u l^^ 1^ ^ % 9^ ^^ -rdJia^ y^c 6/Z ADVANCED COURSE IN ALGEBRA BY WEBSTER WELLS, S.B. PROFESSOR OF MATHEMATICS IN THE MASSACHUSETTS INSTITUTE OF TECHNOLOGY BOSTON, U.S.A. D. C. HEATH & CO., PUBLISHERS 1904 mVS Copyright, 1904, By WEBSTER WELLS. • •- ( c * C C t « . • • • tc • • • «,c PREFACE In preparing the present work, the author has endeavored to meet the needs of Colleges and Scientific Schools of the highest rank. The development of the subject follows in the main the author's College Algebra; but numerous improvements have been introduced. Attention is especially invited to the following : 1. The development of the fundamental laws of Algebra for the positive and negative integer, the positive and negative fraction, and zero, in Chaps. I and II. In the above treatment, the author has followed to a certain extent The Number System of Algebra, by Professor H. B. Fine ; who has very courteously permitted this use of his treatise. 2. The development of the principles of equivalence of equations, and systems of equations, both linear and of higher degrees; see §§ 116-123, 182, 233-6, 396, 442, 470, 477, and 478. 3. The prominence given to graphical representation. In Chap. XIV, the student learns how to obtain the graphs of linear equations with two unknown numbers, and of linear expressions with one unknown number. He also learns how to represent graphically the solution of a system of two linear equations, involving two unknown numbers, and sees how inde- terminate and inconsistent systems are represented graphically. The graphical representation of quadratic expressions, with one unknown number, is taken up in § 465 ; and, in § 467, the graphical representation of equal and imaginary roots. 2S1763 iv PREFACE The principles are further developed for simultaneous quad- ratics, in §§ 482 and 483; and for expressions of any degree, with one unknown number, in §§ 744 and 745. At the end of Chap. XVIII, the student is taught the graphi- cal representation of the fundamental laws of Algebra for pure imaginary and complex numbers. In Chap. XXXVII, the graphical representation is given of Derivatives (§ 751), of Multiple Boots (§ 755), of Sturm's Theorem (§ 762), and of a Discontinuous Function (§ 766). 4. In Chap. VII, there are given the Remainder and Factor Theorems, and the principles of Symmetry. 5. In Chap VIII will be found every method of factoring which can be done advantageously by inspection, including factoring of symmetrical expressions. In this chapter is also given Solution of Equations by Factoring (§ 182). 6. In the earlier portions of Chap. XI, the pupil is shown that additional solutions are introduced by multiplying a fractional equation by an expression which is not the L.C.M. of the given denominators ; and is shown how such additional solutions are discovered. 7. In §§ 264 and 265, the student is taught how to find the values of expressions taking the indeterminate forms ^j — > Ox 00, and oo — oo. 8. All work coming under the head of the Binomial Theo- rem for positive integral exponents is taken up in the chapter on Involution. 9. In developing the principles of Evolution, all roots are restricted to their principal values. 10. In the examples of § 398, the pupil is taught to reject all solutions which do not satisfy the given equation, when the roots have their principal values. 11. The development of the theory of the Irrational Num- ber, and its graphical representation (§§ 399-406). 12. The development of the fundamental laws of Algebra for Pure Imaginary and Complex Numbers (Chap. XVIII). PREFACE V 13. The use of the general form ax^ + 6a; + c = 0, in the theory of quadratic equations (§§ 454-6). 