/(a + h) 12. Extract the mw* root of a^f-F^X v''^X V^^- Ans. a^WX^'V^X'V^. 166 ALGEBRA. 245. Theorems. 1. No irrational quantity can be eayresaed by a fraction. For, if possible, suppose we have (1) Va = ^-. V (1)»=(2) a = ^. . b . 6" . Now, if - is in its lowest terms, — is in its lowest terms, c c" and we shall have an integral number equal to an irreduci- ble fraction, which is impossible. Gar. 1. The root of an imperfect power, if carried out decimally, will be interminate ; for, if terminate, it could be expressed by a fraction. Gor. 2. The interminate root can not be a repetend ; for, if a repetend, it could be expressed by a fraction. 2. A quadratic surd can not be equal to the sum of a rational quantity and a quadratic surd. For, if possible, suppose we have (1) Va = b + Vc. (1)2 = (2) a = 62 _|_ 26 v/T-f- c, a — 6^ — c l/r= 26 That is, an irrational quantity is equal to a rational quan- tity, which is impossible. 3. if two quadratic sv/rds can not be made similar, theil product is irrational.- THEOREMS. 167 Let l/a and Vb be two such surds. Suppose their pro- duct to be rational, and let this product divided by a be equal to r, an entire or a fractional quantity. Then, (1) Vab = or. (1)2 = (2) ah = a^r\ (2)-=- a =(3) b = ar-'. l/(3) = (4) Vb=:rVa: That is, l/a and l/Fcan be made similar, which is contrary to the supposition. Hence, their product can not Be rational, and is therefore irrational. 4. A quadratic 'surd can not be equal to the sum of two dis- similar quadraiie surds. For, if possible, suppose we have (1) i/^= -[/T+ Vl. (1)2 = (2) a = 6 + c + 2l/6^ .-. Vbe = ■h~c Now, if l/^and l/Tare in their simplest form, l/6c will be irrational ; that is, a rational quantity is equal to an irrational quantity, which is impossible. 5. In an eqtiation of which each memher is the sum of a rational quantity and a quadratic surd, the rational quantities of the two members are equal, and also the quadratic surds. Let (1) a + x/b ^ c + \/d. 168 ALGEBRA. If a and c are not equal, let c ^ a + n. Then, (2) a + Vb = a + n + Vd. .-. (3) Vb = n + Vl, which is impossible by Theorem 2. .-. a = G; .-. \/b = V'^. 6. If l^a + 1/5 = a; + -j/y, f/ien l^a — -(/J = x — -j/y. (1) Va + v/5 = a; + v'^. (1)2 = (2) a + vT= x^'+y+2x\/y'. .-. (3) a = a;2 + 2/, and l/6 = 2a; l/^T Th. 5. .-. (4) a — VT^x'^ — 'lxVy^+y. y^=(5) l/a--i/6"=a; — i/^. 7. Ifya + y^=-i/x+i/y,thmya — y'b — x/x — i/y'. (1) V7+V^= l/^+ l/^ (1)2 = (2) a + l/6^a; + y + 2l/^. .-. (3) a^x + y and l/b — 2-\/^. . • . (4) a — Vo = X — 2 Vxy-\- y. y/(4)==(5) ^ "we have Subtracting the (7) from the (6), and substituting l'''^a — |/f for -j/i — -\/y^, we have l/o^TS = ± a/^^±^^ T V^^^"^ ■ 347. Examples. 1. Extract the square root of 16 + 6 v7. C. A. 15. I'^O ALGEBRA. SOLUTION. ^16+6|/7 = Vi6 + y/252^ ,-. a = 16, 6 = 252, l/a^ _ j = ■/256 — 252 = 2. 2. Extract the square root of 21 + 8 1/5. Ans. ± 4 ± v's. 3. Extract the square root of 67 — 16 l/ 3. ^M. ± 8 T 1/3. 4. Extract the square root of 8 + 2 l/l5. ^w. ±1/3 + 1/3^ 5. Extract the square root of 12 — 2 l/35. ^w. ± T/7q= i/X 6. Extract the square root of 131 — 22 l/lO. Am. ± 11 =F l/m 7. Extract the square root of 30 + 12 1/6. Am. ±3\/2'±2VB. 8. Extract the square root of p'^q + pq^ + 2pg l/pj. ^ms. ± p Vq± q Vp. FRACTIOKAL AND NEGATIVE EXPONENTS. 348. Illustrations. (1) a-={y^. m (1)™ = (2) a»= {/a'".' FRACTIONAL EXPONENTS. 171 The numerator of the fractional exponent denotes the power to which the quantity is to be raised; and the denominator, the root to be extracted. It now remains to be proved that we can perform the processes of multiplication, division, involution, and evolu- tion of radical quantities, if we substitute for the radical sign and index the equivalent fractional exponent. lai. In multiplication, we have 1. v^a™ X ycP^'y'cF^^. But 1/0^= a", and p^=ai, m _p mg+np and a'' X a" =a "" =1? a"'«^"''. 2. i/^X\— ==\/— = 1?/^^^^^^=^ But 1^^=: a", and \ — =: — = cr^, and a-:rXaQ=a •>« =",V«^^'- « I 1 . . . / 1 »^ / 1 '^/ „-(mq+np) But „ /T" 1 _™ , ,n~ 1 _p_ " — :r= — =: a ji , and ^ — = — ^a q, \l„n. m^ ' V«P P and a-¥X a~r=a ~S9^= i/a (™5+«p)_ 2d. Similar results can be obtained for division. 172 ALGEBRA. Sd. In involution, we have m flip and (a" )p ^ a » = y^a"'". 4^ (i/^)*".(i/^)'==: l.TZ-^^l/^r 7. (i/^^)*''-*-2= (;/— l)*".(l/"— 1)2= 1 .— 1 = — 1. 8. (l/— 1)*"'-3= (l/— 1)4". (t/— 1)3 = i.^v^rr=_|/_i. By comparing the 5f/i, 6* — 6p2 -j- 1 -I- 4p (p2 _ 1) |/Z:i. EVOLUTION. 262. Theorems. 1. IS Va-\-\/— J = V a; + l/^, then Va — l/— 6 (1) l/a + t/=T= 1/^+ (1)2 = (2) a+y — 6=a; + 2l/— a^ — jr. .•. (3) a^x — y and l/ — 6 = 2 l/ — a;^. .-. (4) a — l/— 6 = a; — 2 V— xy — y. ^^=(5) Va—-\/-b = \/'x--l/- y- EVOLUTION. 179 2. Ifya+ V—b = x + V—y, i/ien v a — V^^^b = x — V — y. (1) '^a-\-V^-b = X + V^. (1)3= (2) a + l/^ = a;3 -|-3a;2i/^— 3ot/ — 2/l/y! .-. (3) a^x^ —2xy &aAV^^^(^^x'^ — y)Vy. .• . (4) a- — V — b^x^ — Zx^V — y — 3xy-\-y'V-—y. f/^= (5) ^a — V^= X — V^^y. 263. Problem. Ho extract the square root of a ±: V — b. Let us assume (1) v^-\- 1/^ =■»/« + 1/^. (2) t/7— l/^ = Va — i/ZTj. (1)2 =(3) a; — i/+2l/— a^ = a + l/^. .-. (4) a; — 2/=:a. Then, (l)X(2)-(5) x^y^Va^^b. ^IMp) = (6) i/^=^±\/ - + ^^ ^(4)^ = (7) V/^=W"--^^- 180 ALGEBRA. Hence, \ 2 ~ ^ 2 and \ 9 \ o ■ 264. Examples. 1. Extract the square root of 6 -|- 6 ]/ — 3. Ans. ± 3 H= l/^^. 2. Extract the square root of — 2V — 1. Ans. ± 1 + I — 1. 3. Extract the square root of 23 — 10 V — 2. Ans. ±5 + "l/— 2. 4. Extract the square root of 1 -)- 56 V — 3. Ans. ± 7 ± 4 v'^^. 5. Extract the square root of a^ — 2a6 + 2(a — b)V — b^. Ans. a — b-\-bV— 1. 365. Miscellaneous Examples in Radicals. 1. Simplify l/Sa^ — Qab + 3b^. Ans. (a — b)VK 2. Simplify l/45a^—60aS63+20a86«. Ans. (3a2 — 263) a l/5a. MISCELLANEOUS EXAMPLES. 181 3. Add V4a^h and VaHK Ans. a (2 + 6) i/^. 4. Subtract l/|| from V' fj. Ans. — ^ i/S. 5. Divide i/2 X f/S by t/4 X v^. Am. iq. 6. Raise js ^-^q to the 9"' power. Am. p^ p^- 7. Extract the 4"' root of ;)' v^. Am. p^ y% 8. Reduce Vftc + 26 l/fee — 6^ 4- v/jc — 26 v'ic - ft^. ^M. + 2 l/6c — 6^. 9. Reduce V^4a2 + 4l/a* — 6* — l/4ffi2_4l/^— 6*. Ans. ± 2 l/2a2 _ 2J2. 10. Reduce ^J^^^?^^/I. ^^. 4 #'3". 12. Reduce Jj (JZiJ^ I \ ^ \ 2l/2(f)* j Am. t/i(i 1/6 + i/"2l). 13. Reduce l/l6 + 30l/^^ + 1^16 — 30i/— 1. J.ns. ± 10. 182 ALGEBRA. INEQUATIONS. 266. Definitions. 1. An inequation is the expression of the inequality of two quantities. Thus, 7 > 5 and 3 < 4. 2. Two inequations subsist in the same sense when the greater is at the left in both, or at the right in both. Thus, 13 > 9 and 10 > 8 ; also, 4 < 6 and 3 < 5. 267. Propositions. 1. If the same quantity be added to both members of an ine- quation, the resulting inequation will subsist in the same sense. Thus, 5 > 3 and 5 + 2 > 3 + 2, or 7 > 5. 2. j^ the same quantity be subtracted from, boA members of an inequation, tJie resulting inequation will subsist in Hie same sense. Thus, 7 < 9 and 7 — 2 < 9 — 2, or 5 < 7. Also, 8 > 5 and 8 — 10 > 5 — 10 or — 2 > — 5. These propositions enable us to transpose. Thus, 2a;— 8>a; + 3, .-. 2a; — a; > 8 + 3, or a; > 11. 3. ^ both members of an inequation be mvMiplied by the same positive quantity, the resulting inequation wiU subsist in the same sense. INEQ UA TlOJSfS. 183 Thus, 8 > 5 and 8 X 3 > 5 X 3, or 24 > 15 ; and — 5< — 3and— 5X2<— 8X2, or— 10< — 5. This proposition enables us to clear an inequation of fractions. Thus, f a; > ^ a, . • . 3a; > 2a. 4. If both members of an inequation be divided by the same positive quantity, the resulting inequation will subsist in the same sense. Thus, 8>4and8^2>4H-2, or4>2; and — 15 < — 12 and — 15 -^ 3 < — 12 ^ 3, or — 5 < — 4. This proposition enables us to reduce an inequation by division. Thus, 3a; > 6a, . • . a; > 2a. 5. J^ both members of an inequation be multiplied or divided by the same negative quantity, the resulting inequation will subsist in a contrary sense. Thus, 8 > 6 and 8 H- — 2 < 6 -^ — 2, or - 4 < —3; and — 5 < 3 and — X — 3 > 3 X — 3, or 15 > — 9. 6. If the signs of both members of an inequation be changed, the sense will be changed ; for changing the signs is equivalent to multiplying or dividing by — 1. ^^^ ALGEBBA. 7. If two inequations whidi subsist in tiie same sense be added member to member, the resulting inequation wiU subsist in the same sense. Thus, 5 < 8 and — 7 < — 3, and 5 — 7 < 8 — 3, or - 2 < 5. 8. If one inequation be subtracted from anoHier which subsisti in the same sense, tlie resulting inequation may or may not subsist in the same sense. Thus, 8 > 5 and 3 > 2, and 8 — 3 > 5 — 2, or 5 > 3. Also, 8 > 5 and 7 > 2, and 8 — 7 < 5 — 2, or 1 < 3. This operation, then, is to be avoided unless the sense can be determined. 9. ^ both members of an inequation of positive quantities be raised to the same power, the resulting inequation will subsist in Hie same sense. Thus, 5 > 3 and 58 > 3^ or 125 > 27. 10. If both members of an inequation of negative quantities be raised to an even power, the resulting inequation wiU subsist in a contrary sense; but if both members be raised to an odd power, the resulting inequation will subsist in the same sense. Thus, — 3>— 5, (— 3)2<(— 5)2 or 9<25, and (— 3)3 > (— 5)^ or — 27 > — 125. 11. If tlie same root be extracted of both members of an ine- quation of positive quantities, or tlie same odd root of an inequa- INJiQ UA TWNS. 185 twn of negative quantities, the resulting inequation will subsist in tJie same sense. Thus, 27 > 8 and f^> 1^8, or 3 > 2 ; — 64 < — 27, and r — 64 < f~ 27, or — 4 < — 3. 268. Examples. 1. Given 2a; + -^ a; — 4 > 6, to find the limit of x. Ans. a; > 4. 2. Given 4 ", X X a — b~a-^~b~' to find the limits of x. Ans. X <^ab and a; >■ a. 3. Prove that a^ -\- b^ > 2ab, if a > 6 or a < 6. 4. Prove that a^ + b'' — 2ab, if a = b. 5. Prove that any positive fraction whose terms are une- qual, plus its reciprocal, is greater than 2. 6. Prove that x^ -\- y^ -\- z'^ ^ xy -\- xz -\- yz, if x, y, z are unequal. 7. Prove that a;^ + 1 > a;^ + •''^) if a; > 1 or a; < 1. 8. Prove that "^+^+" >^ and < ^-" , i{^ (p -\- q — r) (p -^ r — q) (q-^-r — p), if p, q, r are unequal. C. A. 16. 186 ALGEBRA. EQUATIONS OF THE SECOND DEGREE. 369. Definitions and Classification. 1. An equation of the second degree involving but one unknown quantity is an equation in which the greatest exponent of the unknown quantity is 2. 2. A complete equation of the second degree involves both the iirst and the second power of the unknown quan- tity. Thus, ax^ -\- bx:=c is a complete equation. 3. An incomplete equation of the second degree involves only the second power of the unknown quantity. Thus, ax^ = c is an incomplete equation. 4. An equation of the second degree involving two or more unknown quantities is an equation in which the greatest sura of the exponents of the unknown quantities in any of the terms is 2. Thus, xy -\- bz -\- yz = q. 5. Equations of the second degree are also called quad- ratic equations — pure quadratics, when incomplete; affected quadratics, when complete. INCOMPLETE EQUATIONS. 270. Form. Every incomplete equation of the second degree can, by clearing of fractions, transposing, reducing, and changing the signs, if necessary, be placed under the form INCOMPLETE EQUATIONS. 187 1st solution. (1) x^=q. V(l) = (2) x=±Vq. 2d solution. '(1) a;2=:g. .-. (2) ;r2— g=0. Or, (3) (x — Vq) {x + l/g) = 0. Equation (3) will be satisfied ( either x — V^ =0, . • . x = -\- \^ ; .or x-j-l/g^O, .'. x^ — l/^. VERIFICATION. {-\-Vqy = q and {—Vqy= q. 371. Rule. Reduce the equation to the form x^ ^ q, extract the square root of both members, giving to the second member the double "I 373. Examples. g 1. Given 5x^ —-x'^ = 224, to find x. Ana. x~-±l. 25 5 "2. Given 6a;2 — 2a;2 = -^ , to find x. .Atw- a; = ± g • 188 ALGEBRA. 3. Given 5x^ — Tx- =— 18, to find ;r. Ans. x =±3. 4. Given ax'' =^ b, to find .i-. Ans. x= ± -J- . 5. Given -^ x'^ — 4 = "V -"k^ + 9' ^ ^^^ ^■ n f Am.x^±JM±M. ^ adf — bde 6. Given — ; =p, to find x. X -\- n X — n 7. Given ' <^'-P'^' ^ ^^ to find X Ans. a; = ± yp' + q' 8. What two numbers are to each other as 5 to 7, and the diflSerence of whose squares is 216? Ans. 15 and 21. 9. What two numbers are to each other as m to n, and the sum of whose squares is s^ '! , ms , IIS Am. zt ~7=-= :^ and ± —. 1 m^ -|- w^ "l/m^ -|- n^ 10. Divide 21 into two such parts that the quotient of the less divided by the greater shall be to the quotient of the greater divided by the less as 9 to 16. Ans. 9 and 12. 11. Divide /i into two !=uch parts that the quotient of the first divided by the second shall be to the quotient of the second divided by the first as p to q. . nVp , nV q Ans. ---'^-- and 7= — ^■ 1 JJ + V'V Vp+Vq ANALOGOUS FOSMS. 189 12. A rectangular field whose sides are to each other as 3 to 5 contains 6 acres. What are the sides ? Ans. 24 rd. and 40 rd. 273. Analogous Forms. 1. (1) .i;" = fy. ,'nr) = (2) x = i/q. If n is even, the double sign ± must be used. 9 n^ 28x _ 63(a; + 18j ^- ^-^^ .T+18~" 4x (1) X 4a; (.T + 18) == (2) 112a;2 = 63 (x + 18) 2. (2) --7 = (3) 16x2 =9 (a; +18)2. l/(3)=:(4) 4a; = 3 (a; +18). Or, 4x = 3a; + 54. . • . a; = 54. 3. Given (.1; —a)'^—b, to find x. Am. a; = a ± Vb. 4. Given a;^ — 2aa; + a^ = 52, to find x. Ans. x = a ±b. 5. Given a;^ + 2aa; = 6^ _ ^2^ t^ flnd a;. Ans. a; = + & — a. 6. Given px^ +pq = 2fpx-\/q + qx"^, to find x. i^5L V'p-t- l/g Given Ix^ = 875, to find x. Ans. x==5. 190 ALGEBRA. 8. The volume of a rectangular box is 51840 cu. in. ; the length, breadth, and depth are to each other as 5, 3, and 2. Find these dimensions. Ans. 60, 36, and 24. 9. A and B, starting from different places, travel toward each other ; and, on meeting, it appears that A has traveled 30 miles more than B, and that it would take A 4 days to travel B's distance, and B 9 days to travel A's distance. How far has each traveled? Ans. B 60 miles, A 90 miles. COMPLETE EQUATIONS. 374. Forms. Every complete equation of the second degree can, by clearing of fractions, transposing, reducing, and changing the signs, if necessary, be placed under one of the following forms : (1) a;2 + 2px = q. (2) a;2 — 2px = q. (3) a;2 + 2pa; = — q. (4) x^ — 2px = — q. 276. Solution of the First Form. x^ + 2px = q. Adding p'' to both members, we have a;2 + '2px -\- p^ = q -\- pi . COMPLETE EQUATIONS. Extracting the square root, we have Transposing p, we have 191 a; = — p ±V q -\-p''. Adding the square of half the co-efficient of the first power of the unknown quantity to both members is called completing the square, since it renders the first member a perfect square. It may be illustrated geometrically thus : x^ -\- 2px = q px X2 px x'^ -\- 2px -{- p^ ^ q -{- p^ ■ px ?■' x^ px 276. Solution of the Second Form. a;^ — 2px = q. x^ — 2px-\-p^ =q -\- p'^. X — p^^ ± Vq +p^. -p ± Vq-i- p^. 277. Solution of the Third Form. x^ -j- 2px = — q. x^ -\- 2px -\-p^ = — q -\-p^. x -\-p^ ± V — q + p^. x = — p ±: V — q -\-p^. 192 ALGEBRA. 278. Solution of the Fourth Form. x^ — 2px = — q. x^ — 2px -\-p^ ^ — 9 + P'- X — p=^ ±V — g -\- P^- x=p ±1 l/ — q -f p^. By comparing the values of x in each of these forms with the equation from which they are deduced, we have the rule: 279. Rule. Reduce Hie equation to one of the four forms, then write the unknown quantity equal to half the co-efficient of the first power of the unknown quantity with its sign clianged, plus or minus the square root of the sum of the second member and Ike square of half the co-efficient of the first power of the unknown quantity. 280. Examples. 1. Given x'' -\- 6x= 55, to find x. SOLUTION IN FULL. SOLUTION BY THE EULE. a;2 + 6a; = 55. x^ + 6x = 55. a;2 + 6a; + 9 = 64. x = — 3 dz l/55 + 9. a;+3=±8. .1; = — 3 ± 8= 5 or — 11. r = — 3±8 = 5 or — 11. COMPLETE EQUATIONS. 193 2. Given x"- + 8x = 48, to find x. Ans. x^= 4 or — 12. 3. Given x^ — 14a; = 51, to find x. Ans. a; = 17 or — 3. 4. Given a;^ -|- 10a; = — 24, to find x. Ans. x^ — 4 or ^ — 6. 5. Given x^ — 10a; = — 21, to find x. Ans. a; ^ 7 or 3. By comparing the answers with the forms of the equa- tions from which they are deduced, we see that in the first form the values of x have unlike signs, the negative being numerically the greater ; that in the second form, the values of X have unlike signs, the positive being the greater ; that in the third form, the values of x have like signs, both being negative ; and that in the fourth form, the values of x have like signs, both being positive. We also see that in each form the sum of the values of a; is equal to the co-efficient of x with its sign changed, and that the product of these values is equal to the second member with its sign changed. The theory of these interesting facts will shortly be given. 6. Given 12a;2 — 24a; = 420, to find x. Ans. x = 7 or — 5. 7. Given x^ — 35a; = — 300, to find x. Ans. a; = 20 or 15. 8. Given 5x^ -j- 8a; = 21, to find x. C. A. 17. Ans. a; = I- or — 3. 194 ALGEBBA. 9. Given 7x^ — 5x = 522, to find x. Ans. x = 9 or — 8if . 10. Given 4x^ + 3a; = 115, to find a;. Ans. x = 5 or — 5f . 11. Given x^ — ^x = — |f , to find x. Ans. x^ ^ or -t. 12. Given 3.^2 + lOx = — ^, to find x. Ans. x = — I or — f . 13. Given 12a;2 — 187a; = — 481, to find x. Ans. a; = -^ or i/. a^ --I- b^ 14. Given x^ 4^ — a; = — 1, to find x. ab , a b Atis. x= -y- or — a 15. Given ax'' — (a^ — b^x^ ab, to find x. Ans. a; = a or a 16. Given ax'' — ax-\ t ^ 6*^ + bx, to find x. a — , a b Ans. X = r or a — b a — 6 17. Given px'' — pa; = -^ qx^ — gx, to find x. Ans. X = — ir — or p + g p + 1 g2 52 18. Given — j^ a;^ + 2na; = m' -\- w^, to find x. Ans. X = -i-5 y (6n ± l/a^m^+a^n^ — 6^m^"). b^ ^ a- ^ ^ COMPLETE EQUATIONS. 195 281. Hindoo Formula for Solving Quadratics. (1) ax^ ± 6a;= ± c. (1) . X 4a = (2) 4a%2 _,_ 4(^5^ ^ + ^^^ (3) 4(x2a;2 ± 4abx + 62=62 + 4ac. V''(SJ= (4) 2a» ± 6 = ± Vb^ ± 4ac. ' b ± 1/52 ±.4ac (5) X- 2a 282. Rule. 1. Reduce the equation to one of the four forms indicated by the general equation ax^ ± bx = ± c. 2. Write the unknown quantity equal to the co-efficient of the first power of the unknown quantity, witli its sign changed plus or minus the square root of the sum of the square of the co-effi- cient of the first power of the unknown quantity, and four times the co-efficient of the second power of the unknown quantity into the second member, and all divided by twice the co-efficient of the second power of the unknown quantity. 283. Examples. 1. Given ok 2 -\-bx = c, to find x. . ~b± l/62 _j_ 4ac Ans. x^ ■ • 2a 2. Given ax^ — bx^=c, to find x. . b ± Vb^ + 4ac Ans. X = fi — 2a 196 ALOEBRA. 3. Given ax''' -\-hx = — c, to find x. — 6 ± l/62_4ao Alia. X = -; • Aa 4. Given ax'' — hx= — c, to find x. b ± Vb^ — 4ac Ans. x = j; • 2a 5. Given 3a;2 + 7a; = 110, to find x. Am. a; ;:= 5 or — 7^. 6. Given 5x^ — 3a; = 224, to find a;. Ans. a; =^ 7 or — 6f . 7. Given 3a;2 + 25a; = — 8, to find x. Ans. x = — ^ or — 8. 8. Given x^ — 11a; =-■ — 18, to find x. Ans. a; ^ 9 or 2. 9. Given (a — b) x'' — (a + 6) a; =: 1 to find x. a — A a '■ Ans. X == r or a — h a — b 10. Given a; (a; — 1) d = 2s — 2aa;, to find x. ^^ d — 2a±V(d — 2ay+8ds 284. Special Formula. If the co-efiioient of the first power of the unknown quantity is even, the equation will take the form (1) aa;2 ± 26x = ± c. (1) X a = (2) a2a;2 ± 2a6a; = ± ac. (3) 02^.2 + 2abx + 62 = 62 ± ac. COMPLETE EQUATIONS. 197 ■l/(3) = (4) ax±b=± Vh"" ± ac. ... (5) ^ = ±A±y^b!±^, 285. Rule. 1. Reduce the equation to one of the four forms indicated by the general equation ax^ ± 2bx == ± c. 2. Write the unhnovm quantity equal to half the co-efficient of the first power of the unknown quantity with its sign changed, plus or minus the square root of the sum of the square of half the co-efficient of the first power of the unknown quantity, and the co-efficient of the second power of the unknown quantity into the second member, and all divided by the co-efficient of the second power of the unknown quantity. 286. Examples. 