The original of this book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924001498298 Cornell University Library QA 155.P37 1875 An elementary treatise on algebra, to wh 3 1924 001 498 298 ELEMENTARY TREATISE ALGEBRA: TO WHICH ARE ADDED EXPONENTIAL EQUATION8 LOGARITHMS. By BENJAMIN PEIRCE, A. M., PSBKIHB PROFESSOR OF ASTRONOMY AND MATHEMATICS IB HARVARD UNIVERSITY. WITH TABLE OF ERRATA. CAMBKIDGE : CHARLES W. SEVER, Ukivehsity Bookstoke. 1875. /^CORNELiS, JUNI'VERSITY V LIBRARY - Entered accoiiing to Act of Congress. In the year 1864. by Wttxiam il. Dennkt, In the Clerk's Office of the District Court of the District ol Massachusetts CONTENTS. ALGEBRA. CHAPTER 1. FUNDAMENTAL PROCESSES OF ALGEBKb SECTION I. Definitions and Notation. II. Addition. . III. Subtraction. IV. Multiplication. V. Division. 9 14 CHAPTER II. FRACTIONS AND PROPORTIONS. SECTION I. Reduction of Fractions. II. Addition and Subtraction of Fractions. III. Multiplication and Division of Fractions. IV. Proportions. 26 26 40 43 CHAPTER III. EQUATIONS OF THE FIRST DEGREE. 50 SECTION I. Putting Problems into Equations. ... 50 II. Reduction and Classification of Equations. . 59 III Solution of Equations of the First Degree, with one unknown quantity. . . 65 IV. Equations of the First Degree containing two or more unknown quantities. ... 85 CHAPTER IV. NUMERICAL EQUATIONS. 110 SECTION I. Indeterminate Coefficients 110 II. Derivation 112 III. Numerical Equations. . .... 117 IT CONTENTS. CHAPTER V. POWERS AND ROOTS. . ■ 130 SECTION I. Powers and Roots of Monomials. ■ 130 II. Calculus of Radical Quantities 133 III. Powers of Polynomials HI IV. Roots of Polynomials 150 V. Binomial Equations. . .... 155 CHAPTER VI. EQUATIONS OF THE SECOND DEGREE. . 161 CHAPTER VII. PROGRESSIONS. ... . .186 SECTION 1. Arithmetical Progression. .... 186 II. Geometrical Progression 195 CHAPTER VIII. GENERAL THEORY OF EQUATIONS. . 201 SECTION I. Composition of Equations 201 II. Equal Roots 210 III. Real Roots , ... 214 CHAPTER IX. CONTINUED FRACTIONS. . . 248 EXPONENTIAL EQUATIONS AND LOGARI1 HMS. SECTION I. Exponential liquations. ... 263 II. Nature and Properties of Logarithms. 266 III. Common Logarithms and their Uses. 270 ALGEBRA. CHAPTER I. FUNDAMENTAL PROCESSES OF ALGEBRA. SECTION I. Definitions and Notation. 1. Algebra, according to the usual definition, is that branch of mathematics in which the quantities considered are represented by the letters of the alphabet, and the operations to be performed upon them are indicated hy signs. In this sense it would embrace almost the whole science of mathematics, elementary geometry alone being excepted. It is, consequently, subject in common use to some limita- tions, which will be more easily understood, when we are advanced in the science. 2. The sign -f- is called plus or more, or the post' live sign, and placed between two quantities denotes that they are to be added together. Thus 3 -J- 5 is 3 plus or more 5, and denotes the sum of 3 and 5. Likewise a -\- b is the sum of a and b, or of the quantities which a and b represent. 1 ALGEBRA. [CH. I. § I. Signs of Addition, Subtraction, Multiplication, and Division. 3. The sign — is called minus or less, or the negative sign, and placed between two quantities denotes that the quantity which follows it is to be subtracted from the one which precedes it. Thus 7 — 2 is 7 minus or less 2 and denotes the remain- der after subtracting 2 from 7. Likewise a — 6 is the remainder after subtracting b from a. 4. The sign X is called the sign of multiplication, and placed between two quantities denotes that they are to be multiplied together. *A point is often used instead of this sign, or, when the quantities to be multiplied together are represented by letters, the sign may be altogether omitted. Thus 3 X 5 x 7, or 3.5.7 is the continued product of 3, 5, and 7. Likewise 12 x a x b, or 12 . a . b, or 12 a b, is the continued product of 12, a, and b. 5. The factor of a product is sometimes called its coefficient, and the numerical factor is called the numerical coefficient. When no coefficient is writ- ten, the coefficient may be considered to be unity. Thus, in the expression 15 a b, 15 is the numerical co- efficient of a b ; and, in the expression xy, 1 may be re- garded as the coefficient of x y. 6. The continued product of a quantity multiplied repeatedly by itself, is called the power of the quan- tity ; and the number of times, which the quantity is taken as a factor, is called the exponent of the power. Th? srwor is expressed by writing the quantity CH. 1. § I.] DEFINITIONS AND NOTATION. 3 Coefficient, Power. Root. once with the exponent to the right of the quantity, and a little above it. When no exponent is written, the exponent may be considered to be unity. Thus the fifth power of a is written a?; but when a stands by itself, it may be regarded as a 1 . 7. The root of a quantity is the quantity which, multiplied a certain number of times by itself, pro- duces the given quantity ; and the index of the root is the number of times which the root is contained as a factor in the given quantity. The sign V _ i s called the radical sign, and when prefixed to a quantity indicates that its root is to be extracted, the index of the root being placed to the left of the sign and a little above it. The index 2 is generally omitted, and the radical sign, without any index, is regarded as indicating the second or square root. 2 Thus, y/a or ^/a is the square root of a, l/a is the third or cube root of a, t/a is the fourth root of a, n y/a is the reth root of a. 8. The signs -r- and : are called the signs of di- vision, and either of them placed between two quan- tities denotes that the quantity which precedes it is to be divided by the one which follows it. The process of division is also indicated by placing the dividend over the divisor with a line between them. Thus, a ■— b, or a : b, or - denotes the quotient of a di- b vided by 6. ALGEBRA. [CH. I. $ 1 Signs of Equality and Inequality. Algebraic Quantity. 9. The sign = is called equal to, and placed be- tween two quantities denotes that they are equal to each other, and the expression in which this sign occurs is called an equation. Thus, the equation a = b denotes that a is equal to b. 10. The sign > is called greater than, and the =ign < is called less than; and the expression in which either of these signs occurs is called an inequality. Thus, the inequality a^> b denotes that a is greater than b; and the inequality a<^b denotes that a is less than b ; the greater quantity being always placed at the opening of the sign. 11. An algebraic quantity is any quantity written in algebraic language. 12. An algebraic quantity, in which the letters are not separated by the signs + and — , is called a monomial, or a quantity composed of a single term, or simply a term. Thus, 3 a 2 , — 10 a 3 x are monomials. 13. An algebraic expression composed of several terms, connected together by the signs -f- and — , is called a polynomial, one of two terms is called a binomial, one of three a trinomial, &c. Thus, a 2 -f- 6 is a binomial, c-\-x — y is a trinomial, &e. 14. The value of a polynomial is evidently not affected by changing the order of its terms. Thus, a — b — c-\-d is the same as a — c — b -}- d, or a-\-d — b — c, or — b-\-d-\-a — c, &c. CH. I. § I.] DEFINITIONS AND NOTATION. 5 Degree, Dimension, Vinculum, Bar, Parenthesis, Similar Terms. 15. Each literal factor of a term is called a dimen- sion, and the degree of a term is the number of its dimensions. The degree of a term is, therefore, found by taking the sum of the exponents of its literal factors. Thus, 7x is of one dimension, or of the first degree; 5 a 2 b c is of four dimensions, or of the fourth degree, &c. 16. A polynomial is homogeneous, when all its terms are of the same degree. Thus, 3 a — 2 6 — |— c is homogeneous of the first degree, 8 a 3 b — 16 a 2 b 2 -)- b* is homogeneous of the fourth degree. i 17. A vinculum or bar , placed over a quantity, or a parenthesis ( ) enclosing it, is used to express that all the terms of the quantity are to be considered together. Thusj (a -\-'b -)- c) x d is the product of a -\- b -j- c by d, y/x 2 -\-y 2 , or y/ { x * -\-y*) ls 'he square root of x 2 -\- y 2 . The bar is sometimes placed vertically. Thus, a — 26 + 3c is the same as (a— 2 6+3c)a;+(5« 2 T )-3 — 2d)z*+(— 3c-(-4rf— l)* 3 . 18. Similar terms are those in which the literal factors are identical. Thus, 1 ab and — 3 a b are similar terms, and — 5a*b 3 and 3 a 4 b s are similar ; but 2 a 4 6 s and 2 a 3 6 4 are not similar. 1* x-\-oa 2 x 2 —3c + 3 + 4rf — 2d — 1 ALGEBRA. [CII. I. § I. Reduction of Polynomials. 19. The terms of a polynomial which are preceded by the sign + are called the positive terms, and those which are preceded by the sign — are called the negative terms. When the first term is not preceded by any sign it is to be regarded as positive. 20. The following rule for reducing polynomials, which contain similar terms, is too obvious to require demonstration. Find the sum of the similar positive terms by add- ing their coefficients, and in the same way the sum of the similar negative terms. The difference of these sums preceded by the sign of the greater, may be substituted as a single term for the terms from which it is obtained. When these sums are equal they cancel each other, and the corresponding terms are to be omitted. Thus, a 2 6 — 9 a 6 s + 8 a 2 6 + 5 c — 3 « 2 6 + 8 a 6 2 -j- 2a s i-j-c + a6 2 — 8c is the same as 8a a 6 — 2 c. 21. EXAMPLES. 1. Reduce the polynomial 10 a 4 -J- 3 a 4 -}- 6 a 4 — a 4 — 5 a 4 to its simplest form. Ans. 13 a 4 . 2. Reduce the polynomial 5« 4 6-f-3v/a& 2 c — 7 a 6 -|- 17a64-2 v /a6 2 c— 6 a 4 6 — 8 ^/ab 2 c — 10a6-f 9 a 4 6 to its simplest form. Ans. 8 a 4 b — 3 ^/ a b 2 c. 3. Reduce the polynomial 3 a — 2 a — 7/-J- 3/-J-2 a ■•- 4/ — 3 a to its simplest form. Ans. 0. CH. I. § II. J ADDITION. Addition. SECTION n. Addition. 22. Addition consists in finding the quantity equivalent to the aggregate or sum of several differ- ent quantities. 23. Problem. To find the sum of any given quantities. Solution. The following solution requires no de- monstration. The quantities to be added are to be written after each other with the proper sign between them, and the polynomial thus obtained can be reduced to its simplest form by art. 20. 24. EXAMPLES, 1. Find the sum of a and a. Ans. 2 a. 2. Find the sura of 11 a; and 9 x. Ans. 20 x. 3. Find the sum of 1 1 a; and — 9 x. Ans. 2 x. 4. Find the sum of — 11a; and 9 x. Ans. — 2x. 5. Find the sum of — 11a; and — 9 a;. Ans. — 20 a; 6. Find the sum of a and — b. Ans. a — 6, 7. Find the sum of — 6/, 9/, 13/, and — 8/. Ans. 8/. 8. Find the sum of — 12 b, — 4 b and 13 b. Ans. —3 b. 9. Find the sum of y/x -\- a x — a b, a b — y/x~ -\- x y, ax-\-xy — 4 a b, Vi" -}- s/aT — z,and xy -\-xy-\-ax. Ans. 2 ^/x-\-3ax — 4ab-\-4xy — x. ALGEBRA. [CH. I. § III. Subtraction. 10. Find the sum of 7 x* — 6 vT+ 5 z 3 z + 3 — g — xs 3 — y/x —8 — g _a;2_|_ Sx~—3x s z—l+7g — 2x*-\-3 v/J+Sxtz — l—g x i + 8 y/x — 5s 3 z-t-9— g Ans. 4 1 2 + 3 v^ +2 + 5^. SECTION in. Subtraction. 25. Subtraction consists in finding the difference between two quantities. 26. Problem. To subtract one quantity from another. Solution. Let A denote the aggregate of all the positive terms of the quantity to be subtracted, and B the aggregate of all its negative terms; then A — B is the quantity to be subtracted, and let C denote the quantity from which it is to be taken. If A alone be taken from C, the remainder C — A is as much too small as the quantity subtracted is too large, that is, as much as A is larger than A — B. The required re- mainder is, consequently, obtained by increasing C — A by the excess of A above A — B, that is, by B, and it is thus found to be C— A -j- B. The same result would be obtained by adding to C the quantity A — B, with its signs reversed, so as to make it — A-\-B. Hence, To subtract one quantity from another, change the signs of the quantity to be subtracted from -\- to — , CH. I. ^ IV.] MULTIPLICATION. Multiplication of Monomials. and from — to -j-, and add it with its signs thus re- versed to the quantity from which it is to be taken. 27. EXAMPLES. 1. From a take b -\- c. Ans. a — b — c. 2. From a take — 6. Ans. a-\-b. 3. From 5 a take — 5 a. Ans 10 a. 4. From 7 a take 12 a. Ans. — 5 a. 5. From — 19 a take — 20 a. Ans. a. G. From 12 take — 7. Ans. 19. 7. From — 2 take 5. Ans. — 7. 8. From — 11 take —20. Ans. 9. 9. From3a— 176— 10 6 + 13 a — 2a take 6 6 — 8 a — b — 2 a + 3 d + 9 a — 5 A. Xiil 15 a — 32 b — 3 d + 5 h. 10. Reduce 32 a -f- 3 6 — (5 a -J- 17 b) to its simplest form. Ans. 27 a — 14 6. 11. Reduce a-f&— (2 a — 3 5)— (5a-J-7 6) — ( — 13 a + 2 6) to its simplest form. ^rcs. 7 a — 56. SECTION IV. Multiplication. 28. Problem. To find the continued product of several monomials. Solution. The required product is indicated by writing the given monomials after each other with the sign of multi- plication between them, and thus a monomial is formed, which is the continued product of all the factors of the given 10 ALGEBRA. [CH. I. § IV Multiplication of Polynomials. monomials. But, as the order of the factors may be changed at pleasure, the numerical factors may all be united in one product. Hence the coefficient of the product of given mono- mials is the product of their coefficients. The different powers of the same letter may also be brought together, and since, by art. 6, each exponent de- notes the number of times which the letter occurs as a factor in the corresponding term, the number of times which it occurs as a facior in the product must be equal to the sum of the exponents. Hence every letter which is contained in any of the given factors must be written in the product, with an exponent equal to the sum of all its exponents in the different factors. 29. EXAMPLES. 1. Multiply a b by c d e. Ans. abode. 2. Find the continued product of 3 a b, 2 c d, and efg. Ans. 6 a bedefg. 3. Multiply a m by a n . Ans. a m + n . 4. Find the continued product of fi a 3 , a 1 , 7 a 5 , and 3 a e . Ans. 105 a 21 . 5. Multiply 7 a 3 £ 2 by 10 a b 5 c* d. Ans. 70 a* V c 3 d. 6. Find the continued product of 5 a 3 b*, a a b 8 , and 4 a b a c. Ans. 20 cfi b is c. 7. Find the continued product of a m bP c», a n b r c*, and a m + n b. Ans. a 9m + 2 "6P+ r + 1 c«+* 30. Problem. To find the product of two poly- nomials. CII. I. § IV.] MULTIPLICATION. 11 Multiplication of Polynomials. Solution. Denote the aggregate of all the positive terms of one factor by A and of the other by B, and those of their negative terms respectively by C and D ; and, then, the fac- tors are A — C and B — D. Now if A — C is multiplied by B it is taken as many times too often as there are units in D ; so that the required product must be the product of A — C by B, diminished by the product of A — C by D ; that is, (A — C) (B — D) = (A — C) B — (A—C)D. Again, by similar reasoning, the product of A — C by B, that is, of B by A — C, must be (A — C)B = AB — BC, and that of (A — C) by D must be {A — C)D = AD — CD; and, therefore, the required product is, by art. 26, (A — C)(B — D)=AB — BC—AD + CD. The positive terms of this product, AB and CD, are ob- tained from the product of the positive terms A and B, or from that of the negative terms — C and — D ; but the negative terms of the product, as — BC and — AD, are obtained from the product of the negative term of one factor by the positive term of the other, as — C by B or — D by A. Hence, The product of two polynomials is obtained by multiplying each term of one factor by each term of the other, as in art. 28, and the product of two terms which have the same sign is to be affected with the sign +j while the product of two terms which have contrary signs is to be affected by the sign — . The result is to be reduced as in art. 20. 12 ALGEBRA. [cH. I. § IV. Multiplication of Polynomials. 31. EXAMPLES. 1. Multiply x 2 -J- y 2 by x -f- y. Ans. x s -\- a; 2 y -\- x y% -\-y a . 2. Multiply x b -j- x y 6 -|- 7 a x by a a; -J- 5 a a;. Ans. 6 a a 6 + 6 a x* y* -j- 42 a 2 a; 8 . 3. Multiply — a by 6. .4 ns. — a b. 4. Multiply a by — 6. Ans. — a 6.' 5. Multiply — a by — 6. Ans. a b. 6. Multiply —3 a by 14 c. Ans. — 42 a c. 7. Multiply — 6 a 3 6 2 by — 1 1 a b 3 c. Ans. 66 a 4 6 s c. 8. Find the continued product of — a, — a, — a, and — a. Ans. a 4 . 9. Find the continued product of — cPb,c s e, — a, — e 3 z 9 , c, — 2 ax, — 3 a 6 e x, — 7, and b 3 x 3 . Ans. 42 a 5 b s c 3 e 5 a; 7 . 10. Find the continued product oilabx, — ax, — x, b^x", —2 b, — 3, and — 5 a 7 b 3 z 5 . 4«s. — 210 a 9 ft 7 z 16 . ' 11. Multiply a -f- 6 by c -(- a". 4ras. ac-|-aa'-|-&c-}-&d'. 12. Multiply a 3 -f- 5 2 — c by a 2 — 6 3 . 4ns. a5 — a 3 6 3 +a 2 4 2 — a?c — b 5 + b 3 c. 13. Multiply a-\-b-j-c by a-\-b — c. 4ns. a 2 + 2a& f 6 2 — c 2 . 14. Multiply a; 2 — 3a;— 7 by a— 2. 4ns. a; 3 — 5 a; 2 — a; + 14. 15. Multiply a 2 + a 4 -f- a 6 by a 2 — 1 . 4 ns. a 8 — a 2 . 16. Multiply 8 a 9 6 3 + 36 a 8 M + 54 a 1 b s + 27 a 6 6« by 8 a*> b 3 — 36 a 8 6 4 + 54 a> 6» — 27 n^ b&. Ans. 64 a 18 J6 _ 432 a™ b B -f 972 a 14 6" — 729 a 12 fiis. CH. I. § IV.] MULTIPLICATION. 13 Product of Sum and Difference : of Homogeneous Quantities. 17. Find the continued product of 3 x -j- 2 y, 2 x — 3 y, — x-\-y, and — 2z- y Arts. 12 z<— \ h. 9cd8 Ans. 4. Divide — 135 by — 5 a. 5. Divide 7 a 3 x 2 by 21 a 5 %\ 6. Divide a m by a n - 7. Divide —3 a m b n by —4 aP bi c r . Ans. 8. Divide a by — a. 9. Divide — a by a. 10. Divide — a by — a. 37. Corollary. If the rule for the exponents is applied to the case in which the exponent of a letter in the dividend is equal to its exponent in the divisor, when, for instance, tf» is to be divided by a m , the exponent of the letter in the '"• 7be«h' Ans. 27 a' AnS We? Ans. a m — n 3 a m-Pb» ■-* 4c r Ans. - -1. Ans. - -1. Ans, . 1. 16 ALGEBRA. [cH. I. § V. Exponent equal to Zero. Negative Exponents. quotient becomes zero. But the quotient of a quantity di- vided by itself is unity. Whence any quantity with an exponent equal to zero is unity. Thus, a m -*- a m = a° = 1. 38. Corollary. When, in example 6 of art. 30, the ex- ponent n of a in the divisor is greater than its exponent m in the dividend, the exponent m — n in the quotient is nega- tive ; and a negative exponent is thus substituted for the usual fractional form of the quotient. Thus, if m is zero, we have a -=- of* = 1 -=- a n == — = a~ n . a n In the same way we should have afi 2 e 3 -f- a 5 ft 2 e 8 d = a 1 6* e 3 -r- a 5 62 c 8 d 1 = a~ 4 c~ s d~\ Any quotient of monomials may thus be expressed by means of negative exponents without using frac- tional forms. 39. EXAMPLES. 1. Divide 5 a 4 b 3 c 2 d by 15 a b 5 d* d? e*. Ans. Z-ia?l-2d-2e-* 2. Divide 6 a 7 b by 9 a V. Ans. §a 6 6- 6 =2.3- , a 6 6-8. 3. Divide 1 by 8 a 11 b. Ans. %a-" b- 1 = S~ i a~ n b~ l . 4. Divide 3 by a. Ans. 3 a' 1 . 40. Corollary. Quantities, thus expressed by means of fractional exponents, may be used in all CH. I. § V.] DIVISION 17 Division of Polynomials. calculations, and may be added, subtracted, multi- plied, or divided by the rules already given, the signs being carefully attended to. 41. EXAMPLES. 1. Findthesumof7a-3+9a"'6-.P— Gab-^c',— 3a"» Sa m b-P -|-11 ab-^c", a~ 3 — lia m b~P. Ans. 5 a~ 3 -J- 5 a b~ 2 c e . 2. Reduce the polynomial 9 a~ 3 b~ 2 c* — 7ba~ 3 -\- (18a- 3 b — 5a n b m + c* — 3.Z 5 ) — (3a n b m — a- 3 b-2c* -i-3 c" — 5 . 2 s ) to its simplest form. Ans. 10a-3&- 2 c 4 + lla- 3 &— 8a n b m — 2c"-f 2.2 s . 3. Multiply a~ m by a n . Ans. a~ m + n = a n ~ m - 4. Multiply a™ by a~ n . Ans. a m ~ n . 5. Multiply a~ m by a - *. ^Ins. a- m - n = a -(m + M> . 6. Find the continued product of 11 a -2 , —2 a~ s , 4 a 8 , and— 9 a 7 . 4ns. 792 a e . 7. Find the continued product of 2 a -3 , 7 a -9 , and 42 — 3d 6 . Ans. — 42 a -6 = -„. a 6 8. Find the continued product of 5 a 3 b ~ 4 , 10a 2 6 5 c, and _ 3 a 1 . Ans. — 150 a 12 6 c. 9. Multiply — 13 a" 1 b c~ 3 by — 4 a~ 3 6~ 6 e 2 . ^4ns. 52a- 4 6 _s c _I . 10. Divide a _m by a n . Ans. a~ m ~ n = a - <"+"»- 11. Divide a m by - ". .dns. a m + ». 12. Divide a _m by a - ". j4»s. a - ' B +' , = a n ~ m 13. Divide 14 a~ 8 b c 3 d~ 1 e by 2 a b~ 3 c 5 dh. Ans. 7 a~ 9 b* c-s d-* e h-* 14. Divide — 3 a m by 2 a*+" 6 c _1 . .dns. — £a _n 6 -1 e. 8* 13 ALGEBRA. [CH. I. § T. Division of Polynomials. 42. Problem. To divide one polynomial by an- other. Solution. The term of the dividend, which contains the highest power of any letter, must be the product of the term of the divisor which contains the highest power of the same letter, multiplied by the term of the quotient which contains the highest power of the same letter. A term of the quotient is consequently obtained by dividing, as in art. 35, the term of the dividend which contains the highest power of any letter by that term of the divisor which contains the highest power of the same letter. But the dividend is the sum of the products of the divisor by all the terms of the quotient ; and, therefore, If the product of the divisor by the term just found is subtracted from the dividend, the remainder must be equal to the sum of the products of the divisor by the remaining terms of the quotient, and may be used as a new dividend to obtain another term of the quotient. By pursuing this process until the dividend is entirely exhausted, all the terms of the quotient may be obtained. It facilitates the application of this method to ar- range the terms of the dividend and divisor according to the powers of some letter, the term which contains the highest power being placed first, that which con- tains the next to the highest power being placed next, and so on. CH. I. § V.] DIVISION. 19 Division of Polynomials. 43. EXAMPLES. 1. Divide — 16a 3 x 3 -j-a 6 -j-64z 6 by 4x 2 -|-a 2 — 4 ax. Solution. In the following solution the dividend and di- visor are arranged according to the powers of the letter x ; the divisor is placed at the right of the dividend with the quotient below it. As each term of the quotient is obtained, its product by the divisor is placed below the dividend or remainder from which it is obtained, and is subtracted from this dividend or remainder. 64x6 — 16a?x s -\-aP 64x6— 64ax 5 4-16a 9 x 4 4 x 9 ^ — 4 a z -{- a 2 = Divisor. 16x 4 +16ax 3 -}-12a 2 x 2 4-4a 3 z4-a 4 64 a x 5 — 16 a 2 z 4 — 16 a 3 x 3 -\- a 6 = 1st Remainder. 64 a x 5 — 64 a 9 z 4 -\- 16 a 3 z 3 48 a 2 x* — 32 a 3 x 3 -+- a 6 = 2d Remainder 48 a 2 z 4 — 48 a 3 x 3 -f 12 a 4 z 9 16 a 3 x 3 — 12 a 4 z 2 -f- a" = 3d Remainder. 16 a 3 x 3 — 1 6 a 4 z 2 -j- 4 a 5 z 4 e 4 x 2 — 4o 5 i-|-(i 6 = 4th Remainder. 4a 4 z 2 — 4a 5 z-j-a 6 0- 7 ^4 ns. 16 x 4 -f- 16 a x 3 -j- 12 o 2 x 2 -f 4 a 3 x -f- fl 4 2. Divide 6 e 3 — c 3 xbyc 3 . Ans b — x. 3. Divide a 2 -f 2a6-f-6 2 by a-\-b. Ans. a-\-b. 4. Divide — « 8 6 4 -f 15 a" 6 5 — 48 a 14 66 — 20 a". 67 by 10«96 9 — a 66. yins. a 9 6 3 — 5a 5 6 4 — 2a 8 6 5 . 5. Divide 1 — 18z 2 + 8l2 4 by l-|-6z-f 9« 2 . 4/is. 1 — 6 z -f 9 is 9 . 6. Divide 81 a 8 -j- 16 b™ — 72 a 4 66 by 9 a 4 + 12 a 2 63 [- 4 66. ,4ns. 9 a 4 — 12 a 2 6 3 -J-4 6 6 20 ALGEBRA. [CH. I. § \. Division of Polynomials. 7. Divide a; 6m — 3 x* m y 9 " -(- 3 a; 3 "* y 4n — y 6n by x 3m — 3 x im y n -j- 3 x m y** — y 3n . 4ns. z 3m -|- 3 i 9m y" + 3 i m y in -j- y 3 ". 8. Divide — 1 -j- a 3 n J by — 1 -(- a n. .4ns. 1 -J- a n -f a s rfi. 9. Divide 2 a 4 — 13 a 3 6 + 31 a 2 6 2 — - 38 a 6 3 + 24 6* by 2a 2 — 3a6 + 46«. 4ns. a 2 — 5 « 6 + 6 6 2 . 10. Divide a 2 — 6 2 by a — 6. 4ns. a + b. 11. Divide a 3 — 6 3 by a— 6. 4ns. a 2 + a 6 + 6 s . 12. Divide a 4 — b* by a — 6. 13. Divide a 5 — J 5 by a — 6. 4ns. a* + a 3 6 + a 2 6 2 -|- a Z> 3 + 6 4 . 44. Corollary. The quotient can be obtained with equal facility by using the terms which contain the lowest powers of a letter instead of those which con' tain the highest powers. In this case, it is more convenient to place the term containing the lowest power first, and that containing the next lowest next, and so on. This order of terms is called an arrangement ac- cording to the ascending powers of the letter; whereas that of the preceding article is called an arrangement according to the descending powers of the letter. £5. Corollary. Negative powers are considered to be lower than positive powers, or than the power zero, and the larger the absolute value of the expo- nent the lower the power. Thus a 5z-6 — a 4 z-3 + a 3 + a-ia + rt-2i3 > is arranged according to the ascending powers of x, and according to the descending powers of a. CH. I. § V.] DIVISION. 21 Division of Polynomials. 46. EXAMPLES. 1. Divide o 4 + a 2 — a~ 2 — a~* by a? — a-*. Ans. a 2 +l-|-a — 2 . 2. Divide 4 a 4 6-6+12 a 3 6~ 5 + 9 a 2 6—«— 6~ 2 -f 2a'* — a-«6 2 by 2a 2 6" 3 + 3a6~ 3 — 6- 1 + a~ 2 6. ^4n».2a 2 6- 3 + 3 a&~ 2 + &-* — a~ 2 6. 47. In the course of algebraic investigations, it is often convenient to separate a quantity into its fac- tors. This is done, when one of the factors is known, by dividing by the known factor, and the quotient is the other factor. And when a letter occurs as a factor of all the terms of a quantity, it is a factor of the quantity, and may be taken out as a factor, with an exponent equal to the lowest exponent which it has in any term, and indeed by means of negative exponents any mono- mial may be taken out as a factor of a quantity. 48. EXAMPLES. 1. Take out 3 a 2 6 as a factor of 15 a s 6 2 + 6 a 3 b + 9a 2 6 2 + 3a 2 6. Ans. 3a?b (5a 3 6+2a+ 36 + 1). 2. Take out a" as a factor of 3 a m + 1 + 2 a m . Ans. a m (3a + 2). 3. Take out 2a?b 5 c as a factor of 6 a" 6 7 c 2 -f-6a 6 8 e ~2a6 + 2 — cfic. Ant. 2a 3 6 5 c(3a 3 6 2 c + 3a- 2 6 3 — a-^b~*c-^ + a-3 6-5 c -i_2-i a -i6-5. 4. Take out 6 as a factor of a n ~ 1 b — 6". Ans. 6 (a"- 1 — b n ~ l ). 22 ALGEBRA. [CH. 1. ^ V Difference of two Powers divisible by Difference of their Roots. 49. Theorem. The difference of two integral pos- itive powers of the same degree is divisible by the difference of their roots. Thus, a" — 6™ is divisible by a — 6. Demonstration. Divide a" — b n by a — 6, as in art. 42, proceeding only to the first remainder, as follows. a n — b n a — b a n — a n-1 6 a n ~ l 1st Remainder =a n ~ 1 b — 6 n =6(a" -1 — 5 n-1 ). Now, if the factor a" ~ J — b n ~ 1 of this remainder is di- visible by o — b, the remainder itself is divisible by a — b, and therefore a n ~ 6" is also divisible by a — b ; that is, if the proposition is true for any power, as the (n — l)st, it also holds for the nth, or the next greater. But from examples, 10, 11, 12, 13 of art. 43, the propo- sition holds for the 2d, 3d, 4th, and 5th; and therefore it must be true for the 6th, 7th, 8th, &c. powers ; that is, for any positive integral power. 50. Corollary. The division of a 1 * — b n by a — 6 may be continued for the purpose of showing the form of the quotient, a" — b n C — a n_1 b a — b - 1 +a n-2 6+a n -3 6 2 -(- &c + a 6"-«+&»-i a n ~ 1 b — b n a" -1 b — a B ~ a»-2 62 — j» &,c. . . a 2 6»- 2_5» flr2 6"-2_ a &»-» a& n -i — 6» ab n ~i — b n CH. I. (5 v.] DIVISION. 23 Division of Polynomials. that is, a"— 6 n i — = a"" 1 + a"" 2 6 +a»- 3 6 2 + &c +■ a 6»-3+ &"-i, a — o so that each terra of the quotient is obtained from the pre- ceding term by diminishing the exponent of a by unity and increasing that of 6 by unity ; and the number of terms is equal to the exponent n. 51. Corollary. If b is put equal to a in the preceding quotient, each of its terms becomes equal to a n_I , which gives the peculiar result = na n ~ 1 a — a 52. There are sometimes two or more terms in ihe divisor, or in the dividend, or in both, which contain the same highest power of the letter accord- ing to which the terms are arranged. In this case, these terms are to be united in one by taking out their common factor ; and the com- pound terms thus formed are to be used as simple ones. It is more convenient to arrange the terms which contain the same power of the letter in a column under each other, the vertical bar being used as in art. 17 ; and to arrange the terms in the ver- tical columns according to the powers of some letter common to them. 53. EXAMPLES. 1. Divide a 2 x 3 — 6 2 z 3 — 4a6x 2 — 2 a 2 a; + 2 a b x -J- fl2 ja py ax lf X — a — b. 24 ALGEBRA. [CH. I. § V. Division of Polynomials. Solution. a* I v> — Aabx* — 2 a 9 — 68 | +2 a b a 9 I x3_ a 2 — 62 — 2a6 — 62 s + a 9 — 6» a — 6 x — a — 6 + 6 z 2 +a x — a a2 z 2 — 2 a 2 z + a 9 1st Rem. — 2a 6 + 2a6 — 6» + 62 a 9 i 9 — a 2 X — 2a6 + 6 2 + 62 — a 2 xA-cC 2 2d Remainder + 2a6 — 62 -6 2 — a 9 2; + a 9 + 2a6 — 6 2 — 62 3d Remainder 0. In this quotient, the coefficient a + 6 of a 2 , the coefficient a — 6 of x and the term — a + 6 are successively obtained by dividing the coefficient a 2 — J2 f j.3 j n the dividend, the coefficient a 2 — 2 a 6 + 6 2 . of z 2 i n the first remainder, and the coefficient — a a + 2 a 6 — 6 2 of x in the second re- mainder, by the coefficient a -^-6 of x in the divisor. Ans. (a + 6)a;2 + (a_6)a;— (a — 6). 2. Divide (6 6 — 10) a< — (7 6 2 — 23 6 + 20) o? — (3 6° -22 6 2 + 316 — 5)a 9 + (4&3 — 9 6 2 + 56 — 5)a + 6 2 — 26 by(36 — 5)a+6 2 _26. Ans. 2a3— (36 — 4)a 2 + (46 — l)a+l. 3. Divide — — 3 a 2 z 6 — 24 a 5 z 3 . 2 6 2 c a; 4 — 2 a 6 2 c i 3 +4 a 2 6 2 c a; 2 — 6 a 3 6 2 c a!— 20 a 4 6 2 c. The greatest common divisor of these two quantities, found as in the preceding example, is x — 2 a, which, mul- tiplied by the common monomial factor 7 a 2 6, gives 7 a 2 6 (x — 2 a) for the required greatest common divisor. 3. Find ihe greatest common divisor of x 3 — a 3 and a 2 — a 2 . Ans. x — a. 4. Find the greatest common divisor of 5 a 3 — 10a 2 6-(- I56 3 and 3 n 3 -}- 6 a 2 6 + 6 a 6 2 -f- 3 6 3 . Ans. a + b. 5. Find the greatest common divisor of a; 4 -|-a: 3 -p-2; 2 -f-a; — 4 and a; 4 + 2 a; 3 + 3 a; 2 + 4 a;— 10. Ans. x— 1. 6. Find the greatest common divisor of 7 a a 6 -)- 21 a x 4 -\- Uax and 3 a 6 -{- 3 a; 5 -f 3 a; 4 — 3a; 2 . Ans. x 2 -j- x. 7. Find the greatest common divisor of 81 a 4 a; 4 — 24 a 7 a; and 3 a a; 7 — 2 a 2 a; 6 + 3 a 3 a: 3 — 2 a 4 a; 4 . Ans. 3 a a: 2 — 2 a 2 x. 8. Find the greatest common divisor of x 3 -\-x — 10 and a; 4 — 16. Ans.x — 2. 65. When there are several terms in the given polynomials, which contain the same poioer of the letter according to which the terms are arranged, these terms are to be united in one, as in art. 53, and the compound terms thus formed are to be treated as monomials. 34 ALGEBRA. [CH. II. § I. Greatest Common Divisor. 66. EXAMPLES. 1. Find the greatest common divisor of X 3 I y 5_ 2 x* y 4 — 3 x 4 I y 3 + 2 a; 4 I y* —9a + ^ y 3 — 4 x 3 I y 2 + 4 x 3 I y — 12 a: 3 | +12x2 +27 x 2 1 —18 x 2 +18 x 2 — 18 x and x 2 I y* — 3 x 3 + 3x| — 7x 2 + 6z Solution. The factor (x 2 + 3 a:) y is a common factor of all the terms, and is therefore to be suppressed, in order to be multiplied by the greatest common divisor of the re- maining polynomials. The polynomials thus become x I y 4 — 2 k 2 — 3| +8z — 6 and y 3 — 3 a: 2 I y 2 +2z 2 + 9z | — 6x y 2 — 4 x y + 4 x. y 3 — 3z + 2 The suppression of the factor (x — 3) y in the first of these polynomials reduces it to y 3 — 2a: ly 2 — 3zy + 2z, * +2. | ' T by which the second is to be divided, and the rest of the process is as follows : Col. 1. Col. 2. y 3 — 3a; , + * y 3 — 2* +2 y 2 — 4xy + 4x y 2 — 3xy+2x — xy a — xy+2x — y 2 — y+2 — y 2 — 2y y+2 y + 2 y 3 — 2a; + 2 y 3 + y 2 — 3i y 2 -2 y + 2x V — 2x + • — 2x + 1 y 2 — 3z -> + 2 y 2 — 2 a; + 1 y+2x y+4x — 2 X + l| y— 2x y+2 1 — y +2* — l — y i ^«- (y + 2) (x 2 + 3 x) y = (x 2 + 3 *) (y2 + 2y). 3H. II. § I. - ] REDUCTION OF FRACTIONS. 