14. The discussion of the maxima and minima values of quadratic expressions (§ 461). 15. The chapter on Convergency and Divergency of Series (Chap. XXVI). 16. In Chap. XXVIII is given Euler's proof of the Binomial Theorem, for any Eational Exponent. 17. The solution of logarithmic equations (§ 604). 18. The proof of the formula for the number of permuta- tions of n different things, taken r at a time (§ 624). 19. The chapter on Theory of Numbers (Chap. XXXV). 20. In the chapter on Determinants, the double-suffix nota- tion is used only in demonstrations which would not otherwise be complete. Each demonstration of a general principle is preceded by an illustration showing the truth of the principle for a deter- minant of the third order. Multiplication of determinants is taken up only for deter- minants of the second and third orders. 21. In Chap. XXXVII will be found Symmetrical Eunctions of the Eoots (§ 721) ; a shorter proof of Descartes' Rule (§ 735) ; improved methods for finding limits to the roots (§§ 739, 740); the demonstration of two theorems used in the proof of Sturm's Theorem (§ 757) ; and a discussion of Con- tinuous Functions (§ 765). 22. Chap. XXXVIII contains the solution of Cubic Equa- tions by Trigonometry, in Cardan's Irreducible Case (§ 788) ; also, an improved discussion of Newton's Method for deter- mining incommensurable roots (§ 801). The examples and problems have been selected with great care, and include many varieties not found in the College Algebra; no example is a duplicate of any in the College Algebra. The manuscript was read in the most careful manner by Professor George D. Olds, of Amherst College, who offered VI PREFACE many suggestions; these have added materially to the value of the work. The author would be under great obligations to any one who will bring to his attention any error which may be found in the book. WEBSTER WELLS. Boston, 1904. CONTENTS CHAPTER PAGE I. Definitions. Notation. Positive Integers . . 1 II. Rational Numbers 11 III. Addition and Subtraction of Algebraic Expressions. Parentheses 23 N IV. Multiplication of Algebraic Expressions ... 32 V. J^wisiON OF Algebraic Expressions .... 3^ VI. Integral Linear Equations 46 VII. Special Methods in Multiplication and Division . 59 "^ VIII. Factoring 76 IX. Highest Common Factor. Lowest Common Multiple . 99 -^ X. Fractions . 108 XI. Fractional and Literal Equations .... 122 XII. Simultaneous Linear Equations 133 XIII. Discussion of Linear Equations 149 XIV. Graphical Representation 166 XV. Involution and Evolution 173 XVI. lNEQUALmE8_ 206 XVII. Surds. Theory of Exponents 212 XVIII. Pure Imaginary, and Complex Numbers . . . 246 XIX. Quadratic Equations 261 XX. Equations solved like Quadratics .... 292 XXI. Simultaneous Quadratic Equations .... 298 XXII. Indeterminate Linear Equat10"ns 323 v/. XXIIL Ratio and Proportion . 328 XXIV. Variation 338 XXV. Progressions 343 XXVI. CONVERGENCY AND DIVERGENCY OF SeRIES . . . 362 \ vii VIU CONTENTS CIIAPTEK PAGE XXVII. Undetermined Coefficients 378 XXVIII. The Binomial Theobem 392 XXIX. Logarithms . . . .' 399 XXX. Compound Interest and Annuities .... 422 XXXI. Permutations and Combinations .... 428 XXXII. Probability . . 