1. Given okc^ -J- 26a; = c, to find x. , — 6±l/P + ac Ans, X := ■ 2. Given ax'' — 26a; = c, to find x. . b± \/W+ac. Ans. X =^ ■ — a 3. Given ax^ -)- 26a; = — c, to find x. . — 6±l/62 — ac Ans. X = — ' — • • a 4. Given ax'' — 26a; =^ — c, to find x. , b ± l/62 — ac Ans. a; = - a 198 ALGEBRA. 5. Given Sa;^ + 8a; = 115, to find x. Am. « = 5 or — 7f . 6. Given 5x^ — 12a; = 224, to find x. Ans. x = 8 or — 5f . 7. Given Sx^ + 48a; == — 45, to find x. Ans. x^ — 1 or — 15. 8. Given 5a; ^ — 46a; = — 9, to find x. Ans. a; = 9 or ■^. , 9. Given (a + b) x^— 2(a—b)x = -^^ . to find x. CO ~Y~ . 2a 26 Ans. X = — : — - or a -\- b a -\- b 10. Given (p — q) x^ — 2jp l/g a; = — pq, to find x. , pVq±q Vp Am. X =^ ^ — ^ ^ — ^ ■ p — q 287. Problems. 1. What number is that which being subtracted from 10 ■will leave a remainder equal to 25 divided by that number? Ans. 5. 2. What are those two numbers whose difierence is 15, and the cube of the less is equal to 5 times their product divided by 4? Ans. 5 and 20. 3. A drover bought cattle for $1350, and sold all but 20 for $1000, and gained $10 a head on those sold. How many did he buy? Ans. 45. 4. Divide 48 into two such parts that their product may be equal to 70 times their difference. Ans. 20 and 28. COMPLETE EQUA TIONS. 199 5. Divide m into two such parts that their product may be equal to n times their diiFerence. A w — 2n.±T/m^-|-4w- , m-\-2n^'[/m^-'i-4n^ 6. A merchant bought a certain number of pieces of cloth for $375, and sold it at $18 a piece, and gained by the trade 5 times the cost of one piece. How many pieces did he buy? Am. 25. 7. A certain company wished to raise $10000 stock in equal shares ; but after 5 more men joined the company the shares were $100 less. How many men were in the company at first? Am. 20. 8. The base of a right-angled triangle + 5 is equal to the hypotenuse, and the perpendicular -|- 10 i^^qual to the hypotenuse. What are the sides? Am. 25, 20, 15. 288. Formulas for Special Equations. 1. "When the co-efficient of the first power of the unknown quantity is even, and the co-efficient of the second power is greater than unity, the equation is aafl ± 26a; = ± c . ' . x = • 2. When the co-efficient of the first power of the unknown quantity is even, and the co-efficient of the second power is unity, the equation is a;2 ± 26a; = ± c. . - . a; = q= 6 ± l/P~+c. 200 ALGEBRA. 3. When the co-efficient of the first power of the unknown quantity is odd, and the co-efficient of the second power is greater than unity, the equation is 2^7, ^ =fb ±Vb^ ±4ae ax^ ± ox= ± c. . . X = ■ • 4. When the co-efficient of the first power of the unknown :juantity is odd, and the co-efficient of tlie second power is unity, the equation is x' ± ox ^= ± c. . • . x = =^-^^ ^= 289, Examples. 1. Given 3x^ -\- 4x = 95, to find x. Am. 5 or — i^. 2. Given x^ — 6x^= 16, to find x. Ans. 8 or — 2. 3. Given 'Sx'' + llx = — 10, to find x. Ans. — I or — 5. 4. Given x'^ — 9a; = — 14, to find x. Ans. 7 or 2. 290. Analogous Forms. 1. Given x^" -\- 2pa;" = q, to find x. Since x'^" is the square of x", we have a;" = — p ±-\/q+p'^. .-. x=^\/ —p ±-^q-\-'p^. 2. (1) 3a; — 5l/^=12. ... (2) v^^ 5±l/25+ ^_ ^ 6 COMPLETE EQ UA TIONS. 20J 5 ±13 Or, (3) i/^=?-=^==3or-l^ 6 ^• (3)2 = (4) a; = 9orl-J. 3. (1) a;l + 8a;^=128. .-. (2) a;5 = — 4 + 1/16+128. Or, (8) a;4 = — 4±12 = 8 or — 16. f^(3) = (4) xi = 2 or (— 16)*. (4)* = (5) x = 16 or (— 16)t. 4. (1) (3a;2— 8a;)2— 6(3a;2— 8a;)=:7735. .-. (2) 3x2 — 8a; = 3 ± l/9~+7735r Or, (3) 3a;2 — 8a; === 3 ± 88 = 91 or —85. 1st. Let (4) 3x2— 8x = 91. ,.. 4 ± l/l6 + 273 ... (5) .= Or, (6) a;=i±^=7or-4|. 2d. Let (7) 3a;2 — 8x = — 85. ,o. 4 ± 1^16 — 255 (8) x== Or, (9) . = i±J^^. 202 ALGEBRA. (1) 3(2a;2— 7a;)— 5l/2a;2— 7a; + 10:=20. (2) 3(2a;2_7a; + 10)— 5l/2a;2— 7a; + 10=:50. (3) l/ 2.^-7.+10 :^^--^J^f + M. o (4) i/2a;2-7a;+10 = ^-=-^ = 5 or — ^. o o (4)2z=(5) 2x2 — 7a;+ 10 = 25 or ^. l8f. Let (6) 2a;2_7a; = 15. (7) .= 7±l/49 + 120. Or, (8) ^=.lj^ = 5or-l. 4 2 2ci. Let ( 9 ) 2a;2 — 7a; = — . ^ ^ 9 Or, (10) 18a;2 — 63a; = 10. .-..s ^ 63 ± t/3969 + 720 • • ^ ^ * 36 r> /1o^ 63 ± 3 V^52i 21 i l/521 Or, (12) a;= 3^ = j^" ' 6. ( 1 ) 3;=* — 4a;2 + 6a; = 55. ( 2 ) a;* — 4a;'' + 6a;2 — 55a; = 0. (3) (a;2 — 2a;)2 +2a;2— 55a; = 0. COMPLETE EQUATIONS. 203 To make the first member a perfect square, since its first term is (x^ — 2a;) ^, such a quantity must be added as will make its second term of the form n (x^ — 2x), and its third term ( k )• The same quantity must be added to the second member, which must be a perfect square whose first term is either x^, 4x'^, 9x'^. . . , which may be tried in succession till the proper one is found. By a few trials, (3) may be put under the form (4) (x.^-2xy+lKx^-2x)^ (^jy=9x^+Z3x+Qf^' l/(^-(5) x^-2x + ^ = ±(sx+^y 1st. Let (6) a;2 — 5a; = 0. . • . a; = 5 or 0. 2d. Let (7) a;2 + a; = — IL .-. x = — ^^^^ — ^■ 291. Examples. Find the values of x in the following examples. 1. a;* — 12a;2 = 64. Ans. ± 4 or ±2 V—1. 2. a;6 — 3a;' = 40. Ans. 2 or V — 5. 3. 5x + 24 Vx = 176. Ans. 16 or 77^. 4. a;3 + 5xi = 864. Ans. 9 or (— 32)^. 5. a;» — a;«=2. Ans. 2" or (— 1)". 6. a; + 5 — Vx + 5 = 6. .-: ^ns. 4 or — 1. 204 ALOEBBA. 7. (3a;2 — 10a; + 5)2 — 8 (Sa;^ — lOx + 5) == — 12. , 5 ±1/28 o , J.ns. -^=^ , 3 or i- 8. (4a;2 — Sx) ^ — 8 (4a;2 — 3a; — 2) = 36. 3 ± V— 23 9. (3a; — 51/^+1)2 — 7(3a; — 5l/«) = — 5. ^„, 4 4 43±5V^ 10. a;8 — 8a;2 — 28a; + 80 = 0. Ans. 10, — 4, 2. 11. a;3 — 16a;2 + 20x + 112 = 0. Ans. 14, — 2, 4. 12. a;^ — 14a;3 + Sla;^ — 402a; = 448. . -,. 1 1±V— 127 Ans. 14, — 1, ■ • EECUERING EQUATIONS. 292. Beflnition. A recurring equation is an equation in which the co-effi- cients taken in a direct or reverse order are the same numerically. Thus, OKB* — 6a;2 -\- bx — a = 0. 293. Propositions. 1. If the corresponding co-efficients of a reoii/rring equaiion of an. odd degree have like signs, one root is — 1, and the equation RECURRING EQUATIONS. 205 i» divisible by the unknown quantity -(- 1 ; but if these co-effieients have unlike signs, one root -is -|- 1, and the equation is divisible by the unknown quantity — 1. Thus, — 1 is a root of ax^ + bx'^ -|- 6a; -j- a ^ 0, since — 1 substituted gives — a-\-b — 6 + a = 0. Also, + 1 is a root of ax^ — 6a;^ -{-bx — a ^ 0, since -|- 1 substituted gives a — 6 + 6 — -a^O. 2. IJ the corresponding co-effieients of a recurring equation of an even degree, whose middle term is wanting, have unlilce signs, one root is -\- 1 and another root is — 1, and the equa- tion is divisible by the square of the unknown quantity ■ — 1. Thus, +1 and — 1 are roots of ax* — bx*-\-bx — a = 0, since -|- 1 substituted gives a — 6 -(- 6 — as^O, and — 1 substituted gives a-\-b — 6 — a = 0. 294:. Examples. 1. (1) 3a;3 + bx^ -j- 5a; -f- 3 = 0. (2) {x + 1) (3a;2 + 2a; + 3) = 0. Prop. 1. Hence, r either a; -f 1 = 0, .•. x = — 1, lor 3a;2 + 2a; + 3 = 0, .-. a;^ ~^--^"~^ . (1) 5a;3 — 7a;2 + 7a; — 5 = 0. (2) (a; — 1) (5a;2 — 2a; + 5) = 0. Prop. 1. Hence, r either a; — 1^0, •. x = \, lor 5a;2 — 2a; + 5 = 0, .-. a;^ l±2v^=^ 206 ALGEBRA. 3., (1) 2a;* — 5x3 -f- 5a;— 2=: 0. (2) (a;2 — 1) (2a;2 — 5a; + 2) = 0. Prop. 2. either a;^ — 1^0, .•. a; = ± 1, Hence, 'or 2a;2 — 5a; + 2 = 0, .-. a; = 2or4. 4. (1) ax* + hx^ + ca;2 + 6a; -f a = 0. (1) -=- a;2 = (2) aa;2 + 6a; + c + - + 4 = 0. X x^ (3) aa;2 + 2a + -4 + *a; + - + c — 2a =. 0. x^ X (4) „(^ + iy+6(a;+l)=2«-e. .-. (5) !C + - = rP— a; 2a Or, (6) 2aa;2 + (6 =F Vb^+ 8a^—iac) x^ — 2a. — 6dzi/P+8a^— 4ac±v"26^— 8a^— 4ac=F26-|/6^H-8a^-4ac 4a 5. (1) 2a;» — 3a;* + 4a;3 — 4a;2 + 3a; — 2 = 0. (2) (a; — 1) (2a;* — a;^ + 3a;2 — a; + 2) = 0. (3) a; — 1=0, .-. x = l. (4) 2a;*— a;3 + 3a;2— a; + 2 = 0. (4) -4- a;2 = (5) 2x^ — x + 3 — - + -^ = 0. BECURRINO EQUATIONS. 207 6. (6 (7 (8 (9 (lo: (1 (2 (3 (4 (4) H- a;2 = ( 5 (6 (7 ( ( (lo; x' X K'+O "('+:)-'• ,1 1±3 , 1 x-\--^= — - — = 1 or — - • X 4 2 2a;2 + a; = — 2, -1±1/— 15 a;6 _ 2x5 _^ 3^.4 _ 3a;2 _|_ 2a; _ 1 = 0. (a;2 — 1) (a;*— 2a;3 + 4a;2— 2a; + 1) = 0. a;2 — 1 = 0, -•. a;=±l. a;* — 2a;3 + 4a;2 — 2a; + 1 = 0. a;2 — 2a; + 4 — -+ A = 0. a; a;-* a;2 + 2 + i-2a;-?: a;^ X 2. (. + l)-2(. + ^) = -2. : + - = l±l/=T a;2 — (1 ± l/— 1) a; = — 1. 1 ± V^^± ^— 4±2l 208 ALGEBRA. 7. 4x^ +5x^ + 5x + i^ 0. Ans. — 1, B. 5a;S — Ux'^ + 12a; — 5 = 0. A 1 7 ± V — 51 Ans. 1, 10 9. 6a;* — 7x3 _^ 7.^_6==0. ^ws. ± 1, ~ ^^~ • 10. 3a;''' — 4x* f 5x^ — 5a;2 + 4a; — 3 = 0. lil'^ — l±l/^^ ^ns. 1, 2 3 11. 5a;5 + 6a;* + 7x^ + 7a;2 + 6x + 5 = 0. , . 2 ±l/— 21 — 1 ± V— 3 ^^3.-1, ^ , 2 12. 2x6 — 3a;5 + 4x4 — 4x2 + 3x — 2 = 0. Ans. ±1, 3±T/-7±T/-62±6V^=T, 8 BINOMIAL EQUATIONS. 295. Definition. A binomial equation is an equation of two terms of the form x" = a. 396. Examples. 1. (1) x3=a3. (2) x^~a^= 0. (3) (x — a) (x^ + ax + a^) = 0. (4) X — a = 0, . • . X = a. (5) x2 + ax + a2 = 0, .-. x = ~"-"^~^ . 2. (1 (2 (3 (4 (5 3. (1 (2 .-. (3 (3)-a5 = (4 (5 (6 (7 (7) ^2/^ -(8 (9 (lo; (11 (12 (13: (14: C. A. 18 BINOMIAL EQUATIONS. 209 a;* — a* = 0. (a;2 + a2) (a; + a) (a; — a) = 0. X — a ^ 0, . • . x^a. a;-|-a = 0. .". x = — a. a;2-|-a2 = 0, .-. x=±a\/—l. Let X = ay, then x^ = a^y^. a^y^ — a* ^ 0. 2/5_i==o. (2/-l)(2/*+2/'+2/'+2/ + l)=0. 2/-l = 0, .-. y=l. y* + y^ + y^+y + l-=0. y2+y + l + l. + l = 0. 2/^ + 2 + ^+2/ + ^ = !. 1 — ldzl/5 2y2 ^-(1 + -1/5)2/ = -2. ^ = 1 zt 1/5 ± l/_10q:2T/5 :«, a( — 1 ±l/5d=l/— 10q:2l/5' 210 ALOESBA. 4. (1) *«— a6=0. (2) {x^ + a») (a;3 — a^) = 0. (3) (a;+a)(a;2— a.-(; + a2)(a;— a)(a;2+oa; + a2) = 0. (4) a; + a=:0, .'. a;:= — a. (5) a;2-a* + a2 = 0, .-. ^^ g±«>'-3 . (6) a; — a = 0, . • . a; = a. (7) a;2 + aa; + a2 = 0, .-. x = — a±al/— 3 5. a!3=8. 6. a;* = 81. 7. x-5 = 32. ^ns. 2, 8. a;» = 15625. 9. a;«=a8_ ^rw. a; = 2, — 1±1/ — 3. ^ns. ±3, ±8l/^^. 1 ± l/5 d= V^— 10 + 2 l/l -4ns. ±5±5l/^^ ^?i8. ±a, ± d^/ — 1, it al^ — 1, ±ay — l/ — 1. 10. a;i = 1024. Ans. 2, 1 =fc l/5 ± l/— 10+2 1/5 -2, 1 dzT/5 ± 1^— 10 ± 2 1/5 EXAMPLES INVOLVING RADICALS. 211 298. Examples involving Radicals. 1. (1) Va^+x^—a = b. (2) Va'' + x^=a+h. (2)2 == (3") a^ -h a;2 = a2 _|_ 2ab + 6^. (4) a;2 = 2a6 + b"^. '^= (5) a; = ± T/2a6 + bK Multiplying both terms of the fraction by the numerator, we have (2) ^ r^ = n. V72)^C3) rn, + Vm-^-x^ '(2)^(3) — ^^^ - = ±Vn. Clearing of fractions and transposing, we have (4) vw^--x^ = ± a; vn — m. (4)2 = (5) m^ — a;2 = ma;2 q: 2ma;l/n + m^. (6) (\-\-n)x=±2mVn. 1m Vn 1 -{- n 212 ALGEBRA. 3. Given _"' = h, to find x. \ a^ -{- x'^ — X g (6 — 1) Ans. x^ ± 2 >/& , ^. Va + X — Vx J, X c J 4. Given — ^=^= — = b, to find x. Va -\- x-\-y X Ans. X = — ^ ; — ^ . 4o Vx — V X — 2> ^ pq^ Vlc + Vx-^p x—p' Ans.x^P-^^^^^'^-^. l+2g „ ^. a; + p + l'*^ + 2Ba; , ^ , 6. Given — ' ^ ■ — '— = q, to find x. x+p V^q — q"^ GENERAL DISCUSSION OF QUADRATICS. 299. Form of the Equation. a;2 -|- 2px = q. The co-efficient of x is denoted by 2p to avoid fractions ; but 2p is not necessarily even, since p may be a fraction. In this equation, 2p and q may be either -|- or — . 300. Propositions. 1. Evei-y equation of the second degree has two roots. Solving the equation x^ -{- 2px == q, we have x = — p ± Vq-\- p'^. GENERAL DISCUSSION OF QUADRATICS. 213 Separating the values of x, and denoting them respectively by a/ and »", we have x' = — p + Vq-\-p'^. a/'= — j)^ V q -\-p^. Let it be remembered that by the roots we mean the values of x, and not simply the radical parts. 2. An equation of x" = q. . UADBATICS. 216 Q. In the first form, the roots have unlike signs, and the negative root is numericaUy the greater. By Prop. 5, the product of the roots is equal to the second member with its sign changed; but in the first form, the second member is plus ; hence, the product of the roots is minus ; therefore, the roots have unlike signs. By Prop. 4, the sum of the roots is equal to the co- effieient of x with its sign changed ; but in the first form, this co-eflicient is plus ; hence, the sum of the roots is minus ; therefore, the greater numerically is minus. 7. In the second form, the roots have unlike signs, and the positive root is the greater. Let the student prove this proposition as above. 8. In the third form, the roots have like signs, and both are negative. By Prop. 5, the product of the roots is equal to the second member with its sign changed ; but in the third form, the second member is minus ; hence, the product of the roots is plus; therefore, the roots have like signs. By Prop. 4, the sum of the roots is equal to the co- efficient of X with its sign changed ; but in the third form, this co-efficient is plus ; hence, the sum of the roots is minus ; therefore, both roots are minus. 9. In the fourth form, the roots have like signs, and both are positive. Let the student prove this proposition as above. Let Props. 6, 7, 8, and 9 be verified by an inspection of the roots. 216 ALGEBRA. 301. Suppositions and Consequences. 1. p^ >q. Then, in all the forms, the roots are real, unequal, and rational or irrational, as is seen by inspecting the roots. 2. 23^ = q- Then, in the first and second forms, the roots are real, une- qual, and irrational; for, in the first form, x'^ — p-\-p 1/2, re'' = — p — p 1/2 ; in the second form, x' = p -j- p v2, xf' ^p — pv2. In the third and fourth forms, the roots are real, equal, and rational ; for, in the third form, x' = — p, x" = — p ; in the fourth form, x' =p, x" =p. 3. p^ < q. Then, in the first and second forms, the roots are real, unequal, and rational or irrational. In the third and fourth forms, the roots are imaginary, which indicates an impossi- bility ; for, since xf -\- xf' ^^ 2p and a/x" = q, numerically, let x' ^p -\-n, x" ^ p — n; then x'x" ^=p^ — n^ = g, or p^ :== q -\- n^ ; . • . p^ can not be less than q. 4. 3 = 0. Then, in the first and third forms, x' = 0, x" = — 2p ; and in the second and fourth forms, a/ = 2p, x" = 0. This ought to be the case ; for, since ufa^' =: 5 ^ 0, either xf = or a/' = ; but since a/ + x" = 2p, with its sign changed, and since either xf = Q ov x" ^ 0, the other must be equal to 2p with its sign changed. 5. p = 0. Then, in the first and second forms, xf =-{-yq, 0/-"= — l/g"; and in the third and fourth forms, x'=-)-l/ — q, x"^ — l/ — q. GENERAL DISCUSSION OF QUADRATICS. 217 The roots are equal with contrary signs, -which also is evi- dent from the supposition, ^ = ; then a;' + a;" ^ 2p = 0. Since the sum of the roots is 0, they cancel each other in adding ; they must therefore be equal with contrary signs. Indeed, the supposition reduces the equation to x"^ = ± q, an incomplete quadratic ; . ■ . a; = ± V± q. 6. p = and 5 = 0. Then, in all the forms, the equation becomes x^ = 0, and x' = 0, x" = 0. 7. Let us take the equation ax^ -\-bx = c. If a = 0, 9/ = ^, «/'="" ^^ but multiplying both terms of the value of x', which is — h + Vh^ + Aae 2a by / and reducing, we find 2c c .„ „ = — . II a ^ 0. 6 + l/P~+4ac b The infinite value of a/' indicates impossibility; for, since a ^ 0, the equation becomes bx = c, an equation of. the first degree, which can have but one root, a; = — • C. A. 19. 218 ALGEBRA. 302. Problem of the lights. Jb find on the line joining two lights tJie points which are equally iUuniinated by those lights. p//// P'" A P'' P B P' ( a =- the intensity ol'^light A at a unit's distance. Given -{ b ^ the intensity of light B at a unit's distance. [^ d = the distance between the lights. Let X = the distance from A to P, the point equally illuminated. Then d — a; := the distance from B to P. It will be observed that we have assumed that there is a point, equally illuminated, between the lights, as is evi- dently true, and have made the notation to correspond to this supposition. Accordingly, the first value of x, denoted by x', ought to give this point. If there is another point, equally illuminated, not between the lights, as P', the second value of X, denoted by a;", ought to give this point. Since the intensity of the same light, at different distances, varies inversely as the square of the distance, — = the intensity of the light A at P. — = — r= the intensity of the light B at P. Since P is the point equally illuminated, we have a^ b _ d V'a x^~ (d — xy •■ ^~Va±-\/l>' Separating these values of x, and denoting them by a;' and a;", we have d'i/a J „ dVa and x = -^=- l/tf + l/fi V a — 1/6 PROBLEM OF THE LIGHTS. 219 Let us now discuss these values under the following suppositions : 1. a > 6 and d > 0. < 1 and > ^, . • . .- ' — -p^ < d and > | d, Va-\-V% ' ' ]/a + \/b :> 1, .-. — = j=- ^ d, .-. a; =AP. Va — V^b ' ' ' Va — Vh Both points are nearer the weaker light; the first between the lights, the second not between the lights. 2. a < 6 and d > 0. ^^ . 0. l/a , dVa l/a + l/6 ' ' ' l/a + l/6 dl/a = id, .-. a/ = AP" l/a — V^'J ' V^— 1/6 The value oo indicates impossibility. a/'=oo. 220 ALGEBRA. 4. a = 6 and d^O. . = ^, and — = ;= = 0, . . X =0. l-a + y6 l/a + l/ft V a , dVa ,, = 00, and — = ;= ==- ! . • . x - V a — Vl> ' ] a — v1 " " The lights are together, since d = ; and the value a/ = indicates that the point at which the lights are placed is equally illuminated ; and the value x" = -^ indicates that any point of the line is equally illuminated by the lights. 5. a^b, or a < 6 and d ^ 0. "^ :>lor<0, and - ^^^ = 0, .-. ,,"= 0. But shall we conclude that the point at which the lights are placed is equally illuminated by the lights which, by hypothesis, are of unequal intensities? The supposition d ^= reduces the original equation — = ~ to x^ (d — a;) 2 ^ = ^ ' which is false under the hypothesis that a > 6 or a < 6, whatever be the value of x, even if a; ^ ; for then — and — become - and - , which are unequal in- finities. PROBLEM OF THE LIGHTS. 221 303. ProWem of the Lights Generalized. To find all of the points equally illuminated by two lights. 1. a> 6 and d > 0. Finding, as before, the equally illuminated points on the line joining the lights, we have ^ _ ^Ay^ _ AF, and ^' = -AJ^ -_= AP". l/a + l/6 l/a — 1/6 It is evident that there are other points, as the point P, not in the line joining the lights, which are equally illu- minated. Let us find the locus of these points. a AP^ 6 BP^' BP'_6 ' ■ ■ AP' a' BP : AP : : Vb ; y^/b = BP■, y l/'a = AP. BP AP :Va. Vb 'Va Let then Let PE be perpendicular to AE, and n = BE ; then ay'i ~ {d -\- ny = ¥¥.'', &nd by-' ~ n'' = YE'' . . ■ . ay^ — {d-\- ny -—by'' — n^ ; . • . y^ = — ^^ — ' . a — 7ri=r2 1.0 ■> bd^ + 2bdn , " a—b 222 ALGEBRA. EF^AE - AF = d + n— ^^ = -;i^+». l/a + V'6 Va + Vb EF'^AF'-AE =-^t^-(d+n)=^£^-«. Va—Vh Va—Vb \l/a + l/6 /Vl/a-l/6 / A/72 _l_ 9A/7« By reducing, we find EP' X EP" = ^ _7 - »»'■ .-. PE2=EP'XEP", which is a property of a circle. Hence, the equally illuminated points are in the circum- ference of a circle around the weaker light, which, however, is not at the center. ,The diameter of this circle is p,p„ d Va d\ a 2dvab ~-\/a—Vl> Va + Vb^a — b' which is the difference of the two values of x. As P may be any point in the circumference of this circle, it follows that all the points of this circumference are equally illumi- nated by the two lights. Let this circumference be revolved about P'P", as an axis. It will generate the surface of a sphere whose ,. , . 2d Vai) ^ . . , . diameter is j-; and since, in this revolution, the a — points in the circumference of the circle maintain their respective distances from the two lights, all the points of the circumference, throughout the revolution, will be equally illuminated. Hence, all the points in the surface of the generated sphere will be equally illuminated. PROBLEM OF THE LIGHTS. 223 2. a < 6 and d > 0. In this case, all the points in the surface of a sphere about A, whose diameter is -7 < are equally illuminated. 3. OS = 6 and d! > 0. The diameter, ;- > of the sphere increases as a and b a — approach an equality ; and, if a = b, the diameter becomes — - — r= 00, which indicates that the sphere is no longer possible. But, in this case, y V a ^ yv b, or AP =^ BP, or the triangle APB becomes isosceles, and P is any point in the perpendicular to AB at its middle point. Let this perpendicular be infinitely extended and then revolved about AB ; it will generate a circle of infinite radius perpendicular to AB at its middle point, and all the points in the plane of this circle will be equally illu- minated. 