35 Greatest Common Divisor. The third line of col. 1 is the remainder of the division of the 1st line of col. 1 by the 1st line of col. 2 ; and this remainder, reduced by the suppression of the factor % is the 4th line of col. 1. The 5th line of col. 2 is the remainder of the division of the 1st line of col. 2 by the 4th line of col. 1, and this remainder, reduced by the suppression of the factors — x + Lis the last line of col. 2. The 4th line of col. 1 is exactly divisible by the last line of col. 2, and there- fore the greatest common divisor is the product of (}?-\-3x) ybyy + 2. Ans. (z 2 + 3 x) (y 2 + 2 y). 2. Find the greatest common divisor of the polynomials a 2_j_6 2 + c 2 + 2a6-|-2ac-f26c and cp — P — c 2 — 2 be. Ans. a -\- b -j- c. 3. Find the greatest common divisor of the polynomials a 4 — 2 6 2 |a 2 + 6* -2£ 2 | — 2 6 2 e 2 + « 4 and a 3 + 3 b a 2 + 3 6 2 a +63 — 6 c 2 Ans. a 2 -f- 2 a 6 + 6 2 — c 2 . 4. Find the greatest common divisor of the polynomials y 3 -f-a** I y 2 — x 3 \y —x I +x 2 | x\y 5 — 3 a; y*— X s — 1| +3 + 3x — 2 and a&lyi — 3x z \y 3 + x 3 — 1 I +3 I + 2 a s — x — 2 yZ—X 3 -\-x 67. Problem, denominator. Ans.y(y — l)(x—l). To reduce fractions to a common Solution. Multiply both terms of each fraction by the product of all the other denominators. 86 ALGEBRA. [CH. II. § I Common Denominator. For the value of each fraction is, from art. 55, not changed by this process ; and as each of the denominators thus obtained is the product of all the denominators, the fractions are all reduced to the same denominator. 68. But fractions can be reduced to a common denomi- nator which is smaller than their continued product, when- ever their denominators have a common multiple less than this product. For, by art. 55, Fractions may be reduced to a common denomi- nator, which is a common multiple of their denomi- nators, by multiplying both their terms by the quo- tients, respectively obtained from the division of the common denominator by their denominators. 69. Corollary. An entire quantity may, by the preceding article, be reduced to an equivalent frac- tional expression having any required denominator, by regarding it as a fraction, the denominator of which is unity. 70. EXAMPLES. 3 2 5 1. Reduce -, -, — , to the common denominator 24. 8' 3' 6' 9 16 20 AnS - 24' 2T 24' OT ,,a 2 63e D 3a a. lieduce -5-7, r-— 5 , 8 a, -. — , to the common denoim- c 2 a 2 c z led nator 4 c 2 d. ia?b 6ed 32 a c 2 d 3«e Ans. 4c*d'4c2d' 4c 2 d 4c 2 d" CH. II. § I.] REDUCTION OP FRACTIONS. 37 Common Denominator. «t.i o + 6 , 1 c+rf , 3. .Reduce 7 , 1, — ; — =-, — ; — ■ to the common a — 6 a -\- b a 2 — b l denominator a 2 — 6 2 . a 2 +2a5-|- 6 3 a 3 — 6 2 a — 5 c-fri 4ns - a 2ZT62~' a 2 — 6 2 ' a 2 — T 2 ' a 2 — 6 2 ' 71. Problem. To find the least common multiple of given quantities. Solution. When the given quantities are decom- posed into their simplest factors, as is the case with monomials, their least common multiple is readily obtained ; for it is obviously equal to the product of all the unlike factors, each factor being raised to a power equal to the highest power which it has in either of the given quantities. But the common factors can always be obtained from the process of finding the greatest common divisor. 72. EXAMPLES. 1. Find the least common multiple of 2 a 3 6 2 c x, 3 a 5 6 c 3 * 9 , 6 a ex = 2. 3 a ex, 9 c 7 x™ = 3 2 c 7 * 10 , 24 a 8 = 2 3 . 3 a 8 . Ans. 2 3 . 3 2 . a 8 6 2 c 7 x 10 = 72 a 8 6 2 c 7 a; 10 . 2. Find the least common multiple of 16 a x, 40 6 5 a;, 25 a 1 b 3 x*. Ans. 400 a 7 6 5 a 3 . 3. Find the least common multiple of*", a;" -1 , x n ~ 2 , x"- 3 , i. Ans. x n . 4. Find the least common multiple of 6 (a -f- 6) x m , 54(a— 6) 3 , (a+6) 7 , 81 (a—bfx m + 2 ,8(a-{-bfx m - 8 . Ans.64S(a + by(a — b) 3 x m +*. 38 ALU KB It A. [CH. II. § II Sum and Difference of tractions. 5. Find the least common multiple of a 2 -\-2 a b -\- 6 a , fl 2-f 4a6-f 4 6 a , a 2 — 6 a , a? + 3 a b + 2 IP, a? + a*b — a6 9 — tf». Ans. (a + 6) 2 (a — 6) (a + 2 6)2. SECTION n. Addition and Subtraction of Fractions. 73. Problem. To find the sum or difference of given fractions. Solution. When the given fractions have the same denominator, their sum or difference is a frac- tion which has for its denominator the given com- mon denominator, and for its numerator the' sum oi the difference of the given numerators. When the given fractions have different denomi- nators, they are to be reduced to a common denomi- nator by arts. 67 and 68. 74. EXAMPLES. 1. Find the sum of T , — and . b d f . adf+bef—bde Am - bdf • 2. Subtract % from * Am. bc — ad b"""d- ™°-—bd—- 3. Find the sum of i+* and 2=* Ans. a. . „ , a — b „ a 4- 6 4. Subtract — — - from -£_. a„ s , 4, CH. II. § II.] FRACTIONS. 39 Sum and Difference of Fractions. 5. Reduce to one fraction the expression ■ -)- c Ans. b a-\-be ct.j , r *• 2 a , 5 d/ beg 6. Reduce to one fraction — + _£ - ^ 16abc+15cd f— 4beg AnS - 246^ ' .a b 7. Reduce to one fraction s A — ; , £_ 6, ]_ X a;2 ' jj3 . a.T a — bx + l Ans. 5 . x 3 a z 8. Reduce to one fraction — ; 1 . a-\-z ' a — z A ° 2 + ' ;2 Ans. -^-! — ■„. a 3 — z a 9. Reduce to one fraction 3 , 3 1 _ 1— x 4(1-^T8(1-i) + 8(1+i) 4(1+**) 1 + x + x* 10. Reduce to one fraction 3A _ 2/t-fs 5 (A— 2a7p ~"~ (A + a;) (/t-2*) A + x ' 20Az — 22a: 2 .4ns. (A + a:)(A-2x) 2 ' 11. Reduce to one fraction a 3 abb (a + 6) 3 ~~ (a + 6) 2 ~'~a+&' Ans, a 3_|_ a i2_t.fc3 (a +6)3 ' 75. Corollary. It follows, from examples 3 and 4, that the sum of half the sum and half the difference 40 ALGEBRA. [«I. II. § III Product and Quotient of Fractions. of two quantities is equal to the greater of the two quantities ; and that the difference of half their sum and half their difference is equal to the smaller of them. SECTION in. Multiplication and Division of Fractions. 76. Problem. To find the continued product of several fractions. Solution. The continued product of given frac- tions is a fraction the numerator of which is the continued product of the given numerators, and the denominator of which is the continued product of the given denominators. 77. Problem. To divide by a fraction. Solution. Multiply by the divisor inverted. The preceding rules for the addition, subtraction, multi- plication, and division of fractions require no other demon- strations than those usually given in arithmetic. 78. When the quantities multiplied or divided contain fractional terms, it is ge'nerally advisable to reduce them to a single fraction by means of art. 73 79. EXAMPLES. 1. Multiply together r , - and L Am. ^JL o a f bdf 6a3&5 3n 2 rf' 9 a s 2. Mult.ply — — by — — . Ans. 7d* t 9 J 2&5« ""*• 7 obtained from A : B = C : D by transposing its means, arid give A -f B : C+ D = A — B : C— D = m A-\-nB:m C-j- n D = mA — nB:m C — n D = A:C=B:D; and A+ B:A — B = C-\-D:C—D mA-\-nB:mA — nB — mC -\-nD:mC — nD\ that is, the sum of the first two terms of a proportion is to the sum of the last two, as the difference of the first two terms is to the difference of the last two, or as the first term is to the third, or as the second is to the fourth. Likewise, the sum of the first two terms is to their difference, as the sum of the last two is to their dif- ference. Moreover, in finding' these sums and differences, both the antecedents may be multiplied by the same number, and both the consequents may be multiplied by any number. 98. Two proportions, as A : B = C : D and E : F = G : H , CH. II. § IV.] PROPORTIONS. 49 Ratio of Reciprocals. may evidently be multiplied together, term by term, and the result AxE-.B xF=CX G-.DXH is a new proportion. 99. Likewise, all the terms of a proportion may be raised to the same power. Thus, A : B = C : D gives A s : B* = C 2 : D* y/A.y/B — s/C: y/D A m : B m = C m : D m mm mm y/A : s/B = y/C : 24* 25(z + 3) 58 ALGEBRA. L CH> m ' § * Examples or patting Questions Into Equations. 17. Two travellers, A and B, set out at the same time from two different places, C and 1); A, from C to D ; and B, from D to C. When they met, it appeared that A had already gone 30 miles more than B ; and, according to the rate at which they are travelling, A calculates that he can reach the place D in 4 days, and that B can arrive at the place C in 9 days. What is the distance between C and D 1 Ans. If, when they meet, x — the distance gone by A , then, % — 30 = the distance gone by B ; the whole distance = 2 x — 30 ; and the required equation is Ax _ 9(g — 30) x — 30"" x ' 18. Some merchants jointly form a certain capital, in such a way that each contributes 10 times as many dollars as they are in number ; they trade with this capital, and gain as many dollars per cent, as exceed their number by 8. Their profit amounts to $ 288. How many were there of them 1 Arts. If x = the number of merchants, the required equation is tV a 2 (a; -f- 8) = 288. 19. Part of the property of a merchant is invested at snch a rate of compound interest, that it doubles in a number of years equal to twice the rate per cent. What is the rate f interest? Ans. If x = the rate per cent., the required equation is / 100-f a \ 2 x V 100 y ~ CH. III. § II.] REDUCTION OF EQUATIONS. 59 Degree of an Equation. SECTION n. Reduction and Classification of Equations. 104. The portions of an equation, which are sepa- rated hy the sign =, are called its members ; the one at the left of the sign heing called its first member, and* the other its second member. 105. Equations are divided into classes according to the form in which the unknown quantities are contained in them. - But before deciding to which class an equation belongs, it should be freed from fractions, from negative exponents, and from the radical signs which affect its unknown quantities ; its members should, if possible, be reduced to a series of monomials, and the polynomials thus obtained should be reduced to their simplest forms. 106. When the equation is thus reduced, it is said to be of the same degree as the number of dimen- sions of the unknown quantities in that term which contains the greater number of dimensions of the unknown quantities. Thus, x and y being the unknown quantities, the equa- tions a x -\-b = c, 10a:-j-y = 3 » are of the first degree ; ? 2 -f- 3 x -f 1 = 5, xy= 11, are of the second degree, &.C. 60 ALGEBRA. [CH. III. § II. Transcendental Equations; Roots of Equations. 107. But when an equation does not admit of being reduced to a series of monomials, or, when being so reduced, it contains terms in which the un- known quantities or their powers enter otherwise than as factors, it is said to be transcendental ; and the consideration of such equations belongs to the higher branches of mathematics. Thus, a- = 6 (x-\-a)y + i = c, are transcendental equations. 108. An equation is said to be solved, when the values of its unknown quantities are obtained ; and these values are called the roots of the equation. 109. The reduction and solution of all equations depends upon the self-evident proposition, that Both members of an equation may be increased, diminished, multiplied, or divided by the same quan- tity, without destroying the equality. 110. Corollary. If all the terms of an equation have a common factor, this factor may be suppressed. 111. EXAMPLES. 1. If the factor common to the terms of the equation a 2 x 5 + 3 a 3 a 2 = a 2 z 2 is suppressed, what is the resulting equation? Ans. x 3 -|- 3 a = 1. 2. If the factor common to the terms of the equation a* + 3 a' + t x = a x ~ 1 is suppressed, what is the resulting equation ? Ans. a-)-3s s i=l, CH. III. $) II.] REDUCTION OP EQUATIONS. 61 To free an Equation from Fractions. 112. Problem. To free an equation from frac- tions. Sohition. Reduce, by arts. 67 and 68, all the terms of the equation to fractions having a common denominator, and suppress the common denominator, prefixing to the numerators the signs of their re- spective fractions. Demonstration. For suppressing the denominator ol a fraction is the same as multiplying the fraction by its de- nominator ; and, consequently, both the members of this equation are, by the preceding process, multiplied by the common denominator. 113. Corollary. It must be strictly observed that, when the denominator of a fraction is removed, the sign, which precedes the fraction, affects all the terms of the numerator. If therefore this sign is negative, all the signs of the numerator are to be reversed. 114. EXAMPLES. 1. Free the equation a . c a — c , 1 bx ' dx bdz x from fractions. Solution. This equation, when its terms are reduced to a common denominator, is ad be a — c bdhx bd b dx' bdx bdx bd x bdx' 6 *2 ALGEBRA. [CH. III. § II. To free an Equation from Fractions. Suppressing the common denominator, we have ad-^-bc — (a — c)=bdhx — bd, or ad-\-bc — a-\-c = bdhx — bd. 2. Free the equation 3a — 5x , 2a — x a-i-f -, — = —Li — dx a — c a a — c from fractions. Ans. Sad — 5dx-\-2dP — ax — 2ac-\-cx=zad-\- df—ad*x-\-cd*x. 3. Free the equation 8x 20 ■ o = x-\-2 Sx from fractions. A ns. 24 a 3 — 18 x 2 — 36 x = 20 x -f 40. 4. Free the equation 18 -\-x 20z + 9 65 6 (3-x) ~~ 19— 7a ~~ 4 (3— x) from fractions. Ans. 684 — 214 x— 14 ^=612 a; -J- 324— 240 z 2 — 3705 + 1365*. 5. Free the equation x +y *—y = i _i i_ x— y x-\-y x — y x-\-y ' a»— # 2 from fractions. 4ms. 2 2 +2a:y+y2— a2_[-23;y— ^2=3!+y— *4- 6. Free the equation a" _ a" -f- 6* 6* a* — 6" from fractions. Ans. a** — a" b* = a* b* -J- 6 2 * CH. III. "$.*!.] REDUCTION OP EQUATIONS. b3 To free a Fraction from negative Exponents. 115. Corollary. If the given equation contains negative exponents, it can be freed from them by 'irts. 80 and 82. 116. EXAMPLES. 1. Free the equation = a;-i x-\- r~' 1 from fractions and negative exponents. Ans. x 3 -J- x = x% — 1. 2. Free the equation x"-\-x~ a a x — a~* r — a a"-\-a~ x x° — x~ from fractions and negative exponents. Ans. x ia a?* — a 2 * = a ix x* a — z s ". 117. Theorem. A term may be transposed from one member of an equation to the other member, by merely reversing its sign ; that is, it may be sup- pressed in one member and annexed to the other member with its sign reversed from -f- to — , or from — to -f-. Proof. For suppressing it in the member in which it at first occurs is the same as subtracting it from that member ; and annexing it to the other member with its sign reversed is, by art. 26, subtracting it from the other member ; and, therefore, by art. 109, the equality is preserved. 118. Corollary. All the terms of an equation may be transposed to either member, leaving zero in the other member ; and the polynomial thus formed may be reduced to its simplest form, by arts. 20 and 110. 64 ALGEBRA. [ell. III. § II. Equations reduced to their simplest forms. 119. EXAMPLES. 1. Reduce the equation 7x n _ 6s»+ 1 -f-z n 3a"-{-6i n +3 x — 1 — x + l a 2 — 1 to its simplest form in a series of monomials. Solution. This equation, freed from fractions by arts. 112 and 113, is 7z n + 1 +7x n = 6x n +* — 5i"+ 1 — x n — 3z n — 6x n +*, which becomes, by the transposition of its terms and by the seduction of art. 20, 12*"+i-f 11*» = 0, and, by striking out the factor x n , 12x4-11 = 0. 2. Reduce the equation s3-f 1 X — l _ X-\- 1 a; 2_ 1 (x-f-l)2 a: — 1 to its simplest form in a series of monomials. Arts. 2^4-1=0. 3. Reduce the equation a x s -\- b x -\- c ax z — bx — c HF+l — x^—i to its simplest form. Ans. bx-\-c — a = 0. 1. Reduce the equation a x -\-a~ x a" — a,-* x a -\-x~ a x" — x~" to its simplest form. Ans. a z " = z* m : CH. III. § III.] EQUATIONS OP THE FIRST DEGREE. 65 Equations of the First Degree. section im Solution of Equations of the First Degree, with one unknown quantity. 120. Theorem. Every equation of the first de- gree, with one unknown quantity, can be reduced to the form A x + B = ; in which A and B denote any known quantities, whether positive or negative, and x is the unknown quantity. Proof. When an equation of the first degree with one unknown quantity is reduced, as in art. 118, its first mem- ber is composed of two classes of terms, one of which con- tains the unknown quantity, and the other does not. If the unknown quantity, which we may suppose to be x, is taken out as a factor from the terms in which it is contained, and its multiplier represented by A, the aggregate of the first class of terms is represented by Ax; and the aggregate of the terms of the second class may be represented by D ; whence the equation is represented by A x + B = 0. 121. Problem. To solve an equation of the first degree with one unknown quantity. Solution. Having reduced the given equation to the form Ax + B = 0, transpose B to the second member by art. 117, and we have Ax= —B. Dividing both members of this equation by A, gives B 6» A 66 ALGEBRA. [cil. III. § III. Cases in Equations in the First Degree. Hence, to solve an equation of the first degree, re- duce it, as in art. 120, and transpose its known terms to ihp second member, and all its unknown terms to the- first member ; and the value of the un- known quantity is equal to the quotient arising from the division of the second member by the multiplier of the unknown quantity in the first member. 122. Corollary. When A and B are both positive or both negative, the value of x is, by art. 35, nega- tive ; but when A and B are unlike in their signs, one positive and the other negative, x is positive. 123. Corollary. When we have B = 0, the value of x is x = — - = 0. 124 Corollary. When we have the value of x is B — o- But the smaller a divisor is, the oftener must it be con- tained in the dividend, that is, the larger must the quotient be ; and when the divisor is zero, it must be contained an infinite number of times in the dividend, or the quotient must be infinite. Infinity is represented by the sign Co We have, then, in this case, x = — on The given equation is, however, in this case. CH. III. § III.] EQUATIONS OP THE FIRST DEGREE. 67 Cases in Equations of the First Degree. which reduces itself to B=0, an obvious absurdity, unless B is zero. The sign oo is, therefore, rather to he regarded as the expression of the peculiar species of absurdity which arises from diminishing the denominator of a fraction till it becomes zero. 125. Corollary. When we have A = 0, and B — 0, the value of x is _0__0 X ~ ~" 0' which is equal to any quantity whatever, and is called an indeterminate expression. The given equation is, indeed, in this case 0Xi|0 = 0, an equation which is satisfied hy any value Avhatevei of x, and is called an identical equation. 126. EXAMPLES. 1. Solve the equation 8x — 5 = 13 — 7x. Ans. x = 1 J. 2. Solve the equation f + | + "J -7,-718+ f Ans. a; = 116^|f G8 ALGEBRA. [CH. III. § 111. Equations of tile First Degree with one unknown quantity. 3. Solve the equation ax-\- c — -- b x + d. Ans. x ■. d- a- -c -6" 4. Solve the equation dx • = ac-J- ax 1' Ans. X = d r, ' 5. Solve the equation cx m f xn a-J-6 x d -\-ex cd — af af — cd Ans. x = r-j; *- = -± r7 . ' oj — c e ce — bj 6. Solve the equation 3abc a a 6 3 (2 a -f- b) & a x bx a + 6 + (a+6) 3 + a(a + 6) a — dcx + a ' ab Ans. x = — ri- a-j-o 7. Two capitalists calculate their fortunes, and it appears that one is twice as rich as the other, and that together they possess $ 3S 700. What is the capital of each ? Ans. The one has $ 12 900, the other $25 800. 8. To find two such numbers, that the one may be m times as great as the other, and that their sum = a. a , ma Ans. — .— and — 7-^=. OT-j-1 m-\- 1 9. The sum of $ 1200 is to be divided between two per- sons, A and B, so that A'b share is to B's as 2 to 7. How much does each receive ? Ans. A $ 266f , B $ 933$. CH. III. § III.] EQUATIONS OP THE FIRST DEGREE. 69 Equations of the First Degree with one unknown quantity. 10. To divide a number a into two such parts, that the first part is to the second as m to n. ma , n a Ans. : — and ; — . m-\-n m -f-n 11. How much money have I, when the 4th and 5th parts of it amount together to $ 2,25 1 Ans. $5. 12. Find a number such, that when it is divided succes- sively by m and by n, the sum of the quotients = a. mna Ans. tn-j-n' 13. Divide the number 46 into two parts, so that when the one is divided by 7, and the other by 3, the sum of the quotients = 10. Ans. 28 and 18. 14. All my journeyings taken together, says a traveller, amount to 3040 miles; of which I have travelled 3£ times as much by water as on horseback, and 2£ times as much on foot as by water. How many miles did he travel in each of these three ways 1 Ans. 240 miles on horseback, 840 miles by water, and 1960 miles on foot. 15. Divide the number a into three such parts, that the second may be m times, and the third n times as great as the first. a ma na nS ' 1 + m + n' 1 + m + n' 1 + m + n 16. A bankrupt leaves $ 21 000 to be divided among four creditors A, B, C, D, in proportion to their claims. Now A's claim is to B's as 2 : 3; B's claim : C's = 4 : 5; and C's claim : D's =6:7. How much does each creditor T6C61V6 ' Ans. A $ 3200, B $ 4800, C $ 6000, D $ 7000. 70 ALGEBRA. [CH. III. § III Equations of the First Degree with one unknown quantity. 17. Divide the number a into three such parts, that the 1st shall be to the 2d as m to n ; and the 2d part : the 3d = p : q. . mpa npa nqa ' mp+np+nq' mp+np+nq' mp+np+nq' 18. There are two numbers whose sum is 96, and differ- ence 16 ; what are they 1 Ans. 56 and 40. 19. A father gives to his five sons $ 1000, which they are to divide according to their ages, so that each elder son shall receive $ 20 more than his next younger brother. What is the share of the youngest ? Ans. 160. 20. One has six sons, each whereof is 4 years older than his next younger brother; and the eldest is three times as old as the youngest. What is the age of the eldest 1 Ans. 30 years. 21. There is a certain fish whose head is 9 inches ; the tail is as long as the head and half the back ; and the back is as long as both the head and the tail together. What is the length of the fish ? Ans. 72 inches. 22. Five gamesters have lost jointly $ 40$ ; B's loss amounts to £ dollar more than triple A's ; C's loss is $ 2 less than twice B's ; D lost £ dollar less than A and B together; and E twice as much as B less | dollar. How much did each of them lose 1 Ans. A $2,B$6i, C$11, D $8{, J3 $ 12|. 23. A mason, 12 journeymen, and 4 assistants, receive together $ 72 wages for a certain time. The mason re- ceives $ 1 daily, each journeyman £ dollar, and each as- sistant I dollar. How many days must they have worked for this money 1 Am. 9 days. GH. III. § III.] EQUATIONS OP THE FIRST DEGREE. 71 Equations of the First Degree with one unknown quantity. 24. Find a number such that if you multiply it by 5, subtract 24 from the product, divide the remainder by 6, and add 13 to the quotient, you will obtain this number. Ans. 54. 25. A courier left this place n days ago, and makes a miles daily. He is pursued by another making 6 miles daily. In how many days will the second overtake the first ? na , Ans. t days. o — a J 26. A courier started from a certain place 12 days ago, and is pursued by another, whose speed is to that of the first as 8 : 3. In how many days will the second overtake the first? Ans. 7£ days. 27. A courier started from this place n days ago, and is pursued by another whose speed is to that of the first as p is to q. In how many days will the second overtake the first? . nq Ans. p — q 28. Two bodies move in opposite directions ; one moves c feet in a second, the other C feet. The two places, from which they start at the same time, are distant a feet from one another. When will they meet 1 Ans. In -=— -. — seconds. C-f-c 29. Two bodies move in the same direction from two places at a distance of a feet apart; the one at the rate of c feet in a second, the other pursuing it at the rate of C feet in a second When will they meet 1 Ans. In -= seconds. C — c 30. At 12 o'clock, both hands of a clock are together. YVhen and how often will these hands be together in the next 12 hours? 72 ALGEBRA. [CH. III. § III. Equations of Ihe First Degree with one unknown quantity. Arts. At 5j\ minutes past 1, at 10-fJ minutes past 2, at 16/y minutes past 3, and so on, in each successive hour, 5 r 5 T minutes later. 31 . Two bodies move after one another in the circum- ference of a circle, which measures p feet. At first they are distant from each other by an arc measuring a feet; the first moves c feet, the second C feet, in a second. When will those two bodies meet for the first time, second time, and so on, supposing that they do not disturb each other's motion t a p + a 2p + a Arts. In ^ — , j^-i — , -^ , &c, seconds. O — c \*i — — c O ~— c 32. When will they meet if the first begins to move t seconds sooner than the second 1 T a+ct p+a+ct 2p + a+ct . , Ans. In -^ ,— >~ri , &c, seconds. 33. But when will they meet, if the first begins to move t seconds later than the second ? ,. a — ct p\a — ct 2p4-a — ct . , Ans. In -t=j , — ■ ^ , ■ , &c, seconds. 34. When will they meet, if the first, instead of running in the same direction with the second, runs in the opposite direction, and starts at the same time ? Ans - In cT? c¥c> ^7« 3 c$7> &c " seconds - '' 35. When will they meet, if, moving in an opposite di- rection to the second, the first starts t seconds sooner than the second? . T a — ct p + a — ct 2p4-a — ct . . Ans. In w —, c+c , /, c + c , &c, seconds. CH. III. § III.] EQUATIONS OF THE FIRST DEGREE. 73 Equations of the First Degree with one unknown quantity. 36. But when will they meet, if, moving in an opposite direction to the second, the first starts t seconds later than the second 1 a + ct pXaA-ct Zp+a + ct Ans . I n __, -U-, .JL__ & c, seconds. 37. A wine merchant has two kinds of wine ; the one costs 9 shillings per gallon, the other 5. He wishes to mix both wines together, in such quantities, that he may have 50 gallons, arid each gallon, without profit or loss, may be sold for 8 shillings. How must he mix them ? Ans. 37£ gallons of the wine at 9 shillings, with 12£ gallons of that at 5 shillings. 38. A wine merchant has two kinds of wine ; the one costs a shillings per gallon, the other 6 shillings. How must he mix both these wines together, in order to have n gallons, at a price of c shillings per gallon? (a — c)n „ . , . , .... , (c — b\n Ans. r- gallons ol the wine at 6 shillings, and - f- a — 6 & ° a — b gallons of that at o shillings. 39. To divide the number a into two such parts, that, if the first is multiplied by m and the second by n, the sum of the products is b. . b — na , ma — b Ans. and . m — n m — n 40. One of my acquaintances is now 30, his younger brother 20 ; and consequently 3 : 2 is the ratio of his age to his brother's. In how many years will their ages be as 5:4? Ans. In 20 years. 41. What two numbers are those, whose ratio = a :b; but, if c is added to both of them the resulting ratio = m : n F ac(m — n) , bc(m — n) Ans. — r— * and — 7 — -. an — bm an — bin 74 ALGEBRA. [cil. III. § 111. Equations of the Firat Degree with one unknown quantity. 42. Find a number such that 5 times the number is as much above 20, as the number itself is below 20. Ans. 6f. 43. A person wished to buy a house, and in order to raise the requisite capital, he draws the same sum from each of his debtors. He tried, whether, if he obtained $ 250 from each, it would be sufficient for the purpose ; he found, however, that he should then still lack $ 2000. He tried it, therefore, with $ 340 ; but this gave him $ 880 more than he required. How many debtors had he 1 Ans. 32. 44. A father leaves a number of children, and a certain sum, which they are to divide amongst them as follows: The first is to receive $ 100, and then the 10th part of the remainder; after this, the second has $200, and the 10th part of the remainder; again, the third receives $300, and the 10th part of the remainder ; and so on, each succeed- ing child is to receive $ 100 more than the one preceding, and then the 10th part of that which still remains. But it is found that all the children have received the same sum. What was the fortune left 1 and what was the num- ber of children ? Ans. The fortune was $ 8100, and the number of children 9. 45. Divide the number 10 into two such parts, that the difference of their squares may be 20. Ans. 6 and 4. 46. Divide the number a into two such parts, that the difference of their squares may be b. . a 2 +6 3 a*-b Ans. -jr- 1 — and — — — . 2a 2a 47. What two numbers are they whose difference is 5 and the difference of whose squares is 45? Ans. 7 and 2. CH. Ill, § III.] EQUATIONS OF THE FIRST DEGREE. 75 Examples of unknown quantity equal to Zero. 48. What two numbers are they whose difference is a, and the difference of whose squares is 6 1 b — cfi J-fs 8 Ans. — - — and — -i — . 2a 2a 127. Corollary. When the solution of a problem gives zero for the value of either of the unknown quantities, this value is sometimes a true solution ; and sometimes it indicates an impossibility in the proposed question. In any such case, therefore, it is necessary to return to the data of the problem and investigate the signification of this result. 128. EXAMPLES. 1. In what cases would the value of the unknown quan- tity in example 25 of art. 126 become zero 1 and what would this value signify ? Solution. As the value of the unknown quantity of the example is the fraction, which is its answer; it is zero, when b — a or, clearing from fractions, when «« = 0; that is, when n = 0, or when a = ; and, in either case, this value signifies that the couriers are together at the outset; and zero must, therefore, be regarded as a real solution. 2. In what cases would the value of the unknown quan- tity in example 35 of art. 126 become zero? and what wou'd this value signify ? 76 ALGEBRA. [CH. III. § III, Examples or unknown quantity equal to Zero. Ans. When t = — , or = — ! — , or = -^— ' — . &c, c c c ' and either of these equations signifies that the bodies are together when the second ' body starts, the first body hav- ing just arrived at the point of departure of the second, and zero is, therefore, to be regarded as a real solution. 3. In what cases would the value of one of the unknown quantities in example 38 of art. 126 become zero? and what would this value signify ? Ans. When either a =r c, or 6 = c ; and, in either case, these equations indicate that the price of one of the wines is just that of the required mixture, and, of course, needs none of the other wine added to it to make it of the required value ; and zero, must, there- fore, be regarded as a true solution. 4. In what cases would the value of one of the unknown quantities in example 39 of art. 126 become zero? and what would this value signify 1 Ans. When b = n a, or = m a ; and these equations indicate that a is itself such that, multiplied either by m or by n, it gives a product = b; and zero may be regarded as a true solution, expressing that one of the parts is zero, while the other is the num- ber a itself. 5. In what cases would the value of one of the unknown quantities in example 41 of art. 126 become zero 1 and what would this value signify ? Ans. First. When a == 0, or 6 = 0, and, in this case, zero is a true solution by regarding all numbers as having the same ratio to zero. CH. III. § III.] EQUATIONS OP THE FIRST DEGREE. 77 Cases in which the value of an unknown quantity is infinite. Secondly. When c = 0, and, in this case, the problem is impossible, for no two numbers can be in the ratio a : b, and, without having any thing added to or subtracted from them, acquire the different ratio m : n. Thirdly. When m = n, and,, in this case, the problem is impossible, for no two numbers, whose ratio = a : b, and which are therefore unequal, can, by the addition of c to each of them, become equal to each other, as required by the ratio m : n = m : m = 1. 129. When the solution of a problem gives, for the values of one of its unknown quantities, any fractions, the denominators of which are zero, while the numerators are not zero ; such values are, gener- ally, to be regarded as indicating an absurdity in the enunciation of the problem. 