436 "* XXXIII. Continued Fractions ; 449 XXXIV. Summation of Series 459 XXXV. Theory of Numbers 471 XXXVI. Determinants 478 XXXVII. Theory of Equations 502 XXXVIII. Solution of Higher Equations . . . , 545 Appendix. Cauchy's Proof that every Equation has a Root 576 Answers . rM^:xtJ^ ^ -Ti ALGEBRA I. DEFINITIONS. NOTATION. POSITIVE INTEGERS 1. In Algebra, the operations of Arithmetic are abridged and generalized by means of Symbols. SYMBOLS REPRESENTING NUMBERS 2. The symbols usually employed to represent numbers are the Arabic Numerals, and the Letters of the Alphabet. The numerals denote known or determinate numbers. The letters denote numbers which may have any values whatever, or numbers whose values are to be determined. Numbers occupying similar relations in the same investiga- tion are often represented by the same letter, distinguished by different accents; as a', a", a'", read " a prime /^ "a second/^ " a third," etc. They may also be distinguished by different subscript num- bers; as tti, ttg, ttg, read "a sub one," "a sub two," "a sub thi-ee" etc. SYMBOLS REPRESENTING OPERATIONS 3. The Sign of Addition, -f , is read '^plus." The Sign of Subtraction, — , is read ^' minus." The Sign of Multiplication, x, is read "times," "into," or " multiplied by." A point is sometimes used instead of the sign x ; thus, 2.3.4 signifies 2x3x4. The Sign of Division, ^^ is read " divided by." 2 ADVANCED COURSE IN ALGEBRA SYMBOLS OF RELATION 4. The Sign of Equality, =, is read "equals,^^ or 'Hs equal to." The sign =^ is sometimes used for the words " is not equal to.'''' The Signs of Inequality, > and <, stand for "is greater than'^ and "is less than,'^ respectively. The signs ;5^ and 5C are sometimes used for the words " is not greater than'''' and "is not less than^''"' respectively. SYMBOLS OF ABBREVIATION 5. The Signs of Aggregation, the parentheses ( ), the brackets [ ], the braces j \, and the vmculiiyn , indicate that what is enclosed by them is to be taken as a whole. The Sign of Deduction, .-., is read "therefore" or "hence." The Sign of Continuation, •••, is read "and so on." THE POSITIVE INTEGER 6. By the number of things in a group, we mean that attri- bute of the group which remains unchanged however the group may be changed, provided no thing is divided into two or more things, and that two or more things are not merged into a single thing. That is, the number of things in a group is independent of their character, of the order in which they may be arranged, and of the way in which^they may be associated in smaller groups. 7. The numbers of things in two groups are said to be equal when for every thing in either group there is a thing in the other. 8. The number of things in one group is said to be greater than the number in another, or the number in the second group less than the number in the first, when for every thing in the second group there is a thing in the first, but not one in the second for every thing in the first. . . POSITIVE INTEGERS 3 9. The Positive Integer. We define a positive integer as the number of things in a group. A positive integer is also called a whole number. To ensure generality in the results, we represent numbers by- letters. In the remainder of the present chapter, the letters a, h, c, etc., will be understood as representing positive integers. 10. If a and h stand for the numbers in any two groups (that is, for any two positive integers), we use the statement a = 6 to signify that the numbers are equal (§ 7). The statement a = & is called an Equation. Again, we use the statements a> b, and ah-[-Cj then a > 6. If a-\-c —J and =. 26. Equations (2) and (6), § 24, show that a set of subtrac- tions, or of additions and subtractions, can be performed in any order. Equation (4) shows that addition is the inverse of subtraction. Equations (1), (3), and (5), with § 12, II, give complete associative laws for addition and subtraction. Equation (7), with § 14, III, give a complete distributive for multiplication. POSITIVE INTEGERS 9 DIVISION OF POSITIVE INTEGERS 27. We define Division of positive integers as the process of finding one of two positive integers (the Quotient), which when multiplied by another positive integer (the Divisor), gives a third positive integer (the Dividend). Thus, division is the inverse of multiplication. 