4. a = 6 and d^O. On this supposition, the lights are of equal intensities and situated at the same point. Let P be any point in space whose distance from- the lights is x, an indeterminate quan- tity. The intensity of the light A at this point is ~, and of B, — ; but a = b; . • . -^ = -—■, hence, anv point in space x^ x^ X- is equally illuminated by the lights. 5. a > 6 or a < & and d = 0. On this supposition, the lights are of unequal intensities and are situated at the same point. Let P be any point 224 ALGEBRA. in space whose distance from the lights is x, an indetermi- nate quantity. The intensities of the lights A and B at this point are, respectively, — and — ; but a > 6 or a<^b; . • . ~^ > "^ I or — < -^ ; hence, no point in space is equally illuminated by the lights. TWO OR MOEE UNKNOWN QUANTITIES. 304. Examples. f (1) x + y=s.\ 1. Given -l V Required x and ij. (. (2) xy^p. ) (1)2 = (3) x^' + 2xy + y^ = s\ (3) — (2) X 4=(4) a;2 — 2x!/ + 2/2 = s2 _ 4j5_ l/(4)=(5) a; — y=±l/s2_4p_ (1) + (5) _ ,g ^^s±vZE^ 2 ^ ' 2 (l)-(5) __ s+y82_4p 2 '^ ^ ^ 2 f (1) .^ + ^=10. ) 2. Given < ^ Required a; and y. ( (2) a;3 + ys =, 280. ) (2)-(l)=:(3) x2-xi/ + 2/2=28. (1) 2 = (4) x2 + 2x2/ + 2/2 ^ 100. (4) -(3) = (5) 3x2/ = 72. (5)-H- 3 ==(6) a;2/ = 24. (3) -(6) = (7) x2- 2x2/ + 2/2 =4. TWO OR MORE VNKNOWJS qUANTITIES. 225 1/^)^(8) x-y=±2. (lii(8)^(9^ . = 6or4. (1) (8) ^ (10) 2/ = 4or6. 3. Given Required a; and y. (1) a; + y=8. (2) a;* + 2/* = 706. (1)4 = ( 3 ) x*-\-ix^y+ Qx'^y^-^^xy^-\-y*=4:m^. l/(4) = (5) a;2 _|_ ^ _)_ 2,2 ^ 4g_ (1)2 = (6) a;2 + 2a;!/ + 2/2 = 64. (6)-(5) = (7) a;2/ = 15. (5)-(7)X3 = (8) a;2 _ 2.^2, + 2/2 = 4. l/(8) = (9) a;-2/=±2. (1) + (9) = (10) .T = 5 or 3. (1) (9) ^(11-) 2/ = 3 or 5. 4. Griven Required a; and y. (1) a; + y = 5. ( 2 ) »;» + 2/5 = 275. (2) H- (1) = ( 3 ) a;* — a^y + a;22/2 — a^j/' + 2/* = 55. (1)4 = (4) a;''-f4a;3y + 6.T22/2+4a^3_|_y4=625. 226 ALGEBRA. C4) — (3) = ( 5 ) 5x^y + 5x'^y^ + bxy"^ = 570. (5) -f- 5a«/ = ( 6 ) a;2 + a;;/ + 3^2 ^ (1)2= (7) x^ + 2xy + y^=2b. 114 (7)-(6) = (8) a^ = 25- xy (9) a;22^2_25M/ = — 114. .1^^ 25 ± l/625 — 456 „ (10) xy = = 6. (7) — (10)X4 = (11) x^—2xy-\-y^ = l. l/(ir)^(12) x-y=±l. «^ = (14) , = 2 or 3. (1) a;i + 2/i = 5. ) 5. Given -J V Required x and y. Then, .(2) 2/1 + 2,1 = 35.) ( 3 ) Let m = a;7 and n = y^. r ( 4 ) m + n = 5. 1 T m = 3 or 2. 1(5) m3+«,8=35. 3 '" [n,=2or3. (6) a;5=:aor2. .-. a; = 243 or 32. (7) j,i = 2or 3. .-. y= 16 or 81. TWO OB. MORE VNKNOWN QUANTITIES. 227. C 1 ) a;2 + .-n/ + 2/2 = 133. 6. Given (2) a; -)- l/^ + 2/ = 19. (1)^(2) = C3) X — 1/^ + 2/=7. (5)2 =(6) xy = m. (l)-(6) X3 = (7) a;2 — 2a;2/ + 1/2 ^ 25. l/(7) = (8) a; — 2/ = ±5. ^^^ = (9) . = 9or4. (4) -(8) 2 7. Given = (10) y = 4 or 9. (1) .2 + 3., + 2,2 ==48. I jj^^^.^^, ( 2 ) 2a;2 + ajy + j,2 = 44. j a; rnd y. ( 3 ) Let 3/ = nx. ' (4) (l + 3«,+ 2ji2)a;2^48. .-. .2= 48 1 + 3ji + 2»i,2 ( 5 ) (2 + w + «2) .t2 = 44. 44 x^ =- 2 + W + ?l2 8. Given 228 .-. (6 ••• (7 • • (8 .-. (9 .-. (lo; .-. (11 (1 (2 (2) gives ( 3 (3)X2 = (4 (l) + (4) = (5 .-. (6 ••■ (7 (7) in (4) = ( 8 (l)-(8) = (9 l/79) = (lo; 2 ^ (7)-(10)_ (12: ALGEBRA. 48 44 1 + 3w + 2n.2 2 + n + n,2 10«,2 + 21n=13. —21 ±1/441 4- 520 1 ,. = 20^ = *'^'^-^ 44 = 16 or -5^ 2 + »i+n2 "• a; = it 4 or ± y 4^. — — 7 7 2/ z= jia; = ± 2 or T ^ v^f- x2-f 2/2=25. r . J-? ■ . „ ■ « -1^ I a; and y. Required xy—'-^x='-^y. 2xy-^(x + y-)=0. (x + yy-^(x + y} = 25. x + y=\'-±l/25 + W- x + y = 7or—^. 2x2/=24or— ^. (x-yy = lorH^. x—y = ±l or ±f V 73. — 25±5l/73 a; = 4, 3 or 2/ = 3, 4 or — 25=f5i/73 9. TWO OS, MORE UNKNOWN QUANTITIES. 229 ( 1 ) — f— = a. (2) -^^=6. ^ ^ a; + z (3) ^ = c. a; + y (4) (5) (6) xi/z a X -\- z 1 xyz b y + g ^ 1_ a;!/z c (4 ) + (5) + (6) ^ / 7 X x + y + z ^ ab + ae + bo (7)_(4) = (8) Kl/Z 2ffl6c 1 ab -\- ao- — be xy 2abe (11) xy (7)-(5) = (9) (12) xe 2a6e xy ab -]- ac — -be 1 ab -{-be — -ae az 2abc 2abc ab -^ be — ae (7) - (6) = (10) 1 _ ae -{- be — ab yz 2abe (13) yz 2abe ae -\-be — ab 4 (1 1) X (12) (13) 4 (11) X (13) (12) : (14) X (15) y -4 2ah+ys= 5103. a; + 2/ == 7. X* + 2/* = 337. x + y^8. a;5 + 2/' = 3368 a;* + 2/^ = 6. xi + ,/ =.= 72 x"^ +xy + y^ ^ 481. | a; + l/x2/'+2/ = 37. ) a;2 -1-2/2 + 3a; — 32/ = 40. ^ a;2/ = 15. ) a; = 5, Am. Ans. ;=±6. = ±4. y=± Ans. Ans. Ans. Ans. y = 3, C a; = 5 or 3. (2/ = 3 or 5. a; = 15 or 12. 2/ = 12 or 15. Cx^4 or 3. (^ 2/ = 3 or 4. a; ^ 5 or 3. 2/ = 3 or 5. a; = 256 or 16. 2/ ^ 16 or 256. a; = 16 or 9. 2/ = 9 or 16. ■5±l/85 5± V8i x+2/ + 2^12- 1 a; + 2/ — z ^ 2. a;2 -^2/^=»'- a; = 4 or 3. Ans. -j 2/ ^ 3 or 4. z=5. 232 ALGEBRA. 22. 23. r 2a;= + «!/ + 32/2 = 64. 24. J.ns. 4?is. x=±2 01- ±||l/78, (.y = 4 ■ x-\-y 24 7 ■ Xl/Z X + z 4. . 2/ + 2 24 5 • 4 or =F if 1/78. a; = ± 4. ^ns. 3' = ±3. 2 :=±2. a;^ + 2/^ =A.2. x: s:: y.y — s±l/^ + 8^ VV^2- -2s(s ztlZ/i^ + s ^) 2 s±v/7i2 + s2 ■=FV/i2_ -2s (s Tl//i2 + s !) 25. .y- 2 iwa;^ ^ 80. wxz = 100. wi/2 = 200. ^ xyz = 40. Ana. r w = 10. a; =2. 2/ =4. z =5. 305, Problems. 1. Find two numbers whose sum multiplied by the greater will produce 40, and whose difference multiplied by the less will produce 6. Am. ± 3 and ± 5, or ± V^and ± 4 V^. RATIO. 233 2. The product of two numbers is 14, and the sum of their cubes is 351. Required the numbers. Ant. 2 and 7. 3. The product of two numbers is 15, and the sum of their fourth powers is 706. Required the numbers. Ans. 3 and 5. 4. Find three numbers whose sum is 12, the sum of whose squares is 50, and the product of the first and third plus 1 is equal to the square of the second. Ans. 5, 4, 3. 6. There are three rectangles, the length of each is twice its width, the sum of their lengths is 48, the sum of their areas is 400, and the area of the second is equal to the sum of the areas of the first and third. Find the area of each. Ans. 72, 200, 128. 6. If the sum of the squares of two numbers plus their product be divided by their sum, the quotient will be 14; and if the sum of their squares minus their product be divided by their difference, the quotient will be 18. Re- quired the numbers. Ans. 6 and 12. RATIO. 306. Definitions. 1. Ratio is the relation of two quantities of the same kind, expressed by the quotient obtained by dividing the first by the second. A ratio is therefore an abstract number. 2. The terms of a ratio are the quantities compared. The first term is the antecedent, the second the consequent. The antecedent and consequent together form a couplet. C. A. 20 234 ALGEBRA. 3, The sign of a ratio is the colon (:). A ratio i« expressed by placing the sign between the terms. Thus, 4 : 7 expresses the ratio of 4 to 7, and is read the ratio of 4 to 7. Let a denote the antecedent, c the consequent, and r the ratio. Then, by definition, we shall have a r = a: c = - . c 307. Classification of Ratios. 1. A simple ratio is a ratio of integral terms. 2. A complex ratio is a ratio of fractional terms. 3. A compound ratio is the product of two or more ratios. 4. A duplicate ratio is the square of a given ratio. 5. A triplicate ratio is the cube of a given ratio. 6. A suhduplicate ratio is the square root of a given ratio. 7. A subtriplicate ratio is the cube root of a given ratio. REDUCTION OP RATIOS. 308. Case I. To reduce simple ratios to their lowest terms. a a-hn , ,, , a : c = - = = (a -7- m) : (e-i-n). c c -i-n Hence, divide both terms by their g. c. d. MATIO. 235 309. Examples. 1. 120 : 440. Am. 3 : 11. 2. a2_62 . „2_2a6 + 62_ Ans.a + h:a—b. 3. a;3 — 7a; + 12 : a;2 — 9a; + 20. Ans. x — 3:x — 5. 310. Case II. ^0 reduce complex ratios to simple ratios. a a yc n . . a: c^^- = = aXn : c c yc 'II' Xn. Thus, I : 5J = 9 : 64, by multiplying both terms by 12, the 1. c. m. of the denominators 4 and 3. Hence, Multiply both terms by the I. e. m. of the denominators. 311. Examples. '•! . 11 ■ 6 * Ans. 9 : 22 2. a: :6 + V d Ans. ad -.bd-^- c. c d' Ans. ad : be. 312. Case III. To reduce compound ratios to simple ratios. ,v ^ ^^ f C Ct^* 7 7 (a : o) X (c : o) = T X -T = n = ffl" : od. Hence, multiply the several ratios together. 236 ALGEBRA. 313. Examples. 1. (7 : 8) X (8 : 14). Am. 1 : 2. 2. (3^ : 6i) X (4| : ^). Am. 21 : 47. 3. ( m H : m i- n) X (m + w : »r — ?t). \ m — n / ^?is. (m + n)'^ : (m — Ji)^. 314. General Problems. a 1. Given a and c, required /•. Ans. r = -- 2. Given c and r, required a. Ans. a = c X »"• 3. Given a and r, required c. ^ns. c = a r 315. Examples. 1. Given a = x^ — y^, c = a; — y, required r. Ans. r := a;^ -\- xy -\- y^. 2. Given c = x* — x^y^ -\- y*, r = a;^ + y^, required a. Ans. « = »;''-)- y'>. 3. Given a=»i^ — n^, r^(m.-|-«') (l/m+l/n), required i Ans. := Vm — vn. PROPORTION. 316. Definitions. 1, A proportion is an equality of ratios. Thus, (1) a : h :: c : d is a proportion, and is read a is to 6 as c is to d. PROPORTION. 237 By definition of ratio and proportion, (1) is equivalent to 2. The terms of a proportion are the quantities compared. The first and second terms form the first couplet ; the third and fourth, the second couplet. The first and fourth terms are the extremes ; the second and third, the means. The first and third are the antecedents ; the second and fourth, the consequents. The terms of each couplet, the antecedents, and the con- sequents are corresponding terms. 3. A proportion is taken by alternation when antecedent is compared with antecedent, and consequent with conse- quent. If a : 6 : : c : d, by alternation, a : c :: h : d. 4. A proportion is taken by inversion when the conse- quents are taken for antecedents, and the antecedents for consequents. If a : 6 : : c : d, by inversion, h : a :: d : c. 5. A proportion is taken by composition when the sum of the terms of each couplet is compared with either antece- dent or consequent. If a : 6 : : c : d, by composition, a -\- h : a : : c -\- d : c, ov a-\- h : h :: c -\- d : d. 6. A proportion is taken by division when the difference of the terms of each couplet is compared with either ante- cedent or consequent, li a : h : : c : d, by division, a — h : a :: c — d : c, or a — h : h :■. c — d : d. 7. A proportion is taken by composition and division when the sum of the terms of each couplet is compared with their difference. If a : 6 : : c : d, by composition and division, a-\-h : a — b :: e -{- d : o — d. 238 ALGEBRA. 8. Eq^uimultiples of two or more quantities are the prod- ucts of those quantities by a common multiplier. Thus, ma and mb are equimultiples of a and b. 9. Equi-submultiples of two or more quantities are the quotients of those quantities by a common divisor. Thus, — and — are equi-submultiples of a and b. mm 317. Propositions. 1, If the terms of one couplet are equal, ihe terms of the oilier couplet are also equal. If (X) a:b::c:d, then (2) l = t. If a = b, then 7 = 1; hence, - = 1 ; . • . c = d. a 2. if the antecedents are equal, the consequents are also equal, and conversely. If (V) a : h :: c : d, then (2) 1 = ^. If (3) a=:c, then (2) -(3) = (4) i = i. (4) Xbd^{b) d = b. If (6) b = d, then (2) X (6) = (7) a = c. PnOPORTION. 239 3. If one antecedent is greater or less than its conseqiient, the other antecedent is also greater or less than its consequent. If (1) a:b::c:d, then (2) ^ = |. If a > 6, then t > 1 ; hence, j > 1 ; . ' . c^ d. d If a c, then 6 > d ; otherwise, r > -; • b d If a < c, then 6 < (^ ; otherwise, ,-<-;• o d 5. im every proportion, ffie product of the extremes is equal to the product of the means. If (\) a:b::c:d, then (2) 1 = 1- (2) Xhd={S) ad = be. 6. ]ff the product of two quantities is equal to the product of two other quantities, the factors of either product may be made the extremes, and the factors of the other product the means of a proportion. 240 ALGEBRA. If (1) a X d = 6 X c, then (l)--6d=(2) ^=|- .-. (3) a : h :: c : d. 7. If Jour quantities are in fro/portion, tJiey will be in pro- portion by alternation. If (1) a : b :■■ e : d, then (2) a c b~d' (2)X-=(3) c a b c^d' .'. (4) a : c :: b : d. 8. ^ four quantities are in proportion, they wiM be in pro- portion by inversion. If (1) a : 6 : ; c : d, then (2) ^=|. Taking the reciprocal of each member of (2), we have (8) ^ = 1 a c . ■ . (4) b : a :: d : c. 9. If two proportions have a ratio of the one equal to a ratio of the other, the remaining terms are proportional. If (1) a : 6 : : c : d, then (2) 1=1- FBOPORTION. If (3) e:f::g:h, then (4) 1 — 9 f h If (5) a : b :: e : f, then C6) a e •■. (7) d h 241 . • . (8) c : d :: g : h. Cor. If the antecedents of two proportions are in propor- tion, the consequents are in proportion, and conversely. For, by alternation of (1) and (3), we have ( 9 ) a : e :: b : d, and (10) e : g :■./: h. If (11) a : a :: 6 : g, then (12) b : d::f:h. If (13) b : d::f:h, then (14) a : e :: e : g. 10. If four quantities are in proportion, they will be in pro- portion by composition. If (1) a : b :: : d, then (2) ad^be. Prop. 5. (3) ac = ac. (2) + (3) = (4) a(c + d) = c(a + 6). . • . (5) a-\-b : a :: c -[- d : c. Prop. 6. C. A. 21. 242 ALGEBRA. 11. If four quantities are in proportion, they will be in pro- portion by division. If (1) a : b :: c : d, then (2) ad — be. Prop. 5. (3) ac = ac. (3) — (2) =(4) a{c — d)^c(a — b). . ■ . (5) a — b : a :: G — d : c. Prop. 6. 12. If four quantities are in proportion, tliey will be in pro- portion by eomposition and division. If (1) a : b :: G : d, then (2) a -\- b : a :: e -\- d : c. Prop. 10. And (3) a — b:a::e — d : c. Prop. 11. (4) a-\-b : c-\-d :: a — b : c — d. Prop. 9, Cor. (5) a-\- b : a — b :: e-\- d : c — d. Prop. 7. 18. Ikvo quantities liave the same ratio as their equimultiples or equi-submultiples. (1) - = — ^, .-. a : b :: ma : mb. ^ ^ mb am a b (2) r = 7- . • ■ a:b::-:-. b b mm 14. If equimultiples or equi-submultiples be taken of all the terms of a proportion, or of any two corresponding terms, the resulting quantities will be in proportion. PBOPOMTION. If ( 1 ) a : b :•■ c : d, then m 1=1- 243 Hence, ( 3 ) — j- = — ; , . • . ma : mb :: me : md. ntb md (4) m m T~d' • m m abed m m" m m (5) 7na c mb d' ' a • . ma : mb : : c : d. (6) m c m a b , . — : — ;: c : a. m m (7) a mc b~md' ■ e • . a : b :: mc : md. (8) a m b~d' ■' m , e d . a : b :: — : — m m (9) ma me b ~ d' . • . ma : b : : mc : d. (10) a c m m b — d' ■ . ^:b::^:d. m m (11) a e mb md ' . ■ . a : mh :: c: md. (12) a c m m b d a ■.-:■. c: -■ m m 244 ALGEBRA. (13) !li^=?^, .-. ma:mh::nc:nd. mb lid a c (14) ~b H ' ' ' m ' m" n ' n m n (15) ma^me^ .-. ma : nb :: mo : nb. lib lib a c (16) m m . a . b ,, c , d b b ' ' ' m n m n n n 15. If tivo quantities be increased or diminislied by quantities having the same ratio, the resulting quantities will have Hie same ratio. If ( 1 ) a : 6 : : c : d, then ( 2 ) ad = be. ( 3 ) ab^ab. (3) ±(2) = (4) a(b±d)=b(a±c). .-. (5) a : b :: a±:c : b ±d. Cor. 1. If two quantities be increased or diminished by their equimultiples or equi-submultiples, the resulting quan- tities will have the same ratio. Cor. 2. If the antecedents or consequents be increased or diminished by quantities having the same ratio, the results wiU have the same ratio as the consequents or antecedents. PBOFORTION. 245 16. In a continued •proportion, Hie sum of the antecedents is to the sum of the consequents as any antecedent is to its conse- quent. (1) a ; b :: c : d :: e : f :: g : h. (2) ab = ah. (3) ad=hc. (4) of = he. (5) ah = bg. Taking the sum of (2), (3), (4), (5), . . . , we have (6) a(6 + <^+/+/i+...)=H« + c + e + ?-f...)- .-. (7) a + c + e + fif+...:6 + d+/+/i+...::a:&. 17. 27ie products of the corresponding terms of two or more proportions are in proportion. If (1) a : h :: G : d, then (4) r = %- b d If (2) e : f :: g : h, then (5) 1 = |-. If (3) i : j :: k : I, then (6) \ = ^ . J '' Taking the product of (4), (5), (6), . . . , we have aei... _ cgk... ^^ bfj... dhl...' .". (8) aei... : bfj... :: cgk... : dhl... Cor. If a = 6 = i=^ ... , 6 =/=-_/=;.,. , c = g ^=k=^ ... , d^h^l = ... , then (8) becomes (9) a" : h" :: c" : d'\ 246 ALOEBBA. 18. r/ie quotients of the corresponding terms of two propor- tions are in proportion. If (1) a : b :: e : d, then (3) ^ = ^- If (2) e -.f :: g : 7i, then (4) i = |. a c p n abed (3) -. (4) == (5) ^ = |. .-. (6) e-f''g'h- 7 /^ 19. itie powers or roots of the terms of a proportion are in proportion. If (1) a : 6 :: c : d, then (2) «^|. (2)" =(3) |, = |,. .-. (4) rt" : 6" :: c" : d". .-. (6) f a : iyS :: i/S : fZ 318. Examples in Proportion. 1. Given o^ — 6^ : a-\-b : : (a — 6)^ : », required x. Ans. x = a — 6. 2. If a : 6 : : 6 : c, prove that a : c :: a^ : b^. 3. If a : b :: b : e :: c ■■ d, prove that a : d : : a^ : b^. VARIATION. 247 4. What number added to each of the numbers 2, 4, 5, 8 will make the results proportional? Ans. 4. 5. What quantity added to each of the quantities m, n, p, q will make the results proportional ? Ans. '^^-^g m -{- q — n ■ — p 6. It a : b : : c : d, prove that there is no quantity which added to each will make the results proportional. 7. Given r^^ ''' = ''■ ] ««^--<^ ( (2) a;3+ 2/' : a; ■— y^ : : 91 : 37. ) ^ ^^^ V- Ans. a; = 8, y = 6. 8. Given f^x + a : ^x — a :: m : n, required x. Ans.x = ''-^±^. OT,d — n' 9. Find two quantities whose sum, difference, and prod- uct are proportional to m, n, and p. Ans. — i— and — — — m — n m-\- n 10. The product of two numbers is 15, and the cube of their sum is to the sum of their cubes as 64 is to 19. Required the numbers. Ans. 5 and 3. VARIATION. 319. Definitions. 1. One quantity varies directly as another, when it in- creases or decreases in the same ratio as the other. Thus, the space described by a body in uniform motion varies directly as the time, which is thus expressed, s oc t. 218 ALGEBRA. 2. One quantity varies inversely as another when it in- creases in the same ratio as the otlier decreases, or decreases in the same ratio as the other increases. Thus, in uniform motion, if the space is constant, the time varies inversely as the rate, which is thus expressed, to.'. r 3. One quantity varies as two others jointly, when the first increases or decreases in the same ratio as the product of the second and third. Thus, the space described by a body in motion varies as the rate and time jointly, which is thus expressed, s oc rt. 4. One quantity varies directly as a second quantity, and inversely as a third, when it increases or decreases in the same ratio as the quotient of the second by the third. Thus, if a body is in motion, the time varies directly as the space and inversely as the rate, which is thus expressed, toe'-, ■r 320. Propositions. 1. If one quantity varies as another, the first is equal to tlie second, multiplied by a constant. Let X oc y, and let .-);' and y', :;;" and y", . . . , be corre- sponding states of X and y ; that is, if x becomes x', y becomes y', . ■ . Then, by definition, we shall have (1) x: x' :: y : y' . (2) x:x"::y:f. VARIATION. 249 .-. (3) x' : y' :: x" : y". Art. 317, Prop. 9, Cor. Therefore, the ratio of the corresponding states of x and y is constant. Let this constant be denoted by m ; then, (5) m = ^^ = f xl (1) gives (6) x = - y; then (7) X = my. 2. if one variable quantity is equal to another multiplied by a constant, the first varies as the second. Let X and y be two variable quantities, and x^ and y' any corresponding states of x and y. If (1) X = my, then (2) a/ =^ my'. - • . (S) - = m, and (4) -7 = m. 2/ i' •■• (5) y = ^' 0^' (6) x:y::xf:if; or, (7) X : x' r. y : y'. .' . (8) X oc y. 3. if o«e quantity varies as a second, and tlie second as a third, the first varies cts the third. If (1) X oc y, then (2) x = my. If (3) y oc z, then (4) y^nz. 250 ALGEBRA. Substituting the value of xj given in (4) in (2), we have (5) X = mnz. But since m and n are constant, mn is constant, . • . xoc z. Prop. 2. 4. If eaefi of two quantities varies as a third, thdr sum, their difference, or their mean proportional varies as tlie third. If (1) X (X. z, then (2) x = mz. If (3) y -1. ~2/ (2) .':a/':a.l, .