130. EXAMPLES. 1. In what case does the denominator of the fractional value of the unknown quantity in example 25 of art. 126 become zero? and what is the corresponding absurdity in the enunciation of the problem ? Ans. When a = b, and the absurdity is, that, while the couriers are travel- ling at the same rate, it is required to determine the time in which one will overtake the other. 2. In what case do the denominators of the fractional values of the unknown quantity in example 38 of art. 126 become zero? and what is the corresponding absurdity in the enunciation of the problem? 7* fS ALGEBRA. [cH. III. § HI, Cases in which the value of the unknown quantity is indeterminate. Ans. When a = b, and the absurdity is that, while both the wines are of the same value, they should give a mixture of a value differ- ent from their common value. 3: In what case would the denominators of the fractional values of the unknown quantities in example 41 of art. 120 become zero 1 and what is the corresponding absurdity of the enunciation ? Ans. When an = b m, that is, when a : b = m : n ; and the absurdity is, that the ratio of two unequal num- bers should not be changed by increasing them both by the same quantity. 4. In what case would the denominators of the fractional values of the unknown quantities in example 48 of art. 126 become zero ? and what is the corresponding absurdity of the enunciation 1 Ann. When a = 0, and the absurdity is, that the squares of two equal num- bers should differ. 131. Corollary. When the solution of a problem gives for the value of either of its unknown quanti- ties a fraction whose terms are each equal to zero, this value generally indicates that the conditions of the problem are not sufficient to determine this un- known quantity, and that it may have any value whatever. In some cases, however, there are limi- tations to the change of value of the unknown quantity. CH. III. § III.] EQUATIONS OP THE FIRST DKORF.E. 79 Cases in which the value of an unknown quantity is indeterminate. 132. EXAMPLES. 1. In what case would both the terms of the fractional value of the unknown quantity in example 25 of art. 126 become zero 1 and how could this value be a solution 1 Ans. When b = a, and n = ; and these equations signify, that the couriers travel equally fast, and start at the same time; and, therefore, they re- main together, and any number whatever may be taken as the value of the unknown quantity. 2. In what case would both the terms of either of the fractional values of the unknown quantity in example 31 of art. 126 become zero ? and how could this value be a solution 1 Ans. When a = 0, and C = c ; and these equations signify, that the bodies move equally fast, and start from the same place ; they, therefore, re- main together, and any number whatever may be taken as the value of the unknown quantity. But, in this case, all the algebraic values of the un- known quantity but the first become infinite, as they should, because they are obtained on the supposition, that the second body has passed round the circle once, twice, &c., oftener than the first body ; which is here impos- sible. 3 In what case would all the terms of the fractiona. values of the unknown quantities in example 33 of art. 126 become zero? and how could they, then, satisfy the con- ditions of the problem ? Ans. When a = b = c ; and these equations signify, that the wines and the mix- ture are all of the same value ; in whatever proportion, therefore, the wines are mixed together, the mixture 80 ALGEBRA. [CH. III. § m Cases in which the value of an unknown quantity is indeterminate. - , must be of the required value. But the values of the unknown quantities are still subject to the limitation that their sum is n. 4. In what case would the terms of the fractional, values of the unknown quantities in example 39 of art. 126 be- come zero ? and how could they, then, satisfy the con- ditions of the problem ? Ans. When m = n, and i = ns = ns; and these equations signify, that the sum b of the pro- ducts of the parts of a multiplied by m = n is to be equal to the product of a multiplied by n ; and this is, evidently, the case into whatever parts a is divided. 5. In what cases would all the terms of the fractional values of the unknown quantities in example 41 of art. 126 become zero ? and how could they, then, satisfy the con- ditions of the problem ? Ans. First. When a : b = m : n, and c = ; for these equations indicate that the two required numbers are only subject to the condition that their ratio = a : b. Secondly. When m — n, and a:&=m:ra=m:»j = l, that is, a=b; for these equations indicate that the two numbers are to be equal ; and that they are to remain equal, when they are increased by c, which would always be the case. 6. In what case would all the terms of the fractional values of the unknown quantities in example 48 of art. 126 become zero? and how could these values be solutions? Ans. When a = 0, and 6 = 0; and their equations indicate that the numbers are to be equal, and that their squares are to be equal, which i» always the case with equal numbers. CH. III. $ Ml,] EQUATIONS OP THE FIRST DEGREE. SI Cases of negative value of unknown quantity. 133. Corollary. When the solution of a problem gives a negative value to either of the unknown quantities, this value is not generally a true solution of the problem ; and if the solution gives no other than negative values for this quantity, the problem is generally impossible. But, in this case, the negative of the negative value of the unknown quantity is positive ; so that the enunciation of the problem can often be cor- rected by changing it, so that this unknown quan- tity may be added instead of being subtracted, and the reverse. 134. EXAMPLES. 1. In what case would the value of the unknown quan- tity in example 25 of art. 126 be negative? why should it be so? and could the enunciation be corrected for this case? Arts. When a ^> b ; that is, when the second courier goes slower than the one he is pursuing, in which case he evidently cannot over- take him ; and the enunciation does not, in this case, admit of a legitimate correction. 2. In what case would the values of the unknown quan- tities in examples 29, 31, 32 of art. 126 be negative? why should this be so? and could the enunciations be corrected for this case ? Arts. When e > C ; that is, when the first body moves faster than the second, in which case the second cannot overtake it. The enunciation may be corrected for this case by supposing the bodies to travel in the opposite direction to 82 ALGEBRA. [CH. III. § HI Cases of negative value of unknown quantity. that which they are at present taking, that is, by suppos- ing the first body to pursue the second. Examples 31 and 32 are not, however, impossible in this case : for, from the very nature of their circular mo- tion, the first body is necessarily pursuing the second even in their present direction ; the second body must not, however, be considered as a feet or a -\- c t feet behind the first, but as p — a or p — (a -j- c t) feet before it. 3. In what cases would the values of the unknown quan- tity in example 33 of art. 126 be negative? why should this be the case? and could the enunciation be corrected for this case ? Ans. First When C < c, which is subject to the same remarks as in the preceding question. Secondly. When C ^> c, and c t ]> a, or ~^>p -\- a, or ]> 1p -j- a, &c. , that is, when the first body does not start until the second body has passed it once, or twice, or three times, &c. ; and if the bodies were moving in the same straight line, the enunciation would not admit of legitimate correction. As it is, however, the first body is still pursued by the second, and \sp-\-a — ct, 2p-\-a — c t, &,c, feet before the second, when it starts ; so that all the values given for the unknown quantity are correct, except the negative ones. 4. In what cases would the values of the unknown quan- tity in example 35 of art. 126 be negative? why should this be the case ? and could the enunciation be corrected for (his case ? Ans. When ct~> a, or ^>p-f-fl, or ~^>2p-\-a, &lc; that is, when the first body has passed the second once, twice, &c., before the second begins to move. If the bodies were moving in the same straight line, CH. III. § III.] EQUATIONS OP THE FIRST DEGREE. 83 Cases of negative value of unknown quantity. the second body would be obliged to change its direction, and move in the same direction with the first, and even with this change of enunciation the problem is impossible, if the second body moves slower than the first. But as it is, the bodies are still moving towards each other in the circumference of the circle; their distance apart at the instant when the second body starts being p -f- a — ct, or 2 js -J- a — c t, &c, feet ; so that all the positive values of the unknown quantity remain as true solutions. 5. In what cases would the values of either of the un- known quantities in example 38 of art. 126 be negative ? why should this be the case? and could the enunciation be corrected for this case ? Arts. If we suppose, as we evidently may, that a ^> 6 ; one of the values is negative, First. When a <^ c j that is, when the price of the most expensive wine is less than that of the required mixture. Secondly. When b ~^> c ; that is, when the price of the least expensive wine is more than that of the mixture. In either case the problem is altogether impossible, for two wines cannot be mixed together so as to produce a wine more valuable than either of them without a gain, or less valuable than either of them, without a loss. 6. In what cases would the value of either of the un- known quantities in example 39 of art. 126 be negative 1 ? why should this be so? and could the enunciation be cor- rected for this case? Arts. Supposing, as we may, that m^> n; First When n a^> b, that is, when the sum b of the products is less than the product of a by the lea§( pf the numbers m and n. 84 ALGEBRA. [cH. III. § III. Cases of negative value of unknown quantity. Secondly. When m a <^b ; that is, when the sum b of the products is greater than the product of a by the greater of the numbers m and n. In either of these cases, the problem is plainly impos- sible ; and, in the corrected enunciation, a should be the difference of the required numbers, and b the difference of the products obtained from multiplying one of the numbers by m and the other by ». 7. In what cases would the values of the unknown quan- tities in example 41 of art. 12G be negative? why should this be so ? and could the enunciation be corrected for this case? Ans. Fii-st. When m ^> n, and a n <^ b m, or a : b <^ m : n ; that is, when the first ratio is less than the second, and the second is greater than unity. Secondly. When m <^ n, and a : b ^> m : n ; that is, when the second ratio is less than the first, and also less than unity. In either case the problem is impossible, and c is to be subtracted instead of being added in the corrected enunciation. 8. In what case would the value of one of the unknown quantities in example 46 of art. 126 be negative? why should this be so? and could the enunciation be corrected for this case ? Ans. When 6 ]> a 2 ; that is, when the difference of the squares of the parts of a is to be greater than the square of the number itself, which can never be the case; for the greatest possible difference of squares corresponds to the case in which one of the parts is the number a itself, and the other is zero; and CB. III. § IV.] EQUATIONS OP THE FIRST DEGREE. 85 One Equation with several unknown quantities. the difference of the squares is then just equal to the square of a. The enunciation is corrected for this case by stating it as in example 48. 135. Corollary. It follows from example 7 of the preceding section that a fraction or ratio, which is greater than unity, is increased by diminishing both its terms by the same quantity; and a fraction or ratio, which is less than unity, is diminished by di- minishing both its terms by the same quantity ; but the reverse is the case, when the terms are increased instead of being diminished. SECTION IV. Equations of the First Degree containing two or more unknown quantities. 136. In the solution of complicated problems in- volving several equations, it is often found convenient to use the same letter to denote similar quantities, accents or numbers being placed to its right or left, above or below, so as to distinguish its different val- ues. a" a" Thus, a, a', « (l) , a®. «i, a 3 , '«, "«, l a, \ i®, s a, n\ S„. a IV , . . . a<°», &C. a< 4 >, . . . a<">, &c. "t, • •• «». &.c. "a, .. ..'a, &.C. .."a, &.C. 4 «, . . • •»«. &.C. 3„// 1™ )• • V"' &c. "4 ,, 1"3 > 8" may all be used to denote different quantities, though they generally are supposed to imply some similarity between the % 86 ALGEBRA. [CH. III. § 17, Indeterminate Equations referred to the theory of Numbers. quantities which they represent. Care must be taken not to confound the accents and the numbers in parentheses at the right with exponents. 137. Problem. To solve an equation with several unknown quantities. Solution. Solve the given equation precisely as if all its unknown quantities were known, except any one of them which may be chosen at pleasure ; and in the value of this unknown quantity, which is thus obtained in terms of the other unknown quantities, any values whatever may be substituted for the other unknown quantities, and the corresponding value of the chosen unknown quantity is thus-obtained. 138. Corollary. An equation which contains sev- eral unknown quantities is not, therefore, sufficient to determine their values, and is called indeterminate. 139. Scholium. The roots of an indeterminate equation are sometimes subject to conditions which cannot be expressed by equations, and which limit their values ; such, for instance, as that they are to be whole numbers. But their investigation depends, in such cases, upon the particular properties of differ- ent numbers, and belongs, therefore, to the Theory of Numbers. 140. Theorem. Every equation of the first de- gree can be reduced to the form Ax + By-j-Cz + &,c.+M=0; in which A, B, C, Sfc. and M are known quantities, CH. III. § IV.] EQUATIONS OF THE FIRST DECREE. 87 Solution of any Equation of the First Degree. either positive or negative, and x, y, z, 6fc. are the unknown quantities. Proof. When an equation of the first degree is reduced, as in art. 118, the aggregate of all its known terms may be denoted by M. Each of the other terms must have one of the unknown quantities as a factor ; and, by art. 106, only one of them, and that one taken but once as a factor. Taking out, then, each unknown quantity as a factor from the terms in which it occurs, and representing its multiplier by some letter, as A, B, C, &c, the corresponding un- known quantities being represented by x, y, z, &c, the equation becomes A x + B y + C z -f- & c. + M = 0. 141. Problem. To solve any equation of the first degree. Solution. Having reduced the equation to the form Ax-\-By-j-Cz + &,c. + M=0, find, as in art. 137, the value of either of the unknown quantities, as x, for instance, which is, by art. 121, — By— Cz — &c — M x= _ , and any quantities at pleasure may be substituted for y, z, &c. 142. Problem. To solve several equations with several unknown quantities. First Method of Solution called that of Elimina- tion by Substitution. Find the value of either of the unknown quantities in one of the equations in which it occurs, and substitute its value thus found, which is generally in terms of the other unknown quanti- ties in all the other equations in which it occurs. 88 ALGEBRA. [cH. HI. § IV. Solution of Equations. Elimination by Substitution. The new equations thus farmed, together with those in which this unknown quantity does not occur, are one less in number than the given equations, and contain one unknown quantity less, and may, by a succession of similar eliminations be still far- ther reduced in number and in the number of their unknown quantities, until only one equation is finally obtained ; and the solution of all the given equations is thus reduced to that of one equation. 143. Corollary. When there are just as many equations as unknown quantities, the final equation of the preceding solution will, in general, contain but one unknown quantity, the value of which may be thence obtained ; and this value, being substituted in the values of the other quantities, will lead to the determination of the values of all the unknown quantities. 144. Corollary. When the number of unknown quantities is more than that of the given equations, the final equation will contain several unknown quantities, and will therefore be indeterminate ; so that a problem is indeterminate, which gives fewer equations than unknown quantities. 145. Corollary. When the number" of unknown quantities is less than that of the given equations, only as many of the given equations arc required to determine the values of the unknown quantities as there are unknown quantities ; and the problem is therefore impossible, when the values of the uu* CH. III. § IV.] EQUATIONS OF THE FIRST DEGREE. 89 Case in which the roots of two equations are Zero. known quantities determined from the requisite equations do not satisfy the remaining equations. 146. Problem. To solve two equations of the first degree with two unknown quantities. Solution. Suppose, as in art. 140, the given equations to be reduced to the forms Ax-\-By-\-M=Q, A' x -f B' y -j- M ■— : in which x and y are the unknown quantities. The value of x, obtained from the first of these equa- tions, is — By — M X = — i — ; Which, substituted in the second equation, gives -A'Bt,-A'M +Bl!f+MI=0m A The value of y is found from this equation, by art. 121, to be A' 31 — AM' y — AB—A'B' which, substituted in the above value of z, gives BM' — BM X ~ AB—A'B' 147. Corollary. The value of x, obtained by the preceding solution would be zero, if its numerator were zero, that is, if B M'=B' M. But, in this case, if the first of the given equations is multiplied by B', and the second by B, these products be come, by transposition and substitution, 90 ALGEBRA. [CH. III. & IV, Case in which the roots of two Equations are infinite and indeterminate. A B> x = — B B' y — B' M , A'Bx=—BB l y — BM'=—BB'y — B'M; whence AB'x — A'Bx; that is, the given equations involve the condition that two different multiples of x are equal. But this is impossible, unless a; = 0. The value of y would, likewise, be zero, if we had A 1 M = A M', which leads to conclusions with regard to y, similar to those just obtained with regard to x. 148. Corollary. The denominators of the values of both the unknown quantities would be zero, if we had A B' = A' B. But, in this case, if the first of the given equations is multiplied by B' and the second by B, these products be- come, by transposition and substitution, AB'x + BB'y=— B< M, A'Bx-\-BB'y = AB'x+BB'y=—BM>; whence, we must have B> M = B M ' ; that is, they involve (he impossibility that the two unequal quantities B 1 M and B M' are equal. 149. Corollary. Both the terms of the fractional value of x would be zero, if we had B M' = B'M, and A B' = A' B. But, in this case, if the first of the given equations is mul- CH. III. § IV.] EQUATIONS OF THE FIRST DECREE. 91 Equations of the First Degree. tiplied by B' and the second by B, the products become, by substitution, A B 1 x+B B' y+B' M = 0, ABx+BB'y+BM l =AB , x+BB l y+B , M=0; that is, the two given equations are equivalent to but one, and are, as in art. 144, indeterminate. The product of the two equations B M' = B> M, and A B> = A' B, is ABB'M' = A'BB'M, which, divided by B B', is A M> = A' M, so that both the terms of the value of y would also be zero. 150. EXAMPLES. 1. Solve the two equations 3z + 2y=118, x -j- 5 y = 191. Ans. x = 16, y = 35. 2 Solve the two equations X 4- y -8 2 + 3 - 8 ' — _ L — i 3 2 Ans. x = 12, y = 8. 3. Solve the two equations 1^-^ = 3,-5, 5y-7 4s-3 __ + _ 18-5* ./Ires, a; = 3, y=% 92 ALfiEBRA. [CH. III. § IV Equations of the First Degree solved by Elimination by Substitution. 4. Solve the two equations a % = 6 y, x + y=c. . be at Ans. X = r-r, V = r- ;. a-\-V J a-\-k 5. A says to B, "give me $100, and I shall have as much as you." "No," says B to A, "give me rather $ 100, and then I shall have twice as much as you." How many dollars has each 1 Ans. A $ 500, and B $ 700. 6. Said a lad to his father, "How old are we ? " " Six years ago," answered the latter, " I was one third more than three times as old as you ; but three years hence, I shall be obliged to multiply your age by 2£ in order to obtain my own." What is the age of each 1 Ans. The father 36, the son 15 years. 7. A cistern containing 210 buckets, may be filled by 2 pipes. By an experiment, in which the first was open 4, and the second 5 hours, 90 buckets of water were obtained. By another experiment, when the first was open 7, and the other 3£ hours, 126 buckets were obtained. How many buckets does each pipe discharge in an hour? Ans. The first pipe discharges 15, and the second pipe discharges 6 buckets. 8. There is a fraction such, that if 1 be added to its nu- merator its value becomes = -J ; and if 1 be added to its denominator its value becomes = £. What fraction is il 1 Ans. ■&. 9. Required to find two numbers such, that if the first be increased by a, and the second by b, the product of these two sums exceeds the product of the two numbers them- selves by c ; if on the other hand, the first be increased by CH. III. $ V.] EQUATIONS OF THE FIRST DEGREE. 93 Equations of the First Degree solved by Elimination by Substitution. a', and the second by b', the product of these surns exceeds the products of the two numbers themselves by c\ _,, „ a' c — ac'-\-aa'b' — aa'b , , Ans. 1 he first is -7 ; , the second a' — ab' . bc' — b'c+abb'-^a'bb* a' b — a b< ' 10. A person had two barrels, and a certain quantity of wine in each. In order to obtain an equal quantity in each, he poured out as much of the first cask into the second, as the second already contained ; then, again, he poured out as much of the second into the first as the first then con- tained, and lastly, he poured out again as much from the first into the second as the second still contained. At last he had 16 gallons of wine in each cask. How many gal- lons did they contain originally ? Ans. The first 22, the second 10 gallons. 11. 21 lbs. of silver lose 21bs. in water, and 91bs. of cop- per lose lib. in water. Now, if a composition of silver and copper weighing 148 lbs. loses 14f lbs. in water, how many lbs. does it contain of each metal 1 Ans. 112 lbs. of silver, and 36 lbs. of copper. 12. A given piece of metal, which weighs plbs., loses elbs. in water. This piece, however, is composed of two other metals A and B such, that plbs. of A lose albs, in water, and plbs^ of B lose 6 lbs. How much does this piece contain of each metal 1 Ans , (J^klbs. of A, and ^=1^ lbs. of B. b — a — a 13. According to Vitruvius, the crown of Hiero, king of Syracuse, weighed 20 lbs., and lost l£lbs. in water. Assum- ing that it consists of gold and silver only, and that 19,64 lbs. of gold lose lib. in water, and 10,5 lbs. of silver lose lib. in 94 ALGEBRA. fcH. HI. § IV Equations of the First Degree solved by Elimination by Substitution. water. How much gold, and how much silver, did this crown contain? Ans. 14,77... lbs. of gold, and 5,22... lbs. of silver. 151. Problem. To solve any number of equations of the first degree with the same number of unknown quantities. Solution. Let there be three equations with three un- known quantities ; these equations may, by art. 140, be reduced to the forms Ax + By + Cz + M=0, A'x + B'y + C'z-j-M' = 0, A" x -j- B" y -f C" z -f M" = 0. The value of x, given by the first of these equations, is — By—Cz — M which, being substituted in the other two equations, and the resulting equations being reduced, as in art. 140, gives (AB'—A'B)y-\-(AC'—A'C)z+AM'—A'M = 0, (A B"—A"B) y + (A C"—A"C)z+A M"-A"M = 0. These equations, being solved, as in art. 146, give _ (A'C—A"C')M+(A"C—A OM'+jA C—A'C)M" V ~ (A>B"—A"B')e+(A"B—AB")C'+(AB'—A'B)C"' _ (A"B'—A'B")M-r(AB"—A"B)M'+(A'B-AB')M > t 2 — (A'B"— A"B')C+(A"B— AB")C+\aB'— A'B)C"' in which the terms are arranged in groups in order to dis- play the symmetry of the result ; and these values, being substituted in the value of x, give _ (B"C—B l O , )M\(BC" —B"C)M' + (B'C-BC')M" X ~ (A'B"—A"B')C+(A"B—AB")C' + (AB'—A'B)C I CH. Ill, § IV.] EQUATIONS OP THE FIRST DEGREE. 95 Examples to be solved by Elimination by Substitution. If this method of solution be applied to a greater number of equations, it will lead to similar results. 152. EXAMPLES. 1. Solve the three equations x -f- y + z = 6, 2x + 3y -j- 4 z = 20, 3z-j-7y-f5z = 32. Arts. x = l, y =2,z=3 2. Solve the three equations y + ix = 41, * + i * = 20£, y + iz = 34. Ans. x = 18, y = 32, z =r 10 3. Solve the three equations 53— ix — iz = y ^- 109, 5 y = 4 z. .Arcs, a; = 64, # = 80, z= 100 4. Solve the four equations *+ y + z+w=li 16Z-J- 8y+ 4z-j-2a = 9, 81 x -f 27 y -f- 9 z -j- 3 ?* = 36, 256a;+64y-fl6!8 + 4ii= 100. Ans. T = l,y==i,z = i,u = 0. 5. The sums of three numbers, taken two and two, ar« a, b, e. What are they 1 Ans. £(a-f-6 — e), £(a-f-e — b), ^.(6-j-c — a). 6. A, B, C compare their fortunes. A says to B, " give me $ 700 of your money, and I shall have twice as much 96 ALGEBRA. [cil. III. § IV. Examples to be solved by elimination by Substitution. as you retain ; " B says to C, " give me $ 1400, and I shall have thrice as much as you have remaining ; " Csays to A, " give me $ 420, and then I shall have 5 times as much as yau retain." How much has each? Ans. A $980, B $ 1540, C$2360. 7. Three soldiers, in a battle, make $96 booty, which thej wish to share equally. In order to do this, A, who made most, gives B and C as much as they already had; in the same manner, B then divided with A and C; and after this, C with A and B. If, by these means, the intended equal division is effected, how much booty did each soldier make 1 Arts. A $ 52, B $ 28, C $ 16. 8 A, B, C, D, E play together on this condition, that he who !oses shall give to all the rest as much as they already have. First A loses, then B, then C, then D, and at last also E. All lose in turn, and yet at the end of the 5th game they a., have the same sum, viz. each $ 32. How much had each w* in they began to play 1 Ans. A $81, B $41, C$21, D$ 11, E $6. 153. Second Method of solving the Problem of art. 142, called that of Elimination by Comparison. Find the value of either of the unknown quantities in all the equations in which it is contained; place either of the values thus obtained equal to each of the others, and the equations thus formed will be one less in number than those from which they are ob- tained, and will contain one unknown quantity less. By continuing this process on these new equations, the number of equations will finally be reduced to one. CHv HI. § IV.] EQUATIONS OF THE FIRST DEGREE. 97 Examples to be solved by Elimination by Comparison. 154. EXAMPLES. 1. To solve any two equations of the first degree with two unknown quantities. Solution. These equations may, as in art. 146, bo re- duced to the forms Ax-\-By-\-M=0, A> x -J- B' y -f- M! = 0. The values of x, obtained from these equations, are — By — M *= A ' ■ B'y — M 1 x = A' which, being placed equal to each other, give — By — M __ — B' y — M' A A' ' whence _ A< M—AM y ~~ AB' — A' B' and, therefore, B M' — B< M A B' — A' B' being the same values as those obtained in art. 146. 2. Solve the three equations x~ y ~ z 12' i_(_I I — _L 1 """ y z ~ 12' I_I_L I — — x y'T' z~~ \1' 98 ALGEBRA. [CH. III. <$ IT Examples to be solved by Elimination by Comparison. Solution. The values of x, obtained from these equa- tions, are 12 yz X ~ 13yz—l2z—l2y' lyz — 12z + 12y' Wyz _ 5yz+12z — 12 y' the first of which being placed equal to each of the others gives, by reduction, z = 4, y = 3; whence we get, from either value of x, by substitution, 3=2. 3. Solve the two equations 7y= 2x—3y, 19 x = 60 y -f- 621^. .4n.s. a; = 88|, y = 17& 4. Solve the three equations 3 x +5^=161, 7 a; -j- 2 z = 209, 2 y + z = 89. .dras. a; = 17, y = 22, z = 45. 5. Solve the three equations 7 + 7 = °' a; ' z 1 J. 1 y ' x Ans. x = 2 _ 2 _ 2 a-ffc—c'^ — a _6_J_ c ' 2 — 6_j_ c _«' C1I. IU. § IV.] EQUATIONS OF THE FIRST DEGREE 99 Examples to be solved by Elimination by Comparison. 6. Solve the three equations 2 X 3 by +\ -i 1 Ax +7 +' fi 11 = 6 72' 5 Gx 1 ~~ y ^ = 12 ^ Ans. x = 6, y = 9, z = £. 7 A person has two horses, and two saddles, one of which cost $ 50, the other $ 2. If he places the best sad- dle upon the fiist horse, and the worst upon the second, then the latter is worth $8 less than the other; but if he puts the worst saddle upon the first horse, and the best upon the other, then the latter is worth 3f times as much as the first. What is the value of each horse 1 Ans. The first $ 30, the second $ 70. 8. What fraction is that, whose numerator being doubled, and denominator increased by 7, the value becomes §; but the denominator being doubled, and the numerator increased by 2, -the value becomes f 1 Ans. £ . 9. A wine merchant has two kinds of wine. If he mix 3 gallons of the worst with 5 of the best, the mixture is worth $ 1 per gallon ; but, if he mix 3£ gallons of the worst with 8| gallons of the best, the mixture is worth $ 1,03£ per gallon. What does each wine cost per gallon? Ans. The best $ 1,12, the worst $0,80. 10. A wine merchant has two kinds of wine. If he mix a gallons of the first with b gallons of the second, the mix- ture is worth c dollars per gallon ; but, if he mix a' gallons of the first with b 1 gallons of the second, the mixture is worth c< dollars per gallon. What does each wine cost per gallon 1 100 ALGEBRA. [cH. III. § IV. Examples to be solved by Elimination by Comparison. (a + b)b'c — (a>4-b')bc> Ans The first v ' ' — Tl — ^-r 1 dollars, the a b — a b , (a + b)a>c — (a' + b<)ac' „ second v ~ ' ,; — K —rr i dollars. a' b — ab' 11. Three masons, A, B, C, are to build a wall. A and ZJ, jointly, could build this wall in 12 days; J5 and C could accomplish it in 20 days ; A and C would do it in 15 days. What time would each take to do it alone ? A7is. A requires 20, B 30, C 60 days. 12. Three laborers are employed in a certain work. A and B would, togelher, complete it in a days ; A and C require 6 days ; B and C require c days. In what time would each accomplish it singly ? 2abc 2abc Ans. A in - — ; T days, B in = — ; — days, bc-\-ac — ab bc-\-ab — ac „ . 2abc C in —r-r j- days. ao-j-ac — be 13. A cistern may be filled by three pipes, A, B, C. By the pipes A and B, it could be filled in 70 minutes ; by the pipes A and C, in 84 minutes ; and by the pipes B and C, in 140 minutes. In what time would each pipe fill it? Ans. A in 105, B in 210, Cin 420 minutes. 14. A, B, C play faro. In the first game A has the bank, B and C stake the third part of their money, and win. In the second game B has the bank, A and C stake the third part of their money and also win. Then C takes the bank, A and B stake the third part of their money and also win. After this third game they count their money, and find that they have all tlie same sum of 64 ducats. How much had each when they began to play? Ans. A had 75, B 63, C54. CH. III. § IV.] EQUATIONS OF THE FIRST DEGREE. 101 Elimination by the method of the Greatest Common Divisor. 