28. The quotient when a is divided by b is expressed a-i-b, a or-. Since, by the definition of § 27, the product of the quotient by the divisor gives the dividend, we have (1) to c terms, we must ©'■ = a. 29. If ac = be, then a = b. For, if a + aH to c terms = = 6 + 6+... have a = &. 30. Rules for. Division. ^ ^ b' d~bd For by § 14, I and II, a c b' d' bd = b. But if we agree to define an equation as simply a statement that tico symbols, or combinations of symbols, are of such a char- acter that one m,ay be substituted for the other in any operation, then (1), § 21, may be an equation whatever the values of a and b. In this symbolic definition of an equation, it is unnecessary that there should be any real things to which the symbols correspond. We shall attach this meaning to every equation throughout the remainder of the work which does not express the equality of two positive integers. 34. Symbolic Subtraction. If we regard equation (1), § 21, as defining a — b, whatever the values of a and b, we have in this way a symbolic definition of subtraction which holds universally. 12 ADVANCED COURSE IN ALGEBRA This defines subtraction in terms of symbolic addition; for the sign -\- cannot indicate numerical addition, unless the symbols which it connects are positive integers. It is perfectly logical to define an operation by means of an equation. ZERO AND THE NEGATIVE INTEGER 35. In determining the definitions and rules of operation of zero and the negative integer, we make the assumption that the results of §§ 12, 14, 22, and 23 hold for these symbols. If all the letters do not represent positive integers, the results of §§ 12, 14, 22, and 23 are regarded as symbolic statements. 36. Since the results of § 24 are simply formal consequences of §§ 12, 14, and 22, and the definition of subtraction, it follows from § 35 that they hold for the above symbols. If the results of § 24 do not have a positive integral inter- pretation, they are regarded as symbolic statements. In this way they become definitions of symbolic addition, subtraction, and multiplication, and their relations. 37. Zero. Every letter in §§ 37 to 42, inclusive, will be understood as representing a positive integer. Putting 6 = a, in (1), § 28, we have (a — a) -j- a = a. (1) The symbol a — a, if a is any positive integer, is represented by the symbol 0, called zero. Then (1) becomes + a = a. (2) Since the Commutative Law for Addition (§ 12) is assumed to hold if either letter equals (§ 35), we may write equa- ^^o^ (2), a + = a. (3) Again, by definition, a — means a symbol such that when is added to it, the sum is a. That is, (a — 0) -f- = a. RATIONAL NUMBERS 13 Then, by (3), a - = a. (4) Again, by the definition of 0, axO = a(b-b) = ab- ab, by § 24, (7), = 0, by definition. (5) We can use § 24, (7), in the above proof ; for we know from § 36 that it holds, even if the result does not have a positive integral interpretation. From (5), by the Commutative Law for Multiplication, X a = 0. (6) 38. The Negative Integer. Let b be greater than a ; and suppose b = a-{- d, where d is a positive integer. Then, by the definition of subtraction, b — a = d. Then, a — b = a — (a + d) = a — a — d,hj^24:, (1), = — d, by the definition of 0. We can use § 24, (1), in the above, for we know that it holds, even if the result does not have a positive integral interpretation. We then define a — 6, if & is > a, as being equal to — d. It is usual to write — d instead of — d Thus, 0-d = -d. (1) The symbol — d is called a Negative Integer ; in contradis- tinction, the positive integer d may be written + d. 39. The signs + and — , when used in the above manner, are no longer signs of operation; they are called signs of Affection^ Quality, or Opposition. If no sign is written, the sign -}- is understood. 