-. (4) z (5) X z'y a/V y „ == 'i ; or a; = ,■% a;oc " 2/2'' 2/ VARIATION. 253 321. Examples. 1. If 2/ = J) + 9, and p oc x, and q ocx^, and when x = 2, y = 16, and when a; = 3, y = 33, what is the equation be- tween X and y? SOLUTION. Since p oc x, ^J = ww:, and since q oc x'', g = «x^. .•. (1) y = mx-{-nx''. Substituting in (1) first 2 for x and 16 for y, then 3 for x and 33 for y, and transposing, we have (2) 2?a+4n, = 16. (3) Sm + dn = 33. Solving these equations, we find m = 2 and n = 3. Sub- stituting the values of m and n in (1), we have (4) y = 2a; + 3a;2. 2. If 2/ oc X, and when a; — - 2, y ^ 4, what is the equation between x and y ? Ans. y = 2x. 3. If a; oc y, and when a; = 12, y^S, what is the value of X when j/ = 5 ? ^ms. x = 20. 4. If X varies as the sum of two quantities, one of which varies as y and the other as y^, and when y = S, a; ^63, and when y = ^, a; = 140, what is the equation between x and yl Ans. x = ^y -\- 2y^. 5. If X varies as y and s jointly, and a; = 2a when y^l and 2^2, what is the equation between x, y, and z ? Ans. X = ayz. 254 ALiUHjHUA.. 6. If s oc <2 when / is constant, and as / when t is con- stant, and when < ^ 1, /= 2s, what is the equation between s, /, and t ? Am. s = ift^. 7. If X varies as the sum of two quantities, one of which varies directly as y and the other inversely as y, and when y = 1, a; = 4, and when y = 2, x= 5, what is the equation between a; and v? a « , 2 " Ans. x = 2y-}-~. y 8. If a; oc 2/" and y oc 2", and when a; == a, s^b, what is the equation between x and z? . « mn 6""' HARMONICAL PROPORTION. 322. Definitions. 1. Three quantities are in harmonical proportion when the first is to the third as the first minus the second is to the second minus the third. Thus, if a, h, c are in har- monical proportion, we shall have a : c : ; a — b : b — c. 2, Four quantities are in harmonical proportion when the first is to the fourth as the first minus the second is to the third minus the fourth. Thus, if a, b, e, d are in har- monical proportion, we shall have a : d :: a — b : c — d. 323. Problems. 1. Find the first term of the harmonical proportion whose second term is b and third c. . be Ans. . 2c — 6 ARITHMETICAL PROGRESSION. 255 2. Find the second term of the harmonica! proportion whose first term is a and third c. - '2ae a + e 3. Find the third term of the harmonical proportion whose first term is a and second b. . ab 2a— 6 4. Find the first term of the harmonical proportion whose second term is 6, third c, and fourth d. , bd ^' 2d — c' 5. Find the second term of the harmonical proportion whose first term is a, third c, and fourth d. . 2ad — axi Ans. , d 6. Find the third term of the harmonical proportion whose first term is a, second 6, and fourth d. , lad — bd Ans. • a 7. Find the fourth term of the harmonical proportion whose first term is a, second b, and third c. Ans. 2a- ARITHMETICAL PROGRESSION. 324. Definition. An arithmetical progression is a series in which the difference of the consecutive terms is constant. Thus, 3, 5, 7, . . . , is an arithmetical progression of which the common difference is 2, since 5 — 3 = 2 and 7 — 5 = 2...; and a, a -\- d, a -\- 2d, . . . , is an arithmetical progression of which the common difference is d. 256 ALGEBRA. 325. Elements and Notation. 1. The first term, a. 2. The common difference, d. 3. The number of terms, n. 4. The last term, I. 5. The sum of the terms, s. Thus, 5, 8, 11, 14, 17, 20 is an arithmetical progression in which a = 6, d ^ 3, ?i = 6, Z ^ 20, s ;= 75. I denotes the last term considered, but the progression may be re- garded as extending indefinitely. 336. Classification. 1. Increasing, d > 0, as 3, 5, 7, . . . 2. Decreasing, d < 0, as 7, 5, 3, . . . 327. Problems. 1. Given a, d, and n, required I and s. 1st. To find I. iat. 2d. M. ith. lUll. a, a ±d, a± 2d, a±3d, . . . (i ± (n — V) d. .-. (1) l^a±in — l)d. 2d. To find s. (2) 8 = a + (a ± d) + (a it 2d) + (a ± 3d) + . . . + /.. (3) 8 = Z + (Z =P d) + (« =F 2d) + (2 + 3d) + . . . + a. ARITHMETICAL PROGRESSION. 257 (2) + (3)= (4) 2s = (a + + (« + 0+-.. + (« + 0- Or (5) 2s = n(a + l). (6) s = \nia^l). (1) in (6) = (7) s = in [2a ± (ra — 1) d]. Sch. When the double sign ± or + is used, the upper sign will in general apply when the progression is increasing, and the lower sign when the progression is decreasing. Thus, if a = 5, d = 3, m = 10, then, by (1) and (7), Z = 5 + 9 X 3 = 32, and s = 5 (10 + 9 X 3) = 185. Formulas (1), (6), and (7) are used in solving the fol- lowing problems : 2. Given a, d, and I, required n and s. I — a Ans. ^ +1. (a + l)[_d±(l — a)'] 2d 3. Given a, d, and s, required n and 7. Ans. d — 2a± V'8ds+(2a — d)^ 2d I = — d±V8ds+(2a — dy 4. Given a, n, and I, required d and s. Ans. ' C. A. 22. I — a d=± r- n — 1 s = ■|n(a + I). 258 ALGEBRA. 6. Given a, n, and s, required I and d. 2s a. n Am. n (n — 1)' 6. Given a, I, and s, required n and d. 2s w.= - 'a-\-l Ana. (l+a)(l-a-) 28— (a + /) ■ 7. Given d, n, and I, required a and «. ' a = Z =F (n — 1) f^- s=^n[2i=F(»i— i)cri. 8. Given d, n, and s, required a and Z. 28 q: n (to — 1) «Z I tt — ■ ' Am. 2n 2B±n(n — l}d 2n 9. Given d, I, and s, required n and a. 2i + d ± l/(2i + c?)2 /*= 25 d^Vi2i + dy — i 2 ~ 10. Given n, I, and s, required a and (i. 2« — In a = ■ n Ang, 2 (In — s) d = n (n — 1) ARITHMETICAL PB0GBES8I0N. 259 338. Application. The answers of the above general problems are formulas which may be applied in solving numerical examples. 1. Given a = 5, d = 3, and Ji=10, required I and s. Am. 1 = ^2, 8:^-185. 2. Given a = 5, rf = 4, and I = 201, required n and s. A71S. n = 50, s ^ 5150. 3. Given a = 4, d = 3, and 8 =; 714, required n and I. Alls, n == 21, Z = 64. 4. Given a= 5, J = 201, and n = 50, required d and s. ^7is. tZ = 4, s ^ 5150. 5. Given a = 7, n := 10, and s = 205, required I and d. Ans. l = S4, d = d. 6. Given a = 10, I ^ 310, and s= 16160, required n and d. ^m. n = 101, d = 3. 7. Given d = 5, w = 100, and I = 498, required a and s. Ans. a = '6, 8 = 25050. 8. Given cZ = 10, w = 1000, and s = 10003000, required a and I. Ans. a = 5008, I =- 14998. 9. Given d == 5, i = 104, and « = 1134, required n and a. Ans. n ^21, a = 4. 10. Given w = 12, 1= 40, and s = 282, required a and d. Ans. a:=7, d = 3. 260 ALGEBRA. 329. Problem. To insert any number of arithmetical means between any two quantities. The means together with the extremes form an arith- metical progression; and if we knew the common differ- ence, the progression could be written out. The number of terms equals the number of means plus 2. Let m = the number of means, then n =m -\- 2. But according to problem 4, , I — a d = zr. n — 1 Substituting the value of n, we have , I — a m-\-l 1. Insert 8 means between 3 and 30. <50 ^ d^ =3; .-. 3,6,9,12,15,18,21,24,27,30. y 2. Insert 12 means between 12 and 77. 3. Insert 10 means between 82 and 5. 4. Insert 9 means between the consecutive terms of the progression 2, 5, 8, and write out the whole progression. 330. Miscellaneous Problems. In certain problems in arithmetical progression, if the number of terms is odd, it is convenient to represent the middle term by x, and the common difference by y; but ARITHMETICAL PR0ORE8SI0N. 261 if the number of terms is even, to represent the two middle terms by x — y and x -\- y, and the common difference by 2y, as follows : Three terms, x — y, x, x -\- y. Four terms, x — By, x — y, x-\- y, x-]- 3y. Five terms, x — 2^/, a; — y, x, x -\- y, x -\- 2y. 1. The sum of three numbers in arithmetical progression is 24, and the sum of their squares 200 ; required the numbers. Ans. 6, 8, 10. 2. The sum of four numbers in arithmetical progression is 16, and the sum of their squares is 84 ; required the numbers. .4ns. 1, 3, 5, 7. 3. The sum of five numbers in arithmetical progression is 30, and the sum of their squares is 220 ; required the numbers. Ans. 2, 4, 6, 8, 10. 4. The sum of six numbers in arithmetical progression is 86, and the sum of their squares is 286 ; reqfuired the numbers. Ans. 1, 3, 5, 7, 9, 11. 5. The product of four numbers in arithmetical pro- gression is 384, the sum of their squares is 120 ; required the numbers. Am. 2, 4, 6, 8. 6. In the series of odd numbers, find the n"' term and the .sum of n terms. Ans. l^2n — 1, 8 = n^. 7. How many terms of the series 16, 14, 12, etc., must be taken in order that the sum be 60? Ans. 5 or 12. 262 ALGEBRA. GEOMETRICAL PROGRESSION. 331. Definition. A geometrical progression is a series in which the ratio of the consecutive terms is constant. Thus, 2, 6, 18, ... , is a geometrical progression, of which the ratio is 3 ; a, ar, ar"^, ar^, ..., a geometrical progression, of which the ratio is r. 332. Elements and Notation. 1. The first term, a. 2. The ratio, r. 3. The number of terms, n. 4. The last term, I. 5. The sum of Hie terms, s. Thus, 3, 6, 12, 24, 48 is a geometrical progression in which a =: 3, r ^ 2, n = 5, s = 93. « 333. Classification. 1. Increasing, r > 1 ; as 1, 3, 9, . . . 2. Decreasing, r < 1 ; as 21, 7, 2^, . . . 334. Problems. 1. Given o, r, and n, required I and s. 1st. To find I. \al, 2d. 3d. ilh. ulh. GEOMETRICAL PROGRESSION. 263 2d. To find s. (1) 8 = a + ar -j- ar'^ -\- ar^ + • ■ • + a»'"~* . (2) rs = ar -\- ar"^ + ar^ + . . . -f- «?•""' + ar". (2)-(l) = (3) (r-l)8 = a(r"-l). a(r"— 1) .-. (4) 8: r— 1 Thus, if a = 4, r = 3, ?i == 10, then we have Z = 4 X 3» 4 ('310 _i> = 78732, 8 = ^^^-^ — ^ = 118096. 2. Given a, r, and Z, required s and «•. , Zr — a Ans. 8 = ■ r — 1 To find ?i: since i^ar"~i, ^"-1=:-; then raise r to a I "■ power which equals -. and the exponent of this power will be n — 1. Denoting the exponent by e, we have n — l=e, .•. M, = e -(- 1. 3. Given a, r, and s, required I and n. Ans. Z = (?:^l)-Ai«. r After finding I, find w, as in problem 2. 4. Given a, n, and Z, required r and «. Am. -I "~i/r — " v^ 5. Given a, w, and s, required the equations involving r Jim. -{ » Z (8 — «)"-! — a (s — a)"-' = 0. 264 ALGEBRA. 6. Given a, I, and s, required r and n. s — / After finding r, find n, as in problem 2. 7. Given r, n, and i, required a and s. f a = • L = Jir!nll . (r— l)r''-' 8. Given r, w, and 8, required a, and Z. (r — 1) s / a = -^^-^^ ^ • Ans. J ( s = 8(r-l)r'- _ r" — 1 9. Given r, ?, and s, required a and n. J.rw. a^h — (r — 1) s. After finding a, find w, as in problem 2. 10. Given n, I, and s, required the equations involving a and r. r a (8 — a)»-> — Z (s — 0"-' = 0. ( /. L_ /.-I I _L_ ^ 0. 8 — I 8 t 335. Application. 1. Given a = 5, r = 3, and n = 10, required I and 8. JLns. Z = 98415, « = 147620. 2. Given a = 10, r = 10, / = 100000000000, required n and 8. Ans. n = 11, s = 111111111110. GEOMETRICAL PROGRESSION. 265 3. Given o = 1, r = 5, and s = 97656, required n and I. Am. I = 78125, m = 8. 4. Given a —Z, n = 5, and I =^ 768, required r and s. Ans. r = 4, .s ^-= 1023. 5. Given a— .3, n == 5, and s = 1023, required r and I. Ans. r = 4, ^ = 768. 6. Given a = 4, Z = 12500, and s = 15624, required r and n. Ans. r =^ 5, n = 6. 7. Given 7- == 2, «. =^ 10, and I = 2560, required a and s. Ana. a = 5, 8 = 5115, 8. Given r =r 10, n = 9, and 8 = 1111111110, required a and Z. Ans. a = 10, 1= 1000000000. 9. Given r = 10, 1 = 500000, and 8 = 555555, required a and n. Ans. a = 5, w = 6. 10. Given n ^ 6, Z = 12500, and s = 15624, required r and a. ^«.s. r = 5, a ^ 4. 336. Problem. 2b mseri any number of geometrical means between any two quantities. The means together with the extremes form a geometrical progression ; and if we knew the ratio, the progression could be written out. The number of terms equals the number of means plus 2. Let m = the number of means, then n ^ m -\- 2. But according to problem 4, ^"-€ C. A. 28. 266 ALGEBRA. Substituting the value of n, we have m+l ll 1. Insert 5 means between 5 and 3645. r - eJSQ^ ^ 3 . . _ g .^g^ ^g^ ^g-^ ^Qg^ ^215^ 3645. 2. Insert 2 means between 8 and 512. 3. Insert 4 means between 10 and 1000000. 4. Insert 1 mean between the consecutive terms of the progression 6, 150, 3750, and write out the whole pro- gression. 337. Problem. To find the sum of a decreasing geometrical progression having an infinite number of terms. If r < 1 and «,= oo, r"^= 0. Substituting this value of r" in the formula a(r" — 1) r — 1 we have — a r — 1 Changing the signs of both terms of the fraction, we have a 8-- 1 — r GEOMETRICAL PROGRESSION. 267 !• i H" i + "J^ + • • ■ ) ''^ infinitum = 1. 2- i -\- i +2T+---> '^ infinitum = -J. 3. i: + tV + "BT + ■ ■ • ' "'^ infinitum = ^. 4. 1 -,- -| r + • • . , aci infinitum = m ji^ n* 11 — 1 338. Miscellaneous Problems. 1. Find six numbers in geometrical progression whose sum is 252, and the sum of whose extremes is 132. SOLUTION. (1) a-\-ar-\- ar'^ -|- ar^ -\- ar* -\- ar^ = 252. (2) a + ar^ = 132. (1) _ (2) = (3) ar+ ar'^ + ar^ + ar* = 120. (1) = (4) "^ "^^^ ~ ^^ = 252. r — 1 ... (5).= ?5^i^. (3) = (6) «>• ^''*~ -^ = 120. ... w^^ • ,u^ 120(r— 1) 252 (r—n (5) and (7) give (8) 7^^ = ^e _ 1 " 268 ALGEBRA. 10 21 (8) gives (9) r (r2 + 1) r* -f r2 + 1 .-. (10) 10(r2 + l)2 — 10r2 = 21r(r2+l). Or, (11) 10(r2 + l)2 — 21r(r2 + l) = 10r2. ,,„. 2,1 21r±V^441 r^+400r' . Or, (13) r^— frn= — 1. . • . (14) r = 2 or ^. (14) in (2) gives a = 4 or 128. Hence, the progressions : 4, 8, 16, 32, 64, 128, or 128, 64, 32, 16, 8, 4. In certain problems in geometrical progression it is con- venient to represent the ratio by " , and the middle term X by oy, if the number of terms is odd ; and the two middle terms by x and y, if the number of terms is even, as follows : Three terms, thus: x^, xy, y^. Four terms, thus : — , x, y, —■ y " X Five terms, thus: — , a;^, xy, y^ , — • 2. Find 3 numbers in geometrical progression whose continued product is 216, and the sum of their cubes 1971. Am. 3, 6, 12. QEOMETBICAL PROGRESSION. 269 3. Find 4 numbers in geometrical progression the sum of whose extremes is 45, and the sum of whose means is 30. Ans. 5, 10, 20, 40. 4. Find 4 numbers in geometrical progression the sum of whose first and third terms is 40, and the sum of the second and fourth 120. Ans. 4, 12, 36, 108. 5. Find 4 numbers in arithmetical progression which being increased by 1, 3, 7, 14, respectively, the sums will be in geometrical progression. Alls. 7, 9, 11, 13. 6. Find 3 numbers in geometrical progression whose sura is 13, and the sum of whose reciprocals is ■^-. Ans. 1, 3, 9. 7. Find 3 numbers in geometrical progression whose sum is 14, and the sum of whose squares is 84. Am. 2, 4, 8. 8. Find 4 numbers in geometrical progression whose sum is 15, and the sum of whose squares is 85. Ans. 1, 2, 4, 8. 9. Find 6 numbers in geometrical progression whose sum is 126, and the sum of the means 60. Ans. 2, 4, 8, 16, 32, 64. 10. Find 6 numbers in geometrical progression the sum of whose extremes is 488, and of the means 240. A7JS. 2, 6, 18, 54, 162, 486. 270 ALOEBBA. HARMONICAL PROGRESSION. 339. Definition. An harmonical progression is a series in which the first of any three consecutive terms is to the third as the iirst minus the second is to the second minus the third. 340. Propositions. 1. The reciprocals of the terms of an harmonical progression are in arithmetical progression. Let a, b, c, d . . . be in harmonical progression ; then, (1) a.c::a — b:b — c. .'. (3) ab — ac = ac — be. (2) b -.d-.-.b—o: c — d. .-. (4) be—bd = bd — cd. We find that (3) ^ a6c = (5) - — ^ = ^ — ^ ■ c b b a We find that (4) -^- W = (6) \ — - = - — l- d c c b - . - , - , -: . . . are in arithmetical progression. abed 2. The reciprocals of the terms of an ariihrnetieal progression are in harmonical progression. HARMONICAL PROGRESSION. 271 Let a, b, c, d . . . be in arithmetical progression ; then, (1) c — 6 = 6 — a. (I)^a6c=(3) 1(1 l) = i(l_lY a\o c / G \a b / (2) d—e--=c — b. From (3) we have (5) 1 : i : : 1 — i : 1 _ 1 . a c abb e From (4) we have (6) r ■ ~; • ■ 7 ■ i • b a b c c a .'. -. -; — . -J • • • are in harmonical progression. 3. The geometrical mean of tivo quantities is a mean propor- tional between their arithmetical and harmonical means. Let p and q be two quantities, g their geometrical mean, a their arithmetical mean, and h their harmonical mean ; then, '(1) p-g--g-q- ■"■ (4) 9^=pq- (2) a—p = q—a. .-. (5) a^i(p + q). (3) p:q::p-h:h-q. .-. (6) A=^- (5)X(6) = (7) ah=pq. Comparing (4) and (7), we have (8) g'^ = ah. ••■ (9) a:g::g:h. 272 ALGEBRA. 341. Problems. 1. Insert 7)1 harmonical means between a and I. SOLUTION. The problem is solved by inserting m arithmetical means between - and j , and taking the reciprocals. 2. Insert 1 harmonic mean between 20 and 60. Ans. 20, 30, 60. 3. Insert 2 harmonic means between 12 and 30. Ans. 12, 15, 20, 30. 4. Insert 4 harmonic means between 10 and 60. Ans. 10, 12, 15, 20, 30, 60. 5. Continue the harmonic series 10, 12, 15. Ans. 10, 12, 15, 20, 30, 60, oo, — 60, —30, —20, —15, —12, —10. 6. Find the n.* tej:m of the harmonica] progression a, 6, ... . ab Ans. (n — 1) (I — (n — 2) 6 7. If a, b, c are in arithmetical progression, and b, c, d in harmonical progression, prove that a : b :: e : d. 8. Prove that the arithmetical mean of two unequal quantities is greater than the harmonic mean. 9. What quantity added to each of the quantities a, b, c will give results in harmonical proportion? ab — 2ac + be a-~2b-j-c" PEMMUTATIUJS^: 273 10. If a is the arithmetical mean of x and y, and h their harmonical mean, find x and y. r a; = a ± Va^ — Ans. } ah. 11. If jf is the geometrical mean of x and y, and A their harmonical mean, find x and ?/. Ans. J PERMUTATIONS. 342. Befinition. Permutations are the results obtained by placing a certain number of things in every possible order in sets of 1, 2, 3, . . . r,. . . n. 343. Problem. To find the number of permutaMons o/n letters taken 1, 2, 3, . . r, ... n in a set. Let Pj, Pj, P3, . . . Pr, . . . P„, respectively, denote the number of permutations of n letters taken 1, 2, 3, . . . r, . . . »j in a set. n letters taken 1 in a set give n permutations. .-. (1) P,=n. 274 ALOEBBA. For each set of 1 letter there are n — 1 remaining letters which may be annexed singly to the corresponding set of 1, thus giving for each set of 1, n — 1 sets of 2 ; hence, for the n sets of 1, there are n (ji — 1) sets of 2. -•. (2) P^=^n{n—1). For each set of 2 letters there are n — 2 remaining letters which may be annexed singly to the corresponding set of 2, thus giving for each set of 2, « — 2 sets of 3 ; hence, for the 11 (n — 1) sets of 2, there are n (n — 1) (ii — 2) set.:- of 3. .-. (3) P3=n(w — lj(w — 2). Thus far, the following law holds true : The factors are n, n — 1, n — 2, the negaiive number in the last factor being less by unity than the number of letters in a set. To prove the law general, let us assume it true for r — 1 in a set. The last factor will be n — (r — 2) or 71 — r -\- 2. .-. (r— 1) P,._, =n(ro— 1) (n— 2). .. (w — r + 2). For each set of r — 1 letters there are n — (r — 1) 01 n — r -\- 1 remaining letters which may be annexed singly to the corresponding set of r — 1 letters, thus giving, for each set of r — 1, n — r + 1 sets of r; hence, for the )i.(?i — l)(n— 2). . . (n — r + 2) sets of r — 1, there are n(n — l)(n — 2)... (n — r + 2) (w — i- + 1) sets of r. .-. (r) P, = n(m — l)(n— 2)... (w — r+2)(n — r+1). Hence, if the law holds for n letters taken r — 1 in a set, it will hold for n letters taken r in a set. But we have COMBINATIONS. 275 found that it holds for 3 in a set ; hence, it will hold for 4 in a set ; and if for 4, then for 5, and so on up to r. If r =: n, all the letters will be taken in each set ; and by changing the order of the factors, (r) will become (n) P„=1.2.3...w,. Hence, the number oj permutations of n letters taken n in a set is equal to the product of the numbers from 1 up to n inclusive. For the sake of brevity, we shall denote 1.2.3. . .« by \n, which is read factorial n. Then (n) becomes COMBINATIONS. 