15. Five friends, A, B, C, D, E, jointly spend $879 at an inn. This sum is to be paid by one of them ; but, on consultation, they find that none of them had, alone, enough for this purpose. If, then, one of them is to pay it, the others must give him a part of their money. A can pay, if he receives one fourth ; B, if he receives one fifth ; C, if he receives one sixth ; D, if he receives one seventh ; and E, if he receives one eighth of the others' money. How much has each 1 Ans. A $ 319, B $ 459, C $ 543, D $ 599, E $ 639. 155. Third Method of solving the Problem of art. 142, called that of Elimination by the method of the greatest Common Divisor. Solution. This method is generally inapplicable to trans- cendental equations, but can be successfully applied in all other cases to eliminate one unknown quantity after an- other, until the given equations are reduced to one. In order to eliminate an unknown quantity from two equations which contain it, reduce them as in arts. 105 and 118, and arrange their terms accord- ing to the powers of the quantity to be eliminated, taking out each power as a factor from the terms which contain it. It being now recollected that the second member of each of these equations is zero, it will appear evident that, if the first members are divided one by the other, the remainder arising from this division must likewise be equal to zero ; for this remainder is the difference between the dividend and a certain multiple of the divisor, that is, between zero and a certain multiple of zero. 9* 102 ALGEBRA. [cH. III. § IT Elimination by the method of the Greatest Common Divisor. Hence, divide one of these first members by the other, and proceed, as in arts. 60, $*c, to find their greatest common divisor ; each successive remainder may be placed equal to zero. But a remainder will at last be obtained, which does not contain the quan- tity to be eliminated; and the equation, formed from placing this remainder equal to zero, is the equation from which this quantity is eliminated. By eliminating, in this way, the unknown quan- tity from either of the equations which contain it, taken with each of the others, a number of equations is formed one less than that of the given equations, and containing one less unknown quantity ; and to which this process of elimination may be again ap- plied until one equation is finally obtained. 156. Scholium. It sometimes happens, that the first members have a common divisor which contain the given unknown quantity ; and. in this case, the process cannot be continued beyond this divisor. But as the given first members are multiples of their com- mon divisor, they must be rendered equal to zero by those values of the unknown quantities which render the com- mon divisor equal to zero ; that is, the two given equations are satisfied by such values of the unknown quantities ; so that, though they are in appearance distinct equations, they are, in reality, equivalent to but one equation, that is, to the equation formed by placing their common divisor equal to zero. 157. Scholium. Care must be taken that no fac- tor be suppressed which may be equal to zero. CH. III. § IV.] EQUATIONS OF THE FIRST DEGREE. 103 Examples of Elimination by the method of the Greatest Common Divisor 158. KXAMPLES. 1. Obtain one equation with one unknown quantity from the two equations x 3 -(- y x 2 — y 3 -\- 5 = 0, x 3 -f- y s x — 5 = 0, by the elimination of x. Solution. Divide the first members as follows. x 3 -|- y x 2 — y 3 -j- 5 x 3 -f- y 2 x ^— 5 x 3 -J- y 3 a; — 5 1 ' 1st Rem. y a; 2 — y 2 x — y 3 + 10. Divide the preceding divisor by this remainder after mul- tiplying by y to render the first term divisible. yx 2 — y a s — y 3 -f 10 y a;3 -)- y3 z _ 5 y y a; 3 — y 2 x 2 — y s x-|- 10 i y 2 a; 2 +(2y y s x 2 - V 10) a; — 5y a: — y 4 +10y 2d Rem. (3 y 3 — 10) x -\- y 4 — 15 y. Divide the preceding divisor by this remainder after mul- tiplying by (3 y 3 — 10) to render the first term divisible. y3?—y2x—y 3 -\-l0 3y 3 — 10 3y 3 — 10 yx z — 3y 5 + 10y 2 yx*\- y 5 -15y 2 x — 3y 6 - -f-40y 3 X -100 3y 3 |x-|- y* — 10 j — 15y 3y 3 — io yx, — 4y5 + 25y 2 — 4 j/ 5 +25 y 2 x — 3y 6 - -f 40y 3 -100 Multiply by (3y 3 -10), 3y 3 _ 10 (3 y 3_io) (_4y54-25y 2 ) x — 9y 9 -fl50y oy the elimination of a;. .4ns. Either y — 1 = 0, or y a — 3y + 21 = 0. 6. Obtain one equation with one unknown quantity from the three equations x + y + z= a, xz + xy+yz = b, xy z = c, by the elimination of x and y. Ans. z 3 — a « 2 ■} bz — e — 0. CH. III. § IV.] EQUATIONS OP THE FIRST DEGREE. 105 Examples of Elimination by the method of the Greatest Common Divisor. 7. Obtain one equation with one unknown quantity from the three equations x + V + z = a > ** + V* + * = h xy -\-xz-\-y z = c, by the elimination, of x and y. Ans. These three equations involve an impossibility unless a a — b — 2cz=0; and in case this equation is satisfied by the given values of a, b, and c, the three given equations are equivalent to but two, one of thern being superfluous, and, by the elimination of x, they give the indeterminate equation with two unknown quantities i/ 2 + yz-{-z s — ay — a z + c = 0. 8. Obtain one equation with one unknown quantity from the three equations x -f y* = 4, V + z 2 = 2, z + a; 2 = 10, by the elimination of x and y. Ans. z 8 — 8z6 + 16z*+a — 10=0. 9. Obtain one equation with one unknown quantity from the four equations x-\-y+z-\-u = a, xy-\-xz-\-xu-\-yz-\-yu-\-zv=b xyz-{-xyu-\-xzu-\-yzu = c, xy zu = e, by the elimination of x, y, and z. Ans. a 4 — a u 3 -J- 6 w 2 — e « + e — 0. 106 ALGEBRA. [CH. III. § IV. Elimination by Addition and Subtraction. 10. Solve the two equations y x 3 — x 3 -\- x = 3, y x (y x a -j- 1) — x 3 -J- x = 6. Solution. The elimination of x gives 3 y — 3 = 0, or y = 1 ; which, being substituted in the first of the given equations produces x = 3. 11. Solve the two equations x^y i —8y 3 x i + l6x !i =^90xy+60(x—y^)—720{y—l) (y 9 — 4y + 4)z 12 a . 5 x Ans. x=4, y = 2. 12. Solve the three equations xy-j-z = 5, x y z -}- z a = 15, a;y 2 + x 3 j — 2x + 2i8; = 8. -dns. i = 2, t/ = 1 , z = 3. 159. Problem. To solve two equations of the first degree by Elimination by Addition and Subtraction. Solution. The given equations may, as in art. 146, be reduced to the forms A x-{-By-\-M = 0, A' x -J- B' y -j- M< = 0. The process of the preceding article, being applied to these equations in order to eliminate x, will be found to be the same as to Multiply the first equation by A' the coefficient of ▼ in the second, multiply the second by A the co- efficient of x in the first, and subtract the first of these products from the second. CH. III. § IV.] EQUATIONS OF THE FIRST DEGREE. 107 Examples of Elimination by Addition and Subtraction. Thus, these products are A A' x -f- A' B y + A' M = 0, A A 1 x -f A B< y -j- A M> = 0; and the difference is (A B< — A' B) y + A M' — A 1 M = 0; whence _ A' M—A M > y ~AB—A, C= C, &c. 112 ALGEBRA. [CH. IV. § II. A Function ; its Variable, and Rate of Change. SECTION n. Derivation. 164. Definition. When quantities are so connect- ed that their values are dependent upon each other, each is said to be a function of the others : which are called variables when they are supposed to be changeable in their values, and constants when they are supposed to be unchangeable. Thus if y =. a x + b y is a function of the a, b, and x ; but if x is variable while a and b are constant, it is more usual to regard y as simply a function of x. 165. Definition. In the case of a change in the ralue of a function, arising from an infinitely small change in the value of one of its variables, the rela- tive rate of change of the function and the variable, that is, the ratio of the change in the value of the function to that in the value of the variable, is called the derivative of the function. The derivative of the derivative of a function is called the second derivative of the function ; the de- rivative of the second derivative is called the third derivative ; and so on. 166. Corollary. The derivative of a constant is zero. 167. Corollary. The derivative of the variable, regarded as a function of itself, is unity ; and the second derivative is zero. *.H IV. § II.] NUMERICAL EQUATIONS. 113 The Derivative of the sum of any Functions. 168. Theorem. The derivative of the sum of two function& is the sum of their derivatives. Proof. Let the two functions be u and v, and let their values, arising from an infinitesimal' change i in the Value of their variable, be u' and v' ; the increase of their sum will be ("' + »') -(« + ») or u' — u -j- v' — v, and therefore the derivative of the sum is which is obviously the sum of their derivatives. 169. Corollary. By reversing the sign of v, it may be shown, in the same way, that the derivative, of the difference of two functions is the difference of their derivatives. 170. Corollary. The derivative of the algebraic sum of several functions connected by the signs -f- and — is the algebraic sum of their derivatives. 171. Corollary. If, in this sum, any function is repeated any number of times, its derivative must be repeated the same number of times ; in other words, if a function is multiplied by a constant its deriva- tive must be multiplied by the same constant. Thus, if the derivatives of u, v, and w are respectively U, V, and W, and if a, b, c, and e are constant, the deriv- ative of a u -\- b v — c w -j- e a U+b V—c W. 10" 114 ALGEBRA. [cH. IV. § II. The Derivative of a Power. 172. Problem. To find the derivative of any power of a variable. Solution. Let the variable be a and the power a", and let b differ infinitely little from a ; the derivative of a" is then b n — o n b — a ' Now when 6 is equal to a, the value of this quotient is, by art. 51, n a"- 1 ; and this must differ from the present value of this quotient, by an infinitely small quantity, which being neglected gives no»-i for the derivative of a". The derivative of any power of a variable is, therefore, found by multiplying by the exponent, and diminishing the exponent by unity. 173. Corollary. The derivative of mo" when m is con- stant and a variable is nm a" -1 . 174. Problem. To find the derivative of any power of a function. Solution. Let the variable be a, the function u, and the power u n ; let b differ infinitely little from a, and let v be the corresponding value of u ; if U is the derivative of a and V that of u n , we have rv v n — u n v — u V = -, and U= r . o — a o — a But, by art. 51, V n M n V — u ' vhich multiplied by u= v ~ a b — a' CH. IV. § II.] NUMERICAL EQUATIONS. 115 The Derivative of a Power. gives 0" — w" v — u v n — M n _ , TT U' = . , = -; = n u n ~ l U. v — u 6 — a o — a The derivative of any power of a function is, therefore, found by multiplying by the exponent and by the derivative of the function, and diminishing the exponent by unity. 175. EXAMPLES. Find the derivatives of the following functions in which x is the variable. 1. a 2 . Ans. 2 x. 2. x\ Ans. 3 x*. 3. x n -f- a x n -f 6 xP -f &c. Ans. n x n_1 -\-ma x m ~ l -\-p b a;? -1 -{- &c. 4. A-\-Bx-\-Cx*-{-Dx i -\-Ex i -\-Fx5-\-&.a. Ans. B+2Cx+3Dx s -\-4Ex*+5Fx' l -\-&,c. 5. a -j- x. Ans. 1. 6. (a + z) 2 . 4ns. 2(a + z). 7. (a-fz) 3 . 4ns. 3(a + x) 2 . 8. (a -f- a:)". -<4ms. « (a -{- *)*-*. 9. (o-)-6i) a . Ans. 2 6 (a + 6 a:). 10 (o+iit)". -4ws. » 6 (a -fix)"-!. 176. Problem. To find the derivative of the pro- duct of two functions. Solution. Let u and v be the functions, and U and T 7 their derivatives ; then, since the derivative is the rate of change of the function to that of the variable, it is evident 116 ALGEBRA. [CH. IV J 1| The Derivative of a Product. that when the variable is increased by the infinitesimal «", that the functions will become u -\- Ui and v -f- Vi. The product will therefore change from u v to («+ Ui) (v + V i) = u v + v Ui + u Ft + UV$>, and the increase of the product is v Ui + u Vi+UVP; the ratio of which to i is vU+u V+UVi, or, neglecting the infinitesimal U V i, it is v U+u V; that is, the derivative of a product of two functions is equal to the sum of the two products obtained by multiplying each function by the derivative of the other function. 177. Corollary. The derivative of (x — a) n v is, then, »(» — fl)"-ij>-{-(a! — a)" V, because the derivative of (x — a) n is n (x — a) n ~K CH. IV. § III.] NUMERICAL EQUATIONS. 117 Solution of Numerical Equations. SECTION HI. Numerical Equations. 178. Definition. A numerical equation is one all "whose coefficients are given in numbers, so that it involves no literal expressions except those denoting the unknown quantities. 179. Problem. To solve a numerical equation. Solution. Let the equation be reduced as in arts. 105 and 118, to the form m = 0. Find by trial a value of the unknown quantity x which nearly satisfies this equation, and let this value be a ; substitute this value in the given equa- tion, and let the corresponding value of uiem. A correction e in the value of a is then to be found, which shall reduce the value of u from m to zero. Now, if U is the derivative of u, and if M is the value of U which corresponds to x = a, M is, by art. 165, the rate at which u changes in comparison with x, so that when x = a -f- e u = m-\-Me=0, and therefore m . m e=~M> *=" + *=<*- W - By this means a value of x is found which is not 118 ALGEBRA. [CH. IV. § III Rate of Approximation. perfectly accurate, because M is not the rate at which u varies during the -whole interval from x == a to x = a -f- e ; but only while x differs infinitely little from a. Calling, therefore, a' this approximate value of x, we have , m a=a- w , which may be used in the same way in which a was, in order to find a new approximate value a" of x; and if m' and M' denote the corresponding values of u and U, we shall have „ , "»' In the same way, may the approximation be con- tinued to any degree of accuracy. 180. Problem. To determ/ne the rate of approxi- mation in the preceding solution. Solution. This is a most important, practical point, and the determination of it will be found to facilitate the solution. Now, it may be observed, that since e is the correction of a, its magnitude shows the degree of accuracy which belongs to a, and the accuracy of e is, obviously, the same with that of a' = a -f- e. The comparative accuracy of the approximate value of a, and the succeeding approximate value a 1 , is, then, the same with the magnitude of e compared with the error of e. Now, in determining e, M was supposed to be the rate at which u changed throughout the whole interval in the change of i from a to a -\- e. But if the rate of change of CH. IV. § III.] NUMERICAL EQUATIONS. 1 19 Rate of Approximation. M is denoted by N, that is, if N is the derivative of M, the value of 31, at the end of this interval when x is a-j-c, must be increased to M + Ne. In the middle of the interval when * is a + £ e, the value of M is M+lNe, which may be regarded as the average value of the rate of it's increase, throughout the interval. When x, therefore becomes a -\- e, u becomes m + (M + £Nc)e = 0, or whence by transposition and division m N e = " M which differs from e= — m M by the term JV 2M e 2 , which may, therefore, be regarded as the error of e; and its comparison with e gives the rate of approximation. 181. Corollary. If the value of a is accurate to a given place of decimals, as the gth, this will be shown by the magnitude of a, for we shall find and, consequently, ** < We' 120 ALGEBRA. [cH. IV. § III. Rate of Approximation. N If also the value of jrnrj- »"« found to be such that JL < JL 2iB s 10*' then the inaccuracy of e 2 or of a' is N . ^ 1 e 2 < 2ilf ^ 10 2 ff+ fc ' £Aa£ is, a' is accurate to the (2 g -f- &)£/& pfope o/ decimals and the division of ra by M may be carried to this extent. 182. Corollary. When the given equation has the form u = h, in which A is a given number, it may be brought to the form u — h = 0, so that the value of the final member when x = a is m — h, while the value of the derivative is M, because h does not vary, and, therefore, m — h h — m e== W * M ' which is often a more convenient form in practice than that of art. 179. CH. IV. § III.] NUMERICAL EQUATIONS. 121 Solution of Numerical Equations. 183. EXAMPLES. 1. Solve the equation a? — Sx = — 1, which has three roots, the first being nearly 2, the second nearly 0, and the third nearly — 2. Solution. This equation, compared with arts. 179-182, gives u = a? — 3 x, li =z — 1 , U= 3 x s — 3 ; deriv. of U= 6 x. Hencej if a = 2, m = 8 — 6 = 2, M = 12 — 3 = 9, N = 12, = 0-000359232, M'" = 4041072, e>" = 00000S889, g'» = 4, a"' = 1-53208889', which is accurate to 2 g'" = 8 places of decimals. This process may be arranged in the following form, in the first column of which, h is* placed at the top, and the successive values of — m above each horizontal line with those of A — m below it. In the second column are placed the successive values of the divisor M. In the third column the first approximation a is placed at the top of the table, 11 122 ALGEBRA. [CH. IV. § m. Solution of Numerical Equations. and the successive values of e, above each line with those of a -J- e below it. M. h.—l- — m. — 2- 9- 5-67 4 2075 4041072 2- a. — m. — 3- — 03 e. 01 87 17 a —0-813 — 015 0-926125 155 —0073875 — 0018 1000359232 1-532 0000359232 0000085S9 a-fe. In the same way may found, as follows. When x = 0-3 — 1- 0- the 1 -53208889 second and third roots be N _ 180 2 M - 546' °' 0- — 1- 0-873 —0127 0980696 019304 — 100000 9615183 0000009615183 — 3- —273 —2-6532 —2-63814813 0-3 0-3 004 0-34 00073 3473 —00000036446 When — 1- 2- x=— 2 The second root =: -ZV 2 2M 1- 1159 01 59 1004672 0004672 0-3472963554 & = 0. — 2- 01 783 7-6032 The third root = -1-9 002 — 1-88 0000614 — 1 879386 CII. IV. § III.] NUMERICAL EQUATIONS. 123 Solution of Numerical Equations. 2. Solve the equation a; 3 — 12 x = — 132, which has a root nearly equal to — 6. Ans. — 5-87205266. 3 Solve the equation a* + 8 a 2 -j- 16 a; = 440, which has two roots, the first being nearly 4 and the second nearly —4. Ans. 397601 and —4350577. 4. Solve the equation 2 a; 4 — 20 a; = — 19, which has two roots, the first being nearly 1, and the second nearly 2. Ans. 10928 and 1-59407. 5. Solve the equation 5i 3 -6i=-2, which has three roots, the first being nearly 1, the second nearly 0, and the third nearly — 1. Ans. 0-856, 03785, —1-2345. 184. Problem. To find any root of a number. Solution. If the required root is the nth root of the num- ber h, this problem is equivalent to solving the equation x n = h ; so that, if the preceding solution is applied to this case, we have « = x n , Uz=n »"-i. 185. Corollary. When x = a, m=.a n , M=n a" -1 , N=n(n — l)a" -2 N _n(n — l)a"- 2 _w — 1 2M ~ 2raa n -! — 2a ' 186. Corollary. It may be observed, since (10 6 e)" = 10" i e n ; 124 ALGEBRA. [cH. IV. § III. Extraction of Roots. so that if, e<10, 10 i e<10»+ 1 , (10 6 e)»<10"<»+ 1 >; and if e>l, 10 5 e>10 6 , (10 & e)»>10» 1 ; that is, if the root is between 10 6 and 10* + 1 the nth power is between 10™ h and 10 n6 + n ; or, otherwise, if the left hand significant figure of the root is b places from the decimal point, that of the power must be as many as b times n places from this point, and less than 6-j-l times n places from it ; which, combined with the preceding articles, gives the following rule for rinding the root of a number. Divide the given number into portions or periods beginning with the decimal point, and let each por- tion or period contain the number of places denoted by the exponent of the power. Find the greatest integral power contained in the left hand period ; and the root of this power is the left hand figure of the required root, and is just as many places distant from the decimal point as the' corresponding period is removed by periods from this point. Raise the approximate root thus found to the given power and subtract it from the given number, and leave the remainder as a dividend. Raise, again, this approximate root to the power next inferior to the given power, and multiply it by the exponent of the given power for a divisor. The quotient of the dividend by, the div hor gives the next figure or figures of the root. CH. IV. § HI.] NUMERICAL EQUATIONS. 125 Extraction of Roots. The new approximate root, thus found, is to be used in the same way for a new approximation. The number of places to which each division may usually be carried, is so far as to want but one place of doubling the number of places, to which the pre- ceding approximation was found to be accurate. 186. EXAMPLES. 1. Find the fourth root of 5327012345-641. Solution. In the following solution, the columns are the same as the first and second columns in art. 183, except that the top number of the second column is the root which is separated by space into the parts obtained by each suc- cessive division, and the number at the top of the first column is divided by spaces into periods. 2 7* 016 32000000 78732000 Arts. 270-16. % Find the 4th root of 79502005521. Ans. 531. 3. Find the 3d root of 75686967. Ans. 423. 4. Find the 3d root of 128787625. Ans. 505. 5. Find the 3d root of 20548344701. Ans. 5901. 6. Find the 3d root of 512768384064. Ans. 8004. 7. Find the 3d root of 524581674-625. Ans. 806-5. 53 16 2701 2345- 641 37 53 2701 1441 2345- 641 1260 2345 641 * This figure must, in the present case, be found by trial, because the first quotient is so inaccurate. 11* 126 ALGEBRA. [CH. IV, § III. Roots of Fractions. 8. Find the 3d root of 1003003001. Ans. 1001. 9. Find the 3d root of 0^756058031. Ans. 0911. 10. Find the 3d root of 0000003442951. Ans. 00151. 11. Find the 5th root of 418227202051. Ans. 211. 12. Find the 4th root of 75450765-3376. Ans. 932. 13. Find the 5th root of 0000016850581551. Ans. 111. 14. Find the 4th root of 2526 88187761. Ans. 709. 15. Find the 3d root of 12. Ans. 2-289 -|-. 16. Find the 3d root of 28-25. Ans. 3-045 -(-. 187. Corollary. The roots of fractions can be found by reducing them to their lowest terms, and extracting the roots of their numerators and denomi- nators separately. The roots of mixed numbers can be found by reducing them to improper fractions. 188. EXAMPLES. 1. Find the 3d root of §J. Ans. f. 2. Find the 3d root of £}$£§. Ans. ff. 3. Find the 3d root of ,#&. Ans. ^. 4. Find the 3d root of 6-J4 ££. Ans. 1 f f 5. Find the 4th root of 3^f Ans. If 189. Corollary. In the case of the square root, we have u = x Q , U = 2 x, m = a Q , M = 2 a ; and, since the square of a -\- h is CH. IV. § III.] NUMERICAL EQUATIONS. 127 The Square Root of Numbers. (a + hf = a 3 + 2 a h + h s = a 2 + (2 a + h) h it is unnecessary to find the square of the whole root at each successive approximation ; for the square of a being already subtracted, it is sufficient to subtract (2 a -\- h) h from the remainder, in order to obtain the next remainder. In this way, we obtain the fol- lowing rule for the extraction of the square root. To extract the square root of a number, divide it into periods of two figures each, beginning with the place of units. Find the greatest square contained in the left hand period, and its root is the left hand figure of the required root. Subtract the square of the root thus found from the left hand period, and to the remainder bring down the second period for a, dividend. , i 'A ^ Double the root for a divisor, and the quotient of the dividend exclusive of its right hand figure, di- vided by the divisor, is the next figure of the required root ; which figure is also to be placed at the right of the divisor. Multiply the divisor, thus augmented, by the last figure of the root, subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. Double the root now found for a nev) divisor and continue the operation as before, until all the periods are brought dozen. 128 ALGEBRA. [CH. IV. § In. The Square Root of Numbers. 190. EXAMPLES. 1. Find the square root of 28111204. Solution. The operation is as follows ■ 28111204 15302 = Ans.. 25 ! 1st Rem. 311 309 103 1st Divisor. 2d Rem. 3d Rem. 4th Rem 212 1 106 2d Divisor. 21204 10602 3d Divisor. 21204 0. 2. Find the square root of 61009. Ans. 247 3. Find the square root of 57198969. Ans. 7563. 4. Find the square root of 1607448649. Ans. 40093. 5. Find the square root of 48303584-206084. Ans. 6950078. 6. Find the square root of 0000256. Ans. 0016. 7. Find the square root of §£§. Ans. ^f. 8. Find the square root of 1 J. Ans. 1£. 9. Find the square root of 5. Ans. 2-236 -f-. 10. Find the square root of 101. Ans. 10-049 4-. 11. Find the square root of 9-6. Ans. 3098 -f. 12. Find the square root of 0003. Ans. 005477-f. 13. Find the square root of 10. Ans. 3 16227 -f-. 14. Find the square root of 1000. Ans. 31-6227 -f-. 191. Corollary. The roots of vulgar fractions and mixed numbers may be computed in decimals by first reducing them to decimals. CH. IV. § III.] NUMERICAL EQUATIONS. 129 The Square Root of Numbers. 192. EXAMPLES. 1. Find the square root of -^ to 4 places of decimals. Arts. 0-2425 -f. 2. Find the square root of -j^- to 3 places of decimals. Ans. 645-f. 3. Find the square root of If to 2 places of decimals. Ans. 1-32 -f. 4. Find the square root of 11 J-J to 3 places of decimals. Ans. 3-418 -f- 5. Find the 3d root of § to 3 places of decimals. Ans. 873-j- 6. Find the 3d root of £ to 3 places of decimals. Ans. 0941 +. 7. Find the 3d root of 15§ to 3 places of decimals. Ans. 2-502 +. 130 ALGEBRA. [CH. V. § I. Power of a Monomial. CHAPTER V. POWERS AND ROOTS. SECTION I. Powers and Roots or Monomials. 193. Problem. To find any power of a monomial. Solution. The rule of art. 28, applied to this case, in which the factors are all equal, gives for the coefficient of the required power the same power of the given coefficient, and for the exponent of each letter the given exponent added to itself as many times as there are units in the ex- ponent of the required power. Hence Raise the coefficient of the given monomial to the required power ; and multiply each exponent by the exponent of the required power. 194. Corollary. An even power of a negative quantity is, by art. 32, positive, and an odd power is negative. 195. EXAMPLES. 1. Find the third power of 2 a- b s c. ' Ans. 8a e b 15 c 3 . 2. Find the mth power of a n . Ans. a m n . 3. Find the — mth power of a n . Ans. a~ mn . 4 Find the mth power of a~ n . Ans. a~ mn 5. Find the — mth power of a - ". Ans. a nn . CH. V. § I.] P0W1SRS AND ROOTS OF MONOMIALS. 131 Root of a Monomial ; imaginary quantity. G. Find the Cth power of the 5th power of a 3 b c a . Ans. a 90 6 30 c 60 . 7. Find the gth power of the — pth power of the mth power of a _n . Ans. a mn fi. 8. Find the rth power of a m b~" cP d. Ans. a mr b~ nr cr r d r . 9. Find the— 3d power of a~^b 3 c -5 / 6 x ~ 1 - Ans. afi-Sc 15 /- 18 * 3 - a 4 b 5 a 16 6 20 10. Find the 4th power of 5-7-2.. Ans. c>df c™d*f*' cfib 3 11. Find the — 2 mth power of the — 1st power of — r^. cd b a 4mJ6m 12. Find the 5th power of —2 a 2 . Ans. —32 a 10 . 13. Find the 4th power of — 3 6~ 3 . Ans 81 6~ 12 . 14. Find the 5th power of the 4th power of the 3d power of — a. Ans. a 60 . 15. Find the — 5th power of the — 3d power of — a. Ans. — a) 5 . 16. Find the — 4th power of the — 3d power of -. , a 12 Ans. p- 2 . 196. To find any root of a monomial. Solution. Reversing the rule of art. 193, we obtain im- mediately the following rule. Extract the required root of the coefficient; and divide each exponent by the exponent of the required root. 132 ALGEBRA. [cH. V. § I. Fractional Exponents ; imaginary quantities. 197. Corollary. The odd root of a positive quan- tity is, by art. 194, positive, and that of a negative quantity is negative. The even root of a positive quantity may be either positive or negative, which is expressed by the double sign ± preceding it. But, since the even powers of all quantities, whether positive or negative, are positive, the even root of a negative quantity can be neither a positive quan- tity nor a negative quantity, and it is, as it is called, an imaginary quantity. 198. Corollary. When the exponent of a letter is not exactly divisible by the exponent of the root to be extracted, a fractional exponent is obtained, which may consequently be used to represent the radical sign. 199. EXAMPLES. 1. Find the mth root of a mn . Ans. a n . 2. Find the mth root of a~ mn Ans. a~ n . 3. Find the square root of 9 a 4 J9/-i2g— 8», Ans. ±3o 2 J/-6f- 4 ". . t,- , , .^ n a 8 6 20 e 4 , a?b s c • *'" 16 d™Z™' -*"- 4dW 5. Find the 9th root of — 2 36 « 45 6 9 . Ans. — 2 4 a 5 6. 6. Find the mth root of a". R Ans. a™. 7. Find the mth root of — . a" 1 i Ans. -n= a ~" CH. V. § H.] CALCULUS OF RADICAL QUANTITIES. 133 Calculus of Radicals. 8. Find the J»th root of -. Ans. n £. _JL JL b m c m e m . , 9. Find the 5th root of — a 3 . Ans. — a". 10. Find the square root of a. Arts. a?. 11. Find the 3d root of — a. Ans. — a'. 12. Find the mth root of a. Ans. am. 200. Corollary. By taking out — 1 as the factor of a negative quantity, of which an even root is to be extracted, the root of each factor may be extracted separately. 201. EXAMPLES. 1. Find the square root of — a 9 . Ans. a^/ — 1. 2. Find the 4th root of— a 4 b s c 3 . A , 9 | * , Ans. oJ 2 c 4 y/ — 1. 3. Find the 8th root of — a. - 4 8 , Ans. a 8 y/ — 1 SECTION II. Calculus of Radical Quantities. 202. Most of the difficulties in the calculation of radical quantities will be found to disappear if frac- tional exponents are substituted for the radical signs, and if the rules, before given for exponents, are ap- plied to fractional exponents. In the results thus obtained, radical signs may again be substituted for the fractional exponents ; 12 134 ALGEBRA. [CH. V § H Examples in the Calculus of Radical Quantities. but, before this substitution is made, the fractional exponents in each term should be reduced to a com- mon denominator, in order that one radical sign may be sufficient for each term. When numbers occur under the radical sign, they should be separated into their factors, and the roots of these factors should be extracted as far as pos- sible. Fractional exponents greater than unity should often be reduced to mixed numbers. 203. EXAMPLES. 1. Add together 7 1/ 54 a 3 6 5 c 3 and 3 1/ 16 a 3 ¥> c 3 Solution. We have 7 1/ 54 a 3 b 5 c 3 = 7 1/ 2 . 3 3 . a 3 6 5 c = 7 . 2 s " . 3 . a b % e = 21. 2*. ab 1 %c = 2l.2$abbh = 21 abc 1/1 6 2 . 3 1/ 16 a 3 6 5 c 3 = 3 ^2* a? &c — Z. 2*. ab%c whence = 3 . 2 . 2^ a b $ c = 6 a b c ^ 2 IP, 7^54a 3 65 c 3 + 3^ 1 6 a 3 6 s c 3_2 la 6 c ^262+6a5c^263 = 27a6c^/2 6 2 . 2. From the sum of ^24 and ^/54 subtract y/6. Ans. 4^6. 3. From the sum of ^/45c 3 and v'5a 2 c subtract v/SOe 3 . 4«s> [a — e) or y 12 = a + x > we have or y« + iiy» + By +C=0; the solution of which gives the value of y, which, being substituted in % = y™ — a, gives that of x. 2. Free the equation ( x + j3)i _i_ ( x -f-a;3)-J-(ar-{-a:3)^ = a from radical quantities. Ans. From the equation y 6 -\- y 3 -\- y* = a obtain the value of y, and substitute it in a?-\-r — y 6 . CH. V. § III.] POWERS OF POLYNOMIALS. 141 Binninial Theorem. 3. Solve the equation (2 a; + 13)^ = 3 (2 a; +13)^. Ans. a; = 358. 4. Solve the equation ( 5 + (3 + «)*)4 = l+(8 + *)*- Ans. x = S. SECTION m. Powers of Polynomials. 