40. Rules for Addition, Subtraction, and Multiplication, involv- ing Negative Integers. (1) a-^(-b) = a-b. For, by the definition of § 38, a + (-&) = « + (0 - Z>) = a + - &, by § 24, (5), = a-b,hy ^ 37, (3). 14 ADVANCED COURSE IN ALGEBRA (2) -a+(-b)=-(a + b). For, —a+{—b)= — a+{0 — b)=—a + — b = - a - 6, by § 12, I, = - (a + ?^), by § 24, (1), = - (a + ^>), by § 38. (3) a-(-b) = a-\-b. For, a - (-5) = a - (0 - 6) = tt - + &, by § 24, (3), = a + 6, by §37,(4), (4) -a-(-b) = -a + b = b-a. For, - a - (- 6) = _ a - (0 - 6) = - a - + &= - a+6-0 = -« + &, by §37, (4), = 6 - a, by § 12, I. Putting b for a in (1) and (4), we have (5) 6-f (-&) = 0. (6) -j)^b = 0. (7) _^,_(_6)=0. (8) a(-6) = -a6. By § 37, (5), = a X = a[6 + (- &)], by (5), = o6 + «(-&), by § 14,111. Then, ab — ab = ab -\-a(— b). Whence, by § 22, - a6 = a (- 6). (9) (-b)a = -ab = -ba. This follows from (8) by § 14, I. (10) (-a) xO = 0. For, by the definition of 0, (- a) X = (- a) (b-b) = (-a)b- (- a)b, by § 24, (7), = (_a6)-(-a6),by(9), = 0,by(7). (11) Ox(-a) = 0. This follows from (10) by § 14, I. RATIONAL NUMBERS 15 (12) (-a)(-b) = ab. By (10), 0=(-a)x0 = (-a)[& + (-&)],by(5), = (- a) 6 + (- a) (- b), by § 14, III, = _a&4-(-a)(-6), by (9). Then by (6), - a& + a& = - a6 + (- a) (- 6). Then by § 22, ^6 = (- a) (- 6). 41. If 6 is > a, + 6 > + a, by § 37, (2). Then, _a + a + 6>-5 + & + a, by§40, (6). Then by §23, -a>-b. (1) In like manner, if 6 is < a, — a is < — b. These may be regarded as defining greater and less inequality *in negative integers. 42. If a and b are positive integers, a a, a — b = —(b — a)', for the result of § 38 also holds when either a or 6 is a positive fraction (§ 47). Then, a-}- (-b) = - (b-a). (2) From (1) and (2), we have the following rule : To add a positive and a negative number, subtract the smaller absolute value (§ 53) fro7n the greater, and place before the result the sign of the number having the greater absolute value. Thus, 5i + ( - 3i) = If; 2 + ( - 5) = - 3. 68. Addition of Negative Numbers. If a and b represent positive numbers, we have by § 40, (2), (-a) + (-6) = -(a + &). We then have the following rule : To add two negative numbers, add their absolute values, and put a negative sign before the result. Thus, (-lJ) + (-2^)=-3i|. 69. Addition of Monomials. The sum of a and b is a-{-b', and by § 40, (1), the sum of a and —bisa — b; hence. The addition of monomials is effected by uniting them with their respective signs. 24 ADVANCED COURSE IN ALGEBRA Thus, the sum of a, —b,c,— d, and — e is a — b-\-c — d — e. Since the Commutative Law for Addition holds for any- rational numbers (§§ 47, 51), the terms may be united in any order, provided each has its proper sign. 70. Definitions. If two or more numbers are multiplied together, each of them, or the product of any number of them, is called a Factor of the product. Thus, a, b, c, ab, ac, and be are factors of the product abc. 71. If a number be expressed as the product of two factors, each is called the Coefficient of the other. Thus in 2 ab, 2 is the coefficient of a6; 2 a of 6; a of 2 &; etc. 72. If one factor of a product is expressed in Arabic numerals, and the other in letters, the former is called the numerical coefficient of the latter. Thus in 2 ab, 2 is the numerical coefficient of ab. If no numerical coefficient is expressed, the coefficient unity is understood. Thus, a is the same as 1 a. By § 47, the result (9), § 40, also holds when 6 is a positive fraction. 