344. Definition. Combinations are the results obtained by placing a certain number of things in sets of 1, 2, 3, ... r, ... ti, any two sets differing from each other by at least one of the things. 345. Problem. To find the number of combinations of n letters taken 1, 2, 3, ... r, ... n in a set. Let C,, Cj, Cg, . . . C,., . . . (/„, respectively, denote the number of combinations of n letters taken 1, 2, 3, ... r, ... n in a set. 276 ALGEBRA. Combinations, Permutations, < To ascertain the law of relation between combinations and permutations, let us take 3 letters, a, b, c, two in a set. Then, ah. ha. ac. ca. be. [cb. The combinations are converted into permutations by permut- ing the letters of each combination. Now, suppose the C,. combinations formed, then each of these combinations will give [r permutations; hence, the Q. combinations will give |r X Cy permutations ; hence. ab. ae. I be. Hence, (1) 6V \r X a = P, ■ I\_n a Lr /ON f< _ -^2 __ n (n — 1) (3J C'3- -' P3_ )t(w — 1) (w — 2) [3' "~1 . 2 . 3 ' M p _ ^r »t (^ - 1) Qi - 2) ■ ■ . (n~r + 1) (?i — 1) «, uoMniJSJiTiujsa. 277 346. Proposition. Tlie number of comMnations of n things taken r in a set is equal to the number of combinations of n things taken n — r in a set. This is evident from the fact that for each combination of n things taken r in a set there is a complementary com- bination taken n — r in a set. 347. Problems. 1. To find the number of permutations of n letters taken all in a set, if p of these letters are of one kind, q of another hind, s of another, and so on. The p letters which are alike would, if diiferent, give for each position of the other letters [^ permutations ; but since these letters are alike, \2_ permutations reduce to one ; hence, there will be ,— as many permutations when the p letters are alike as when they are different. In like manner, there will be ,— as many permutations when the q letters are alike as when they are different. Also, there will be p as many permutations when the s letters are alike as when they are different. But if the p letters were different, also the q letters and the 8 letters, . . . the number of permutations of the n letters would be \n. Hence, when the p letters are alike, also the •2i8 ALGEBRA. q letters and the s letters, ... the number of permutations of n letters taken all in a set is (n 2. To find the number of sets of n different letters wlien each may occur 1, 2, 3, ... r times. The number of sets of 1 in a set is n. Placing each letter before each set, the number of sets of 2 in a set is n^. Placing each letter before each set, the number of sets of 3 in a set is n^. In general, the number of sets of r in a set is n\ MS. Examples. 1. How many permutations can be made of 6 letters taken 1, 2, 3, 4, 5, 6 in a set? Ans. 6, 30, 120, 360, 720, 720. 2. What is the total number of permutations that can be made of 10 letters? Am. 9864100. 3. How many combinations can be made of 8 letters taken in sets of 1, 2, 3, 4, 5, 6, 7, 8? Ans. 8, 28, 56, 70, 56, 28, 8, 1. 4. What is the total number of combinations that can be made of 12 letters ? Am. 4095. 5. How many permutations, taken all in a set, can be made of the letters in the word Algebra? Ans. 2520. INDETERMINATE GO -EFFICIENTS. 279 6. Prove that the product of any r consecutive numbers is divisible by [r. 7. Prove that the number of combinations of n things, taken r in a set, is equal to factorial n divided by factorial r into factorial n — r. INDETERMINATE CO -EFFICIENTS 349. Theorem I. If Ax' f 5a;'' + Gi;= + . . . =-- ^V + B'x^' + CV + . . . , for all possible values of x, the co-effieients and exponents being finite quantities, and independent of x, and the terms arranged according to the ascending powers of x, tlien a = a' , 6 = 6', c = c',... and A = A!, B^B', G^C... Since the co-efficients and exponents are independent of x, if we find their values for one value of x, we shall have their values for all values of .r. (1) Aaf + Bx'' + Cx'-+...^ ^V+ jBV+ C'x''' + . . . 1. Suppose a < a'. Dividing (1) by a;", we shall have (2) A+Bx'^" + Cx"-" + . . . ^^ V'-^-f £V-"+ C V-<'+ . . . The exponents of x are all positive ; and making a;= 0, we find A^O, which is contrary to the hypothesis that the co-efficients are finite quantities. Therefore, it is not true that a < a'. 280 AIjUKJiUA. 2. Suppose a > a'. Dividing (1) by a^', we shall have (3) .4a:<'-»'+Ba;''-"'+ &=-"'+ . . . r=4'+BV'-"'+Cy-"'+. . . The exponents of x are all positive ; and making x = 0, we find = A! , which is contrary to the hjrpothesis. Therefore, it is not true that a > a'. Since neither a<^a' nor a > a', it follows that (4) a = a'. .-. (5) X = a;"'. Dividing (1) by (5), member by member, we have (.6) A + J5x'-« + Crr'-" + ...=A'+ B'x^-"' + V'-"' + . . . Making a; = 0, we shall have (7) A = A'. (5) X (7) = (8) Asfl = A'x'^'. Subtracting (8) from (1), member from member, we have (9) Bx'' +Cx'= +...= B'x^+ OV + . . . Since (9) is the same in form as (1), by a similar process it can be proved that 6 = 6' and B =^ B' • and, in like manner, that c = e' and (7= C", and so on. 350. Theorem II. If Aaf + jBx'' + CxP + Dafl + . . . = 0, for all possihk values of X, the co-efficients and exponents involved in the equation being independent of x, and the terms arranged according INDETERMINATE CO -EFFICIENTS. 281 to the ascending powers of x, fken J. = 0, £ = 0, C = 0, Z) = 0, ajid so on. (1) ^a;" + Ba;'' + 0);<= + Da^ + . . . = 0. Dividing (1) by a;", we shall have (2) A -\- JBs^-" + Gxf=-"- + Da?'--+ . . . = 0. Making x = 0, we shall have (3) ^ = 0. .-. (4) ^a;" = 0. Subtracting (4) from (1), member from member, we have (5) Bx" -{- CxP -{- Dsfl + . . . = 0. Since (5) is the same in form as (1), by a similar process it can be proved that 5 = 0; and, in like manner, that 0=0, jD = 0, and so on. 351. Expansion of Fractions into Series. 4 1. Expand ~ into a series. ^ 2 + 3a; 1st solution. — BY DIVISION. * : 2 — 3a; + I a;2 — ^ a;3 + . . . 2 + 3a; '2 4 2d SOLUTION. — BY INDETERMINATE CO-EPPICIENTS. It is evident by division that the first term of the devel- opment will be independent of a; ; . • . assume (1) ^^^ = A + Bx + Gx^+Dx^+... C. A. 24. 282 ALGEBRA. Clearing of fractions and transposing, we have (2) = '2A +2B x+ 2C a;^ + 2D — 4+3^ +35 +30 Hence, by Theorem 2, we have 2JL — 4 = 0. .-. A = 2. a;8 + , 2B + 3A = 0. 20 + 3B = 0. 2D + 3C = 0. •. B = — ^A = — 3. •■ G = -iB = l •. B = — ^C = — !^. Substituting these values in (1), we have (3) —^ = 2 — 3a; + I a;2 — ^ a;3 + , 2 + 3a; ^2 4 ^ Any term can be obtained by multiplying the preceding by — fa;, which is the scale of the series. 2. Expand into a series. 3a;2 — 4a;^ We find by division that the 'first term of the expansion is 3a;2 3" ^= -a;~^. Let us therefore assume (IJ JLa:-2+ 5a;-' + Ca;»+I>a; +£S(;2+ . .3a;2 — 4a;3 Proceeding as before, we shall find (2) --^ = A + 20 80 3_20 1280^,,_, ^^ :!.r- — 4a;3 3a;2 ^ 9a; ^ 27 81 ^ 243 ' ' INDETERMINATE CO-EFFICIENTS. 283 3. Expand into a series. 1 -\- X Ans. 1 — a; -|- x^ — a;^ -(- a;* — x^ -{- . . . 4. Expand into a series. 1 — X . 1 + a; + a;2 + a;8 + a;* + a;5 + . . . 2 5. Expand into a series. 3H-4x A 2 8 , 32 „ 128 , , Atis. - — —x-\--z.x x^ + . , 3 9 ^27 81 6. Expand :; into a series. h-\- ex 1 -\- X 7. Expand ^_n , o 2 ^^^ ^ series. 1 ~j~ ^X ~\~ OX Atis. 1 — X' — a;^ + 5a;' — 7a;* - . . , 353. Decomposition of Rational Fractions. 1. Let us take the fraction „ "*",„ • x^ — 0^ Since x'^ — h^ ^= (x -\-li) (x — 6), we may assume ,^^ ax -\- c A . B ^^^ x-i — h^"' x-\-h^ x~b' Clearing of fractions and reducing, we have (2) ax -[-c={A^B)x — hA-\- bB. 284 ALGEBRA. Hence, by the principle of indeterminate co-efficients, . . ah — c ::! I — hA + hB=G.) I R — ah + c ~ 26 Substituting the values of A and B in (1), we have ax-\- c ah — c ah -\- c x2 _ 62 — 26 (x + 6) + 26 {x — 6) ' 2. Decompose /^ + ^, • Ans. ^ ii- . a;2 — 7a;+12 x — 4 a; - 3 3. Decompose „ ^~ , „ ■ Am. -I • , -n 5a; + 32 . 2,3 4. Decompose — — ^ — ■ Ans. A -■ ^ a;2 + 13x + 42 a; + 6 ^ a; + 7 THE BINOMIAL THEOREM. 353. Lemma. — I =na;''~i, for all values of m. ^U „,Jt The above is read — , when ii=^.(, r=nx"''*. 1. When n is a positive integer. By division, we shall have ^" ~y" = x"^' + a;"-2 y + a;"-3 2/2 4- ... 4. y»-i. a; y BINOMIAL THEOREM. 285 There are n terms in the second member, since there are n — 1 terms involving y and one term involving x only; but when y =- x, each term = a;"~ ' . ^ 1 ^=nx^'~^, when n is a positive integer. X y /y = x 2. When ri= ", a positive fraction. Let x^v XQ:^r'', and r''~^ = Xg''^. Let y^sP, . ■ . 2/9 = s'', and if s ^ r, y = cc. ' I* . Xq — yq r" — s" X — y r' — s' P rP-9 = £a;9 '; r ' — s r« — s" r — s a;9 p p^p qfi q q ( x—y 3. When n is a negative integer. ^x" . 9 a:~" — ?/ x — y — a;-"2/-" = — a;-2"x wx" / x" — y" \ j__ V .T — y )v=x \ X — y /y==o 4. When n = — — , a negative fraction. 9 _p p g 9 — y~g x — y — ^P _£ •a; « ar~(j (_p p x—y Jy^x -^- X Q . 286 ALGEBRA. 354. Application. Let us now find the development of (a + xT, when n is integral or fractional, positive or negative. Let us assume (1) (a + a;)" = 4 + £a;+ai;2+X>a;3 + ... Making a; ^ 0, we find a" = J. ; hence, (2; (a + a;)" = a" -\- Bx + Cx'' ^ Dx^ -\- . . . Substituting y for x, we have (3) (a + 2/)" = o" + %+Q,2 + Z)2,M---- (4) (a-\-x) — {a + y)=x — y. Subtracting (3) from (2), and dividing by (4), {a + xr~{,a + yY ^^(x-y\ ^ ^ (a + x)-(a-\-y) \x-yj \ X — y J \ X — y J But, for all values of n, when y = x, (5) becomes (6) nCa + x)"-^=B + 2Cx + BDx^ +... Multiplying both members of (6) by (a + «), we have (7) n(a + xy = aB + 2aC + B x + 3aD + 2C x^ +... BINOMIAL THEOREM. Multiplying both members of (2) by n, we have (8) n, (a + a;)" = nO" + nBx + nCk'^ + . Equating the second members of (7) and (8), 287 (9) aB-\-2aG x + ZaD + 20 a;2 +... =na"+n£a; + nGi;2 +.. Hence, by the principle of Indeterminate Co-efficients, 2aC+ B=nB,.-. C=C?L=llg ^ »C^-1^ a-''. 2a 1.2 ^ ' 3a 1.2.3 Substituting the values of B, G, D, . . . in (2), we have, for any exponent, the Binomial Formula, (a + xy = a" + na"- ' a; + ^'i'^ — '^l a"" ^ a;^ _l_ ri,(w — l)(w — 2) ^„_3 3.3 _j_ _ _ 2 ^2 355. Examples. 1. (a + x)i = ai + ^ J-' X + ^1^^^=^ ai- ^1.2.3 ^ i , a; a;^ , a;' •'b* 1 2a^ 8al 16a^ 128a* 288 ALGEBRA. 3. Expand fl—x^{l— x)s. ^ws. 1 — ~x — -x"^ a;' 3 9 81 243 4. Expand Va'^—h'^. , 6^ J4 J6 4ws. a — 2a 8a3 IGct^ 5. Expand \ = = at (a^ + h'^yi. , 1 262 554 4060 ^'^- -1 - ri + ^u - ^7^ + • ■ ■ aa das 9aT ovan DIFFERENTIAL METHOD OF SERIES. 356. Definitions. 1. The first order of differences of a series is the series obtained by subtracting the first term of the given series from the second, the second from the third, and so on. 2. The second, third, etc., orders are the series obtained from the first, second, etc., orders in the same manner as the first order is obtained from the given series. DIFFERENTIAL METHOD OF SERIES. 289 Thus, if the given series be 1, 4, 9, 16, 25, . . . The 1st order of differences is 3, 5, 7, 9, . . . The 2d order of differences is 2, 2, 2, . . . The 3d order of differences . is 0, 0, . . . 357. Problem I. To find the first term of any order of differences and any term of the series. The series and orders of differences will be as follows : beries, tj, c2j *sj *4j ^sj • ■ • *m, • • • 1st order, tj— tj, tg— 'tj, f^ — tg, tg — ti, . . . 2d order, tg — 2f ^ + f , , t^ - 2t3 + t^, tg — 2t4 + tg, . . . 3d order, t^ — 3t3 + Stj — tj, t^ — 3*4 -|- 3f3 — t^, . . . ith order, tj — 4t^ -\- Qt^ — it^ -\- t^, . . . Let the 1st terms of these successive orders be denoted by dj, dj, dg, d^, . . . d,„ . . . Then, dj^^ — ti+ tj, -•. t2=*i+ <^i- dj^ ti— 2t2+ ig, .-. t3=f, +2di +d2. d3 = — f , + 3t2 — Sfj + t4, . • . <4 = f 1 + 3d, + 3d2 + £^3- ,^. , , n (n — 1) , (1) d„ = -t- t, ± ntj -H J 2 *3 , ii0i-l)(» -2), ± 1.2. " 3~" ** ^ • ■ ■ The upper sign applies if n is odd ; the lower, if ii is even. C. A. 25. (2) f„ =t, + (,n-l)d, + (.n-inn-2) ^^ (n-1 ) (n-2)CH-3) + r~: 2 ^ 3~ ''3 + • ■ • 358. Examples. 1. Find d,, (^2, dg, . . . and t„ of the series P, 2^, 3^, . . . ^718. dj ^3, 6^2= 2, dg = 0, t„ = )i^. 2. Find d^, d^, d^, . . . and t„ of the series 1', 2', 3', . . . Ans. di =7, dj = 12, d^ ^ 6, d^ = 0, t,, = n^. 3. Find d,, d2, dg, . . . and <„ of the series 1, 3, 6, 10, . . . Ana. d] =2, dj = 1, dj = 0, i„=: ^7" • It 4. Find d,, d2, dj, . . . and t„ of the series 1.2.3, 2.3.4, 3.4.5, ... Am. di=18, d^=\'ii, dg='6, d^=^, «„=n(?i2 + 3n+2). 5. Find d,, d2, dg, . . . and i„ of the series 1 (m -)- 1), 2 (m + 2), 3 (m + 3), . . . Ans. d, = ?rt -|- 3, d2 = 2, dg = 0, i„= '^ (™ + '0- 359. Problem II. To find the sum of n terms of the series. Let the series be (]} t], C2; *3> ^41 ■ ■ ■ '11' • • • Let us assume the series (2) 0, t„ t,+ t^, t, + t^+t^, «, + (!,+ tg+ t,+ ... PIFFEBENTIAL METHOD OF SERIES. 291 It is evident that the {n + 1)"' term of (2) is equal to the sum of n terms of (1), and that «, of (1) = d^ of (2), d-^ of (1) = dj of (2), d^ of (1) = d^ of (2), . . . But the (w + 1)''' term of (2), -which is the sum of n terms of (1), can be found from formula (2) of the pre- ceding problem, if we substitute n -f 1 for w, for <,, i, for dj, d] of (1) for d^, d^ of (1) for dg, . . . and s for t„^i, which give (3) s^wti + ^^ ^ ^ di + j-^ ^ '^ g ^ da + ■ ■ ■ 360. Examples. 1. Find the sum of n terms of the series 1, 3, 5, 7, 9, . . , Ans. s = n^. 2. Find the sum of n terms of the series 1, 2, 3, 4, 5, . . . It 3. Find the sum of n. terms of the series 1^, 2^, 3^, 4^,... 7l2 (n + 1)2 '^ 4 • 4. Find the sum of n terms of the series 1*, 2^3*, 4*,... 5. Find the sum of n terms of the series 1.2, 2.3, 3.4, 4.5,... ^i„,, >iC»+l)('^ + 2) 6. Find the sura of n terms of the series 1.2.3, 2.3.4, 3.4.5, 4.5.6, . . . ^^^^ ^ ^ w(» + l)(n + 2)C» + 3) _ ALOEBRA. APPLICATION TO PILES OF BALLS. 361. Problem I. To find tiie number of balls in a triangular pile. In the 1st course the number of balls =1. "■ " " " =1+2=3. " " =1+2+3=6. " " =1+2+3+4=10. " " 2d " " 3d iC l( itJi nth " " " " " =1+2+3+... + n, ^ >i-(w+l) ~ 2 Hence, the number o^ balls in a triangu- lar pile of n courses is equal to the sum of n terms of the series 1, 3, 6, 10, . . . By formula (3), Art. 359, we find n (n + 1) (n + 2) *~1 . 2 . 3 ■ n(n + l) 2 363. ProWem II. To find the number of baUs in a square pile. In the 1st course the number of balls = 1. " " 2d " " 3d (i ith = 22 = 4. = 32 = 9. = 42 = 16. nth " " =«.2 DIFFERENTIAL METHOD OF SERIES. 293 Hence, the number of balls in a square pile of n courses is equal to the sum of n terms of the series 1, 4, 9, 16, ... n^. By formula (3), Art. 359, we find __ n {n + 1) (2n + 1) 1.2 . .3 3«3. Problem III. To find the number of baUs in an oblong pile. Let n = the number of courses = the number of balls in breadth of base. Let m = the number of balls — 1 in the top row = the number of balls in the length of base — the number of balls in the breadth of base. The rectangular pile = a square pile of n courses -f- m triangular strata, each = — ~^ — - • But the number of balls in the square pile is n (n + 1) (2ra + 1) 1.2 . 3 ' and the number of balls in the m triangular strata is mn (w. -|- 1) 2 294 ALGEBRA. Adding and reducing, we have for tlie number of balls in an oblong pile of n courses ^ n(n + l) (l+2it + 3m) ""1.2 . 3 364. Examples. 1. Find the number of balls in a triangular pile of 12 courses. Am. 364. 2. How many balls in a triangular pile of 15 courses, and how many will remain after 5 courses are removed? Am. 680, and 645. 3. In an incomplete triangular pile, the number of balls on one side of the lower base is 20, and on one side of the upper base 10; how many balls in the pile? Am. 1375. 4. How many balls in a square pile of 12 courses? Ans. 650. 6. How many balls in a square pile of 15 courses, and how many will remain after 6 courses are removed? Am. 1240, and 1149. 6. In an incomplete square pile, the number of balls on one side of the lower base is 30, and on one side of the upper base 15 ; how many balls in the pile ? Ana. 8440. 7. How many balls in a complete oblong pile of which the length of the base is 45, the breadth 15 ? Am. 4840. LOGARITHMS. 295 8. In an incomplete oblong pile, the length and breadth of the lower base are respectively 50 and 25, the length and breadth of the upper base 35 and 10 ; how many balls in the pile ? Ava. 12240. 9. The number of balls in a triangular pile is to the number in a square pile of the same number of courses as 9 to 17. How many balls in each pile? Am. 2925, and 5525. LOGARITHMS. 365. Definition and Principles. 1. A logarithm of a number is the exponent denoting the power to which a fixed number, called the base, must be raised in order to produce the given number. Thus, in the equation a!'^n, x = logan, which is read x = the loga- rithm of n to the base a. 2. Any positive number, except 1, may be assumed as the base ; but when assumed, it remains fixed for a system of logarithms. 3. There may be an infinite number of systems of loga- rithms, since there may be an infinite number of bases. 4. The logarithm is dependent on both the base and the number, since a change of either involves a change of the logarithm. -96 ALGEBRA. 366. General Properties of Logarithnis. 1. In any system, the logarithm of 1 is 0. For, a" = 1 ; .'. by def., log„l = 0. 2. In any system, the logarithm of tlie base is 1. For, a> = a; .-. by def., logatt := 1. 3. In a system tvhose base is greater than 1, the logarithm of is — -00. For, a— = - = 0, if a>l; .-. by def, log„>,0 = -oo. 4. In a system whose base is less than 1, the logarithm of is -|- oc. For, a°° =^ 0, if a < 1 ; . • . by def , log„ ^ j = + oc. 367. Brigg's Logaritlims. Brigg's Logarithms, so called from their inventor, are the logarithms of numbers in the system whose base is 10. This is the common system, and it possesses advantages for computation superior to all other systems. 368. logarithms of the Powers of 10. 100 = 1; .-. logiol=0. 101=10; .-. Iogiol0 = l. 102 = 100; .-. log, 100 = 2. 103 = 1000; .-. log, 1000 = 3. Hence, the logarithm of an exact power of 10 is a whole number, equal to the exponent of the power. LOGARITHMS. 297 Cor. 1. If 71 > 1 and < 10, logn, n > and < 1, or 4- a decimal. Cor. 2. If n > 10 and < 100, log, o « > 1 and < 2, or 1 + a decimal. Cor. 3. If w > 100 and < 1000, log, o « > 2 and < 3, or 2 + a decimal. Hence, if the number is not an exact power of 10, its logarithm, in the common system, will consist of two parts, an entire part and a decimal part. The entire part is called the characteristic, and the decimal part the mantiisa. 369. Problem. To find the laws for the characteristic. Take the equations (1) 10^ =n. (2) 10' =10. (l)-(2) = (3) 10-'-^- •■• Iog,oYo = ^— l=l"gTO«— 1- Hence, the logarithm of the quotient of any number by 10 is less by 1 than the logarithm of the number. Let us now take the number 8979, and its logarithm 3.953228, as given in a table of logarithms, and divide the number successively by 10, and for each division subtract 298 ALGEBRA. 1 from the logarithm of the dividend, and we shall have the following: Log. 8979 = 3.953228. 897.9 = 2.953228. 89.79 = 1.953228. 8.979 = 0.953228. .8979 = 1.953228. .08979 = 2.953228. .008979 = 3.953228. The minus sign applies only to the characteristic over which it is placed. The mantissa is always positive, and is the same for all positions of the decimal point. An inspection of the above will reveal the following laws : 1. if the number' is integral or mixed, tlie charaeteristie is positive and is one leas than the number of integral figures. 2. if tJie number is entirely decimal, the characteristic is neg- ative and is one greater, numerically, than the number of O's immediately following the decimal point. 370. Propositions and Rules. 1. The logarithm of the product of two numbers is equal to the sum, of their logarithms. Let (1) a''=m; then, by def., log„m = a;; and (2) a'' = n; then, by def., log„?i ^ i/. (1) X (2) = (3) a''*!' ~mn, . • . by def. , log, mn = a; + y. logo mn = log^ m -f- log, n. LOUAKITUMS. 