211. Problem. To find any power of a binomial. Solution. This power might be obtained directly by mul- tiplication, but the operation is long and tedious, and can be avoided by a process invented by Newton. To obtain this process, let the given binomial be a-\-x, let n be the exponent of the power, and let the product be arranged according to the powers of x, so that (a -f xf =A + B x + Cx* + D a* + E x* -f &c, in which the coefficients, A, B, C, &a, are to be deter- mined ; none but positive integral powers of a; are written in the second member, because the product could, evidently, give no others, and all the positive integral powers of x are included, because the coefficients of any which are super- fluous must be found to vanish. First. To find the value of A. Let x = and the development becomes a" = A. 142 ALGEBRA. [CH. V. § III. Binomial Theorem. Secondly. To find the form in which a enters into the development. Let a = 1, x = x 1 , and let the corresponding values of a", B, C, &c, be 1, B', C", &c, and we have (l-fx') n = 1 + B' x' + O z' 2 + D' z' 3 -f &c. ir. which A 1 , B\ C, &c, must be independent of a. The product of this equation by a n is a n (l+z') n = (a + ax') n :=a n -|- J B'a n z'-|-C l 'a n z l2 -j-&c. > in which, if we put a x' = x, or a;' = — , we have a a n x 1 = o B_1 x, a n x'2=a n ~ !l x s , &c, and (a-\-x) n = a n -{-B'a n -^x-\-C'a n -^x ;i -{-D'a n - 3 T i -}-6i,ii. Thirdly. To find the coefficients. The derivative of the last equation is, by examples 4 and 8 of art. 175, »(a-|-s)«-i = J5'a n - 1 -|-2C"«' , - 2 3;4-3I?'a n - 3 a: 2 +4 J E'a ,, - 4 a; 3 -|- &c, which, multiplied by (a-J-z), gives n(a-(-a;)"= J B'a n -f 2C"a n -' x-j-3 D' a"" 2 z 2 -f 4£' a"" 3 z 3 -j-&c + B'a" 1 z-f2Ca n - 2 z 2 -|-3D a"- 3 x 3 +&c. The product of (a + z)* 1 by n gives, also, n(a-f x)» = n a n -\-n B'a n ~ J x-\- «C"a n - a z 2 -|-n.D'a n - 3 z 3 + &c. which, compared with the preceding equation, gives, by art. 163, B' — n, 2C+B' = nB', or2C* = (n — 1)B', or C'=\{n — \)B'; 3D+2C — nC, 3D = (n— 2) C", />= \ (n — 2)C"; 4.E'+ 3 D = nD', 4E'=(n—3)D', E'=\(n — 3)D'; &c. &c. &,c. CH. V. § III.] POWERS OF POLYNOMIALS. 143 Binomial Theorem. The combination of these results gives Sir Isaac Newton's Binomial Theorem. The first term of any power of a binomial is the same power of the first term of the binomial. In the following terms of the power the expon- ent of the first term continually decreases by unity whereas the exponent-of the second term of the bine* mial, which is unity in the second term of the powef continually increases by unity. The coefficient of the second term of the power if the exponent of the power. If the coefficient of any term is multiplied by the exponent which the first term of the binomial has in that term, and divided by the place of the term, the result is the coefficient of the next fol- lowing term. 212. Corollary. The equations of the preceding article give B' = : n C' = n (re 1 . 2 1) D' — n (n — 1)(» 2) 1. 2 . 3 &c. Hence ffl + j)» = S»-fllll ,l - 1 I+ n n « B-2 z 3 -t- At H (,-l)(«-2) an _ 3a 3 +W (n-l)(»-2)( W -3) an . 4a4+&c> 144 ALGEBRA. [cB. V. § III, Binomial Theorem. 213. Corollary. If x is changed into — a; in the pre- ceding formula, it becomes (a — x) n =z a n — n a"- 1 x + I~ ' a"- 2 a 3 — «(n — l)(ra — 2) _, « , „ -^- '-+— — I a n ~° z 2 -f &c. the signs of every other term being reversed. 214. Corollary. The preceding formula, written in the reverse order of its terms, gives (x + a)" = x n -f re a x n ~i -f "^"7" ' a 2 a;" -2 -f &c. whence it appears that The coefficients of two terms which are equally distant, the one from the first term, and the other from the last term, are equal. 215. EXAMPLES. ! a c ~W 1. Find the 6th power of -^ — J 6 e 2 d. Solution. Place x = — nr- = 2 a b~* c. y = ±bc 2 d; and we have But, by the above formula, (z — y)o = z« — 6z 5 y + 15a; 4 y 2 — 20 a 3 y + lSa^y 4 — 6xy s + y«; in which, if we substitute the values of x and y, we have CH. V. § III.] POWERS OF POLYNOMIAL. 145 Binomial Theorem. (2 a 6" 2 c — i b c* a')8 = 64 a 6 b~™ ^—18 a 5 6~ 9 c 7 d-\- 15a*6 •8c 8 rf2_J a 3J-3 e 9rf3_)_J| a 2 c 10d* 2. Find the 10th power of a -j- 6. Ans. aio+10a 9 6+45a8 6 2 +120a 7 6 3 + 210o6 6 4 -|- 252aH 5 + 210a 4 6 6 -{-120a 3 6 7 4-45a 3 6 < 4-10a6 9 -j- 3. Find the 11th power of 1 — x. Ans. 1— 1 1 x+55 x 2 — 1 65 x 3 +330 z 4 — 462 z E +462 x« — 330 a; 7 + 1 65 x- 8 — 55 x 9 + 1 1 a; 10 — x". 4. Find the 4th power of 5 — 4x. Ans. 625 — 2000 a: -|- 2400 x 2 — 1280 r>-f 256 x 4 . 5. Find the 7th power of £ x -j- 2 j. ,4ns. ^x 7 +£. ««y + V *» ^24. y ^.J. 703: 3j4 4- 168 x 2 j/ 5 + 224x^6+ 128 1/ 7 . 6. Find the 4th power of5a 2 c 2 rf — label*. Ans. 625 a 8 c 8 d 4 — 2000 a 7 6 c^ rf 5 + 2400 a" 6 2 e 4 d« — 1280 a 5 6 3 c 2 d 7 + 256 a 4 6 4 d*. 216. Problem. To find any power of u poly- nomial. Solution. Suppose the terms of the given polynomial to be arranged according to the powers of any letter, as x, as follows ; a _1_ b x + c x 2 + d x* -f- e x 4 + &c, in which the successive coefficients are denoted by the suc- cessive letters of the alphabet. The following is Arbogasfs rule for finding any power of the poly- nomial. 13 146 ALfiEBUA. [CII. V. § IH Polynomial Theorem. The first term of the power is the same power of the first term a of the given polynomial. The coefficient of x in the second term is b times the derivative of the first term taken with reference to a. To obtain any other coefficient from the preceding coefficient; let r be the letter farthest advanced in the alphabet which is contained in any term of the given coefficient. Then r times the derivative of this term with ref- erence to q, is a term of the new coefficient ; but this process is obviously inapplicable, or at least useless, when q is the last letter of the given polynomial so that r is zero. If the given term contains the preceding letter p as well as q, q times its derivative with reference to p, divided by the increased exponent of q, is also a term of the new coefficient. Thus the term T p" q i gives, in the following coefficient, the two terms 6 Tp" q'+ l r and ^7 T p*- 1 tf+\ Proof. First. The value of the first term is obtained precisely as in the binomial theorem by putting x = 0. Secondly. Let V denote the given polynomial, so tha V = a -{- b x -\- c z 2 . . . -j- p x* -f- q a; s + 2 -|- &c The derivative of V with reference to p is, then, x", and tnat of V" is, by art. 172, n V"- 1 x' , CH. V. § III.] POWERS OF POLYNOMIALS. 147 Polynomial Theorem. the derivative of V with reference to q is and that of V n is n F n_1 s"+i. Let, now, the required power be the nth and let V n = A-\-Bx + Cx* ... +Px* -f.Q^+i-f.&c. and let the derivatives of A, B, C, ... P, Q, &c, with reference to p be A', B 1 , C", ... P', Q', &c, and with reference to q, A ', B" , C", . . . P", Q", &c. ; the deriva- tives of the preceding equation with reference to p and q are n V n -ix* = A'+B'x + C'x2... _(. p> x * -\- Q' x*+i + &c. nV n - 1 x-+ l —A"+B"x+C'x z . . . 4 P"x*+Q" xH +*+ &c. the first of which, multiplied by x, is n V»-ix*+ 1 = A , x + B'x s +Cx*... fF'i4i + Qv+2+ &c, which, compared with the other, gives, by art. 163, for the coefficient of x* + 1 , P< = Q", that is, the derivative of any coefficient with reference to any letter p, is the same with that of the succeeding coefficient with reference to the succeeding letter q. Thirdly. Every term of Q, in which q is the letter far- hest advanced in the alphabet, such as T p" q* , must give n Q" a similar term 6 T p" q*~ l , or else, if 6=1, a term T p" , in which there is no letter so far advanced as q. Every such term, as it belongs also lo P', must be the derivative of a similar term in P, that is, of a term in which p is either the farthest advanced letter, or the next to the farthest advanced letter. The terms of P' = Q" 148 ALGEBRA. [CH. V. § HI. Polynomial Theorem. are, then, obtained from those of P by derivation, while those of Q are obtained from those of its derivative Q" by a process, which is the reverse of derivation, and which, by art. 172, consists in multiplying by q, that is, in increasing its exponent by unity, and dividing by its exponent thus in- creased. This process is identical with the last paragraph of tin rule in this article, and the three preceding paragraphs refer merely to those cases in which the increased exponent of q is unity, so that the division by it is superfluous. 217. Corollary. If x is put equal to unity in the value of (a-l-ii-fc^-f&c.)", we have the value of (a -\- b -\- c -\- &.c.) n , so that any power of a polynomial, the terms of which contain no common letter, is readily found by multiplying the successive terms, after the first, re- spectively by x, 2; 2 , x 3 , x*, &c, obtaining the power of the polynomial thus formed, and putting x= 1 in the result. 218. EXAMPLES. 1. Find the 5th power of 1 -f 2 x -f 3 x a -f- 4 a 3 . Solution. Represent the successive coefficients 1, 2, 3, 4 by a, b, c, and d, so that a=\, 6 = 2, c = 3, d=i; and the given polynomial becomes a-\-bx-\-cx s -\-daP, CH. T. § III. | POWERS OF POLYNOMIALS. 149 Polynomial Theorem. the fifth power of which, is found by the rule to be o 5 -f5a 4 Aj;4- 5a 4 c |a; 2 4- 5 a 4 d | a: 3 4- 20 a 3 6 d + 10a 3 6 s | +20«3fic + 10 a 2 A 3 20 a 3 c d 4- 30 a 2 62 d 30 a 2 6 c 2 --20 ab 3 c 65 4- 30 a 2 cd 2 30a6 2 rf s (Mab&d 5 a c 4 --206 3 erf -- 10 6 2 c 3 10 a 3 d* --Mr,? bed 10 a 2 e 3 -j- 20 a 6 3 d 30 a 6 2 c 2 5 6" c 10 a 2 d 3 -- 60a6cd 2 - - 20 a c 3 d 10 6 3 d 2 30 6 2 c 2 d 4- 5 6 c 4 --10a 3 c 2 - - 30 a 2 6 2 e -- 5 a 6 4 --30a 2 6d 2 --30a 2 t 2 d --60a6 2 ed --20 a be* --5 6 4 d -- 10 6 3 c 2 20 a b d 3 30 ffi r? tf 2 30 6 2 e d 2 20 6 c 3 d C 5 i' 3 -)-5cd 4 2; 14 +rf 5 ii5. --20acrf 3 x»+ 5 ad 4 x™ + 5bd* --106 3 d 3 +20 6 erf 3 + 10c 2 d 3 --306<: 2 d 2 + 10c 3 d 2 -- 5c 4 d Now, if we substitute for a, b, c, d, their values, the pre- ceding expression becomes (1 _|_ 2 x + 3 a: 2 + 4 j: 3 )5 — i 4. 10 a: + 55 a: 2 + 220 a* -f- 690 a; 4 + 1772 a. 5 -f 3830 a; 6 -f- 7040 a? -f 1 1085 a; 8 + 14!)70 a: 9 + 17203 x i0 + 1G()60 a 11 4 13280 a;' 2 -j- 8320 a; 13 + 3840 a; 14 + 1024 a:". 2. Find the 3d power of a -)- 6 x 4 c x 3 . Ans. a 3 4- 3 a 2 6 x + 3 a 2 c 3 a 6 2 6 a b c 6 3 4- 3 a c 2 a; 4 4- 3 6 c 2 a; 5 + c 3 f 6 3. -f 3 6 2 e Find the 6th power of a 4 * + c - Ans. a 6 4- 6 n 5 6 4 6 a 5 c -f 15 a 4 6 2 4 30 « 4 6 c -f 20 a 3 6 3 + 15 a 4 c 2 4 CO a 3 6 2 c 4 15 a 2 6 4 + 60a 3 6 c 2 -\- 13* 150 ALGEBRA. [CH. V. § IV. Root of a Polynomial. 60 a 2 ft3 c _|_6 a 6 5 -f 20 a 3 c*-\- 90 a 2 6 2 c*-\- 30 a b 4 c -|- 6 6 +60a 2 6e 3 -j-60a6 3 c 2 -|-66 5 c-fl5a 2 c' 1 -)-60a6 2 c3 -f- 15 6< e 2 _j_ 30 a b c* + 20 63 1 3 _|_ 6 „ c 5 _|_ 15 52 e 4 _j_ 6 6 c 5 + c«. 4. Find the 4th power of a 3 — a 9 x -j- a a; 2 — a: 3 . .4res. ai 2 — 4 a 11 Z+ 10 a 10 re" — 20 a? x 3 -j- 31 a* z* — — 40 a 7 a; 5 -|- 44 aots ' Polynomials. 219. Problem. To find any root of a polyno- mial. Solution. If the root is arranged according to the pow- ers of either of its letters as x, whether ascending or de- scending, it is evident from the rule of art. 216, that The term of the required root which contains the highest power of x, is found by extracting the root of the corresponding term of the given polynomial. If, now, R is the first portion of the root, and R' the rest of it, and if P is the given polynomial of which the nth root is to be extracted, we have P = (R+R')" = R» -f n JJ»-i R' -f- &c. CH. V. § IV.] ROOTS OF POLYNOMIALS. 15] Root of a Polynomial. or and P — R n — reiJ n -iK'-f&o. P—R n n Rn-l T and if, in P — R n and n.R n_1 , only the first term is re- tained, the first term of the quotient is the first term of R' ; and a new portion of the root is thus found, which, com- bined with those before found, gives a new value of R and of P — jR", which, divided by the value of n R n ~ l already obtained, gives a new term of the root, and so on. Hence to obtain the second term of the root, raise the first term of the root to the power denoted by the exponent of the root, and subtract the result from the given polynomial, bringing down only the first term of the remainder for a dividend. Also raise the first term of the root to the power denoted by the exponent one less than that of the root, and multiply this power by the exponent of the root for a divisor. Divide the dividend by the divisor, and the quo- tient is the second term of the root. The next term is found, by raising the' root al- ready found to the power denoted by the exponent of the required root, subtracting this power from the given polynomial, and dividing the first term of the remainder by the divisor used for obtaining the sec- or.d term. This divisor, indeed, being once obtained, is to be used in each successive division, the successive divi- dends being the firs< terms of the successive remain' ders. 152 ALGEBRA. [CH. V. § IV Root of a Polynomial. 220. EXAMPLES. 1. Find the 4th root of 81a; 8 — 2 16 a: 7 + 336 z 5 — 56*« - 224z 3 + 64a:+16. Solution. The operation is as follows, in which the root is written at the left of the given power, and the divisor at the left of each dividend or remainder; and only the first term of each remainder is brought down. 8 1 s 8 -2 1 6a: 7 +336i 5 — 56*"— 224z 3 + 64i+ 16 1 3z 2 — 2x— 2. 81 2)8 1st Rem. —216 r 7 I 108x 6 = 4x (3 a; 2 ) 3 81 1 8 — 216^+216^ — 96 x^-j-ldx 4 = (3 z 2 — 2 x)* 2d Rem. — 2l6z6 | 108x6 8 Ii8_2 16i 7 +336i5_56i"_224x 3 +64aH-l 6=(3a: 2 -2ai-2) 4 3d Rem. 0. 2. Find the 3d root of a 3 + 3 a 2 6 + 3 a 2 e + 3 a 6 2 + 6a 6c + 3ac 2 + 6 3 + 3 6 2 c+ 36 c 2 +c 3 . Ans. a-f"6-f~c. 3. Find the 3d root of a 3 + 6a 9 6 — 3a 2 c+l2a6 2 — 12 ab c + 3 a c 2 + 8 6 3 — 12 6 2 c + 6 6 c 2 — c 3 . .dns. a -|- 2 6 — c. 4. Find the 3d root of 343 x« — 441 x 5 y + 777 x* y* — 531 z 3 y 3 + 444 a: 2 y 4 — 144 x y" + 64 ij e . Ans. 7x s — 3zy + 4y*. 5. Find the 4th root of 81 a 4 — 540 a 3 b — 72 a 3 c + 1350 a 2 6 2 + 360 « 2 6 c + 24 a 2 c 2 — 1500 a 6 3 — 600 a 6 2 e — 80 a b e 2 — ^ 2 a c 3 + 625 b*+ i^p 6 3 c + £$£ 6 2 c 2 -f- W 6c 3 + 4f A ylns. 3a — 56 — § c CH. V. § IV.] ROOTS OF POLYNOMIALS. 153 Root of a Polynomial. 6. Find the 5th root of 16807 a w b$ — ^-^ a e b 4 -\- Jjyj. a 6 fi3 _ ZAfrU) „4 £7 c _ 2_|J> 4 fc2 _ ftj^LP a 2 S c _j_ ff a 2 6-f-245 6 5 e— •& + Lapp a - 2 6 9 c 2 — V" _2 6 4 c — ap a~ 4 6 8 c 2 + ^ a -4 6 3 c + ^ a -e J7 c a_|_ ajf s> a-8 fcii c 3 — f a -8 66 c 2 — ^ 7 ° a -10 6 10 e 3 -f- f$ a~ 12 6 £ c 3 + ^0 a -14 J13 c 4 _ |ft ^-16 fcl2 c 4 -f^. a" 20 b™ C 5 . .4ns. 7a 2 A — ^- + |a- 4 6 3 *e. 7. Find the 9th root of y a '+27i/ 95 + 324y 23 +2268 i / 21 -f- 1020G y 19 + 30618 y" -f 61236 j/ 15 + 78732 # 13 + 59049 y n + 196S3 y 9 . Ans. y z -(- 3 y. 221. Corollary. When the preceding method is applied to the extraction of the square root, it admits of modifica- tions similar to those of art. 189, and we have the following rule To extract the square root of a given polynomial. Arrange its terms according to the powers of some letter, extract the square root of the first term for the first term of the root. Double the part of the root thus found for a divi- sor, subtract the square of this part of the root from the given polynomial, and divide the first term of the remainder by the divisor ; the quotient is the second term of the root. Double the terms of the root already found for a new divisor ; subtract from the preceding remainder the product of the last term of the root multiplied by the preceding divisor augmented by the last term of the root. Divide the first term of this new remain- der by the first term of the corresponding divisor, and the quotient is the next term of the root. 154 ALGEBRA. [CH. V. § IT Square Root of a Polynomial. Proceed in the same way, to find the other terms of the root. 222. EXAMPLES. 1.. Find the square root of x<3 -f 4 x 5 -\- 20 « a — 16 x + 16. Solution. In the following solution, the arrangement is similar to that in the example of art. 190. tfl+i X s + 20 a 2 — 16 x + 16 4i5_(_20a; 2 — 4z5_|_ 4^4 16a;+16 — 4a; 4 + 20 s 2 — — 4 a; 4 — 8x 3 -\- 16* + 16 4 a; 2 8 x? +16 a; 2 — 8z 3 + 16a; 2 — 16a; + 16 16a; -f- 16 a; 3 + 2 a: 2 — 2a; + 4. Ans. 2*3 2 x 3 -f 4 x s 2 a? + 4 a; 2 — 4 a: 0. 2. Find the square root of 25 a 4 — 30 a 3 6 + 49 a 2 J 9 — 24a6 3 +166 4 . .4ras. 5a 2 — 3a6 + 46 2 . 3. Find the square root of 4 x e + 12 a; 5 + 5 a; 4 — 2 a? -f 7j 2 -2z+1. 4ns. 2x 3 + 3a; 2 — a; + 1. 4. Find the square root of a 4 — 2 a 3 x + 3 a 2 x z — 2ax 3 -\-x*. Ans. a 2 — ax + xK 5. Find the square root of £ + 6 x — 17 a; 2 — 28 x> + 49^- 4»s. f + 2 a; — 7 a; 2 . CH. V. § V.] BINOMIAL EQUATIONS. 155 Solution of Binomial Equations. SECTION V. Binomial Equations. 223. Definition. When an equation with one un- known quantity is reduced to a series of monomials, and all its terms which contain the unknown quan- tity are multiplied hy the same power rf the un- known quantity, it may be represented by the gen' eral form A x n + M = 0, and may be called a binomial equation. 224. Problem. To solve a binomial equation. Solution. Suppose the given equation to be' A x n + M = 0. Transposing M and dividing by A, we have M x - A , the nth root of which is M * = v- T Hence, find the value of the power of the unknown quantity which is contained in the given equation, 'precisely as if this power were itself the unknown quantity, and the given equations were of the first degree. Extract that root of the result which is denoted by the index of the power. 225. Corollary. Equations containing two or more un- known quantities will often, by elimination, conduct to binomial equations. 156 ALGEBRA. [CH. V. § V Examples of Binomial Equations. 226. EXAMPLES. 1. Solve the two equations *^-f O y 7_4j / 3_ 8a ;-|-16=0, zsy: _ 4 y? _ 4 ,£ y 3 _^_ g #3 _|_ 32 Z _ 64 = 0. Solution. The elimination of y between these two equo> tions, by the process of art. 155, gives 8 x 2 — 32 = 0, whence we have 12=4, x = ±2. Now the value of x, x = + % being substituted in the first of the given equations, pro- duces 4 y* — 4 y* = o, which is satisfied by the value of y, y = 0; or if we divide by 4 y 3 , we have y 4 - i = o, y 4 =h y = vl = ±loi -: = ±\/ — 1, as will be shown when we treat of the theory of equations Again, the value of x, x = -2, being substituted in the first of the given equations, pro- duces — 4 y + 32 = 0, whence we have y = 2or =■— li^ — 3, as will be shown in the theory of equations. CH. V. § V.] BINOMIAL EQUATIONS. 157 Examples of Binomial Equations. 2. Solve the equation Ans. x = ± 3. 3. Solve the equation 2x—7 _ i-fl x—1 ~ 2x + 7' Ans. z = ± 4. 4. Solve the equation , 1 26 1 z a; 3 — a; ii»s. z = 3. 5. Solve the equation aP + z + 8 , a?+x— 8 x3_j_4 + j;3_4 — *• Ans. x = ± 2. 6. Solve the equation V/ (2 z + 2) = z -f 1. ylns. X = ± 1. 7. Solve the equation V (x 5 — 81 * + 1) = 1. Ans. z =z 0, or i = ± 3 8. Solve the two equations z 3 + y s _ 2 n> a; 3 — # 5 = 2 6, 3 5 Ans. x = y/ (a -J- 6), y = ^/ ( a — 6). 9. Solve the two equations y 6 — 33 y 3 _|_ a,4_ 17 ^2 — 0) y 6 + 17 y 3 -f x* — 33 z 2 = 0. Ans. x = 0, and y = 0; or a; = ± 5, and y = 2 14 158 ALGEBRA. [on. V. § V Examples of Binomial Equations. 10. What number is it, whose half multiplied by its third part, gives 864 ? Arts. 72. 11. What number is it, whose 7th and Sth parts multi- plied together, and the product divided by 3, gives the quotient 298| 1 Ans. 224. 12. Find a number such, that if we first add to it 94, then subtract it from 94, and multiply the sum thus obtained by the difference, the product is 8512. Ans 18. 13. Find a number such, that if we first add it to a, then subtract it from a, and multiply the sum by the difference, the product is b. Ans. \/ (a 2 — 6). 14. Find a number such, that if we first add it to a, then subtract a from it, and multiply the sum by the difference, the product is b. Ans. \/(a 2 -4-6). 15. What two numbers are they whose product is 750, and quotient 3£ 1 Ans. 50 and 15. 16. What two numbers are they whose product is a, and quotient ft ? a Ans. \/ab and \/ — . 17. What two numbers are they, the sum of whose squares is 13001, and the difference of whose squares is 1449? Ans. 85 and 76. 18. What two numbers are they, the sum of whose squares is a, and the difference of whose squares is b 1 Ans. v 'H a + 6 ) ar) d Vh ( a_ *)• 19. What two numbers are to one another as 3 to 4, the sum of whose squares is 324900? Ans. 342 and 456. 20. What two numbers are as m to n, the sum of whose squares is a? m \/ a n \/ a CH. V. § V.J BINOMIAL EQUATIONS. 159 Examples of Binomial Equations. 21. What two numbers are as m to n, the difference of whose squares is a? m %/ a , n*/« Ans. — . „ — -„. and , ^ — „ T . ^/{rrfi — rfi) v (™ 2 — rfi) 22. A certain capital is let at 4 per cent. ; if we multiply the number of dollars in the capital, by the number of dollars in the interest for 5 months, we obtain 117041|. What is the capital t Ans. $ 2650. 23. A person has three kinds of goods, which together cost $ 5525. The pound of each article costs as many dol- lars as there are pounds of that article ; but he has one third more of the second kind than he has of the first, and 3| times as much of the third as he has of the second How many pounds has he of each ? Ans. 15 pounds of the first, 20 of the second, and 70 of the third. 24. Find three numbers such, that the product of the first and second is 6, that of the first and third is 10, and the sum of the squares of the second and third is 34. Ans. 2, 3, 5. 25. Find three numbers such, that the product of the first and second is a, that of the first and third is b, and that of the second and third is c. ab a c , be Ans. \/ — , v/ -r-i and \/ — • cb a 26. What number is it, whose third part, multiplied by tts square, gives 1944 ? Ans. 18. 27. What number is it, whose half, third, and fourth, multiplied together, and the product increased by 32, gives 4040? Ans. 48. 28. What number is that, whose fourth power divided by £th of it, and 167 subtracted from the quotient, gives the remainder 12000? Ans. 11 J. 160 ALGEBRA. [CH. V. § T. Cases of imaginary Solution. 29. Some merchants engage in business ; each contrib- utes a thousand times as many dollars as there are partners. They gain in this business $2560; and it is found that this gain is exactly half their own number per cent. How many merchants are there 1 Ans. 8. 30. Find three numbers such, that the square of the first multiplied by the second is 112; the square of the second multiplied by the third is 588 ; and the square of the third multiplied by the first is 576. Ans. 4, 7, 12. 227. Corollary. When the solution of a problem gives for either of its unknown quantities only imag- inary values, the problem must be impossible. 228. EXAMPLE. In what case would the value of the unknown quantity in example 13 of art. 226 be imaginary 1 and why should the problem in this case be impossible 1 Ans. When b > a a , that is, when the product of the sum and difference i,i required to be greater than the square of a. Now if th# required number is x, this product is (a -f x) (a — x) = a 9 — z?; and, therefore, less than o 2 CH. VI.] EQUATIONS OF THE SECOND DEGREE. 161 Equations of the Second Degree. CHAPTER VI. EQUATIONS OF THE SECOND DEGREE 229. It may easily be shown, as in art. 120, that „ „ ,™ (10a: 2 — 66a: + 62)y + 60:r (y-5)y 2 + 40x 2 -100 = ^ z _ 2 Solution. If we proceed to eliminate y between these two equations, by the process of art. 155, the remainder of the first division is (a;2_ 6 x _j_ 5)y 2 — ( 10 a; 2 — 60 a; + 50)y+24 a; 2 — 144 a; +120, in which a; 2 _ 6 x -+- 5 is a factor of each of the coefficients of y, and y z , and of the terms which do not contain y. Before suppressing this factor, we must see whether, as in art. 157, it may not be equal to zero, in which case we have aS_ 6 a: + 5 = 0, the roots of which are x ■=. 5, and x = 1. Now if the value x = 5 is substituted in the given equations, each of them becomes y 8 -— 5 1/ 2 + 6 y = 0, which is satisfied by the value or, dividing by y, we have j,2^5j + 6 = 0, the roots of which are y = 2, and y = 3. But if the value z = l is substituted in the given equations, each of them becomes 168 ALGEBRA. [CH. VI Examples of Equations of the Second Degree. which is the same as the preceding equation, and gives therefore the same values of y. Having thus obtained all the roots of the given equation corresponding to K 2 _ (j x _|_ 5 _ o, we may omit this factor of the above remainder, and it becomes ya— 10y+24; and as this does not contain x it is unnecessary to proceed farther in the elimination of y, but we may obtain the roots of the equation y* — 10y + 24 = 0, which are y = 4, and y = 6, and substitute them in the given equation to obtain the corresponding values of z. Thus, if the value y = 6 is substituted in the given equations, each of them becomes 5x2 — 48^4-61 =0, the roots of which are * = i(24± k « 2 . that is, that is, the product of two numbers cannot be greater than the square of half their sum. 23. What two numbers are they, whose difference is 5, and product 24 ? Ans. 8 and 3 ; or — 3 and — 8. 24. What two numbers are they, whose difference is a, and product b ? Ans. £a± v '(&-f-4a 2 ), and — £ a -± \/(b-\- J a 2 ). 25. Find a number, whose square exceeds it by 306. Ans. 18, or — 17. 26. A person being asked his age, answered, " My mother was 20 years old when I was born, and her age multiplied by mine, exceeds our united ages by 2500." What was his age 1 Ans. 42. 27. A person buys some pieces of cloth, at equal prices, for $60. Had he got three more pieces for the same sum, each piece would have cost him $ 1 less. How many pieces did he buy? Ans. 12. 28. A person dies, leaving children, and a fortune of $ 46800, which, by the will, is to be divided equally among them. It happens, however, that immediately after the death of the father, two of his children also die. If, in consequence of this, each remaining child receives $ 1950 more than it was entitled to by the will, how many children were there? Ans. 8. 29. Twenty persons, men and women, spent $ 48 at an inn ; the men $24, and the women the same sum. Now, on inspecting the bill, it is found that the men have to pa/ 172 ALGEBRA. [CH. VI. Examples of Equations of the Second Degree. ijS 1 each more than the women. How many men, there- fore, were there in the company 1 Ans. 8. 30. What two numbers are they, whose sum is 41, and the sum of whose squares is 901 1 Ans. 15 and 26. 31. What two numbers are they, whose sum is a, and the sum of whose squares is b 1 Ans. \a-\-{ v (26 — a 2 ), and \a — \ (2 V 6 — a=). In what case would the values of these unknown quan- tities be imaginary 1 Ans. When we have a 2 > 2 6 ; that is, the square of the sum of two numbers cannot be greater than twice the sum of their squares. 32. What two numbers are they, whose difference is 8, and the sum of whose squares is 544 ? Ans. 12 and 20 ; or —12 and —20. 33. What two numbers are they, whose difference is a, tnd the sum of whose squares is b 1 Ans. ^±^(26 — a 2 ),and— Jnd=^V(26— a 9 )- In what case would the values of these unknown quan- tities be imaginary ? Ans. When we have a 2 > 2 b ; that is, the square of the difference of two numbers can- not be greater than twice the sum of their squares. 34. Divide the number 39 into two parts, such that the Bum of their cubes may be 17199. Ans. 15 and 24. 35. A person being asked about his yearly income, answered, " My income is such, that if I add $ 1578 to it, and also subtract $ 142 from it, and extract the cube roots CH. VI.] EQUATIONS OP THE SECOND DEGREE. 173 Examples of Quadratic Equations higher than the Second Degree. of the numbers thus obtained, the difference between the roots is 10." What was his income? Ans. $150. 36. Find two numbers whose difference added to the dif- ference of their squares, makes 150, and whose sum added to the sum of their squares is 330. Ans. The one is 15, or — 16; the other is 9, or — 10. 37. What two numbers are they, whose sum, product, and difference of their squares, are all equal to each other } Ans. J (3 ± v'5), and \ (1 ± ,/ 5). 38. Find a number consisting of three digits, such, that the sum of the squares of the digits, without considering their position, may be 104 ; but the square of the middle digit exceeds twice the product of the other two by 4 ; farther, if 594 be subtracted from the number sought, the three digits are inverted. Ans. 862. 234. Corollary. The preceding method is not only applicable to equations of the second degree, but to all equations of the form Ax* n + B x n + M = 0, in which there are two terms multiplied by different powers of x, the highest exponent being the double of the lowest ; and n may be either integral or frac- tional. 235. EXAMPLES. 1. Solve the equation A z 2n + B x n + M = Solution. If the square is completed, as in the preceding irticle, and the square root extracted, the result is A x n + i B = ± v" (— A M + \B a) ; Jn* 174 alc;ebra. [en. vi Examples of Quadratic Equations higher than the Second Degree. from which we obtain, by art. 224, x = i -\B±y/(-A M+\B*) \L 2. Solve the equation z 4 — 74 z 2 = — 1225. Ans. x = ± 5, or = ± 7. 3. Solve the equation 3 z" + 42 x* = 3321. 3 Ans. x = 3, or = — ^ 41. 4. Solve the equation 5 s/ x— y/x = 6. Ans. x = 16, or 81. 5. Solve the equation ( x _|_ 12)* + (x + 12)* = 6. Ans. x = 4, or 69. 6. Solve the equation z + 16 — 7 (z+ 16)* = 10 — 4 (a;-}- 16)i Ans. x= 9, or — 12. 7. Solve the equation J* — a: = 2 A .dns. x = 0, or 1, or 4. 8. Solve the equation* a? — 2# = 56. .4 ns. x = 4, or ( — 7)'. 9. Solve the equation i^-fi^= 756. 4ns. a; = 243, or (_— 28')-'. CH. VI.]: EQUATIONS OF THE SECOND DEGREE. 175 Examples of Substitution of Unknown Quantities. 10. Solve the equation (z 2 -f 5) 2 — 4 z 2 = 160. Ans. x — 3, or \/ — 15. 11. What two numbers are they, whose product is 255, and the sum of whose squares is 514 ? Ans. 15 and 17, or — 15 and — 17. 12. What two numbers are thej, whose product is a, and the sum of whose squares is 6 ? and±^[4 6— • (J ft»— <*«)]. 13. What number exceeds its square root by 20 ? Ans. 25. 14. What number is it, the ex«ess of whose square above its square root is equal to 56 divided by the number 1 3 Ans. 4 or y/ 49. 236. There are equations of higher degrees, which can be reduced to equations of the second degree by introducing other unknown quantities instead of those contained in them. Thus if the same algebraic ex- pression is involved in different ways, it will often be found successful to consider this expression as the unknown quantity. 237. EXAMPLES. 1. Solve the two equations fz 2 — 23y)3 + (z 2 — 23y) a +(s 2 — 23y)(x— 2y) = 18, 176 ALGEBRA. [CH. VI. Examples of Substitution of Unknown Quantities. Solution. Consider («■ - 23 y), and (z - 2 y), as the unknown quantities, making x' = a:2 _ 23 y, and the equations become a;/ 3 -|- X ' 2 -f z' y' = 18, a' 9 -j- V' = 7. Hence, by the elimination of y', we have i' s -f7s ; = 18, aiid, therefore, a;' = 2 ) or = — 9; and the corresponding values of y 1 are y' = 3,or = — 74; that is, » a — 23 y = 2, or = — 9, x — 2 y = 3, or = — 74. The solution of these equations gives % = 5, y = 1 ; or, * = G£, y = If; or, z = |(23±v'14001),y = 4(319±V'14001), 2. Solve the equation * + (* + c )^ = 2 + 3 {x -f 6)*. .dfls. a; = 10, or — 2. 3. Solve the two equations (* + y) + (* + y)* = 12, x 3 + y 3 = 189. .4ns. z = 5, or = 4 ; y = 4, or = 5. CH. VI.] EQUATIONS OP THE SECOND DEGREE. 177 Examples of Substitution of Unknown Quantities. 238. Corollary. When there are two unknown quantities which enter symmetrically into the given equation, the solution is often simplified by substi- tuting for them two other unknown quantities, one of which is their product and the other their sum. 239. EXAMPLES. 1. Find two numbers whose sum is 5, and the sum of whose fifth powers is 275. Solution. Let the numbers be x and y, represent their product by p, and we have ■*■ + V = 5. a 5 _f_ y h _ 275. But we also have (x-\- y)$ = x 5 + 5 x i y +10 x 3 y* -f-10z s y* + 5xy4-\-y s _ x 5 +1J & +5x y (tf + y 3) + 10 x*y* {x + y) ; and & + 3,3 = (x _|_ yf _ 3 x 2 y — 3 X 1J S = (x -\- yf — 3 x y {x + y) = 125— 15 p. Hence (^4-^)5= 275 +5p (125— 15p) + lOp* X 5 ==55; or, by reduction, p2_25;> = — 114, p = 19, or = 6 ; and s = 2, or=3,or=i (5±v/ — 51), j [ ==3 j or=2, or = j[ (5Tv/-51). 