2 That is, — 3 a is the product of — 3 and a, and —-ab is the 2 ^ product of — - and ab. Then, — 3 is the numerical coefficient of a in — 3 a, and 2 . . .2 — -is the numerical coefficient of ab in —-ab. o o Thus, in a negative term (§ 61), the numerical coefficient includes the sign. 73. Similar or Like Terms are those which either do not differ at all, or differ only in their numerical coefficients ; as 2 x^y and — 7 x^y. Dissimilar or Unlike Terms are those which are not similar ; as 3 x^y and 3 xy"^. ADDITION 25 74. Addition of Similar Terms. 1. Find the sum of 5 a and 3 a. Since the Distributive Law for Multiplication (§ 14, III) holds for any rational numbers (§ 47), we have 5 a 4- 3 a == (5 + 3) a = 8 a. 2. Find the sum of 5 a and — 3 a. By §72, 5a + (-3a) = 5a + (-3)a ^ = [5 + (_3)]a ' (§14,111) = 2 a. (§ 67) 3. Find the sum of — 5 a and 3 a. ( - 5) a + 3 a = [( - 5) + 3] a = - 2 a. (§67) 4. Find the sum of — 5 a and — 3 a. (_5)a + (-3)a=[(-5)4-(-3)]a = -8a. (§68) Therefore, to add tvjo similar terms, find the sum of their numerical coefficients (§§ 67, 68, 72), and affix to the result the common letters. 5. Find the sum of 2 a, — a, 'da, — 12 a, and 6 a. Since the additions may be performed in any order, we may add the positive terms first, and then the negative terms, and finally combine these two results. The sum of 2 a, 3 a, and 6 a is 11 a. The sum of — a and — 12 a is — 13 a. Hence, the required sum is 11 a + (— 13 a), or —2 a. 6. Add 3(a-6), -2(a-h), 6(a-b), and -4(a-6). The sum of S(a-b) and 6(a-b) is 9(a-b). The sum of — 2 (a — 6) and — 4 (a — 6) is — 6 (a — b). Then, the result is [ 9 + (- 6)] (a - b), or 3 (a - b). 75. If the terms are not all similar, we may combine the similar terms, and unite the others with their respective signs (§ 69). Ex. Required the sum of 12 a, —5 a;, —Sy^, —5 a, Sx, and — 3 a;. 26 ADVANCED COURSE IN ALGEBRA The sum of 12 a and — 5 a is 7 a. The sum of —^x, S-a:, and —3 a; is 0; for the result, (5), § 40, holds for any value of b (§ 47). Hence, the required sum is 7 a — 3 y^. 76. A polynomial is said to be arranged according to the descending powers of any letter, when the term containing the highest power of that letter is placed first, that having the next lower immediately after, and so on. Thus, x* + 3oc^y-2 xY + 3 xy^ -4.y^ is arranged according to the descending powers of x. The term —4 ?/*, which does not involve x at all, is regarded as contain- ing the lowest power of x in the above expression. A polynomial is said to be arranged according to the ascend- mg powers of any letter, when the term containing the lowest power of that letter is placed first, that having the next higher immediately after, and so on. Thus, a;* + 3a;3?/-2a^/-f-3a;/-4?/* is arranged according to the ascending powers of y. 11. Addition of Polynomials. It follows from § 12, II, and § 24, (5), that the addition of polynomials is effected by uniting their terms with their respective signs. 1. Required the sum of 6 a — 7 a^, 3 a;^ — 2 a + 3 2/^ and 2 a^ — a — mn. We set the expressions down one underneath the other, similar terms being in the same vertical column. We then find the sum of the terms in each column, and write the results with their respective signs ; thus, 6a-7a^ -2a + 3a;2_^3^3 — a-\-2Qi? — mn 3a — 2 a;^ 4-32/^ — mn. ADDITION. SUBTRACTION 27 2. Add 4a;-3aj2-ll + 5aj3, 12aj2 _ 7 _ 8a^- 15a;, and U-\-(ja^-\-10x-9x\ It is convenient to arrange each expression in descending powers of x (§ 76) ; thns, 5x^- Sx^-^- 4a;-ll 6a^_ 9a;2 + 10« + 14 and 3oi? - X- 4:. 3. Add 9(a + 6)- -8(6 + c), _3(6 + c)- -7(c + a), 4(c-f -a)- 5(a + 6). 