299 Hence, to multiply by means of logarithms, find the logarithms of the factors and take their sum, which will be the logariilvm, of the product; find the number corresponding, which will be their product. 2. The logarithm of the quotient of two numbers is equal to the logarithm of the dividend mimis the logarithm of the divisor. Let (1) a" =m; then, by def., log<,m = a;; and (2) a" ^n; then, by def., log„n ^y. (1)^(2) = (3) «--"= ^ ; then, by def. , log,. - = x — y. .-. log«— = logam — log„Ji. Sence, to divide by means of logarithms, find the logarithms of the numbers, subtract the logarithm of the divisor from the logarithm of the dividend, and the remainder toill be the logarithm of the quotient ; then find the number con-esponding, which will be the quotient. 3. The logarithm of any power of a number is equal to the logarithm of the number multiplied by the exponent of the power. Let (1) a-' :=n; then, by def., log^u^a;. (1)" = (2) av'^n^; then, by def. , logan" =px. .-. l0gaU''=2) (log„»l). Hence, to raise a number to a given power by means of loga- rithms, find the logarithm of the number and multiply it by the exponent of the power, and the product will be tlie logarithm of tJie power; find the number corresponding, which will be the power. 300 ALGEBRA. 4. The loganthm of any root of a number is equal to the log- arithm of the number divided by the index of the root. Let (1) a' =^n; then, by def., log^n^^^x. yJT) = (2) a'' -^ y'n ; then, by def., log,, {/'n --= - ■ , r, — log„« .-. log„ v/ti =- -^ ■ Hence, to extract a given root of a number by means of loga- rithms, find the logarithm of the number, divide it by the index of ike root, and the quotient will be the logarithm of the root; then find the number corresponding, which mil be the root. 5. The logarithm of a number to Hie base b is equal to the logarithm of the same number to the base a, divided by the loga- rithm of b to the base a. Let (1) a^=«; then, by def., log„n^a;; and (2) 6"=^; then, by def., logi,n^y. Hence, a^=b", .-. a« = b; then, by def., \og^b = -- . • . l0g„6 = z^^S^, or logiW X logafe = logaM. log,,»i , loff„ n .-. log,,W = r-^. Hence, to find the loganthm of a number corresponding to a given base, divide the logarithm of the number to any base by the logarithm of the given base ti the same hose. LOGARITHMS. 301 371. Table of Logarithms from 1 to 100. N. Log. N. Log. ^N. Log. N. Log. 1 0.000000 26 1.414973 ■51 1.707570 76 1.880814 2 0.301030 27 1.431364 52 1.716003 77 1.88649) 3 0.477121 28 1.447158 ,53 1.724276 78 1.892095 4 0.602060 29 1.462398 64 1.732394 79 1.897627 5 0.698970 30 31 1.477121 ! 65 1.740363 80 1.903090 6 0.778151 1.491362 56 1.748188 81 1.908485 7 0.845098 32 1.505160 57 1.755875 82 1.913814 8 0.903090 33 1.518514 58 1.763428 83 1.919078 9 0.954243 34 1.531479 69 1.770852 84 1.924279 10 I.OOOIIO'J 35 36 1.544068 60 1.778151 85 1.929419 U 1.041393 1.556303 61 1.785330 86 1.934498 12 1.079181 37 1.568202 62 1.792392 87 1.939519 i:i 1.113943 38 1.579784 63 1.799341 88 1.944483 14 1.146128 39 1.591065 64 1.806180 89 1.949390 15 1.176091 40 1.60206;) 65 1.812913 90 1.954243 10 1.204120 41 1.612784 66 1.819544 91 1.959041 ir 1.230449 42 1.623249 67 1.826075 92 1.963788 18 1.255273 43 1.633468 68 1.832509 93 1.968483 19 1.278754 44 1.643453 69 1.838849 94 1.973128 20 1.301033 45 1.653213 70 1.845098 95 1.977724 21 1.322219 46 1.662768 71 1.851258 96 1.982271 22 1.. 342423 47 1.672098 72 1.857333 97 1.986772 23 1.361728 48 1.681241 73 1.863323 98 1.991226 24 1.380211 49 1.690196 74 1.869232 99 1.S95635 25 1.397940 50 1.698970 75 1.875061 100 2.000000 373. Examples. 1. Multiply 8 by 7 by means of logarithms. 2. Multiply 9 by 6 by means of logarithms. 3. Divide 99 by 9 by means of logarithms. 4. Divide 35 by 7 by means of logarithms. 5. Square 6 by means of logarithms. 6. Cube 4 by means of logarithms. 302 ALGEBRA. 7. Extract the square root of 81 by means of loga- rithms. 8. Extract the cube root of 27 by means of logarithms. 9. Find the logarithm of 9 in a system whose base is 8. Ans. 1.05664. 10. Find the logarithm of 100 in a system whose base is 15. Am. 1.70055. 11. Find the logarithm of mn Ans. log a -)- log 6 + log c — log m — log n. 12. Find the logarithm of • Ans. m log a -f 71 log 6 + p log c — q log d. 13. Find the logarithm of a'^—b'^. Ans. log (a -f 6) -1- log (« — &)• 14. Find the logarithm of (a + by Ans. I log (a — 6) — I log (a + 6). 15. Given lOO"-' = 80, to find x. Ans. a; = 0.951545. 16. Given (t/5)^ = 10, to find x. Ans. x-^ 2.861353. 373. Logarithmic Series- Let us endeavor, by the method of indeterminate co-effi- cients, to develop logo;, to any base, into a series arranged according to the ascending powers of x. Let us assume the series, (1) log X = M + Nx -\- Px^ -\- Q.r3 + . . LOGARITHMS. 303 The co-efficients M, N, P, Q, . . ■ are independent of x and dependent on the base. If X ^ 0, log = M; but log ^ q; oo, . • . =p oo = Jf , a finite quantity, which is absurd ; hence, the assumed series will not give the development required. Let us assume the series, (2) log x=:Mx + Nx'- + Px3 + Qx* + . . . If a; = 0, log =: ; but log = =F oo, . • . q: oo = 0, which is absurd; hence, the assumed series will not give the development. Let us assume the series, (3) log (1 -h a;) == M; + Nx^ + Px^ + Qa;* + . . . If X ^ 0, log 1^0, which is true. Let us also assume, (4) \og(l +y)= My ^ Ny^ + Py^ -^ qy* +... Subtracting (4) from (3), we have (5) log (1 + x) — log (1 + 2/) = M(x — y) + N(x^— 2/2) + P (x^ — 1/3) + . . . But log (l+x)-log(l+?/)=log (^) =log(l+|^) + Q(pzyY+...=.M(x-y) + Nix^-y') + P(x^ -y^) + Q (x* -y') + . . ■ •^04 ALGEBRA. Dividing (6) hj x — y, we have (7) M -\- N 2// 0-+yy ^—y , p (a=— y)' , n (^—2/) +Q (l+2,)«^'*(l+2/)4 Making y ~ x, we have (8) -M'( j^~) = if + 2iVa; + 3Prr2 + 4§.r3 + . . . Clearing of fractions and transposing M, we have (9) = M+ 2N — M+ M x^ + 4Q + 3P x» +. x + SP + 2N Then, from the principle of indeterminate co-efficients, M— M=0, .-. M=M. 2N+ M=0, .-. N= — IM. 3PH-2iV"=0, .-. P= — |iV"=p£ 4Q + 3P-.0, .-. q=-^P=-\M. Substituting the values of N, P, Q, . . . in (3), we have (10) log (1 + x) = if (x — -1x2 + Ja;3 — :|x* + . . .)■ Equation (10) is called the logarithmie series. Hence, the logarithm of a number is the product of two factors; one a series which is dependent on the number, the other independent of the number, and therefore dependent on the hose; for, since the logarithm varies with _ the base, if the base elutnges, x remaining the same, M must change. 305 374. The Modulus. The modulus of a system of logarithms is the factor dependent on the base. If we denote the bases of two systems, respectively, by a and b, and their moduli by M and M', then (1) log, (1 + a;) = M(x — ix^ + ^x^ _ J^a;* + . . .). (2) log, (1 + s) = M'(x - ix^ + ix^' -ix^+.. .). .-. (4) log,, (1 + a;) : log, (1 + a;) : : 1/ : if'. Hence, ihe logarithms of the same number in two systems are proportional to the moduli of those systems. 375. The.Naperian System. The Naperian system of logarithms, so called from the name of the inventor, is the system whose modulus is 1. The base of the Naperian system is usually denoted by e. If the base of another system be denoted by a and its modulus by M, we shall have, by the preceding Article, (1) log,(l+a;) : log,(l+a;) :: M:l. .-. (2) \og,(l+x)=log,il+x)XM. Hence, the logarithm of a number in any system is equal to Ike Naperian logarithm of the same number, multiplied by the modulus of. that system. C. A. 26. 306 ALGEBRA. Dividing (2) by M, and changing the members, we have (3) log,(l + .)=i2gii^). Mence, tlie Naperian logarithm of any number is equal to Ike logarithm of the same number in any other system, divided by the modulus of that system. Dividing (2) by log^ (1 + x), and changing the members, we have ^^^ ^ log.(l + ^) Heme, the modulus of any system is equal to the logarithm of any number in that system, divided by the Naperian loga- rithm of the same number. 376. Problem. To transform the logarithmic series into a converging series. Since, in the Naperian system the modulus is 1 and the base e, the logarithmic series in this system will become (1) log,(l + x)=x-ix^+ix»-^* + ... Substituting — x for x, (1) becomes (2) log, (1 — ic) == — a; — ^2 — ^a;3 - Ja;" - ... Subtracting (2) from (1), we have (3) log,(l+.r)-logXl-a;) = 2(a; + ^8 + ^j,5^^7^,..). LOQARTTHMS. ' 307 But log, (1 + x) — log„ (1 — a) = log, ( _ ) ; lience, (4) log. ( J3^) = 2 (a; + ia;3 + ^x^ + |x-' + ...)• Let = , then x = -^ — r— r . and 1—x y 2t/ + 1 •og« (.rirty = ^"s. r' — j = loge (y + 1 ) — log, y. .' . (4) becomes (5) log, (2/ + 1) — log, 2/ "" ^ LvFT ^ 3(2y+l)« + 5(22/ + l)« + ■■•]■ This series is converging for all values of y > ; and the greater the value of y, the more rapidly will the series con- verge, and the less will be the number of terms necessary to be taken to give the approximation to the required degree of accuracy. 377. Problem. To compute a Table of Naperian logarithms. Formula (5) of the preceding Article, by transposing — log, y, becomes (1) log,Cy+l) =log,2/ + 2 [^q^^ + 3(2^^1)8 + 6 (22/ + 1)5 +••■]• Eemembering that log, 1 = 0, that the logarithm of the product of two numbers is equal to the sum of their loga- rithms, and that the logarithm of any power of a number 308 ALGEBBA. is equal to the logarithm of the number multiplied by the exponent of the power, we shall have, by making in (1) 2/ = 1, 2, 3, ... in successioa, log, 1=0.000000. log, 2 = 0.000000 + 2(1 + 3!^ + ^ + ...) = 0.693147. log. 3 = 0.693147 + 2 Q + 3l3+_^+...) = 1.098612. log, 4 = 2 X log, 2 = 1 . 386294. log, 5 = 1.386294 + 2 (i + 3l^ + g-lj + ...) = 1 .609438. log, 6 = log, 2 + log, 3 =1.791759. log, 7 = 1.791759 + 2 (l + 3-i3^ + ^{3p+...) = 1.945910. log, 8 = 3 X log, 2 = 2. 079442. log, 9 = 2 log, 3 = 2.197225. log, 10 = log, 2 + log, 5 = 2.302585. 378. Problem. To find the modulus of the common ayatem. By Art. 375, formula (4), we have 309 379. Problem. To compute the Table of common logarithms. By Art. 375, formula (2), we have log, (1 + a;) = log, (1 + a;) X M. Hence, to compute common logarithms, multiply tlie Naperian hgarithms by .4342944819, the modulus of the common system. 380. Problem. To find the numerical value of e,'the Naperian base. Iog,oe = log,e X M=l X .4342944819 == .4342944819. From a table of common logarithms, we find 6=2.71828128.. . 381. Propositions. 1. j^ tlie base is greater than 1, and the number greater than 1, the logarithm is positive. For, let a'^ = n, then x = log^ n, in which a > 1 and n > 1. Then, if x is negative, we have — := n, ; but — < 1 and n. > 1 ; hence, we have a quantity less than 1 equal to a quantity greater than 1, which is absurd. .•. x can not be negative ; . • . a; is positive. 310 ^ij^\jr jjjjjja,.a.t 2. 7/" tlie bane is greater than 1, and the number less Hian 1, ti>e logarithm is negative. For, let a''^=n, then x = log„n, in which a > 1 and n < 1. Then, if x is positive, a'" > 1 ; but n < 1 ; hence, we have a quantity greater than 1 equal to a quantity less than 1, which is absurd. . • . x can not be positive ; - • . a; is neg- ative. 3. Jff the base is less than 1, and the number greater titan 1, the logarithm is negative. For, let a'^n,, then a; = log„n, in which a< 1 and n>l. Then, if a; is positive, a'' < 1 ; but m > 1 ; hence, we have a quantity less than 1 equal to a quantity greater than 1, which is absurd. . ■ . x can not be positive ; . • . a; is neg- ative. 4. if ^'^ ^"^^ ^ ^^ *''™* 1> "'^'^ *^ number less than 1, tlie logarithm is positive. For, let a'^=n, then a;^log„H, in which a 1 and w ■< 1 ; hence, we have a quantity greater than 1 equal to a quantity less than 1, which is absurd. .". x can not be negative; .•. a; is positive. From Props. 1, 2, 3, 4 we have If a; = log„ n, then ( 1. If a > 1 and «, > 1, a; > 0. 2. If a > 1 and ?» < 1, a; < 0. 3. If a < 1 arid n > 1, a; < 0. ,4. If « < 1 and »i < 1, a; > 0. LOGARITHMS. 311 1st. if the base and the number are both greater or boHi less than 1, t/ie logarithm is positive. 2d. If the base is greater than 1, and the number is less than 1, or the reverse, the logarithm is negative. 5. If the base is greater than 1, the modulus is positive; but if the base is less than 1, the modulus is negative. For, by Art. 375, formula (4), we have ^^log^i_ log„ n Then <^ If a > 1 and If a < 1 and ( n>l I »<1 )i> 1 1«<1 log„ w > 0. log, n > 0. loga n < 0. loge n < 0. loga n < 0. log, n > 0. log„ re > 0. log, »i<0. , M>0. M<0. 382. Examples. 1. Given (10)"* = 50, to find re. Ans. a; = 1.3034. 2. Given (10)2^ — 2(10)"= t=: 24, to iind x. Ane. a; ==.778151. 3. Given a*"= + 2pa'' = q, to find x. Ans. loga 4. Given aP' bi" = r, to find X. Ans. X = logr jploga + glogi 6. Given x" = f and a;*" = i/", to find x and y. 6. Given a^ '■''* ^"'^^ — iha,""- 2'"',= e^, to find x. Am. X = — p ± \— ^ ^^!^ ^ + p2. log a ^ THEORY OF EQUATIONS. 383. Form of the General Equation. Every equation of the w* degree involving but one unknown quantity may, by clearing of fractions, trans- posing, reducing, and dividing by the co-efiicient of »", be placed under the form x" + ax"-" + haf-'' + ea;"-8 + ... + kx + l = 0. In this equation, the co-efficients may be integral or fractional, positive or negative, rational or irrational. The exponents may all be considered positive ; for if any of the exponents are negative, let — p he numerically the greatest negative exponent. Then the equation can be freed from negative exponents by multiplying every term by 3f. THEORY OF EQUATIONS. 313 384. Problem. To find the numerical value of Hie first member, when a par- ticular number is substituted for the unknown quantity. Take, for example, the cubic equation, a;3 + aa;2 + fa + c = 0. Substituting n for x, the first member becomes n^ + an^ + 6n, + c ^ 0. When n is a particular number, this result may be most conveniently obtained by detaching the co-efficients, multi- plying the co-efficient of the highest power of the unknown quantity by n, adding the product to the next co-efficient, mu]ti])lying the sum by n, adding the product to the next co-efficient, and so on. The last sum will be the value sought. Thus, in the above equation, we shall have 1, a, b, c. n n'^ -j- are n^ -\- an^ -j- bn n-\- a, n^ -\- an -\- b, m* -f- an^ -\- bn -|- c. The last sum is the result obtained by direct substitution. 385. Illustration. Let us find the numerical value of the first member ol' a;'' -\- 4x^ — 5x — ■ 174 = 0, when 2 is substituted for x. 1. + 4, + 0, - 5, — 174. + 2, + 12, + 24, + 38. + 6, -f 12, + 19, — 136. (' A. 27. 314 ALGEBRA. Hence, when 2 is substituted for x, the first member becomes — 136. Let us now substitute 3 for x. 1, +4, +0, — 5, —174. + 3, +21, +63, +174. + 7, +21, +58, 0. Since 3, when substituted for x, reduces the first member to 0, 3 verifies the equation, and is therefore a root. 386. Examples. 1. Find the numerical values of the first member of the equation x'^ + 3a;* — Ix — 16 =: 0, when 2, 3, and 5 are substituted for x. Am. 10, 125, 949. 2. Find the numerical values of the first member of the equation Zx^ — 2a;* + 4a;2 — a; + 7 = 0, when 1, 2, and 6 are substituted for x. Am. 11, 85, 20881. 387. Proposition. If V ii a root of the general equation, the first member is divisible by x — r. 1st demonstration. Assuming the division performed, denoting the quotient by q, and the remainder, which is independent of x, by r', we shall have x" + ax"-' +...+kx + l=(x — r)q+ /. THEORY OF EQ UA TIONS. 3 1 5 Substituting r for x, the first member will become 0, since r is a root, and {x — r) q will become 0. Then 0=0+/, . • . / = 0. Hence, the first member is divisible by a; — r. 2d demonstration. (1) x" + oo;"-' + . . . + te + i := »" + ax"-' + . . . + fo -f Z. (2; = r"+ ar"-' + •• • 4-fc'- + '- (Ij — (2) = (3) X" + aa;"-i + . . . + te + / = a;" — r" + a (s;"-' — r"-') -\- . . . -\- k (x — r). But the second member of (3) is divisible by x — r ; therefore the first member, which is the first member of the general equation, is divisible by x — r. 388. Problem. To find the quotient when, the first member of the equation, x" + ax"~' -j- . . . -\- hx -\- I ^ 0, is divided by the tmknown quantity, minus a root of the equation. Since x" + ax"-' + bx"-^ + . . . + kx + I = x" — r- + aCx"-'— r"->) + b (a"-2— r'-^) + . . . + i(x— r). The quotient of the first member divided by a; — r can be obtained by dividing the second member by x — r, which gives ^n-i _j_ _^u-2 ^ _(. a;«-3 ^2 _|„ _ _ _ _f. xr"-^ + r"-' + a (x"-2 + a;"-3 r _)_ a;«-4 ^2 ^ . . . _|_ ^f'-^ + r'-^) + 6 (x"-^ + X"-* r + «""' r2 + . . . + .rr"-" + r'-^) Jt. 316 ALGEBRA. Arranging the terms with reference to the descending powers of x, we have a;"-' +(r + a) rB"-^ + (r^ -\- ar + b) x"-^ + ... The co-efficients of the powers of x in the quotient can evidently be obtained by the process employed in finding the numerical value of the first member, when a particular number is substituted for the unknown quantity. Art. 384. 389. Illustrations. 1. Thus, 3 is a root of the equation, x^ 4-4a;» — 5a;— 174 = 0, the first member is divisible by x — 3, and the quotient is obtained thus : 1, +4, +0, - 5, -174. + 3, + 21, + 63, + 174. + 7, + 21, + 58. .-. a;' + 7a;2 + 21a: + 58 = the quotient. ... a;4 _i_ 4a;S _ 5x — 174 = (x— 3) (a;^ + 7a;'^ + 21a; + 58). 2. Again, 2 is a root of the cubic equation a;3_9a;2 -f 26a; — 24 = 0, what are the other roots? Proceeding as above, we shall have 1, -9, +26, -24. + 2, — 14, + 2 4. - 7. + 12. THE OB Y OF EQ UA TIONS. 317 . ■ . x^ — 7a; + 12 = the quotient. .-. a;^ — 9a;2 + 26a; — 24 = (a; — 2) (a;^ — 7a; + 12) == 0. f either a; — 2 = 0, - • . a; = 2. (or a;2 — 7a; + 12 = 0, .-. x = 4or3. 390. Examples. 1. A root of the equation, x" — 12x3 + 48a;2 — 68a; + 15 = 0, is 5 ; factor the first member. Ans. (a; — 5) (a;^ — 7.t2 + 13a; — 3) = 0. 2. One root of the equation, a;3_2a;2_ 23a; +60=0, is 4 ; what are the other roots ? Ans. 3 and — 5. 3. Two roots of the equation, a;* + 2a;3 — 25a;2 — 26a; + 120 = 0, are 4 and — 3 ; what are the other roots? Ans. 2, — 5. 4. Two roots of the equation, X* — 8a;3 — lla;2 ■+ 198a;— 360 = 0, are 3 and — 5 ; what are the other roots ? Ans. 4, 6. 391. Proposition. if the first member of the general equation is divisible by x — r, I is a root of the equation. For, let q denote the quotient; then x" + aa;"-' + bx"-'' + . . . + te + Z = (a; — r) 5. 318 ALGEBRA. Substituting r for a; in both members, we have r" + ar"-^ + 6r"-2 + .. . +kr + l = 0. Hence, r is a root of the equation. 392. Problem. To find ihe remainder when the first member of the general equation is divided 'by x — p, ■i/" p is not a root of the equation. Denoting the quotient by q^ and the remainder, which is independent of x, by /, we shall have X" + as"-' + 6x"-2 -\-... + kx + l=(x — p)q+r'. Making x=p, we shall have p" + op"-' + bp"-^ +...+kp + l=r-r'. Hence, r' is what the first member of the equation becomes when p is substituted for x, which is foimd by the process of Art. 384. 393. Examples. 1. Find the remainder when the first member of the equation x* — 3x^ — 15a;2 + 49a; — 12 :== is divided by 3. 5_ Am. 108. 2. Find the quotient and remainder when the first mem- Ijgj. of j;4__8a;8 — lla;2 + 198a; — 360 = is divided by a; _ 7. Ans. x^—x^ — 18x + 72, 1 44. THEOR Y OF EQ VA TfONS. 319 DEKIVED FUNCTIONS. 394. Beflnition and Notation. A function of a quantity is any expression containing that quantity. For the sake of brevity, let us denote the equation, x" + ax"-' + bx"-' +...+kx + l = 0, by /(s) = 0, which is read a funetion of x is equal to zero. In this expression, / is not a factor, but a symbol denoting that / (x) is an expression involving x. Let / (p),f (q), / (r) . . . denote what / (x) becomes when p, q, r... are sub,";tituted for X. 395. Problem. To find the co-efficients of the powers of y, when x -|- y is svbstituled for x in the equation f (x) = 0, and the result is arranged according to the ascending powers of y. f(x) = x" + ax"-' + bx"-^ + ...