2. Solve the two equations ( x -y)(x2-y*) = 7, (x + y) (s» + # a ) = 175 - 178 ALGEBRA. [CH. VI. Examples of Substitution of Unknown Quantities. Solution. These equations become, by development, x 3 — a; 2 y — x y' z -\- y 2 = 7, z 3 + z 3 y -f x y s -f- y 3 = 175 ; and, by the substitution of x + V = s » * y=p; they still farther become, s 3 — 4 s p = 7, s 3 — 2 sp = 175. If we eliminate p we have s 3 = 343, whence s = 7; and this value gives, by substitution, 343 — 14 p = 175, p ■= 12. Hence x = 3, or = 4 ; y = 4, or = 3. 3. Solve the two equations z + y + * 2 +0 a =i2. .dns. a:=2, or =J(— 3± v/21); 0=2, or =4 (— 3^=^/21). 4. Solve the two equations i 3 _f_ y 3 = 189, z*y-{-zyZ= 180, ,/lras. a; = 4, ov = 5; y = 5, or = 4. 5. Solve the two equations *a + y> = 5, * y = 2. j4ns. a; = ± 2, or = J; 1 ; y = ± 1, or = ± 2. CH. VI.] EQUATIONS OP THE SECOND DEGREE. 179 Examples of Substitution of Unknown Quantities. G. Solve the two equations a: 2 y -f- x y 3 = 6, x s f _j_ j;2 y 3 _ 12. Ans. x = 1, or = 2; y = 2, or = 1. 7. Solve the two equations 4 x y = 96 — s 3 j9, a; -f- y = 6. 4ns. x = 2, or 4, or 3 ± %/ 21 ; ?/ = 4, or 2, or 3 =F %/ 21. 8. Find two numbers such, that their sum and product may together be 34, and the sum of their squares may ex- ceed the sum of the numbers themselves by 42. Ans. 4 and 6 ; or J(— H-J-y — 59,) and ^(—11 — v/ — 59). 9. What two numbers are they, whose sum is 3, and the sum of whose fourth powers is 17? Ans. 2 and 1 ; or £(3+^ — 55), and |(3— ■• — 55). 10. What two numbers are they, whose product is 3, and the sum of whose fourth powers is 82 1 Ans. ± 1, and ± 3; or ± /—S: or x = ± | v> — 5, and y = 1. 4. Solve the two equations x 3 + 2 a: y 9 = 3, .4ns. a; = 1, and y = 1. CH. VI.] EQUATIONS OF THE SECOND DEGREE. 185 Examples of Substitution of Unknown Quantities. 5. What two numbers are they, twice the sura of whose squares is 5 times their product, and the sum of whose sixth powers is 65. Ans. 2 and 1, or — 2 and — 1. 6. What two numbers are they, the difference of whose fourth powers is 65, and the square of the sum of whose squares is 169. Ans. ± 2, and ± 3. 186 ALGEBRA. [CH. VII. § I To find the last Term. CHAPTER VII. PROGRESSIONS. SECTION I. Arithmetical Progression. 244. An Arithmetical Progression, or a progres- sion by differences, is a series of terms or quantities which continually increase or decrease by a constant quantity. This constant increment or decrement is called the common difference of the progression. Throughout this section the following notation will bo retained. We shall use a = the first term of the progression, I = the last term, r = the common difference, n = the number of terms, S = the sum of all the terms. 245. Problem. To find the Ian. term of an arith- metical progression when its fir si <,erm, common dif- ference, and number of terms are known. Solution. In this case a, r, and n, are supposed to be known, and / is to be found. Now the successive terms of the series if it is increasing are a, a-\ r, a-^-lr, a -\- 3 r, a-\-ir, &c. : Oil, VII. <§ I.] ARITHMETICAL PROGRESSION. 1S7 Sum of two Terms equally distant from the extremes. so that the nth term is obviously I = a -\- (n — 1) r. But if the series is decreasing, the last term must be I = a — (n — 1) r. Both these cases are, however, included in one, if we suppose r to be negative when the series is decreasing. 246. Corollary. In like manner any other term, such as the wth, is a -\- (m — 1) r. 247. Corollary. By writing the series in an inverted order, beginning with the last term, a new series is found, of which the first term is I, and the common difference — r. Hence the rath term of this series, that is, the mth term counting from the last of the given series, is J — (m — 1 ) r. 248. Corollary. The sum of the mth term and of the mth term from the last is, therefore, [ a -\-(m — l)r] -f- [/— (w — 1) r] = a + 7; that is, the sum of any two terms, taken at equal distances from the two extremes of an arithmetical series, is equal to the sum of the two extremes. 249. Problem. To find the sum of an arithmeti- cal progression when its first term, last term, and number of terms are known. Sulution. In this case, a, I, and n are supposed to be known, and S is to be found. 1S8 ALGEBRA. [CH. VII. § I. To find the Sum of the regression. Suppose the terms of the series to be written as follows first in the regular order, and then in an inverted order: a, b, c, . . . . i, k, l\ I, k, i, c, b, a. The sum of the terms of each of these progressions being S, the sum of both of them must be 2 S, that is, 2S=(a+l)+(b+k)+(c+i) . . ,+(i+c)+(k+b)+{l+a). But by the preceding corollary, we have a-\-l=b-\-k=c-\-i = &c. Hence 2 S is equal to as many times (a -j- /) as there are terms in the series, that is, 2S= (a + Z)rc; or S = \ (a -j- /) n ; that is, the sum of a progression is equal to half the sum of the two extremes, multiplied by the number of terms. 250. Corollary. From the equations /=«+(»-])r, S=\{a + l)n; either two of the quantities a, I, r, n, and S can be deter mined when the other three are known. 251. EXAMPLES. 1. Find the 100th term of the series 2, 9, 10, &c. Am. 695. 2. Find the sum of the preceding series. Arts. 34850. 3. Find S, when a, r, and n are known. Ans. S= \ [2a-f(»- l)*]n. CH. Vil. ^ I.] ARITHMETICAL PROGRESSION. 189 Examples in Progression. 4. Find n and S, when a, I, and r are known. A l a I 1 Ans. n = r- 1 ; r ' 72 a 2 5. Find the number and sura of terms of the series of which the first term is 6, the last term 796, and the com- mon difference 10. Ans. The number of terms = 80, the sum = 32080. 6. Find r, when a, I, and n are known. I — a Ans. r ==. -. n—1 7. Find the common difference and sum of the series, of which the first term is 75, and the last term 15, and the number of terms 13. Ans. The common difference = — 5, the sum = 585. 8. Find r and n, when a, I, and S are known. 2S Ans. n = — j— ,, Z2— a^ ~" 2S— (a+Z) 9. Find the common difference and number of terms of a series, of which the first term is 2, and the last term 345, and the sum 8675. Ans. The number of terms = 50, the common difference = 7. 10. Find I and n, when a, r, and S are known. Ans „_ ^VrS+(a-try]-(a- r /=VprS+(a-ir)»]-Jr. 190 ALGEBRA. [cil. VII. § I. Examples in Progression. 11. Find the last term and number of terms of a series, of which the first term is 3, the common difference 4, and the sum of the terms 105. Ans. The last term = 27, the number of terms = 7. 12. Find a and n, when /, r, and S are known. Ans. n = / +ir=FVl(*+*O a -g'-fl] > r 13. Find the first term and the number of terms of a series, of which the last term is 13, the common difference 3, and the sum of the series 35. Ans. The first term = 1, the number of terms = 5. 14. Find / and r, when a, n, and S are known. A 7 * S Ans. I = a, n __ 1{S—an) n(n — 1) 15. Find the last term and common difference of a series, of which the first term is §, the number of terms 12, and the sum 100. Ans. The last term = 16, the common difference = 1£|. 16. Find a and r, when /, n, and S are known. n _ 2(ln—S ) n(n — 1 ) ' Ans. a = /, n CH. VII. § I.] ARITHMETICAL PROGRESSION. 191 Examples in Progression. 17. Find the first terra and common difference of a series, of which the last term is 50, the number of terms 20, and the sum 600. Ans. The first term = 10, the common difference = 2^-. 18. Find a and S, when I, r, arid n are known. Ans. a = l — (re — 1) r, S = %[2 1— (re— \)r]n. 19. Find the first term and sum of the terms of a series, of which the last term is 100, the common difference \, and the number of terms 51. Ans. The first term = 75, the sum of the terms = 4462 \. 20. Find a and /, when r, n, and S are known. Ans. a — £ (» — 1) r, 21. Find the first and last terms of a series, of which the common difference is 5, the number of terms 6, and the sum 321. Ans. The first term = 41, the last term = 66. 22. Find the sum of the natural series of numbers V, 2, 3, &.c. up to n terms. Ans. \ n (n -J- 1). 23. Find the sum of the natural series of numbers from 1 to 100. Ans. 5050. 24. Find the sum of the odd numbers 1, 3, 5, &c. up to n terms. Ans. n a 192 ALGEBRA. [cH. VII. § I. Examples in Progression. 25. Find the sum of the odd numbers from 1 to 99. Ans. 2500. 26. Find the sum of the even numbers 2, 4, 6, &c. up to n terms. Ans. n (n -j- 1). 27. Find the sum of the even numbers from 2 to 100. Ans. 2550. 28. One hundred stones being placed on the ground, in a straight line, at a distance of 2 yards from each other ; how far will a person travel, who shall bring them one by one to a basket, placed at 2 yards from the first stone 1 Ans. 11 miles, 840 yards. 29. We know, from natural philosophy, that, a body which falls in a vacuum, passes, in the first second of its fall, through a space of lG^- feet, but in each succeeding second, 32£ feet more than in the immediately preceding one. Now, if a body has been falling 20 seconds, how many feet will it have fallen the last second ? and how many in the whole time 1 Ans. 627£ feet in the last second, and 6433J feet in the whole time. 30. In a foundery, a person saw 15 rows of cannon-balls placed one above another, and asked a bombardier how many balls there were in the lowest row. " You may easily calculate that," answered the bombardier. " In all these rows together, there are 4200 balls, and each row, from the first to the last, contains 20 balls less than the one immediately below it." How many balls, therefore, were there in the lowest row ? Ans. 420. 252. The arithmetical mean between several Ci:- VII. '^ I.J ARITHMETIC .L PROGRESSION. 193 Arithmetical Mean. quantities is the quotient of their sum divided by their number. Thus the arithmetical mean between the two quantities a, and 6 is half their sum, or ^ («-{-&); that between the four quantities 1, 7, 11, 5 is 6. 253. Problem. To find the arithmetical mean between the terms of an arithmetical progression. Scholium. It is, by the preceding definition S or, since it is *(« + *); that is, half the sum of the extremes, and also, by art. 248, half the sum of any two terms at equal distances from the extremes. 254. Problem. To find the first and last terms of a progression of which the arithmetical mean, the number of terms, and the common difference are known. Solution. If we denote the arithmetical mean by M, we have which, substituted in the result of example 20, in art. 251, gives a = M—\ (n — \) r, I = M + I (» — 1) r. 255. Scholium. In very many of the problems involving arithmetical progression, it is convenient 17 194 ALGEI1RA. [CH. VII. <§ 1 Examples involving Arillmielicr.l Progression. to use for one of the unknown quantities '.he arith- metical mean. 256. EXAMPLES. 1. Find five numbers in arithmetical progression whose sum is 25, and whose continued product is 945. Solution. Denote the arithmetical mean by M, and the common difference by r, and we have, by art. 254, and the first term = M — 2 r = 5 — 2r, the second term = M — r =5 — r, the third term = M = 5, the fourth term = M -j- r = 5 -J- r, the fifth term =M-j-2r = 5-\-2r; and the continued product of these terms is (5-2r)(5— »-)5( e H-r)(5+2r)= 3125-625 )-2+20H =945. Hence we find r = ± 2, or = ± v 54 J ; and the only rational series satisfying the condition is, there- fore, 1, 3, 5, 7, 9. 2. Find four numbers in arithmetical progression whose sum is 32j and the sum of whose squares is 276. Ans. 5, 7, 9, 11. 3. A traveller sets out for a certain place, and travels 1 mile the first day, 2 the second, and so on. In five days afterwards another sets out, and travels 12 miles a day. How long and how far must he travel to overtake the fiist? Ans. 3 days and 36 miles CH. VII. § II.] GEOMETRICAL PROGRESSION. 195 Examples involving Arithmetical Progression. 4. Find four numbers in arithmetical progression whose sum is 28, and continued product 5S5. Ans. 1, 5, 9, 13. 5. The sum of the squares of the first and last of four numbers in arithmetical progression is 200, and the sum of the squares of the second and third is 136; find them. Ans. 2, 6, 10, 14. 6. Eighteen numbers in arithmetical progression are such, that the sum of the two mean terms is 31 1, and the product of the extreme terms is 85 J. Find the first term and the common difference. Ans. The first term is 3, the common difference is 1J. SECTION II. Geometrical Progression. 257.' A Geometrical Progression, or a progression by quotients, is a series of terms which increase or decrease by a constant ratio. a, I. n, and S will be used in this section as in the last, to denote respectively the first term, the last term, the num- ber of terms, and the sum of the terms; and r will be used to denote the constant ratio. 258. Problem. To find the last term of a geomet- rical progression when its first term, ratio, and num- ber of terms are known. J96 ALGEBRA. [cH. VII. $ II To find the last Term and Sum. Solution. In this case a, r, and n are given, to find / Now the terms of the series are as follows : a, ar, ar' 2 , ar 3 . . . &.C. ... ar n — 1 , so that, the last or nth term ia that is, the last term is equal to the product of the first term by that power of the ratio whose exponent is one less than the number of terms. 259. Problem. To find the sum of a geometri- cal progression, of which the first term, the ratio, and the number of terms are known. Solution. We have S=za-\-ar-\-ar fi -\-&.a. .. . -j- a r n ~ 2 -\-a r"" 1 . If we multiply all the terms of this equation by r, we have rjS=ar-|-ar 2 -(-a r 3 -|-&c. ,.,4-ar*" 1 4-ar~ n ; from which, subtracting the former equation, and striking out the terms which cancel, we have r S — S = a r n — a, or (r — 1) S = a r n — a = a (r n — 1) ; whence ~ ar n — a a (r n — 1) r — 1 r — 1 Hence, to find the sum, multiply the first term by the difference between unity and that power of the ratio whose exponent is equal to the number of terms,, and divide the product by the difference between unity and the ratio. CH. VII. § II.] GEOMETRICAL PROGRESSION. 197 Examples in Geometrical Progression. 260. Corollary. The two equations l = a r" _1 . (r— l)S==a(r n — 1) give the means of determining either two of the quantities a, I, r, n, and S, when the other three are known. But it must be observed, that, since n is an exponent, it can only be determined by the solution of an exponential equation. 261. EXAMPLES 1. Find the 8th term and the sum of the first 8 terms of the progression 2, 6, 18, &.C., of which the ratio is 3. Ans. The 8th term is 4374, the sum is G560. 2. Find the 12th term and the sum of the first 12 terms of the series 64, 16, 4, 1, \, &lc, of which the ratio is J. Ans. The 12th term is ^-j-J-j^, the sum is 85/ 9 V¥oV 3. Find S, when a, I, and r are known. rl — a Ans. S = — . r — 1 4. Find the sum of the geometrical progression of which the first term is 7, the ratio |, and the last term 1|. Ans. 12£. 5. Find r and S, when a, /,'and n are known. <■— 1 l n — 1 n— 1 «-' I AllS - r== V a' S =~^Tl S a 1 y/l — y/a 198 ALGEBRA. [cil. VII. § II. Examples in Geometrical Progression. 6. Find the ratio and sum of the series of which the first term is 160, the last term 3S880, and the number of terms 6. Ans. The 'ratio is 3, the sum is 58240. 7. Find r, when a, I, and 8 are known. S — a Ans - r = s=T 8. Find the ratio of the series of which the first term is 1620, the last term 20, and the sum 2420. Ans. J. 9. Find a and S, when I, r, and n are known. . I Ans - a = ^rrr. Sr _ ___/(r-i) f n __ ™. u — 1 " 10. Find the first term and sum of the series of which the last term is 1, the ratio \, and the number of terms 5. Ans. The first term is 16, the sum is 31. 11. Find I, when a, r, and £ are known. Ans. 1 = 8 . r 12. Find the last term of the series of which the firpl term is 5, the ratio \, and the sum 6^. Ans. 2^. 13. Find a, when I, r, and S are known. Am. a = S—(S—l)r. 14. Find the first term of the series of which the last term is ^ s , the ratio £, and the sum 6^ Ans. 5. CH. VII. § II.] GEOMETRICAL PROGRESSION. 19& Infinite Geometrical Progression. 15. Find a and /, when r, n, and jS are known. An, a - < r -V S 1 = »■» — 1 ' r n — 1 16. Find the first and last terms of the series of which the ratio is 2, the number of terms 12, and the sum 4095. Ans. The first term is 1, the last term 2048. 262. An infinite decreasing geometrical progres- sion is one in which the ratio is less than unity, and the number of terms infinite. 263. Problem. To find the last term and the sum of the terms of an infinite decreasing geometrical progression, of which the first term and the ratio are known. Solution. Since r is less than unity, we may denote it by a fraction, of which the numerator is 1, and the denomi- nator r' is greater than unity; and we have 1 r'°° co Since, then, the number of terms is infinite, the formula* for the last term and the sum become I = a r"- 1 = a X = 0, r I — a — a S = r—1 ~ r — V a ar 1 I — r — r 1 — V 200 ALGEBRA. [(III. Til. § II Examples in Geometrical Progression. that is, the last term is zero, and the sum is found by dividing the first term by the difference between unity and the ratio. 264. Corollary. From the equation 8 : ' 1 — r' either of the quantities a, r, and & may be found, when the other two are known. 265. EXAMPLES. 1. Find the sum of the infinite progression, of which the first term is 1, and the ratio £. Arts. 2. 2. Find the sum of the infinite progression of which the first term is 0-7, and the ratio 0-1. Ans. J-. 3. Find r, in an infinite progression, when a and S are known. Ans. r = 1 ^-. 4. Find the ratio of an infinite progression, of which the first term is 17, and the sum 18. Ans. *fa. 5. Find a, in an infinite progression, when r and S are known. Ans. a — S (1 — r). 6. Find the first term of an infinite progression, of which the ratio is §, and the sum 6. Ans. 2. CH. VIII. § I.] COMPOSITION OF EQUATIONS. 201 Form of any Equation. CHAPTER VIII. GENERAL THEORY OF EQUATIONS. SECTION I. Composition of Equations. 266. Any equation of the nth degree, with one unknown quantity, when reduced as in art. 1 IS, may be represented by the form i^ + Bi'-'+C^'H^ +M = 0. If this equation is divided by A, and the coefficients B C M —, — , &.C., — represented by a, b, &c, m, it is reduced to Jl A. A x n -\-ax n ~ 1 -{- b x n ~ 2 -\- &.c. -\- m = 0. 267. Theorem. If any root of the equation a n_|_ aa; ii-l_j_ft i"-2-j-&c. -{-m = is denoted by x', the first member of this equation is a polynomial, divisible by x — x', without regard to the value of x. Proof. Denote x — x' by x™, that is, x i" = x — x', or x = x< -f- zW. If this value of x is substituted in the given equation, if P zM is used to denote all the terms multiplied by at 1 ', or 202 ALGEBRA. [CH. VIII. § I. Form of any Equation. by any power of a* 11 , and Q used to denote the remaining terms, the equation becomes />a;ii]_|_ Q = 0. Now the given equation is, by hypothesis, satisfied ay the value of x. or * =r x', a;[i) = 0; by which the preceding equation is reduced to Q=z0. The terms not multiplied by a;! 1 ', or a power of a: I", must, therefore, cancel each other ; and the first member of the given equation becomes Parr", which is divisible by x t'J, or its equal x — %'. 268. Corollary. The preceding division is easily effected by subtracting from the polynomial the expression a:'" + aa:'"-i + 6a;"—2-|.&c. -|- m — 0, which does not affect its value, but brings it to the form x n — z'»-j-a(ar»-i_ z'*~i)-{-b(z n -2 — x' n -2)-\-&,c., of which each term is, by art. 49, divisible by x — x'. The quotient is, by art. 50, x«-i -L x '\ x n-3 __ + a ■ a x' 4-6 2-/3 -bx' ■ c 1 + &C 2G9. Corollary. The first term of the preceding quotient is x n ~ l ; and if the coefficients of x n " s , x n ~ 3 , &c, are denoted by a', b', &c, the quotient is a;"-! -f „' zn-a -j_ j/ j.n-3 _|_ &, c . . CH, Oil, § I.] COMPOSITION OP EQUATIONS. 203 Number of the Roots of an Equation. and the equation of art. 2G7, is (i — x') (i»-i-(- a' i"- 2 -f 6' x n ~ 3 + &c.) = ; which is satisfied either by the value of x, x = x', or by the roots of the equation z »-i _|. a ' z"- 2 -±b'x n - 3 -\-&,c — 0. If now x" is one of the roots of this last equation, we have in the same way a:»-i -|- a 1 Z"- 2 -(- &.c. = {x — x"){x n - i +-a" x»~ 3 -\- &c.) = 0, and the given equation becomes (x — x 1 ) (x — a;")(s"- 2 + a"a;''- 3 + &,c.) = 0; which is satisfied by the value of x", x = x" ; so that x" is a root of the given equation. By proceeding in this way to find the roots x"', x™, = 1. Ans. x=l, = —l,=l(—\± K / — 3), and=J(l±V — 3). We might proceed in the same way to higher equations, such as the 8th, 9th, 12ih, &c. ; but, since much more simple solutions are given by the aid of trigonometry, this subject will be postponed to a more advanced part of the course. 273. Corollary. Before proceeding farther, we may remark, that the method of solution used in the last example of the preceding article may be applied to any equation of an even degree, in which the successive coefficients of the different powers of x are the same, whether the equation is arranged ac- cording to the ascending or according to the descend- CH. Till. § I.] COMPOSITION OF EQUATIONS. 207 Solution of Equations of a peculiar Form. ing powers of x, as is the case in the following equa- tion. A x* n + B a: 2 "" 1 + C 2? n -* + &c. + Cx s + Bx + A = 0. 274. EXAMPLES. 1. Solve the equation Ax*S~Bx3-\-Cx*-\-Bx-\-A = 0. Solution. Divide by x 2 , and we have and, if we make y = x +- x > we have and Ay* + By+C-2A= we have and the equation becomes, by substitution, y 3 -\- 3 y* — 10 y = 0. The roots of this equation are y = 0, = 2, and == — 5 ; and, therefore, the values of z are % = ±y/— 1, = l,or = £(— 5±V21). 3. Solve the equation sB _^_ 2 2 6 _ 6x i -f 2 s 2 -f 1 = 0. 4ns. i=± 1, or =±£ v (— 2±\/3). 4. Solve the equation 2 a; 4 — 3 a? _ x a _ 3 j _j_ 3 _ . 4ns. ar = 2, or = ^, or=r|( — 1±\/ — 3). 275. Corollary. It follows, from art. 269, that an equa- tion of the second degree has two roots, both of which are given by the process of art. 230 ; and if the equation is re- duced to the form x 2 -f- a x -f- b = 0, and the roots denoted by x' and x", we have z*-\-az-±-b = (x — x')(x — x") = 0. OH. VIII. § I.] COMPOSITION OP EQUATIONS. 209 Values of Coefficients in Equations. But the product (x — x 1 ) (x — x") being arranged according to powers of x, is x s — (x'-\-x")x + x'x"; which, being compared with its equal, x 2 -f- a x -(- b, gives — (x 1 -f- x") = a, x 1 x" = 6 ; that is, the coefficient of x is the negative of the sum of the roots of equation, and the terra which does not contain x is the product of the roots. 276. Corollary. If the roots of the general equation of the third degree x*-\-ax2-\-bx-\-c = Q are denoted by J. j A/ j Ui j we have X 3+ ax 2 + b x + c = (x —x')(x — x")(x^x"') = 0. But the product (z — x') (x — x") (x — x 1 ") is, when arranged according to powers of x, a? — {x< -f x"-\-x"' ) x* + {x 1 x" -f x> x'" -f x" x"') x — x' x" x'" ; whence, by comparison with the given equation, we have « = — (*' + x" -f a-'"), b = x' x" + *' r"' -f- *" x">, c = — T,' cc" e'" ; that is, the coefficient of x 2 is the negative of the sum of the roots, the coefficient of x is the sum of the products of the roots multiplied together two and two, and the term which does not contain x is the negative of the continued product of the roots 18* 210 ALGEBRA. [CH. VIII. § II. To find the Equal Roots. 277'. Corollary. It may be shown in the same way that, in the equation x n -\-ax n - 1 +bx n - 2 -{-cx n - 3 -j-&c. = 0, the coefficient of x n-1 is the negative of the sum of the roots; the coefficient of #"~ 2 is the sum of the products of the roots multiplied together two and two; the coefficient of x n ~ 3 is the negative of the sum of the products of the roots multiplied together three and three ; and so on, the last term being the pro- duct of the roots when n is even, and the negative of this product when n is odd. SECTION n. Equal Roots. 278. Problem. To find the equal roots of an equation. Solution. Let x' be one of the equal roots which occurs n times as a root of the given equation, the first member of which is therefore divisible by {% — x') n . If the quotient is a function of/f denoted by X, the equation is, then, {x — x'Y x=o. The derivative of this first member is, as in art. 177, „ (x — r')*-» X -f (x — x'Y Y, provided that Y is the derivative of X. The factor x — x' occurs, then, (re — 1) times in this derivative of the first member, that is, once less than in the first member itself. The greatest common divisor of the first member and its derivative must, therefore, consist of the factors (x — x') of CH. VIII. § II. J EQUAL ROOTS. 211 Examples of finding Equal Roots. the first member, each being repeated once less than in the first member. No one of them is, then, a factor of the common divisor, unless it is more than once a factor of the first member, that is, unless it corresponds to one of the equal roots. The equal roots of an equation are, therefore, ob- tained by finding the greatest common divisor of its first member and its derivative, and solving the equation obtained from putting this common divisor equal to zero. 279. Corollary. The common divisor must, it- self, have equal roots, whenever a root is more than twice a root of the given equation. 280. EXAMPLES. 1. Find all the roots of the equation a;3 _ 7 32 _|_ 16 x _ 12 — which has equal roots. Solution. The derivative of this equation is 3 z 2 — 14 * + 16, the greatest common divisor of which and the given first member is z — 2. The equation x — 2 = 0, gives 8 x = 2. Now since the given equation has two roots equal to 2, it must be divisible by (x — 2) a = z 2 — 4 x +4, 212 ALGEBRA. [cH. VIII. $ II. Examples of finding Rqunl Roots. and we have a? — 732+16*— 12=(z — 2) s (:k — 3) = 0; whence x = 3 is the other root of the given equation. 2. Find all the roots of the equation x i _ 9 x s _j_ 6 x* + 1 5 z3 — 1 2 z 2 — 7 x -f 6 = which has equal roots. Solution. The derivative of this equation is 7 i 6 — 45 z 4 -f- 24 a;3 -(- 45 z2_ 24 x — 7, the greatest common divisor of which and the given equa- tion gives z 3 — x * _ x + 1 = 0, which is an equation of the third degree, and we may con- sider it as a new equation, the equal roots of which are to be found, if it has any. Now its derivative is 3^ — 2 a— 1, and the common divisor of this derivative and the first mem- ber gives x — 1 = 0, or a; = 1. Hence the first member of a? — a? — a; -j- 1 = must be divisible by and we have indeed s 3 — x 2 — x + 1 = (x — 1)2 ( x + 1) = 0. The equal roots of the given equation are, therefore, x = 1, and = — 1 ; and its first member is divisible by (*_l)3( x _|_l)t f CH. VIH. § II.] EQUAL ROOTS. 213 Examples of finding Equal Roots. and is found by division to be (x — 1)3 (x + l)2(x? + x — 6). The remaining roots are, therefore, found from solving the quauralic equation x 2 -f x — 6 = 0, which gives x = 2, or = — 3. 3. Find all the rools of the equation X34-3X 2 — 9x — 27 = which has equal roots. Arts, x = 3, or = — 3 4. Find all the roots of the equation a? _ 15 x 2 -|- 75 x — 125 = which has equal roots. Arts, x = 5 5. Find all the roots of the equation a; 4_9 a; 3_^29x 2 — 39x-f 18 = which has equal roots. Ans. x = 1, or = 2, or = 3 6. Find all the roots of the equation a,4 _ 2 a .3 _ 59 & 4. 60 x -)- 900 = which has equal roots. Ans. x = 6, or = — 5. 7. Find all the roots of the equation x* — 6x* — 8x — 3 = which has equal roots. Ans. x = 3, or = — 1. 8. Find all the roots of the equation x 4 + 12 x 3 + 54x a 4-108x+ 81=0 which has equal roots. Ans. x = — 3. 214 ALGEBRA. [CH. VIII. § III. Number of Real Roots. 9. Find all the roots of the equation ) X Y becomes (p — x 1 ) (p—x") (p — x'") X Y', wnen we substitute p for x, and denote the corresponding 216 ALGEBRA. [CH. VIII. § HI. Number of Real Roots between two given Numbers. value of Y by Y 1 ; and when we substitute q for x, and de note the corresponding value of Y by Y" , it becomes (q — x') {q — x") (q — x">) X Y" The quotient of these two results is (p — x') (p — x") (p — x'") Y< lq—x')(q— x") {q — x'") Y" which can be written p — x 1 p — x" p — x" < Y> q — x 1 q — x" q — x"' Y" Now since each of the roots x', x", x'", &c, is included between p and q, the numerator and denominator of each of the fractions p — x' p — x' p — x" &c, q — x 1 ' q — x"' q — %"' must be affected with contrary signs, and therefore each of these fractions must be negative. But since Y' and Y" must, by art. 282, have the same sign, the fraction XL Y 7 ' is positive. The product of all these fractions is therefore positive, when the number of the fractions p — x' p — x' 1 „ q — x' q — x" is even, that is, when the number of the roots, x', x'.', x"' t &c, is even ; and this product is negative, when the num- ber of these roots is uneven. The values which the given first member obtains by the substitution of p and q for x roust, consequently, be affected with contrary signs in the latter case ; and with the same sign in the former case. CH. VIII. § III.] REAL ROOTS. 217 Number of Real Roots of an Gqunlion of an Odd Degree. 284. Theorem. Every equation of an uneven degree, has at least one real root affected with a sign contrary to that of its last term, and the number of all its roots is uneven. Proof. Let the equation be x n -j- a x n ~ l -J- &.C. . . . -J- m := 0, in which n is uneven. First, to prove that there is a real root, and that the number of real roots is uneven. Every real root must be contained between -|- CD and — go. Now the substitution of X r= CD, gives the value of the first member a n_|_ aQ o n-1 -|- 6co n-2 + &c - ... + '»; the first term of which is infinitely greater than any other term, or than the sum of all the other terms. The sign of this result is therefore the same as that of its first term, or positive. Again, the substitution of x = — co gives, since n is uneven, Q0 n_|_ ncc n-1 6 0D n-2 -f- &C. . . . -f- OT, which may be shown by the above reasoning to be negative. The given equation must then, by art. 2S1, have at least one real root, and by art. 283, the number of its real roots must be uneven. Secondly. To prove that one, at least, of the real roots is affected with a contrary sign to that of the last term. The substitution of x — Q, reduces the given first member to its last term m. 19 218 ALGEBRA. [CH. VIII. § IIL Number of Real Roots of Equations. Comparing this with the above results, we see that, if m is pnsitioe, the given equation must, by art. 281, have a real root contained between and — od, that is, a negative root; but if m is negative, there must be a real root contained between and -j- oo, that is, a positive root ; so that there must always be a root affected with a sign contrary to that of m. 285. Theorem. The number of real roots if there arc any, of an equation of an even degree must be even, and if the last term is negative, there must be at least two real roots, one positive and the other negative. Proof. Let the equation be z n -f- a x n ~ l -f b z n_2 + &c -}- m = 0, in which n is even. First. To prove that the number of real roots is even. The substitution of x = on gives for the value of the first member oo n -f- nOD" -1 -\-b go" " 2 -|- &c. . . . -\-m. which is positive. The substitution of x = — oo gives for the value of the first member 00" — acD"~ 1 -\-b od" - " 2 -J- &c. . . . -f- m, which is also positive. Hence, if the given equation has any real root, ibere must, by art. 283, be an even number of them. Secondly. To prove that when m is negative, there must be two real roots, the one positive, the other negative. The «abstitution of x = CH. VIII. § III.] KEAL ROJTS. 219 Number of Imaginary Roots ; of Real Positive Roots. reduces the given first member to its last term m, and this result is therefore negative in the present case. Comparing this with the above results, we see that there must be a real root between and -}- go, and also one be- tween and — co : that is, the given equation has two real roots, the one positive and the other negative. 286. Corollary. Since the number of real roots of an equation of an uneven degree is uneven, and that of an equation of an even degree is even, the number of imaginary roots of every equation, which, has imaginary roots, must be even. 287. Theorem. The number of real positive roots of an equation is even, when its last term is posi- tive ; and it is uneven, when the last term is nega- tive. Proof. The substitution of x = co gives, for the first member of the given equation, a positive result ; while the substitution of x = reduces the first member to its last term. Hence if this last term is positive, the number of real roots contained between and oo, that is, of positive roots, must, by art. 2S3, be even ; and if this last term is negative, the number of these roots must be uneven. 288. Theorem. If a function vanishes, that is, is equal to zero for a given value x' of its variable x ; the function and its derivative must have like signs for a value of the variable which exceeds x' by 220 ALGEBRA. [CH. VIII. § III. Variation and Permanence. an infinitely small quantity, and unlike signs for a value of the variable which is less than x' by an in- finitely small quantity. Proof. Let the given function be u, and its derivate U, and, as in art. 176, when the variable is increased by the infinitesimal i, the function becomes- u-\- Ui. This value of the function, when w = ij reduced to Ui, which has, obviously, the same sign with U. In the same way when the variable is decreased by i, the function becomes w — Ui, which, when u = 0, is reduced to — Ui, having the opposite sign to U. 289. Definition. A pair of two successive signs in a row of signs, is called a permanence when the two signs are alike, and a variation when th^y are unlike. 290. Sturm's Theorem. Denote the first member of the equation X* + a x n ~ x + &c. = by u and its derivative by U. Find the greatest common divisor of u and U, and, in performing this process, let the several remainders which are of con- tinually decreasing dimensions in regard to x, be de- noted, after reversing their signs, by U, W, U", = m U" — U"> ehes V = ; and, in the same way, it is shown that U= and u = 0, so that the function and the derivative are both zero at the same time, which, by art. 278, corresponds to the case of one of the equal roots of the equation. Secondly If any term of the series, except the first oi last, has a different sign in the row corresponding to the value p of the variable from that which it has in the row 19" 222 ALGEBRA. [CH. VIII. § III. Sturm's Theorem. corresponding' to the value q of the variable, it must, by art. 281, vanish for some value of the variable contained be- tween p and q. But for this value of the variable, the pre- ceding term must have a different sign from the succeeding term ; thus, when U"=0 the equation IP = m" U" — IP" gives V = — IP". By the change of sign which the term undergoes in vanish- ing, therefore, it can only change from forming a perman- ence with one of its adjacent terms to forming one with the other of these terms, and the change of sign of a term, which is neither the first nor the last of the series, does not increase or diminish the number of permanences of the row of signs. Thirdly. When the first term u of the series, in chang- ing its sign, vanishes, while the second term U does not »anish, the corresponding value of the variable is, by art. 278, a root of the equation which is not one of the equal roots. If, moreover, the variable is decreasing in value, the signs of these two terms constitute a permanence before the change and a variation after the change. When the. vari- able, therefore, in decreasing passes throvgh a value which is one of the unequal roots of the equation, the number of per- manences in the row of signs is increased by unity. Fourthly. When the given equation has no equal roots, u and U can, by art. 278, have no common divisor, and therefore the last term of the series will not contain the variable ; it must, therefore, be of a constant value and no change of sign can arise from it. In this case, then, the number of permanences must by the preceding division of the proof be greater in the row which corresponds to the greater CH. VIII. § III.] REAL ROOTS. 223 Sturm's Theorem. of the tico limits p and q, than in the row which corre- sponds to the less of these ttoo limits, by a number which is exactly equal to the number of real roots contained -between p and q. Fifthly. When the given equation has equal roots, u and U must, by art. 278, have a common divisor which will be the last term of the series. This divisor must also, by art. 59, be a divisor of all the other terms of the series ; and if the series is divided by it a new series », V, F', V", &c, is obtained, which has in all cases either the same signs as the given series or t'he reverse signs, so that each pair of successive signs is of the same .name, whether perma- nence or variation, in each series. And by dividing the equations before found by this same common divisor, they become v = m V — V' V = m' V — V" &.C The first term of the new series has, by art. 278, the same roots with the given series .except that it has no equal roots, and the last term is unity. The reasoning of the preceding portion of this article may, therefore, be applied to the new series ; and it follows that the theorem is applicable to the case of an equation which has equal roots, as well as to one width has unequal roots. 291. Corollary. If infinity is substituted for p and negative infinity for q in the series of divisors, the resulting rows of signs show at once the wholt number of real roots of the given equation. 224 ALGEBRA. [til. VIII. § III Sturm's Theorem. 292. EXAMPLES. 1. Fir.d the number of real roots of the equation 2 x* — 20 x -f- 19 = and also the number contained between 1 and 2. Solution. In this case, we have u = 2 a: 4 — 20 x -f 19, U = 8 a: 3 — 20 ; and by the method of the common divisor -2A — 38* — 722 — 3157 U> = 15 x — 19 U" =3157. When, therefore, x = oo the row of signs is +.+.+. + ; and when x = — CD, it is there are then two real roots. Again when x =. 2, the row of signs is +.+.+. + ; and when x = 1, it is the two real roots are therefore both between 1 and 2. 2 i 4 — 20 x -f- 19 2 z 4 — 5 i 2I 3 - 5 — 15 i 4- 19 3 x* — 75 - 30 x 3 — 38 a? - 38 x* — 75 -570 a:- 2 — 1125 570 x* — 722 a; 722 z— 1125 10830 a; — 16875 10831) x — 13718 CJI. VIII. § III.] REAL ROOTS. 225 Number of Real Roots. 2. Find the number of real roots of the equation i 3 -)- a x -j- 6 = 0. Solution. In this case, we have u = x 3 -j- a x -\- b U = 3 x* + a V = — 2 ax — 3 6 U" =z — 4 a? — 27 6 3 , First ease. When a is such that U" is negative, that is when — 4a3<27 6 2 , or— 3 Va3<^6 2 ,or(-ia)3<(J6)2 the row of signs when x = ao is -j-, -j-, =F (the reverse of a), — ; and when x = — cc it is — , +, ± (like a), — , so that there is only one real root. The row of signs when x = is, when b is positive, +, ± (like a), — , — , so that, in this case, the real root is negative. This row, when 6 is negative, is — , ± (like a), -f , — , so that, in this case, the real root is positive, which agrees with art. 284. Second case. When a is negative and of such a value that (h bf = - (i «) 3 > that is U" = 0, in which case the equation has the equal root, obtained from the equation V' = — 2 ax — 36 = 36 x== -=2a'' 226 ALGEBRA. [CH. VIII. § III Number of Real Roots. and in this case the row of signs when x = oo is +.+. + ; and that when x = — oo is so that the two different roots of the equation are, in this case, real. The row of signs when x = is ± (like b), —, q= (unlike 6) ; so that one of the roots is positive and the other negative. Third case. When a is negative and of such a value that U" is positive or (- I «) 3 > ih *) 2 in which case, the row of signs when x = oo is and when x = — oo it is so that all three of the roots of the equation are real. The row of signs when x = is ± (like 6), — , =p (like 6), — . If, then, b is positive the equation has one positive real root and two negative ones; and if 6 is negative, it has two posi- tive real roots and one negative one. 3. Find the number of real roots of the equation x n -f- a = 0. Solution. In this case, we have « = x n -f- a U = rax"- 1 U'= — a. First case. If n is even, the row of signs, when z = go is, then, +. +» =F f "(unlike a) ; CH. VIII. § III.] REAL ROOTS. 227 Number of Real Roots. when x = — co it is -j-, — , 1= (unlike a) ; so that there is no real root when a is positive, and two real roots when a is negative, which agrees with art. 2S5. Second case. If n is odd, the row of signs when x = on is -f, +, =F (unlike a) ; when x = — oo it is — , +, =F (unlike a) ; so that, in either case, there is only one real root, which is, by art. 284, of a sign unlike that of a. 4. Find the number of real roots of the equation x n -f a x -f- b = 0. Solution. In this case, we have u = ,x n -j- a x -\- b, U = n z n_1 -(- a . £/' = — (n — l)az — ra 6, U"= — a n (ra— I)" -1 — n n (— 6)" _1 . JFVrsf case. When n is even and greater than 2, and J7" positive, that is, when 6 is positive, and C)' +> T (unlike a) ; when a; = — oo it is +>— > ± (like a); so that in either case there is no other real root than the above equal root. Third case. When re is even and greater than 2, and U" negative, that is, (J)">(iP the row of signs when x = qd is +» +» =F (unlike n), — ; when x = — it is +. — , ± (like a),—; so that when a is negative there is no real root, and when a is positive there are two real roots. In the latter case, the row of signs when x = is ± (like 6), -f, zp (unlike b), — ; so that when b is positive, both the roots are negative, and when b is negative, one of the roots is positive and the other negative, which agrees with art. 2S7. Fourth case. When n is odd, and U" positive, that is, when a is negative and (-!)"> C4,y '- CH. vrn. § III.] REAL ROOTS. 229 Number of Real Roots. in which case, the row of signs when x = go is +>+>+> + ; when x = — go it is so that .the equation has three real roots ; the row of signs when x = is ±. (like 6), — , =F (unlike 6), -f ; so that when 6 is negative, one of the real roots is posi- tive and the other two negative ; and when b is positive, one of the real roots is negative and the other two posi- tive. Fifth case. When n is odd, and U" zero, that is, when a is negative and n-l K)"=C.-^T- in which case, there is the equal root nb _"~j_JL * — — ( n — l)a ~ \ "»' the row of signs when x = co is +.+. + ; when x = — go it is so that there is another real root besides the above equal root. The row of signs when x = is ± (like b), — , =p (unlike b) ; so that one of the roots is positive and the other negative. Sixth case. When n is odd, and V" negative, that is, Kr<(^r- 20 230 ALGEBRA. [cH. VIII. § III. Number of Real Roots. in which case, the row of signs when x = oo is +, -h =F (unlike a), — ; when x = — it is — . +. ± (like a), — , so that there is only one real root, which, by art. 284, Las i sign contrary to that of its last term. 4. Find the number of real roots of the equation x 3 _ 6 a;2 -|- 19 x — 44 = 0. Ans. It has one positive real root. 5. Find the number of real roots of the equation x* — 10 a? -f 35 z 2 — 50 x + 24 = 0. Ans. It has four positive real roots. 6. Find the number of real roots of the equation x i 4- a? _ 24 3? + 43 x — 21 = 0. Ans. It has three positive real roots and one negative one. 7. Find the number of real roots of the equation x 4 + 8 I s + 16 x — 440 = 0. Ans. One positive root and one negative root. 293. Sturm's theorem is perfect in always giving the number of real roots, but often requires so much labor, that theorems, which are much less perfect, may be used with great advantage. 294. Stern's Theorem. Denote the first member of the equation x n + a x n ~ l + & c - = by u and its first, second^ &c. derivatives by U, U', &c. CH. VIII. $ III.] REAL ROOTS. 231 Stern's Theorem. Find the row of signs corresponding to the values of u, U, U', U",'&,c, for any value p of the variable, and also for a value q of the variable. The number of real roots of the equation, com- prised between p and q, cannot be greater than the difference between the number of permanences of the first row of signs and that of the second row. Proof. First. It may bo shown as in the third division of the proof of art. 290, that one 'permanence at least is always lost from the row of signs when the variable in de- creasing passes through a value which is one of the roots of the equation. Secondly. "When any term of the series except the first or the last, vanishes, it passes, by art. 288, with the de- creasing variable, from having the same sign with its deriv- ative, which is the next term of the series, to having the reverse sign of it. Even then, if it had before the change the reverse sign of the preceding term and after the change the same sign, it introduces a permanence which is only sufficient to take the place of the other permanence which is lost. The number of permanences of the row of sig?is is not, therefore, augmented by the vanishing of such an inter- mediate term. Thirdly. The last term of the series must be constant, for the number of dimensions is diminished by each deriva- tion ; and, therefore, as x decreases from a value p to a smaller value q, the number of permanences of the row of signs must be diminished by as large a number at least as the number of roots comprised between p and q. 232 ALGEBRA. [cH. VIII § III. Descartes' Theorem. 295. Definition. An equation x n -\-ax n - 1 -f-«fcc. + &i a +lx-(-i=0. is said to be complete in its form, when it contains terms multiplied by every different power of x from the highest to unity, and also a constant term, such as I. 296. Descartes' Theorem. A complete equation cannot have a greater number of positive roots than there are variations in the row of signs of its terms, nor a greater number of positive roots than there are permanences in this row of signs. Proof. If the equation is that of art. 295, the values of u, U, V, &c. in art. 294, are « = x n -f- a z*-i -|- &c. + h x s -f- k x -f- 1 U = nx"- ! + (» — l)az M - 2 + &c.-f 2Az-f-& U'=n(n-l)x n -*+(n— l)(»-2)ax—3-j-&.c.+2A &.C. The row of signs when x == oc is +> +. +. +» +, &c-> consisting wholly of permanences. When x = — oo it is ±. =F, ±, q=, &c, in which the upper row of signs is used when n is even, and the lower row when n is odd. In either case, this row consists wholly of variations. The row of signs when x = is ± (like I), ± (like k), ± (like 2 h) that is, it is the same as the row of signs formed by the terms of the equation taken in the inverse ordpr. CH. VIII. § III.] REAL ROOTS. 233 Zero substituted for a Term which is wanting. The limit of the number of positive roots is, therefore, by art. 21)4, equal to the expess of the whole number of pairs of successive signs of the lerms, over the number of per- manences ; that is, it is equal to the number of variations ; and in the same way, the number of negative roots cannot exceed the number of permanences. 297. Corollary. The whole number of successions of signs of an equation, that is, the sum of the permanences and variations, is one less than the number of terms, or the same as the degree of the equation, that is, the same as the number of roots. If, therefore, all the roots are real, the number of positive roots must be the same as the. number of variations, and the number of negative roots must be the same as the number of permanences. 298. Scholium. Whenever a term is wanting in an equation, its place may be supplied by zero, and either sign may be prefixed. 299. Corollary. When the substitution of + for a term which is wanting gives a different num- ber of permanences from that which is obtained by the substitution of — 0, and consequently a differ- ent number of variations also, the equation must have imaginary roots. 300. Theorem. When the sign of the term which precedes a deficient term is the same with that whic\ follows it, the equation must have imaginary root! . Proof. For if the terms which precede and follow the deficient term are both positive, the substitution of -f- gives two permanences; while the substitution of — ' gives 20* 234 ALGEBRA. [CH. VIII. § III. Imaginary Roots, when Terms are wanting. two variations. The reverse is the case when both these terms are negative. The equation must, therefore, in either case, have imaginary roots. 301. Theorem. When two or more successive terms of an equation are wanting, the equation must have imaginary roots. Proof. For the second deficient term may be supplied with zero affected by the same sign as that of the term pre- ceding the deficient terms ; and the first deficient term is then preceded and followed by terms having the same sign, so that there must, by the preceding article, be imaginary roots. 302. Theorem. When an uneven number (m) of successive terms is wanting in an equation, the number of imaginary roots must be at least as great as (m -f- 1), if the term f receding the deficient terms has the same sign with the term following them ; and the number of imaginary roots must be at least as great as (m — 1), if the term preceding the deficient terms has the reverse sign of the term following them. Proof. First. If the sign of the term preceding the de- ficient terms is the same with the sign of the term following them ; supply the place of each deficient term with zero affected by this same sign. All the (ra -|- 1) successions, dependent upon the deficient terms, must in this case be permanences. But if the sign of every other zero beginning with the first is reversed, namely, of the first, third, fifth, &c, all these permanences are changed into variations ; so that (m -f- 1 ) roots can be neither positive nor negative, and are, consequently, imaginary CH. VIII. § III. J REAL ROOTS. '235 Superior Limit of Positive Roots. Secondly. If the sign of the term preceding the deficient terms is the reverse of the sign of the term following them ; supply the place of the two last deficient terms with zero affected by the same sign as that of the term preceding the deficient terms. This case becomes the same as the pre- ceding with (m — 2) deficient terms, and there must there- fore be [m — 1) imaginary roots. 303. Theorem. When an even number of suc- cessive terms is wanting in an equation, the number of imaginary roots must be at least as great as the number of these deficient terms. Proof. Let the place of the first deficient term be sup- plied by zero affected with the same sign as that of the term which follows the deficient terms. The number of deficient terms is thus reduced to the uneven number m — 1; and, as the term preceding the deficient terms is now of the same sign with that of the term following them, the number of imaginary roots of the equation must, by the preceding article, be at least as great as (m— ]) 4-l = /«. 304. A number, which is greater than the greatest of the positive roots of an equation, is called a su- perior limit of the positive roots ; and one, which is less than the least of the positive roots, is called an inferior limit of the positive roots. In the same way, a superior limit of the negative roots is a number which, neglecting the signs, is greater than the greatest negative root ; and an infe- rior limit of the negative roots is a number which is less than the least negative root. 236 ALGEBRA. [CH. VIII. § III Superior Limit of Positive Roots. 305. Problem. To fold a superior limit of the .positive roots. Solution. The sum of all the negative terms being equal to the sum of all the positive terms, must exceed each positive term. Let, then, — S be the greatest negative co- efficient of the equation of the nth degree, and m the ex- ponent of the highest negative term ; the sum of the nega- tive terms, neglecting their signs, must evidently be less than ihnf of the series S + Sx+Sx*+&c + Sx m , for each term of this series is greater than the correspond- ing negative term of the equation. But this series is a geometrical progression of which S is the first term, Sx m the last term, and x the ratio ; so that its sum is, by example 3, of art. 261, x — 1 ' and must be greater than any positive term, as x n , or ^Sx m + l — S ^ Sx m + i ^ x — l ^ x—1 Hence (x— l)z n , and, therefore, y > L" or •>h- so that -=- is an inferior limit of the positive roots of the Li given equation. 307. Problem. To find the limits of the negative roots of an equation. Solution. Substitute for x x = — y, 23S ALGEBRA. [ell. VIII. § III. Limits of Real Roots. and the positive roots of the equation thus formed are the negative roots of the given equation ; and, therefore, the limits of its positive roots become, by changing their signs, the required limits. 308. Corollary. By the substitution of different numbers for p and q, in arts. 290 or 294, the limits between which each root is obtained can be narrowed to any extent which may be desired, until they may be adopted as the first approximations to the roots in the method of art. 179. Thus, it is easy to obtain the first left hand significant figure. 309. EXAMPLES. 1. Find the left hand significant figure of the real roots of the equation 5^ — 6a:-f2 = 0. Solution. First. In this case, 6 is the greatest negative coefficient and — 6 # is the first negative term, so that, by art. 305, 1 -f- s/ 6 = 3-5 is a superior limit of the positive roots. To find the limit of the negative roots, let * = — y, and the equation becomes, by reversing its signs, 5y 3 — 6y — 2 = 0; so that — (I+V6) = -3-5 is the superior limit of the negative roots, and the roots are all contained between 4 and — 4. CH. VIII. § III.] REAL 239 Limits of Heal Roots. Secondly. Sturm's theorem gives u = 5 x 3 — 6a-J-2 U = 15 a 2 — 6 U' = 2x—1 U"=S; so that the row of signs when x = 4 is +>+.+, + ; when x = — 4 it is 80 that the equation has three real roots. The row of signs when x = is +, -,-. + ; so that two of the roots are positive and one is negative. The substitution of positive integers, gives for the rows of signs when x = 1 +. +,+. + ! so that both the positive roots are contained between and 1. The substitution of the positive decimals 0-1, - 2, - 3, &c, gives the following rows of signs. x = 0-l * = 02 x= 3 x = 0-4 x = 0-5 1 = 0-0 * =0-7 x = 0-8 a: = 0-9 + : so that one real root is contained between 03 and 0-4, and the other between 0-S and 9 ; their first approximate val ues are, then, 0-3 and 0-8. The substitution of the negative integers gives, in the 240 ALGEBRA. [cH. VIH. § in. Limits of Heal Roots. same way — 1, for an approximate value of the negative root 2. Find the left hand significant figures of the roots of '.lie equation x i _|_ 8 & -f 16 x — 440 = 0. Ans. 3 and — 4. 3. Find the first approximation to the roots of the equa- tion a 5 — 15 x 3 + 132 a; 3 + 3G x + 396 = 0. Ans. 1, — 1, — 5. 4. Find, by Stern's theorem, the greatest possible num- ber of real roots which the equation a-io _ io s? — x* -f x — 1 1 = can have between -(- 1 and — 1. Solution. In this case we have, by art. 294, u = * 10 — IOje 8 — z 4 + a: — 11 U = 10 x 9 — 80 x> — 4 x 3 -f- 1 V = 90 a; 8 — 560 x 6 — 12 z 2 U" = 720 x> — 3360 x* — 24 * U"> = 5040 x 6 — 16800 it* — 24 U" = 30240 z 5 — 67200 Z 3 U> = 151200 x* — 201600 a 9 If" = 604800 »» _ 403200 x U™ = 1814600 x* — 403200 17»»' = 362800 x W* = 3628800 ; the row of signs when x = 1 is • - — > > > > i "■" "» i~» r» ("• i » when x = — 1 it is bo that the number of these roots cannot exceed 8. CH. VIII. § III.] REAL ROOTS. 241 Integral Root. Again, when x is- infinitely little greater than zero, for which value some of the differential coefficients vanish, the row of signs is I |~ I ' 3 > ~~ > J > ~ | > » so that there cannot be more than three roots between and 1 ; and since the sign of the first term is the same when x = 0, that it is when x= 1, there cannot, by art. 283, be an odd number of real roots between and 1, and consequently there cannot be more than 2. The row of signs when x is less than zero by an infinitesimal is i "T~i > i > > i > i> > > ~T~ > so that there can be no real root between and — 1. 5. Find, by Stern's and Descartes' theorems, the greatest possible number of real roots of the equation comprised between and 1. Ans. 2. 310. A Commensurable Root is a real root, which can be exactly expressed by whole numbers or frac- tions. 311. Problem. To find the commensurable roots of the equation i'Ioi*-' +&x n - 2 +&c. + lx + m = 0, in which a, b, Sfc. are all integers, either positive or negative. Solution. Let one of the commensurable roots be, wl a reduced to its lowest terms, x=P-. 9 31 242 ALGEBRA. [CH. Till. § III. Method of finding Integral Roots. As this root must verify the given equation, we have ?? _|_ a ?2ll _L 6 £11 _L &c. + m = ; whence, multiplying by q n ~ 1 , and transposing, we obtain — =z — ap n ~ l — bp n ~ z q — &.c. . . . — uij" -1 ; and, therefore, as the second member is integral, the first member must also be integral, or we must have whence x = p; that is, every commensurable root of the given equa- tion must be an integer. Again, the substitution of x=p, in the given equation, produces p n -|-ap n-1 -J-fcc \-lcpi-\-lp-\-m = 0; whence, dividing by p, and transposing, we obtain — = — I — kp — &c. ... — ap n ~ 2 — p n ~ 1 ; and, therefore, as the second member is integral, the first member must be so likewise; that is, every integral root must be a divisor of m. If, now, we denote by nv, m' = — + I, P the preceding equation gives, by transposing and dividing by p, mf , _ . — = — k — tp — hp* — gp 3 — &c. — ap n ~* — p*-*. so that this integral root must likewise be a divisor of m' CU. VIII. § III.] REAL ROOTS. 243 Method of finding Integral Roots. In the same way, if we use m", m'" t my, &c. as follow m" = — + h, P r m" m i" — _ _l_ i P n m!" m" = — + A, P &c, &c. ; tAis integral root must be a divisor of m", m'", m ,T , &x. , and the last condition to be satisfied is njt»-i] _j- p = 0, or ffl[ m-1 l = — jp. Hence to find all the commensurate roots of the given equation, write in the same horizontal line all the integral divisors of m, which are contained between the extreme limits of the roots. Write below these divisors all the corresponding values of m', m", Sfc, which are integral, remembering that a divisor cannot be a root, when the value which it gives for either m', m", m'", fyc, is fractional. Proceed in this way till the values ofwf -11 are obtained, and those divisors only are roots which give — p for the value of this quantity. 312. EXAMPLES. 1 . Find the commensurable roots of the equation x 5 — 19^ + 34x 2 4-12a; — 40=0. Solution. The extreme limits of the real roots are -J- 7-4, and — 6-9. Hence we have 244 ALGEBRA. [ell. VII). § in. Commensurable Roots of any Equation. m =—40; p = 5, 4, 2, 1,_ l,_ 2,— 4,— 5; m! = 4, 2, — 8, —28, 52, 32, 22, 20 : m" = , , 30, 6,-18, 18, , 30; m"'= , , — 4, —13,— 1,-28, ,—25; m"= , .— 2, — 13, 1, 14, , 5; and, therefore, 2, — 1, and — 5 are roots of the given equation, and its first member, divided by the factor (x — 2)(ar + l)(* + 5) = z 3 + 4 z 2 — 7 x — 10, gives the quotient a;2_4a;-{-4; and, therefore, the remaining roots are those of the equation a: 2 — 4 x + 4 = 0, which are equal to each other, and each is x = 2. 2. Find the commensurable roots of the equation a 8 — 3 x7_l0 i6_ 2a; 4 +6 i 3 +21ar 2 — 3 z— 10= 0. Ans. 5, 1, — I, and — 2. 3. Find all the roots of the equation K 4 + 2:3 — 21a; 2 + 43u. 21 = which has commensurable roots. Ans. 1, 3, — |±J%/52. 4. Find all the roots of the equation x 3 — 6 x s + 19 x — 44 = which has a commensurable root. Ans. 4, and 1 ± \/ — 10. 5. Find all the roots of the equation x* — 1 z 3 + 35 a; 2 _ 50 x -j_ 24 = which has commensurable roots. Ans. I 2, 3, 4. CH. Till. § III.] REAL ROOTS. 945 . Commensurable Roots of any Equation. 6. Find all the roots of the equation % f> _ 3 mi _ g x 2 _|_ 24 x * _ 9 x _f_ 27 = which has commensurable and equal roots. Ans. 3, — 3, and ± y/ — 1 7. Find all the roots of the equation x 6 _ 23 x 4 — 48 x3 -f 95 x 2 -f 400 x -\- 375 = which has commensurable and also equal roots. Ans. 3,5, and— 2±%/ — 1. 313. Problem. To find the commensurable roots of an equation. Solution. Reduce the equation to the form A x n + B x"" 1 + &c + L x -f M = 0, in which A, B, &c, are all integers, either positive or negative. Substitute for x the value x-1- x ~ A' and the equation becomes J^ + ^I 1 + ^ =2 + &c... + ^ + M = 0i ;p=l-r- A nr-i -r A *-^ 1 «"<-••-. -r ^ -r-» which, multiplied by ^ n_1 , is 2/ «+By'>-*+ACy n - ;i +&c....+A n -*Li/+A n -iM=0. The commensurable roots of this equation may be found, as in the preceding article, and being divided by A, will give the commensurable roots of the re- quired equation. 314. Scholium. The substitution of V x= A 21 • 246 ALGEBRA. [CH. Till. § III. Commensurable Roots of any Equation. • is not always the one which leads to the most simple result. But when A has two or more equal factors, it is often the case that the substitution y X = -A! leads to an equation of the desired form, A' being the product of the prime factors of A, and each fac- tor need scarcely ever be repeated more than once. 315. EXAMPLES. 1. Find the commensurable roots of the equation 64 x* — 328 a* + 574 x* — 393 x -)- 90 = 0. Solution. We have, in this case, A = 64 =26; hence we may take A' equal to some power of 2 ; and it is easily seen that the third power will do, so that we may make * = iV- Hei.ce the given equation becomes y i _ 41 yZ _|_ 574 ,,2 _ 3144 y _j_ 5760 _ o. The commensurable roots of which are found, as in art 311, to be y = 4, 6, 15, and 16 ; eo that the roots of the given equation are * = i, i, 1J, and 2. 2. Find the commensurable roots of the equation 8 a? -f 34 x* — 79 x -f 30 = 0. Ans. £, £, and — 6 CH. VTII. § III.] REAL ROOTS 247 Commensurable Roots of any Equation. 3. Find the commensurable roots of the equation 24 ti> — 26 s2 _|- 9 x — i = o. Ans. |, §, and \. 4. Find the commensurable roots of the equation 3 a? — 14 x s -f 21 x — 10 = 0. Ans. 1, f , and 2. 5. Find the commensurable roots of the equation 8 s 4 — 38a^4-49»2 — 22 k -{-3 = 0. -Aws. |, |, 1, and 3. G. Find all the roots of the equation 6 a* -f- 7 z 2 -f 39 x + 63 = which has a commensurable root. Ans. — f, and £±£v/ — 251. 7. Find the commensurable roots of the equation 9x6 _|_ 30 a* _|_ 22z 4 + 10x3 + 17 z 2 — 20 a: + 4 = 0. Ans. | and — 2. 243 ALGEBRA. [CII. IX. Vaiue of Continued Fractions. CHAPTER IX. CONTINUED FRACTIONS. 316. A continued fraction is one whose numeratoi is unity, and its denominator an integer increased by a fraction, whose numerator is likewise unity, and which may be a continued fraction. Thus, and _1_ l i 1 d -j- &c. are continued fractions. 317. Problem. To find the value of a continued fraction which is composed of a finite number of fractions. Solution. Let the given fraction be 1 . 1 »+i • + i Beginning with the last fraction, we have successively . 1 cd+1 c + ¥ = -ir- 1 d , l~cd + l C + d OH. IX.] CONTINUED FRACTIONS. 249 Value of Continued Fractions. i d - l>( cdJ r i ) + d = 6 + cd-j-1 """ c d -f- 1 cd+1 crf-f- 1 b (c d + 1) + d ~ (bc-{-\)d+~b 6 1 * 1 6 1 l c +d | X a •■; [ " 1 1 , » a r 1 i. i 6 + 1 _ ad(bc-\-\)-\-ab + cd+\ (6 c +])+ K'\ N _ M q + L ~N> ~ Wq~+L» that is, the value required to satisfy the theorem. If, therefore, it can be shown that the proposition is true for any approximate value, it follows that it must be true for every succeeding value. But the comparison of the values given in the preceding article shows that it is true for the third value, and therefore for every succeeding value. 