+kx+l. f(.x + y) = (x + yy + a{x + yy-^ + ...+k(x + y)+l. Expanding by the Binomial formula, and arranging the result according to the ascending powers of y, f (x -\- y) becomes 3f+ ax"-' + bx"-'' + .. . + kx + l+ \nx"- ' + (?i — 1) ax"-"^ + {n — 2) bx"-^ +. . . fc] y + [re(n— l)af-2 + (n — 1) {n — 2) ax"-^ + {n — 2) (w — 3) 6a;"-* -f . . . J |- + [K(n.— 1) (n— 2)a;"-3 + (n-1) (7i-2) {n—Z)ax"-^+...]-^ 320 ALGEBRA. The co-efficients of 2/" and y" are /(a;) and 1, respectively. Denoting the co-efficients of y, ^ , f^ , . . . by /' (a;), /" (a;), /'" (x), . . . , respectively, then f(x-^y) ^f(x) +f (x) y +/" (x) g +/'" (x) ^ +. . . In this development, /(a;), the co-efficient of y'^, is the original function of x; f (x), the co-efficient of y, is the first derived function of /(a;); /" (a;), the co-efficient of ^ , is the second derived function of /(a;), and so on. By an inspection of the values of /(.r), /' (a;), /" (a;) . . . , we see that any derived function is obtained by multiplying each term of the preceding function by the exponent of x in that term, and diminishing the exponent of x by unity. Thus, let it be required to find the derived functions of x* + 3a;3 -I- 5a;2 — 7a; — 8 = 0. / (x) = a;* + 3a;3 -f 5a;2 — 7a; — 8. /' (a;) = 4x» + 9a;2 + 10a; — 7. /" (x) = 12x2 j^ i8x + 10. /'" (a;) = 24x + 18. /"" (x) = 24. /""'(x) = 0. 396. Proposition. In f (x) involving different powers of x, a value sufficiently great may he assigned to x as to cause any term which occurs to contain the sum of the terms involving the inferior "powers of x any finite number of times, and a value suffieientJ/ij small may he THEOR Y OF EQ UA TIONS. 321 assigned to x as to cause any term which occurs to contain the mm of the terms involving the superior powers of x any finite number of times. Letf(x) = x" + ax"-' + . . . + daf-'*' +...-\-kx + l. 1st. Let dx"^''*' be the r"" term, and let g be the numer- ical value of the numerically greatest coefficient of any of the terms involving powers of x inferior to the (n — r.-|-l)'* power. The sum of the terms which follow the r'* term can not exceed g (af ^' -|- x"~''~' -|- . . . -j- a; -)- 1) which is equal to ^-^ ; , since the series within the parenthesis is a X — 1 geometrical progression; but ^.,.-r., _ gCx-^'-l) _ d{x-l)x"-'*' _ djX-V, x-l -^(a;"--^'-l) -^ g ■ As X increases the numerator increases, and the denomi- nator approaches the limit g, which it can not exceed. Hence, by making x sufficiently great, the numerator may be made to contain the denominator any finite number of times. Since this fraction expresses the quotient of the r"' term divided by the sum of the terms involving the inferior powers of x, and since the r"' term may represent any term, any term which occurs, by assigning a value sufficiently great to x, may be made to contain the sum of the terms involving the inferior powers of x any finite number of times. 2c?. Let X ^ -; then / (x) becomes 322 ALGEBRA. Reducing to a common denominator and factoring, we have i (1 + a^ + ...+%•■-'+... + kf-> + ly% Now, a value sufficiently great may be assigned to y, or, which is the same, a value sufficiently small may be assigned to x, as to cause d'f~' to contain the sum of the terms within the parenthesis, involving the inferior powers of y, any finite number of timies, according to the first part of the proposition; hence, -^ times. this term will contain — times this sum any finite number of times; that is, da;"^'"' will contain the sum of the terms involving the superior powers of x any finite number of times. 397. Proposition. Jf f (s) and f(t) he values of f(x), corresponding to the values X ^ s and x ;= t, respectively, then if x changes from s to t, passing through every intermediate value, f (x) will change from f (s) to f (t), and. will pass through every intermediate value. Let x = v, then f(x) —f(v). If in f (v) we substitute V -\- h for V, we shall have, by Art. 395, f(y+h) =f{v) +/' {V-) h +/" (y) ^ +/'" (t;) ^ + . . . Now, by the preceding proposition, we may make the first term which occurs of the second member of the equation, /(« + h) -/ w =/' w h +/" w ^ +./'" («) I + . . . , THEOR Y OF Eq VA TIONS. 323 contain the sum of tlie following terms any finite number of times. Then, by making h small enough, the second member can be made less than any assignable quantity; therefore, f(v -\- h) — / (v) will be less than any assignable quantity. Hence, as x changes from s to t, passing through every intermediate value, f (x) will change from/(s) to/(t), and will pass through every intermediate value. Scholium 1. It is not asserted that f(x) always increases or always decreases from/(s) to/(<), but that any two con- secutive values of / (x) differ insensibly from each other. Scholium 2. It is to be observed that s, t, /(s), f (t) are not restricted to positive quantities. 398. Proposition. If two numbers substituted for x in the equation, f (x) = 0, give results with contrary signs, then the equation, f (x) =:= 0, has at least one real root between these numbers. Let p and t be these numbers ; then, by hypothesis, / (p) and f{t) will have contrary signs. But as a; changes from p to t, passing through every intermediate value, f(x) will change from f{p) to /(i), and will pass through every intermediate value. But since f(p) and f(f) have contrary signs, is one of these intermediate values ; hence, as x changes from p to t, a certain value of x, which we shall denote by r, will make /(a;) ^ ; . • . r is a root of / (a;) = 0. 399. Proposition. Any equation of aw odd degree, the co-efficients of which are all real, has at least one real root, whose sign is contrary to that of the last term. 324 ALGEBRA. Let f(x) = x'' -\- ax"~' -{-... -\- kx -\- 1=^0, in which n is odd. If we substitute for x a value, v, sufficiently great to make a;" greater than the sum of the remaining terms, f(x), which becomes f(v), will have the same sign as v, since n is odd. If we substitute for x, f (x), which becomes /(O), reduces to I, and, of course, has the same sign as I. 1st. Let I be negative. If a; = 0, / (a;) becomes / (0) = — /, a negative quantity. If a; = + V, f(x) becomes f{+v), a positive quantity. Hence, the equation / (a;) = has at least one real root between and + v, which root is positive, or has a sign contrary to that of I. 2d. Let I be positive. If a; =: 0, / (a;) becomes / (0) = Z, a positive quantity. If a; := — V, f (x) becomes / ( — v), a negative quantity. Hence, the equation / (a;) ^ has at least one real root between and — v, which root is negative, or has a sign contrary to that of I. 400. Proposition. Any equation of an even degree, the co-efficients of which are all real, and whose last term is negative, has at least two real roots with contrary signs. Let f(x) =x" -\- aa;"-i -\- . . . -\-kx — l = 0, in which n is even. Let V bo a quantity sufficiently great to make ■w" > av"-> + 6'y"-2+ ...-\-hv — l. THEORY OF EQUATIONS. 325 Then, since n is even, v'\ and consequently /(d), is positive, whether v is positive or negative. If x= -\- V, f(x) becomes /(-|- ■«), a positive quantity. If a; = 0, / (x) becomes / (0) ^ — I, a negative quantity. If a;^ — V, fix) becomes /( — v), a positive quantity. Hence, the equation / (a;) = has at least one real posi- tive root between -\- v and 0, and at least one real negative root between and — v. 401. Proposition. If the signs of the terms of the equation, f (x) = 0, are all positive, the equation has no positive root. For every positive number substituted for x would make the first member positive, which would not verify the equa- tion. 402. Proposition. If in the equation, f (x) = 0, tlie last term and all the terms containing the even powers of x have the same sign, and all the terms containing the odd powers of x have the contrary sign, the equation has no negative roots. For the substitution of a negative quantity for x will make the signs all plus, if the terms containing the even powers of x are plus; otherwise the signs will all become minus. 403. Proposition. Jf the equation, f (x) = 0, involves only the even powers of x, and all the terms have the same sign, the equation has no real roots. For if the signs are all alike, the substitution of any real quantity, positive or negative,, for x, will render the signs 326 AljijtJiXSIiA. of all the terms alike, which will not verify the equation. In this case, therefore, the roots are all imaginary. 404. Proposition. jff the equation, f (x) = 0, involves only the odd powers of x, a)id tlie terms all have the same sign, and all contain x, the equation has no real root except x = 0. For, by factoring, the equation becomes (1) X (x"-> + 6x"-8 + &"-« + . . . 4- ifc) = 0, which may be satisfied by making a; ^ 0, or (2) a;"-^ + bxf-^ + rfa^-« + ...-\-k^0. But equation (2) involves only the even powers of x, and the terms all have the same sign ; therefore, it has no real root ; hence, equation (1) has no real root except a; = 0. 405. Remarks on tlie Existence of a Root. 1. We have found that any equation of an odd degree of the general form, involving only real co-efficients, has at least one real root ; and that any equation of an even degree of the general form, involving only real co-efficients, whose last term is negative, has at least two real roots. 2. It now remains to be demonstrated that any equation of the general form has a root. 3. It may, indeed, be granted that every equation which is the statement of a determinate problem has at loast one root; for, if not, then the problem is not determinate. But the question is, Has any function of x of the general THEORY OF EQUATIONS. 327 form a" + oa;"~' -\- ...-{- kx -\- I, written at random, and placed equal to 0, a root? It will not do to say, "If the two members of an equation are equal, they must be so for at least some one value of the unknown quantity, real or imaginary;" for the members have only been assumed equal; and now the question is. Can they be made equal for any value assigned to x : 4. The subjoined demonstration of the proposition. Any equation of the general form has at least one root, is, in substance, Cauchy's ; but as it is somewhat difficult, it may be omitted at the discretion of the teacher. 406. Proposition. Any equation, f (x) =; 0, of the general form, in which the co-efficients are real or imaginary, Jias at lead one root, real or imaginary. Let a-f- bV — 1 be substituted for a; in / (x) ; then / (x) will become P -\- Q V — 1, P denoting the algebraic sum of the real quantities, and Q,V — 1 the algebraic sum of the imaginary quantities. If P ^ and Q = 0, then P -\- Q V — 1 =^ 0, and a -\- b l/ — 1 is a root of the equa- tion / (x) = 0. Let us suppose that P and Q are not both 0. Let, now, a-\-bv — l-{-h be substituted for a; in / (x), which may be done by substituting a + 6l/ — 1 for x in the development, f(x + h)=f(x)+f' (x)h+f" (x)l^+f"' ix)^+... Then f(x) will become P+ Ql/ — 1, as above. Some of the co-efficients may vanish, but all can not vanish, since the last term is h"- 328 ALGEBRA. Let h" be the lowest power of h, whose co-efficient does not vanish. Denote this co-efficient by E -f Si' — 1 ; then f(a+bv'^ + h) = P+QV^ + {R + SV^-)h''+... Let P'+Q'V'^ = P + QV^^l+(R+Sl^^)hP+... Let h = Id, t" being — 1 or + 1 ; then P'+ Q'l/^^=:P+ Qr^^ + (i2 + iSl/^^) F + ... P'=P=pRk''+... Q'= Q zp Si" + . . . .-. P'2 + Q'2=P2 + Q2iF.2(PP-f QS)]^^-... ... p'2 _|. Q'3 _ p2 _ Q2 =, q: 2 (^PB 4- QS) fcP 4- . . . Now, if PP + QS is not 0, the sign of the second mem- ber, by making k sufficiently small, will be the same as that of =F 2 (PP -|- QS) k", which may always be made minus by making V = — 1 or + 1 , according as PR + QS is positive or negative. .•. P'2-|-Q'2 — P^ — Q"^ can always be made negative. .-. P'2+ Q'2 + bx"-^ + ...+kx + l^O, into another whose roots shall be less by q than the roots of the given equation. Let y^x — q, then x = y -\- q. THEORY OF EQUATIONS. 345 Substituting this value of x in (1), developing and arranging with reference to the descending powers of y, the equation will take the form, (2) f + ay-' + 6y-2 + . . . + fc'y 4- i' ^ 0. Substituting x — g for ?/ in (2), we have (3) {X- g)"+ a'{x -- g)-' + . . . + fc' (x-q)^l'=Q. Equation (3), when developed and reduced, must be identical with (1) ; for (2) was obtained from (1) by sub- stituting 2/ + 3 for X, and (3) was obtained from (2) by substituting x — -q for y, which must reproduce (1). Our object is to find the co-efficients in (2). If we divide (3) by X — q, the quotient will be (4) (x-qr-^+a'{x-q)-^ + ...^V, and the remainder V, the last term of (2). Dividing (4) by a; — q, the quotient will be (5) (x — gi)'-2 -f (Z (« — 5)»-3 + . . . , and the remainder ¥, the co-efficient of jf in (2). In a similar manner, the co-efficients of jf^, y^ . . . , in (2) can be found. But since (3) and (1) are identical, we can obtain the co-efficients in (2) by operating on (1) as we have on (3), which need not be obtained. The operation of di- vision is performed as in Art. 389, and the remainder is found as in Art. 392. 438. Illustration. Let us transform the equation, a;4 _ 3a.3 _ i5a;2 j^ 49a; — 12 = 0, into another whose roots shall be less than the roots of the given equation by 3. 346 ALGEBRA. We are to divide the first member of the equation by X — 3, the quotient by a; — 3, and so on ; the remainders ■will be the co-efficients of the transformed equation. OPERATION. 1—3—15+49 — 12 + 8 + — 45+12 + - 15 + 4 + + 3+ 9- 18 + 3— 6 — 14 + 3+18 + 6 + 12 + 3 + 9 Performing the first division, we have, for the co-efficients of the quotient, 1, +0, — 15, +4, and the remainder 0, which is the last term of the transformed equation. As the co-efficients of the quotient thus obtained are already detached, we proceed with the second division, and obtain — 14 for the second remainder, which is the co- efficient of y in the transformed equation. In a similar manner, we find that 12 is the co-efficient of y^, 9 the co-efficient of j/*, and 1 the co-efficient of y*. Hence, the transformed equation is y4 ^ 9^,3 ^ 122,2 _ 143, + = 0. 439. Examples. 1. Transform the equation, x^ — 27a; — 36 = 0, into another whose roots are less than the roots of the given equation by 3. Am. y^ + 9y^ — 90 = 0. THEOR Y OF EQ UA TIONS.^ 347 2. Transform x* — ISa;^ — 32a;2 + 17a; + 9 = into an equation whose roots are less by 5 than the roots of the given equation. Ans. y* + 2y^ — lb2y-^ — 11532/— 2331 = 0- 3. Transform x^ -\- 2a;* — Qx"^ — 10a; ^ into an equation whose roots are less by 2 than the roots of the given equa- tion. Ans. y^ + lOy* + 42yS + 8%^ + 70y + 4 = 0. 4. Transform x^ — 4a;^ — 17a; -(- 60 ^ into an equation whose roots are greater by 4 than the roots of the given equation. Ans. y^ — 1%^ + 63y ;= 0. 430, Fifth Transformation. To transform one equation into another, whose second term is warding. (1) x" + aa;"-> +...+hx + l = 0. Let y = x -\ — . then x^y Substituting y for x, we shall have Developing, we shall find that the co-efEcient of t/"""^ is 0, or the second term disappears. This transformation is accomplished by successive divis- ions of (1) by a; -j — , as in the preceding Article. n 431. Examples. 1. Transform a;* — 6a;2 + 8x — 2 = into an equation wanting the second term. Ans. y^ — Ay — 2^0. 348 ALGEBRA. 2. Transform x^ — G.-b^ + Ix — 2 = into an equation wanting the second term. Ans. y^ — by — 4 = 0. 3. Transform the equation x^ — dx^ + 7.4x3 ^ 7.92a;2 _ i7.872a; — .79232 = 0, into another wanting the second term. Ans. y^ —7y^ + 2y~8 = 0. LIMITS OF THE ROOTS. 432. Definitions. 1. The limits of the roots of an equation are the quanti- ties between which the roots are situated. 2. A superior limit of the roots is a quantity greater than any of the roots. 3. An inferior limit of the roots is a quantity less than any of the roots. 433. Proposition. A superior limit of the roots of an equation, or any quantity greater, wUl, when substituted for the unknown quantity, render the first member positive. If a superior limit made the first member 0, it would be a root and not a limit. Therefore, a superior limit can not reduce the first member to 0. If a superior limit made the first member negative, then, since a positive value, greater than the limit, may be as- signed to X which will make a;" numerically greater than the sum of all the other terms, in which case the first member is positive, there would be a root included between this value and the superior limit, and this root would be greater than the limit, which is impossible. Therefore, a superior limit renders the first member positive. THEOB Y OF EQ UA TIONS. 349 If any positive number greater than the superior limit makes the first member negative, then there would be a root included between this number and the superior limit, and this root would be greater than the limit, which is impossi- ble. Therefore, any positive quantity greater than a supe- rior limit will make the first member positive. 434. Proposition. Unity plus that root of the greatest negative co-efficient, whose index is equal to the number of terms pr'eeeding the first negative term, is a superior limit of tlie positive roots. Let g be the numerically greatest negative co-efiicient, and r the number of terms preceding the first negative term. To take the most unfavorable case, suppose all the terms after the first negative term to be negative, and each nega- tive co-efficent to be — g. Then the equation becomes (1) a;" + aa;"-' -\- . . . — g (x"-"- + . . . + a; + 1) = 0. Since the series within the parenthesis is a geometrical pro- gression, (1) becomes (2) a;" + aa;"-' + ...— ?^ T^ = 0- Dropping all the positive terms in (2) but .t", we have (3) ,.._g(^-"-l)^o. ^ ^ X — 1 Now, a superior limit of x in (3) will, for a stronger reason, be a superior limit in (2), since the first member of (2) is the same as that of (8), with the addition of posi tive terms. But the superior limit of x in (3) will be found if we find the superior limit in (4) x"--^-^-0. ^ .r — 1 350 ALGEBRA. For the positive parts of (3) and (4) are the same, and the negative part of (3) is less, numerically, than that of (4). Clearing (4) of fractions, dividing by a;""""'"', and trans- posing g, we have (5) x'-'(x-l)=g. Since x — l<:x, (a— l)'-'■ r, r >■ s' > s, s > t' > t, . . . , be substituted for x in the equa- tion, the results will he alternately positive and negative. By factoring, the equation becomes (1) {X — r) {x — s) {x — t) . . . = 0. Substituting /, s' , tf,..., in succession for x in (1), we have (2) (/ — r) (/ — s) (r' — ■ . . > 0, since the factors are all positive. (3) (s'-r)(s'-s)(s'-0---<0, since one factor is negative and the rest positive. (4) {t'-r)(f-s)(f-t)...>0, since two factors are negative and the rest positive. Cor. If, when two numbers are substituted for x, the results have unlike signs, there is an odd number of roots included between those numbers ; but if the results have like signs, there is no root or an even number of roots included between them. 439. Problem. To find the integral roots of an equation. Find, by Art. 385, the value of the first member when the integral divisors of the last term which are included within the limits of the roots are substituted for the un- known quantity. Those divisors which reduce the first member to are roots. 352 AljbfJillUA. 440. Examples. 1. a;* + 4x3 _ a;2 _ iQ^ —12 = 0. Avs. 2, — 1, — 2, — 3. 2. x^ — 9x2 _j_ 23a; _ 15 =. 0. Ans. 1, 3, 5. 3. a;3 — 3x2 _ lOx + 24 = 0. Ans. 2, 4, — 3. 4. X* — 13x2 + 36 = 0. Ans. 2, 3, — 2, — 3. 5. x= — X* — 13x3 + 13x2 + 36x — 36 = 0. Ans. 1, 2, 3, —2, —3. 6. x" — 2x' — 11x2 + 42a; _ 40 = 0. Ans. 2, — 4, 2 H- l/^, 2 — l/^. STURM'S THEOREM. 