321. Theorem. If two succeeding approximate values are reduced to a common denominator equal to the product of their denominators, the difference, of their numerators is unity. Demonstration* Let the (n — 2)nd, (« — l)st, and ntb approximate values be Ll JUL j^_ m i+ l JV M 1 ' N< ~ JU'q + L'' the difference between the (n — 2)nd and (n — l)st is LW—L'M =fc TTW' ; and that between the (n — l)st and nth is M'N—MN' _ (M M' —MM')q + M>L—ML' ^ WW "~ M N> LM—L'M ~ ± M> N' ' of both which differences the numerators are the same; and, therefore, this is always the case. CH. IX.] CONTINUED FRACTIONS. 253 Approximate Value compared with True Value. Now the first and second approximate values, as given in art. 319, are, when reduced to a common denominator, a b -j- 1 ah a (a b -fl) a(aft+l) ; the difference of the numerators of which is 1 ; and, there- fore, unity must always be the difference of two such nu- merators. 322. Theorem. The approximate values of a con- inued fraction are alternately larger and smaller 'han its true value, the first being larger, the second smaller, and so on alternately. Demonstration. Since, in the preceding demonstration, the subtraction of the (n — l)st value from the (re — 2)nd, gave a fraction having the same numerator as that obtained, by its subtraction from the nth ; we see that if the (re — I )st value is larger than the (re — 2)nd, it must also be larger than the nth ; and if the (n — l)st is smaller than the (n — 2)nd, it is also smaller than the nth. But the true value is, by art. 319, nearer the (n — l)st value than the (re — 2)nd, and nearer the reth than the (re — l)st; so that when the (n — l)st value is larger than the (re — 2)nd, the true value must likewise be larger than the (re — 2)nd, and smaller than the (n — l)st, and so on alternately ; but when the (re — l)st value is smaller than the (re — 2)nd, the true value must be smaller than the (re — 2)nd, and larger than the (re — l)st, and so on alternately. Now the first value is, by the preceding article, larger than the second, and therefore the true value is smaller than the first, larger than the second, and so on alternately. 323. Theorem. Each approximate value of a continued fraction differs from the true value by a 22 25 4 ALGEBRA. [CH. IX. Transformation of a Quantity to a Continued Fraction. quantity less than the fraction whose numerator is unity y and whose denominator is the square of tht denominator of this approximate value. Demonstration. Let the denominator of the two succea sive approximate values be M 1 and N' ; N' must, by art. 320, be larger than M' ; and the difference between these two values must be 1 M 1 N>' But, by the preceding article, the true value is contained between these two approximate values, and therefore differs from either of them by a quantity less than their difference Now, since M' < N', we have and — > 1 Jl#'8 "* M'*^ M'N 1 ' bo that the true value must differ from the approximate value, whose denominator is M', by a quantity less than 1 JK'8* that is, less than a fraction whose numerator is 1, and de- nominator M ' 2 . 324. Problem. To transform any quantity into a continued fraction. Solution. Let X be the quantity to be trans- formed. Find the greatest integer contained in X, and denote it by A, and denote the excess of X above A by the fraction —j ; and we have ctf. XX.] CONTINUED FRACTIONS. 255 Approximate Value compared with True Value. A + ±=X, and 1 3/ = X—A' From this value of x', find the greatest integer con- tained in x', and denote it by a, and the excess of x' above a by —r, ; whence * x" 3/— a' from which the greatest integer contained in x" is to be found, and so on ; so that we have 1 X=A . 1 a ■ a'+&c. 325. EXAMPLE. Transform |ff into a continued fraction. Solution. We have, in this case, successively, A = 2; — = m. a = 1 ; *" = w. a' = 2; x»'=|f, a" = 1 ; ^ — 87 = 0"'* 256 ALGEBRA. [CH. IX Approximate Values of Fraction or Ratio. and the required continued fraction is *tt = 2 + r l +7TT 326. Corollary. The values of a', a", Sfc, in the ease of a vulgar fraction, are evidently the quotients which would be obtained by the process of finding the greatest common divisor of the numerator and denominator of x'. The preceding process might therefore be performed as follows : 263 ! 351 1 l = a 263 88|263|2: = a' 176 87188 1 1 = a" 87 1 1 87187 = 87 0' a"' 327. Corollary. If a fraction or ratio is trans- formed into a continued fraction by the preceding process, the approximate values of this continued fraction are also approximate values of the given fraction or ratio, which are often of great practical use. Thus the approximate values of §|-£j are 2 3 8 U • of which the last differs from the true value by only rl ^ 4 CH. IX.] CONTINUED FRACTIONS. 257 Approximate Values of Fraction or Katio. 328. EXAMPLES. 1. Find approximate values of the fraction f |i. Arts, i, |f, ff, and Itf. 2. Find approximate values of the fraction T f§£j. Arts. 2V, T 2 r , F VV. rih- tsWs' and i&W 3. Find approximate values of the fraction ^W^fV Ans. jV, A. s&ir; sVV slis. &c - 4. Find approximate values of the fraction 0-245. Ans. J and £f 5. Find approximate values of the fraction 127. Ans. |, I, If, ft, and |f 6. The lunar month consists of 27-321G61 days. Find approximate values for this time. Ans. 27, %?, V¥> 3 T W> &c. days, which show that the moon revolves about 3 times' in 82 days; or with greater accuracy, 28 limes in 765 days; and with still more accuracy, 143 times in 3907 days. 7. The sidereal revolution of Mercury is 879G9255 days. Find approximate values for this time. • Ans. 88, 2 -&p, &c. 8. The sidereal revolution of Venus is 224700817 days. Find approximate values for this time. Ans. 225, *$*. Mp, *f§J, «ff p, &c. 9. The ratio of the circumference of a circle to its diameter is 3141592G535. Find approximate values for this ratio. Ans. 3, V 2 , m, Hi, &0 - 258 ALGEBRA. [CH. IX. Approximate Roots of Equation. 329. Corollary. The process of art. 324 may be applied to finding the real roots of an equation, the approximate values of which, obtained by this pro- cess, can easily be reduced to decimals. 330. EXAMPLES. 1. Find the real root of the equation a? — 3 x — 8 = 0. Solution. We have, in this case, A = 2, and if we substitute •= a + y. in the given equation, we obtain 6^3 — 9a' 2 — 6a'— 1 = 0, whence we have a = 2; and the substitution of gives a"S — 30z"2 — 27*" — 6 = 0; whence we have a' = 32, and so on. The approximate values of x are, therefore, 2, 2J = 2-5, 2|| = 2-492, &c. 2. Find the real root of the equation z 3 — 12 x — 28 = 0. Ans. x = 4-30213. CI1. IX ] CONTINUED FRACTIONS. 259 Approximate Roots of Equation. 3. Find the real root of the equation a 3 — 12 a 2 -f 57 x — 94 = 0. Ans. x — 3-36216. 331. Corollary. If the given equation is a bino- mial one, as in art. 223, we can obtain, by this pro- cess, a root of any degree whatever. 332. EXAMPLES. 1. Extract the square root of 5 by means of continued fractions. Solution. Representing this root by x, we i have a 2 =5, whence 4 = 2; and the substitution of X =,+ • gives ar' a — 4 *' — 1 = ; whence we have a = 4 ; and the substitution of x' = 4 + 1 gives a;" 2 — 4x" — 1, which, being precisely the same with the equation for xf, we may conclude that 4 = a = a' = a" = a'" = &c. 260 ALGEBRA. [CH. IX Approximate Roots of Equation, and the approximating values are and the value in decimals is 2-23600. 2. Extract the third root of 46 by means of continued fractions. Ans. 3-58305. 3. Extract the third root of 35 by means of continued fractions. Ans. 3271. 4. Extrar' the square root of 2 by means of continued fractions. Ans. 1-4142136. EXPONENTIAL EQUATIONS AND LOGARITHMS. EXPONENTIAL EQUATIONS AND LOGARITHMS. SECTION I. EXPONENTIAL EQUATIONS 1. An Exponential Equation is one in which the unknown quantity occurs as an exponent. 2. Problem. To solve the exponential equation b" = m. Solution. This equation is readily solved by means of continued fractions, as explained in Alg. art. 324. 3. EXAMPLES. 1. Soke the equation 3" = 100. Solution. Since we have 3 4 = 81, and 3 s = 243, 2C4 EXPONENTIAL EQUATIONS. [§ I. Solution of Exponential Equations. the greatest integer contained in x must be 4. Substituting then ' %>' we have 3 4 r " = 100, or 3 4 3 I ' = 81 x3"= 100; and 3^= Vr°; which being raised to the power denoted by x', is /I00\x' [w) ■ By raising igOp to different powers, the greatest integer contained in x' is found to be 5. Substituting then x> = 5 + — ~ x" we have _ /lOOX 5 ^ /100\ s /100\^ 3 -U; = \sr) Vsi) : or Hence _ 10000000000 . /100\^ ' " 3486784401 X V~87/ ' fl 0460353203V" = ~, from which the greatest integer contained in x" is found to be 4 ; and in the same way we might continue the process The approximate values of x are, then, 4, 4i, 4^, = 4-19, &c. $ I.] EXPONENTIAL EQUATIONS. 265 Solution of Exponential Equations. 2. Find an approximate value for x, in the equation 3* = 15. Arts, x = 2-46. 3. Find an approximate value for x, in the equation 10* = 3. Arts, x = 0-477. 4. Find an approximate value for x, in the equation Ans. x= 0-53. 4. Corollary. Whenever the values of 6 and m are both larger or both smaller than unity, the value of x is positive. But when one of them is larger than unity while the other is smaller, the value of x must be negative ; for the positive power of a quantity larger than unity must be larger than unity, and the positive power of a quantity smaller than unity is smaller than unity ; whereas the negative power, being the reciprocal of the corresponding positive power, must be greater than unity, when the positive power is less than unity, and the reverse. ■ Hence to solve the equation b x = m, in which one of the quantities, b and in, is greater than unity, while the other is smaller than unity make x = — y, which gives or b~y = m, tt)'~ which may be solved as in the preceding article. S3 2G6 LOGARITHMS. [$ II. Positive and Negative Logarithms. 5. EXAMPLES. 1. Solve by approximation the equation Ans. x — — 0*25 2. Solve by approximation the equation 2- = *. Ans. xz= — 1-58. SECTION II. NATURE AND PROPERTIES OF LOGARITHMS. 6. The root of the equation b x = m is called the logarithm of m ; and since, by the pre- ceding section, this root can be found for any value which m may have, it follows that every number has a logarithm. The logarithm of a number is usually denoted by log. before it, or simply by the letter I. 7. But the value of the logarithm varies with the value of b, and therefore the. value of b, which is called the base of the systetn of logarithms, is of great importance ; and the logarithm of a number may be denned as the exponent of the power to which the base of the system must be raised in order to produce this number. § II.] NATURE AND PROPERTIES OF LOGARITHMS. 267 Logarithm of Product and of Power. 8. Corollary. When the base is less than unity, it follows, from art. 3, that the logarithms of all num- bers greater than unity are negative, while those of all numbers less than unity are positive. But when, as is almost always the case, the base is greater than unity, the logarithms of all numbers greater than unity are positive, while those of all numbers less than unity are negative. 9. Corollary. Since 6° = 1, it follows, that the logarithm of unity is zero in all systems. 10. Theorem. The sums of the logarithms of several numbers is the logarithm of their continued product. Proof. Let the numbers be m, m', m", &.C., and let 6 be the base of the system ; we have then b lo =- m = m, the product of which is, by art. 28, J log. m + log. mr + log. ,»"-(- ic. __ m ,„/ m „ &<._ Hence, by art. 7, log. m m! m" &c. = log. m -\- log. m! -\- log. m" -|- &c. 11. Corollary. If the number of the factors, m, ml, && is n, and if they are all equal to each other, we have log. mmm &c. = log. m -\- log. m -f- log. m -\- &c. or log. m n = n log. m ; 268 LOGARITHMS. [§ II. Logarithm of Root, Quotient, and Reciprocal. that is, the logarithm of any power of a number is equal to the logarithm of the number multiplied by the exponent of the power. . 12. Corollary. If we substitute p =. m n , or m= y/p, in the above equation, it becomes n log. p = n log. y/ p, or log- \/ p = — — ; n that is, the logarithm of any roof, of a number ts equal to the logarithm of the number divided by the exponent of the root. 13. Corollary. The equation log. m m' = log. m + log. m', gives log. ml = log. mm' — log. m ; that is, the logarithm of one factor of a product is equal to the logarithm of the product diminished hy the logarithm of the other factor ; or, in other words, The logarithm of the quotient is equal to the loga- rilhm of the dividend, diminished by the logarithm of the divisor. 14 Corollary. We have, by arts. 13 and 9, ] °g- — — log- 1 — log. n — — log. n ; that is, the logarithm of the reciprocal of a number is the negative of the logarithm of the number. § II.] NATURE AND PROPERTIES OF LOGARITHMS. 2G9 Logarithms in different Systems. 15. Corollary. Since zero is the reciprocal of in- finity, we have in log. = — log. qo = — oo ; that is, the logarithm of zero is negative infinity. 16. Corollary. Since we have 6 1 = 6, the logarithm of the base of a system is unity. 17. Theorem. If the logarithms of all numbers are calculated in a given system, they can be ob- tained for any other system by dividing the given logarithms by the logarithm of the base of the re- quired system taken in the given system. Demonstration. Let b be the base of the given system, and b' that of the required system ; and denote by log. the logarithms in the -given system, and by log. 1 the logarithms in the required system. Taking, then, any number m, we ■ have, by art. 7, 6 ,0 * " = m, and 6"°s-"» = m; whence If we take the logarithms of each member of this equation in the given system, we have, by arts. 11 and 16, log.' m X log. b' = log m X log. 6 = log. m, or, dividing by log. 6', log. m log.' m = .-^-r. 6 log. b' 23» 270 LOGARITHMS. [§ III. Logarithms of a Power of 10. SECTION III. COMMON LOGARITHMS AND THEIR USES. 18. The base of the system of logarithms in com- mon use is 10. 19. Corollary. Hence in common logarithms, we have by arts. 16 and 9, log 1 = 0, log. 10 = 1, log. 100 = log. 10 2 = 2, log. 1000 = log. 10 3 = 3, log. 10000 = log. 10 4 = 4, also, log. 01 = log. 10- 1 = — 1, log. 01 = log. 10~ 2 = — 2, log. 0001 = log. 10-s = — 3, &.c, &c. ; that is, the logarithm of a number, which is com- posed of a figure 1 and cyphers, is equal to the num- ber of places by which the figure 1 is removed from the place of units ; the logarithm being positive when the figure 1 is to the left of the units' place, and negative when it is to the right of the units' place. 20. Corollary. If, therefore, a number is between 1 and 10, its log. is between and 1, if between 10 and 100, it3 log. is between 1 and 2, if between 100 and 1000, its log. is between 2 and 3, and so on. § III.] COMMON LOGARITHMS. 271 To find the Logarithm of a given Number. But if between O'l and 1, its log. is between — 1 and 0, if between 00 1 and - 1, its log. is between — 2 and — 1 , and so on. Hence, if the greatest integer contained in a loga- rithm is called its characteristic, the characteristic of the logarithm of a number is equal to the num- ber of places by which its first significant figure on the left is removed from the units' place, the characteristic being positive when this figure is to the left of the units' 1 place, negative when it is to the right of the units' place, and zero when it is in the units' place. 21. Logarithms have been found of such great practical use, that much labor has been devoted to the calculation and correction of logarithmic tables. In the common tables they are given to 5, 6, or 7 places of decimals. In almost all cases, however, 5 places of decimals are sufficiently accurate ; and it is, therefore advisable to save unnecessary labor, and avoid an increased liability to error, by omitting the places which may be given beyond the first five. 22. Problem. To find the logarithm of a given number from the tables. Solution. First. Find the characteristic by the rule of art. 20. The characteristic is the most important part of the loga- rithm, and jet the unskilful are very apt to err in regard to 272 LOGARITHMS. [§ III. Finding a Logarithm. it, not appearing to consider that an error of a single unit in its value will give a result 10 times as great or as small as it should be. If the characteristic thus found is negative, Ike negative sign is usually placed above it, that this sign may not be referred to the decimal part of the logarithm, which is always positive. But calcula- tors are in the habit of avoiding the perplexity of a negative characteristic by subtracting its absolute value from 10, and writing the difference in its stead ; and, in the use of a logarithm so written, it must not be forgotten that it exceeds the true value by 10. Secondly. In finding the decimal part of the loga- rithm, the decimal point of the given number is to be wholly disregarded, and any cyphers which may precede its first significant figure on the left, or fol- low its last significant figure on the right are to be omitted. When the number thus simplified is contained within the limits of the tables, which we shall re- gard as extending to numbers consisting of four places, the decimal part of its logarithm is found in a horizontal line with its three first figures, and in the column below its fourth figure ; the second, third, and fourth figures, when wanting, being supposed to be cyphers. When the number consists of more than foui places, and is therefore, beyond the limits of the tables, point off its first four places on the left and $ III.] COMMON LOGARITHMS. 273 Finding a Logarithm. consider them as integers, regarding the other places as decimals. Care must be taken not to confound the decimal point thus introduced with the actual decimal point of the num- ber, of which it is altogether independent. Find, in the tables, the decimal logarithm corre- sponding to the integral part of the number thus pointed off ; and, also the difference between this logarithm and the one next above it, that is, the logarithm of the number which exceeds this integral part by unity ; this difference is often given in the margin of the tables. Multiply this difference by the decimal part of the number as last pointed off, and omit in the product as many places to the right' as there are places in this decimal part of the number. The product, thus reduced, being added to the decimal logarithm of the integral part of the num- ber, is the decimal part of the required logarithm. 23. Corollary. This process for finding the decimal part of the logarithm of a number, which exceeds the limits ol the tables, is founded on the following law, easily deduced from the inspection of the tables. If several numbers are nearly equal, their dif ferences are proportional to the differences of their logarithms. 24. EXAMPLES. 1 Find the logarithm of 00325787. 274 LOGARITHMS. [§ III Number corresponding to Logarithm. Solution. The characteristic is — 3, instead of which may be written 10 — 3 = 7. For the decimal part, the number is to be written 3257 87; aid we have log. 3258 — log. 3257 == 13 now, multiplying by -87 and omitting two places 91 on the right, 104 we have 1 1 which, added to log. 3257 = 51282 gives 51293; and the required logarithm is log. 000325787 = F51293, or, it may be written, 7-51293. 2. Find the logarithm of 1-8924. Ans. 027701. 3. Find the logarithm of 757-823000. Ans. 887956. 4. Find the logarithm of 000041359. Ans. TeiCS?, or G 61657 5. Find the logarithm of 0- 12345. Ans. T09149, or 909149. 6. Find the logarithm of 99998. A ns„ 4-99999. 25. Problem. To find the number corresponding to a given logarithm. Solution. First. In finding the figures of the required number, the characteristic is to be neg- lected. § III.] COMMON LOGARITHMS. 275 Number corresponding to Logarithm. When the decimal part of the given logarithm is exactly contained in the tables, its corresponding number can be immediately found by inspection. But when the given logarithm is not exactly con- tained in the tables, the number, corresponding to the logarithms of the table which is next below it, gives the four first places on the left of the required number. One or two more places are found by annexing one or two cyphers to the difference between the given logarithm and the logarithm of the tables next below it, and dividing by the difference between the logarithm of the tables next below and that next above the given logarithm. When tables are used in which the logarithms are given to five places, the accuracy of the corresponding numbers is never to be relied upon to more than 6 places, and rarely to more than 5 places; so that in finding the last quotient, one place is usually sufficient. Secondly. The position of the decimal point of the required number depends altogether upon the characteristic of the given logarithm, and is easily ascertained by the rule of art. 20 ; cyphers being prefixed or annexed when required. 26. EXAMPLES. 1. Find the number, whose logarithm is 8 - 19325. Solution. We have for the logarithm of the tables nea below the given logarithm •19312 = log. 1560. 276 LOGARITHMS. [§ III, Multiplication of Logarithms. Hence the diff. between given log. and log. 1560 = 13, also log. 1561 — log. 1560 = 28, and the quotient *f|P = 46 gives the two additional places; so that the six places of the required number are 156046 ; and the number is, therefore, 156046000. 2 Find the number, whose logarithm is 2-13511. Ans. 136493. 3. Find the number, whose logarithm is 1-76888. Ans. 58-7328. 4. Find the number, whose logarithm is 0-11111. Ans. 1-29153. 5. Find the number, whose logarithm is 2 T 98357. Ans. 00962875. 6. Find the number, whose logarithm, when written 10 more than it sho.ild be, is 9-35846. Ans. 22828. 27. Problem. To find the product of two or more factors by means of logarithms. Solution. Find the sum of the logarithms of the factors, and the number, of which this sum is the logarithm, is, by art. 1 0, the required product. When the logarithm of any of the factors is writ- ten, as in art. 22, 10 more than its true value, as many times 10 should be subtracted from the result as there are such logarithms. $ HI.] COMMON LOGARITHMS. 277 Involution by Logarithms. 28. EXAMPLES. 1. Find the continued product of 78-052, 0-6178, 341000, 100-008, and 00009. Solution. We find, from the tables, log. 78052 = 1-89238 10 -f- log. 0-6178 = 9-79085 log. 341000 = 5-53275 log. 100-008 = 2 00003 10 + log. 0-0009 = 6-95424 log. 1479960 frl7025 and the required product is 1479960. In the sum of the preceding logarithms 20 was neglected, because two of the logarithms were written 10 more than they should be. 2. Find the continued product of 00001, 7,9004, 56, 032569, and 17899- 1. Arts. 0-257792. 3. Find the continued product of 3-1416, 0-559, and 6401. Arts. 112-41 4. Find the continued product of 3-26, 00025, 25, and 0003. Arts. 0-00000611257. 29. Problem. To find any power of a given number by means of logarithms. Solution. Multiply the logarithm of the given number by the exponent of the required power, and 24 278 LOGARITHMS.. [§ III Evolution by Logarithms. the number, of which this product is the logarithm, is, by art. 11. the required power. When the logarithm of the given number is writ- ten 10 more than it should be, as many times 10 must be deducted from the product as there are units in the given exponent. 30. EXAMPLES. 1. Find the 4th power of 98573. Solution. We have, by the tables, 10 -f- log. 0-98573 = 9 99375 multiply by 10 + log. 0-94406 = 9 97500 and the required power is 0-94406. In the above product, 40 should have been neglected, but in order to avoid a negative characteristic, only 30 was neglected, leaving the exponent 10 too large. 2. Find the 3d power of 25. Ans. 0015625. 3. Find the 7th power of 3-1416. Ans. 3020-28. 4. Find the square of 0031422. Ans. 0-00000987325. 31. Problem. To find any root of a given num- ber by means of logarithms. Solution. Divide the logarithm of the given number by the exponent of the required root, and the number, of which this quotient is the logarithm, ii., by art. 12, the required root. § III.] COMMON LOGARITHMS. 279 Evolution by Logarithms. When the logarithm of the given number has a negative characteristic, instead of being increased by 10, it shoiild be increased by as many times 10 as there are units in the exponent of the root, and the quotient will in this case exceed its true value by 10. 32. EXAMPLES. 1. Find the fifth root of 0028 145. Solution. We have, by the tables, 50 + log. 028145 = 48-44940, which, divided by 5, gives 10 + log. 0-48964 = 9-68988, and the required root is 0-48964. 2. Find the cube root of 0-002197. Ans. 0-13. 3. Find the 10th root of 0-000000001. Ans. 12589. 4. Find the square root of 238149. Ans. 154317. 33. The arithmetical complement of a logarithm ib the remainder after subtracting it from 10. 34. Corollary. The arithmetical complement of the logarithm of a number is, by art. 14, and the preceding article, the logarithm of its reciprocal in- creased by 10. 35. Corollary. The most convenient method of finding the arithmetical complement of a logarithm is to subtract the first significant figure on the right from 10, and each figure to the left of this figure from 9. 280 LOGARITHMS. [$ III. Arithmetical Complement. 36. EXAMPLES. 1. Find the arithmetical complement of 9-62595. Ans. 0-37405. 2. Find the arithmetical complement of the logarithm of 6. . Ans. 9 22185. 3. Find the arithmetical complement of the logarithm of 007. Ans. 1115490. 4. Find the reciprocal of 0-01115. Solution. We have, by the tables, log. 0-01115 (ar. co.) 11-95273 subtract 10- log. 89-0S6 1-95273 and the required reciprocal is 89-686. 5. Find the reciprocal of 2330. Ans. 000042918. 6. Find the reciprocal of 68-99. Ans. 0014494. 37. Problem. To find the quotient of one number divided by another by means of logarithms. Solution. Subtract the logarithm of the divisor from that of the dividend, and the number, of which the remainder is the logarithm, is, by art. 13, the required quotient. Or, since, by art. 81, multiplying by the reciprocal of a number is the same as dividing by it, add the logarithm of the dividend to the arithmetical comple- ment of the logarithm of this divisor, and the sum diminished by 10 is the logarithm of the quotient. § III.] COMMON LOGARITHMS. 281 Division by Logarithms. Wheti the logarithm of the dividend is written 10 more than its true value, 20 must be subtracted from the sum, instead of 10. 38. EXAMPLES. 1. Divide 0-01478 by 0-9213. Solution. We have, by the tables, 10 -f log. 001478 8-1G967 log. 0-9243 (ar. co.) 10-03419 10 -f log. 01599 8-203S0 and the required quotient is 0-01599. 2. Divide 00815 by 00025. Ans. 3-26. 3. Divide 40-32 by 2240. Ans. 0018. 4. Divide 0-875 by 25. Ans. 035. 5. Divide 0013 by 013. Ans. 01. 39. Corollary. The value of any fraction may be found by adding together the logarithms of all the factors of the numerator and the arithmetical complements of the logarithms of all the factors of the denominator, and subtracting from the sum as many times 10 as there are arithmetical complements plus as many times 10 as there are logarithms of the factors of the numerator, which are written greater than their true value by 10 ; the remainder is the logarithm of the fraction. 282 LOGARITHMS. [§ 111. Various Examples of the use of Logarithms. 40. EXAMPLES. 1 Find the value of the fraction (0-32?)" x x/ 1981 (1-23)4 X (0 005) 2 Solution. We have, from the tables, 10 -j- log. (0-327)7 6-60185 log. y/ 19-S1 0-64844 log. (1-23)4 (ar. co.) 9-97003 log. (0 005) 2 (ar. co.) 14 60206 log. 66-433 1-82238 and the required value is 66-433. 2. Find the value of the fraction •S / 365 X x/ 2 \ k 788 )• Ans. 0-2308. 3. Find the value of the fraction V ( 347 X 6276 and we have x= 11555. 2. Find thu fourth term of the proportion 00138 : 0-319 = 765 : x. Ans. x = 1768-3. 43. Problem. To solve the exponential equation a 1 = m, by means of logarithms. Solution. The logarithms of the two members of this equation give x log. a = log. m ; hence log. m log. a or log. x = log. log. m — log. log. a ; that is, the root of this equation is equal to the logarithm of m divided by the logarithm of a, and this quotient may be obtained by the aid of loga- rithms. 284 LOGARITHMS. [§ III Exponential Equations. 44. EXAMPLES. 1. Solve the equation 625" = 3125. Solution. We have, from the tables, log. 3125 = 3-49485, log, 625 = 279588; and also log. log. 3125 = log. 3 49485 = 0-54343 log. log. 625 = log. 2-79588 = 0-44652 log. x = log. 1-25 61)9691 ; hence x= 1-25. 2. Solve the equation' — — — — 3* = 15. 3. Solve the equation 10» = 3. Arts, x = 2-464. Ans. x = 0-477 ERRATA. Page 8, line 2, for — x 2 3 — Va: read . — X 2 — 3 Vx. 5> » )) "» ,, " S' » Zg. „ 11, ,, 2, „ those „ that. » 16, „ 6, „ art. 30 „ art. 36. „ „ last line, „ fractional „ negative- ,, 17, ex. 11, „ by-» by a — n . „ 22, line 12, „ a n ~b n „ a n — J» „ 32, „ 2, after algebraic work, ; interchange « Col. 1 " and " Col. 2." „ 33, ex. 7, for -f- 3 a 3 x 3 read -f 3 a? x 5 . „ 37, line 10, omit unlike. „ 43, last line, for A X -B ,, AXD. „ 59, art. 106, 1. 4, for greater „ greatest. Page 60, art. 109, add: But if a factor which may equal zero is multiplied into an equation or taken away from it, the possible solutions of the equation are increased or diminished. Thus x — 1 = has only one root, namely 1. If the equa- tion be multiplied by x, we obtain x 2 — x = 0, which has two roots, 1 and 0. If we next divide by a; — 1, we obtain x = 0, which no longer has the root 1. Page 60, art. 110, add : after taking account of any solu- tion of the equation which would be furnished by putting that factor equal to zero. Page 64, ex. 2, ans. for 2 x 2 read 3 x 2 . „ 120, line 3, „ of e 2 „ of e. „ ;, art. 182, 1. 7, „ final „ first. 286 • ERRATA. Page 123, ex. 2, the answer should be — 5.871947344. „ „ ex. 3, the second answer should be — 4.350415. „ „ ex. 4, the answers should be 1.092411 and, 1.591428. „ „ art. 186, for e read c. „ 124, lines 8, 9, „ decimal point „ unit's place. n 125, ex. 5, „ 20548344701 „ 205483447701. » 126, art. 189, 33 " 33 e " n 132, ex. 4, ans. „ 4 d? z 4 „ 2 d 3 z 4 . j? 134, line 17, „ 3 ^2* a 3 ¥ c „3^2*a3 55 c 3 » 143, „ -2, for »(*-l) rf '-2a? read n ^~^ a»- 2 a: 2 . 'ag e 144, line 4, for a™ — 3 a£ read a" — 3 a?. 3) 146, „ 7, 33 »• 33 ?• 33 v » 21> „ dTpvqft + ir „ QTpiq -ir. 33 9 3) 33 u t „ 172 „ 174. 5) 152, ex. 4, „ — 144 x y* „ — 144 xy 5 . 3) 153, ex. 6, 33 — 24 § 1 - a a 4 i7 c „ _|_ aigjji a 4 ii e# 33 162, line 16, „^^ „ ^iK 33 164, lines 19, , 20, for h „ e. 33 165, line 1, „ fi „ e. „ „ -5, for 6 X 128 + 3 X 1 read 128+3 X 1. 167, „ 3, „ -f 40z 2 „ -f 40*. 168, „ -11,-2, for Vl71 „ V271. 170, ex. 17, the second answer should be -|- .2641. „ ex. 20, ans. for Vl5 read */ — 15. 172, ex. 31, ans. for (2 *Jh — a?) „ V(25 — a 2 ). 181, ex. 5, ans. „ or =-\- 4 „ or = — 4. 182, line 13, „ +6xtf „ —6xf. „ „ „for(s« — 5y + 4 „ (2,2 — 5^ + 4). 196, „ 15, „ ar- n „ ar n . 198, ex. 9, aw*, for I (r™ — *) „ / (r» — 1). EEEATA. 287 Page 206, line 12, for £[_i_V5± V(— 10 + 2-v/o)] read £ [— 1 -f- s/h ± / -\ , there are two real roots. Page 232, line 10, for positive read negative. „ 240, ex. 3, „ + 396 „ .— 396: „ „ line -8, „ 1814600 „ 1814400. „ „ „ -7, „ 362800 „ 3628800. » 241, „ 4, read —.+,—,—,—,—,—,—,— ,+,+. „ „ „ 5, for three read two. „ „ lines 6-9, omit " and since more than 2." 255, line -7, for x" read x'. 267, » 2, „ art. 3 „ art. 4. ?) » 13, „ sums „ sum. 274, ex. 3, ans. „ 8.87956 „ 2.87956. 276, ex. 4, arcs. „ 1.29153 „ 1.29155. 277, ex. 2, „ 7,9004 „ 7.9004. „ 0.257792 „ 0.25792. 278, ex. 4, a»s. „ 987325 „ 987375. . ■ ;'■' ■■■■■, " " mm