441. Statement. Let f (x)=0 be an equation of the n"' degree' freed from equal roots; let fi(x) be its first derived function; let the opera- tion for finding the g. c. d. be applied to f (x) and fi(x), with tlie modification that the signs of Hie remainders be changed; let the operation be continued till a remainder is obtained which is independent of x; let the modified remainders be denoted by fjCx), fjCx), . . . f,(x), . . . f„(x); then, if in the series f (x), f, (x), faCx), . . . fr(x), . . . f„(x), we make x = p, and arrange the signs of the results in a line, and if we make x = v, a number algebraically greater than p, and arrange the signs of the results in a line, the number of variations of signs in the first arrangement minus the number of variations in the second arrangement will be equal to the number of i/ie real roots of f (x) = compreheiided betiveen p and v. THEORY OF EQUATIONS. 353 442. Preliminary Relations. Call/(.i;),/i(a;),/2(a;), . . ./^(rc), . . ./„(a;) Sturm's functions; and call fi(x), f^ix), . . .fr(_x), . . ./„(a;) auxiliary functions. Let 5), 321 93v3n"-?»-i be the quotients obtained in applying the process for finding the g. c. d. to /(x) and fi(x). Then we shall have Ji{x) = q^f^{x) —f^Qjo). /n-2 W = ?"-l/«-l W ~fn(x). 443. Lemmas. 1. fn(x) is not ; for, by hypothesis, it is independent of x, and if it were 0, then / (x) and /, (x) would have a common divisor involving x, and f (x) would have equal roots; but, by hypothesis, f (x) has been freed from equal roots ; hence, f„{x) is not 0. 2. No two consecutive functions can reduce to for the same value of x ; for suppose that a certain value substi- tuted for X should make, for example, fiix) = and /2(a;)=0: then, since fi{x) =q^J^{x) —f^ix), we shall have /3(a;) = 0; then, smce J ^i^) = Is f&i^) —fii^)^ we shall have/4(a;) = 0, and so on, till we should find/„(a;) = 0, which is impossible by Lemma 1 ; hence, no two consecutive functions can reduce to for the same value of x. 3. If any of the functions reduce to when a certain value is substituted for x, the function preceding and the function following will have unlike signs for the same value. C. A. 30. 354 ALGEBRA. for, if a certain value substituted for x makes fr(,x) = 0, then fr-i(x) = — /r+, (x) ; that is, /r-, (a;) and /•+,(«;) have unlike signs. 4. As X increases from p toward v, the number of varia- tions will be neither increased nor diminished till x passes through a root of / (x) = 0, when a variation will be lost. There will be no change of sign in any of Sturm's functions, except when x passes through a root of that function. Let g be a root of /,.(»;) ; then /■(?) = 0, and /r^i(g) and /..,., (5) will have contrary signs. Then, just before and just after x^^q the three functions, /r_, (a:), /(a;), /,.,., (x), will give one permanence and one variation of signs, so that no vari- ation will be either lost or gained in passing through a root of any of the auxiliary functions. Now, let r be a root of f{x) = 0, and let r — h and r-f- h be substituted for x, and the results be developed and ar- ranged according to the ascending powers of h ; then, since /(r)=0 and /, (r) =/'(r), we have. Art. 395, Kr-h) =/,(r) C-h) +/"(r) ^^ +/'"(^) i=^ + . . . /(r + 70 =/i (r) h +/" (r) ^ +/"'(r) ^ + • • - Now, so small a value may be assigned to h as to make the term containing the first power of h greater than the sum of all the following terms; hence, the sign of /(r — h) will be the same as that of /i(r) ( — /i), and the sign of / (r + A) will be the same as that of /, (r) h. Since r is not a root of /, (x) = 0, h may be taken so small that no root of /, (x) = shall lie between r — h and r -\-h; then the signs of /i (r — 7i), /j (r), /i (r + h) will be the same. THEORY OF EQUATIONS. 355 But the sign f(r-~K) is like the sign of /i(r) ( — A), or unlike the sign of /, (r), since h is negative ; but the sign of /i(r — /i.) is like the sign of /i(r); .-. f(r — h) and /j (r — h) have unlike signs. Again, the sign of / (r + '0 is like the sign of /i(r) h, or like the sign of /i(r), since h is positive ; but the sign of /i (r -(- h) is like the sign of /i W ; • ' ■ / (*■ + ^) ^iwi /i (*■ + '0 have like signs. There- fore, in passing from r — h to r -j- h, Sturm's functions lose one variation of signs. Now, since Sturm's functions never lose or gain a varia- tion of signs except when x passes through a root oif(x)=Q, when a variation is always lost, the number of variations lost in passing from p to v will denote the number of real roots between p and v. Cor. 1. The number of variations lost in passing from — (x> to -\- (X will be the number of real roots of the equa- tion, since — co and -|- oo are limits of the roots. In sub- stituting — 00 and +00 for a; in Sturm's functions, it will be sufficient to determine the sign of the first term, since this term will become greater than the sum of all the other terms, and will determine the sign of the result. Cor. 2. In passing any root of /(a;) = to the root next in order, we must pass a root of /, (x) := 0; for, just after passing any root of / (x) = 0, / (x) and /, (x) have like signs, and just before reaching the next root of /(x) ^0, / (x) and /i (x) have unlike signs ; but since the sign of f(x) has not changed, the sign of /i (x) has changed ; hence, a root of /i (x) = has been passed ; hence, between any two consecutive roots of/(a;) = there is a root of/](a;)=^0. Cor. 3. The limits of the roots can be found by substitut- ing 0, 1, 2, 3, . . . in Sturm's functions, till a number is found which gives the same number of variations as -|- oo ; and by substituting 0, — 1, — 2, — 3, . . . till a number is 356 ALGEBRA. found which gives the same number of variations as — oo. The situation of the roots can be found by substituting 0, 1, 2, 3, ... , — 1, — 2, — 3, . . . , and observing when a variation is lost. 444. Application. Find the limits and situation of the roots of a;* — 8x3 _^ j4_^2 _^ 4a;_ g ^ 0. OPERATION. / (x) = X* — 8x» + Ux^ -j- 4a; _ 8. /j (a;) = a;3 — 6a;2 + 7a;+ 1. f.^{x)=5x^- — nx+6. /3(a;) = 76x— ]03. /,(x) = 15125. In deriving /, (a;), fiix), . . . , reject positive factors. /W> .fi(P), f.,(x), fs(x), f^(x). X = — 00 -\- — + — + 4 variations. X = -\- CD + + + + +0 variation. X ^^ — + + — +3 variations. a; = l + -\- — — +2 variations. X = 2 -\- — — -|- +2 variations. x^3 — — ± + + 1 variation. a; = 4 — — + + +1 variation. X = 5 — + + -|-+1 variation. X = 6 + + + + +0 variation. X = — 1 + — + — 4-4 variations. THEORY OF EQUATIONS. 357 Since + 6 gives the same number of variations as + oo, and — 1 the same number as — co, -)- 6 and — 1 are limits of the roots. Since 4 variations are lost in passing from — oo to -|- oo, the equation has 4 real roots. Since 3 variations are lost in passing from to -)- 6, the equation has 3 positive roots, and hence one negative root. Since 1 variation is lost in passing from to -)- 1, one root lies between and -|- 1 ; ■ ' • this root ^ -|- a fraction. Since 1 variation is lost in passing from -j- 2 to + 3, one root lies between -\-2 and +3 ; . • . this root ;= 2 + a fraction. Since 1 variation is lost in passing from -)- 5 to + 6, one root lies between -|-5 and -|-6; . ■ . this root ^b -\- & fraction. Since 1 variation is lost in passing from — 1 to 0, one root lies between — 1 and 0; .'. this root = a negative fraction. If two or more variations should be lost in passing from one integral number to the next consecutive number, the roots lying between these numbers can be separated by increasing the less of these numbers by .1, .2, .3, . . . , or, if necessary, by .01, .02, .03, ... , till a variation is lost. 445. Examples. 1. Find the integral part of the roots of the equation x3_7a; + 7 = 0. ^«s. 1, 1, — 3. 2. Find the integral part of the real roots of the equation X* + a;3 + a;2 + 3a; — 100 = 0. Am. 2, — 3. 3. Find the integral part of the roots of the equation x^ — 12a;2 + 12* — 3 = 0. Ans. 0, 0, 2, — 3. 4. Find the integral part of the real root of the equation x3 + 2x2 _J 3j; ^ 9. Ans. 1. 5. Discuss x^ — Ix* + IZx^ + x"^ — 16a; + 4 = 0. ^ns. The roots are 2, 2, — 1, 2 + t/S, 2 — 1/3. 358 ALOEBBA. HORNER'S METHOD OF APPROXIMATION. 446. General Statement. 1. The integral parts of the roots are first found by Sturm's Theorem. 2. The equation is then transformed into another whose roots are less than the roots of the given equation by the integral part of a certain root of the given equation. One root, at least, of the transformed equation will then be the decimal part of that root of the given equation. 3. The first figure of this decimal root is found by sub- stituting ,1, ,2, ,3, . . . in the transformed equation, till two successive substitutions give results with contrary signs, in which case the less will be the tenths of the root. 4. This transformed equation is then transformed into another whose roots are less than the roots of the first transformed equation by the tenths just obtained. One root of this second transformed equation will be the remain- ing part of the root of the first transformed equation, and so on. 447. Illustration. Let us find one root of the equation a* — 8a;» + 141;^ + 4x — 8 = 0. By Sturm's Theorem we find that 5 is the integral part of one root. We next transform the equation into another whose root." are less than the roots of this equation by 5, thus : THEORY OF EQUATIONS. 359 — 8 4- 5 + 14 — 15 + 4 — 5 — 8 — 5 — 3 + 6 — 1 + 10 — 1 + 45 — 13 + 2 + 5 + 9 + 35 + 44 + 44 + 7 + 5 + 12 The transformed equation then is yi _|_ i2j,3 -j_ 443,2 _j_ 44y _ 13 = 0. Let us now substitute .1, .2, .3, . . . , in succession, in this equation till two successive substitutions give results with contrary signs, thus : 1+12 +44 +44 — 13 + .1 + 1.21 + 4.521 + 4.8521 + 12.1 + 45.21 + 48.521 — 8.1479 The result of this substitution is — 8.1479. 1+12 +44 +44 — 13 + .2 + 2.44 + 9.288 + 10.6576 + 12.2 + 46.44 + 53.288 — 2.3424 The result of this substitution is — 2.3424. 1+12 +44 +44 - 13 + .3 + 3.69 + 14.307 + 17.4921 + 12.3 + 47.69 + 58.307 + 4.4921 The result of this substitution is + 4.4921. 360 ALGEBRA. Since .2 and .3 give results with contrary signs, one root is between .2 and .3, or .2 is the tenths of the root. We next transform the first transformed equation into another whose roots are less than the roots of this equation by . 2, thus : 1+12 +44 +44 — 13 + .2 + 2.44 + 9.288 + 10.6576 + 12.2 + 46.44 + 53.288 — 2.3424 + .2 + 2.48 + 9.784 + 12.4 + 48.92 + 63.072 + .2 + 2.52 + 12.6 + 51.44 + .2 + 12.8 Therefore, the second transformed equation is z* + 12.8z' + 51.44«2 + 63.0722 — 2.3424 -= 0. Since z is a small fraction, we can determine the hun- dredths by neglecting, for the present, the terms containing powers of z higher than the first. We shall then have 63. 072z — 2.3424 = 0; .-. z = .03+, which is the hundredths. We next transform the second transformed equation into another whose roots are less than the roots of this equation by .03, and so on. These successive transformations may be combined as in the following condensed form : THEORY OF EQUATIONS. 361 — 8 5 + 14 -15 + 4 — 5 — 8 (6.2360679 — 5 — 3 5 — 1 10 49 49 58 )7 37 — 1 + 45 44* 9.288 — 13* 10.6576 2 5 9 35 — 2.3424* 1.93880241 7 5 44* 2.44 53.288 9.784 — .40359759* .39905490 12* .2 46.44 2.48 63.072* — .00454269 1.554747 .00400954 12.2 .2 48.92 2.52 64.626747 — .00053315 1.566321 .00046777 12.4 .2 51.44 .38 66.19306 .31608 8* — .00006538 .00006014 12.6 .2 51.82 .38 66.50915 .31656 .00000524 12.8* .03 52.21( .38( 66.825|71 12.83 .03 52.59 .08 74* 12.86 .03 52.68 .08 12.89 .03 52.76 112.92* After obtaining two decimal places, the work can be contracted by omitting unnecessary decimal places at the right ; and after three decimal places have been found, the remaining figures of the root can be found with sufficient C. A. 31. 362 ALGEBRA. accuracy by division. The other positive roots can be found in a similar manner. The negative roots can be found by substituting — y for x, and finding the positive roots of the resulting equation. 448. Rule. 1. Find the integral parts of the roots, by Sturm's Theorem. 2. Transform the equation into another whose roots are less than the roots of the given equation by the integral part of one of the positive roots. 3. Substitute .1, .2, .3, . . . in Hie transformed equation till two successive substitutions give results with contrary signs, and the less wiU be the tenths of the root. 4l. Transform, this transformed equation into another whose roots are less than the roots of this equation by the tenths just obtained. 5. Disregard all the terms containing higher powers of the unknown quantity than the first, and find the hundredths from the resulting equation. 6. Transform the second transformed equation into anoUier whose roots are less than the roots of this equation by the hun- dredths just obtained. 7. Disregard all the terms containing higher powers of the unknown qtumtity than the first, and find from the resulting equation the remaining figures of the root. 449. Examples. 1. Given a;' — 7a; + 7 = 0, to find x. Ans. 1.356896, 1.692021, — 3.048917. 2. Given x* — Vlx"^ + 12a; — 3 = 0, to find x. Ans. 2.858083, .6060183, .443278, —3.907378. THEORY OF EQUATIONS. 363 3. Given x^ —•Sx^ — 2x + 5 = 0, to find x. Ans. 8.128419, 1.201639, —1.830059. 'i. Given x^ = 41068625, to find x. Ans. 845. 6. Extract the cube root of 43614208. Ans. 352. 6. Extract the fifth root of 86936242722357. Ans. 517. CUBIC EQUATIONS. 450. Cardan's Formula. Causing the second term to disappear, the cubic equation will take the form (1) x^ +bx + c = 0. Let y -\-e = x; then we shall have (2) (y + zy +b(y + z) + c--=0. ... (3) yB^^B^^Syz + b-)(y + ^')+c = 0. Let Syz + b = 0; then z = — ^ , and (3) becomes (4) y3 j^z^ +e = 0. Substituting the value of z, (4) becomes (5) ,3_^ + c = 0. (6) y' + cy^=^ 63 27" (7) 2,3^__±^^+_. (8) 2/ = V-^±^4+27- 30 i ALGEBRA. Substituting the value of y^ in (4), we find (9) ,' = -t^^t.+ '^. But x^y -\- z; hence, we shall have \''4 "^ 27 ■ If — -|- -=- is negative, the formula fails, since the root can not be obtained. In this case, contrary to appearance, the roots are all real and unequal, as is thus proved : / (a;) = x^ -\- bx -]- c. f,(x)^ Sx^ + b. /j (a;) = — 26a; — 3c. f^(x) = — 463 — 27c2. In order that the roots be all real, — oo substituted in Sturm's functions must give three variations, and + 00 three permanences, which can be the case only when — 463 _ 27c2 > 0. -.2 7,3 ■•• V+2T <'■ In this case 6 is negative, and sif > -7 ' The roots are unequal, since fi(x) is not 0. By combining the signs in Cardan's formula, nine values are obtained ; but a cubic equation can have but three roots, and these roots must satisfy the condition 1/2 = — „ • THEORY OF EQVATIOm. 365 451. Examples. 1. Given x^ — 12x— 16 = 0, to find x. Am. 4, — 2, — 2. 2. Given x^ -^x^~?,x= 12, to find x. Am. 3, — 2, — 2. 3. Given x^ + 6a;2 _ 32 = 0, to find x. Am. 2, — 4, — 4. 4. Given a;'' — 27a; + 54 = 0, to find x. Am. 3, 3, — 6. 5. Given x^ + Sx^ — 24x — 80 -= 0, to find x. Ans. 5, — 4, — 4. 6. Given x^ — 6x^ + 13x — 10 = 0, to find x. Am. 2, 2 + l'''^^, 2 — 1/^. 452. ETans's Solution of Cubics. (1) x^ ^^ax -\- b. .-. (2) a;2=a + -. ,-. (3) x^ + yja + l- ■. (4) x=±^a- ±^a- ±V^- ^V« + 4 366 ALGEBRA. 453. Rule. 1. Find the square root of a -|- b, divide b by this root, add a to the quotient, extract the square root of the sum, divide b bij this root, and so on; tlie sitceessive roots wiU afford nearer and nearer approximations to the true root of the equation. 2. Jff h is positive, there wUl be one positive root, which may be found by considering the radicals all positive. Z. If h is negative, there viUl he one negative root, which may be found by considering the radicals all negative. 4. To find the other roots, transpose all the terms of the equa- tion to the first member, which divide by the unknown quantity minus the root obtained, and the quotient placed equal to will be a quadratic which will give the other roots. Scholium. This method is applicable to the failing case of Cardan's formula, and Cardan's formula is applicable to the failing case of Evans's method. 454r. Examples. 1. Given o;^ = 6a; + 2, to find x. Ans. 2.601679, —.339877, —2.261802. 2. Given a;' —lx = — 7, to find x. Ans. — 3.048. BIQUADRATIC EQUATIONS. 455. Descartes's Method. Causing the second term to disappear, the biquadratic equation will take the form (1) x'^ -{-bz"^ -\-cx-\-d = 0. THEORY OF EQUATIONS. 367 Assuming the first member resolved into quadratic factors, the co-efficients of x will be equal with contrary signs, since there is no term involving «*, and we shall have (2) a;* + 6a;^ + ca; + d = (a;^ -{- mx -\- n) (x^ — mx -\-p). We are now to find values for m, n, and p which will satisfy the equation. Developing, we have (3) x*-\-bx^-l-cx-\-d=^x*-\-(p-\-n--m^)x^-\-m(p—n)x-\-p>i. Equating the co-efficients of the like powers of x, we have (4) p -\-n — m^ = 6. (5) m (p -:— fi) ^ c. (6) pn^ d. (4) gives (7) p -{- n ^ m^ -\- b. (5) gives (8) G p — w = — . (7) + (8) _ 2 .(9) P = l(r.^ + h + ^. (7) - (8) 2 = (10) . = i(.^ + 6-;). Substituting the values of p and n in (6), and reducing, we shall have (11) m^-f 26m^+(62 — 4d)m2— c2 = 0. Taking m^ as the unknown quantity, this equation may be regarded as a cubic. It will have one real root, since the last term is negative ; and this root may be found by Horner's method. Cardan's formula, or Evans's method. "We may therefore regard m^, and hence m, as known. 368 ALGEBRA. Now, equation (1) will be satisfied x^ -\-mx-\-n-=^Q. \ — m ±: Vm'^ — in X= 7: ■ If , I ■ • 1 _x^ — mx-j-p = 0.) \ m±:\/m^ — 4p 2 456. Examples. 1. Given x* + 4x^ + Sx^ — 44a; — 84 = 0, to find x. Ana. 3, — 2, ^ 2. Given x* — 6x^ —8x — 3 = 0, to find x. Ans. — 1, — 1, — 1, 3. 3. Given a* — 12a!» + 'j^x'^ — 78* + 40 = 0, to find x. Ans. 1, 2, 4, 5. (x^+y = 7. ) 4. Given < [- Required a; and y. (y'^-^x^U.) Cx = 2, 3.131313, — 1.848126, or — 3.283186. Ans. I (y = 3, — 2.805118, 3.584428, or — 3.779309. Two equations of the m'* and n"' degree respectively, will, by elimination, give one equation of the mn"' degree. ECLECTIC EDTJC^'1'IOJirA.Xi SERIES. THE Eclectic Series of Geographies, By A. VON STEINWBHR. The Primary Geography) ^o. 1. The plan of this book is natural, the language simple, and the definitions and descriptions exact. Illustrated, small 4to. The Intermediate Geography, iN'o. 2; for more advanced classes. Contains tlie leading principles of the science, so arranged as to give correct ideas to pupils without requiring the constant aid of the teacher. Full instructions in Map Drawing. Illustrated, large 4to. TJte School Geography , No. 3; embraces a full Mathemat- ical, Physical, and Political description of the Earth, and is intended for the highest classes in this branch of study. Illustrated, large 4to. GRADED SCHOOL ARITHMETICS: A New Series, on an Entirely New Flan, BY E. E. WHITE, M. A. White's Primary Arithmetic. The distinguishing feature of this book, as well as of the series of which it forms a part, is the complete union of Mental (oral) and Written Arithmetic. This is secured by making every oral exercise preparatory to a written one, and by uniting both as the essential complements of each other. Illustrated, 16mo., 144 pages. White's Intermediate Arithmetic. This treatise possesses three very important characteristics: 1. It is specially adapted to the grade of pupils for which it is designed, and is not an abridgment of the Complete Arithmetic. 2. It combines Mental and Written Arith- metic in a practical and philosophical manner. 3. It faithfully embodies the Inductive Method. Illustrated, 16mo., 192 pages. White's Complete Arithmetic. This, like the preceding books of tlie series, combines Mental and Written Arithmetic. It embraces a particularly full aud clear presentation of the subject of Percentage, ijvvith its various applications. Abridged Methods of operation are given under most of the rules. Illustrated, 12mo., 320 pages. K^CH SERIES compi>i;te iw three books. ECLECTIC JS,JDU