9(a + 6) - ■ 8(6 + c) - ■ 3(6 + c)-7(c + a) — 6(c t + 6) + 4(c + a) 4(a + 6)- -ll{b-{-c)-S(c + a). 4. Add 3 , 2, 4" + 5''- loandia-|6 + |o. !- i"-\' ^»- ¥*!' 11 14^, 8 „ SUBTRACTION 78. Subtraction of Monomials. By § 47, the result (3), § 40, holds for any values of a and 6. Hence, to subtract a monomial, we change its sign and add the result to the minuend. 1. Subtract 5 a from 2 a. Changing the sign of the subtrahend, and adding the result to the minuend, we have 2a-5a = 2a + (-5a)=-Sa (§74). 28 ADVANCED COURSE IN ALGEBRA 2. Subtract — 2 a from 5 a. 5a — (—2a) = 5a-\-2a = 7a. 3. Subtract 5 a from — 2 a. — 2a — 5a = — 7 a. 4. Subtract — 5 a from —2 a. -2a-(-5a) = -2a + 5a = 3a. 5. From — 23 a take the sum of 19 a and — 5 a. It is convenient to change the sign of each expression which is to be subtracted^ and then add the results. We then have — 23 a — 19 a + 5 a, or — 37 a. * 79. Subtraction of Polynomials. By § 36, (1) and (3), § 24, hold for any values of the letters. Hence, to subtract a polynomial, we chayige the sig^i of. each of its terms, and add the result to the minuend. 1. Subtract 7ab^-9 a'b + 8 6^ from 5a^-2a^b + 4. ab\ It is convenient to place the subtrahend under the minuend, so that similar terms shall be in the same vertical column. We then change the sign of each term of the subtrahend, and add the result to the minuend ; thus, 5a3-2a26 + 4a&2 -9a^6 + 7a?>^ + 8?>^ 5a^-{-la^b-'6ab''-%b\ The student should perform mentally the operation of changing the sign of each term of the subtrahend. 2. Subtract the sum of ^x^ — %x-\-x^ and 5 — a?-\-x from 6a^-7x-4. We change the sign of each expression which is to be sub- tracted, and add the results. Qx^ -7x-4: ~x^-9x^ + Sx + x^ — X — 5 SUBTRACTION. PARENTHESES 29 80. By § 78, subtracting + a is the same thing as adding — a, and subtracting — a the same thing as adding + a. That is, — (h- a) = 4- (— a), and — (— a) = + (+ a). In these results, the signs within the parentheses are signs of affection (§ 39), and those without signs of operation. Comparing the results with those of § 58, where all the signs are signs of affection, we see that the signs + and — , when used as signs of affection, are subject to the same laws as when used as signs of operation. Thus the meaning attached to the signs + and — , in § 39, is consistent with their meaning as symbols of operation. PARENTHESES 81. Removal of Parentheses. It follows from § 12, II, and § 24, (5), that: ' Parentheses preceded by a -\- sign may he removed without changing the signs of the terms enclosed. Again, it follows from § 24, (1) and (3), that : Parentheses preceded by a — sign may be removed if the sign of each term enclosed be changed, from -{- to —, or from — to -{-. The above rules apply equally to the removal of the brackets, braces, or vinculum (§ 5). It should be noticed, in the case of the latter, that the sign apparently prefixed to the first term underneath is in reality prefixed to the vinculum. Thus, + a — & and — a — b are equivalent to + (a — 6) and — (« — 6), respectively. Parentheses often enclose others ; in this case they may be removed in succession by the rules of § 81. Beginners should remove one at a time, commencing with the inner- most pair ; but after a little practice they should be able to remove several signs of aggregation at one operation, in which case they should commence with the outermost pair. Ex. Simplify 4:X — \3x-]-{—2x — x-a)}. We remove the vinculum first, then the parentheses, and finally the braces. 30 ADVANCED COURSE IN ALGEBRA Thus, 4:X — l3x-{-(—2x — x — a)l = 4: X — \S X -\- (— 2 X — X -{- a)l = 4:X — [3 X — 2 X — X -{- a] = 4:X — 3x-\-2x-{-x — a==4:X — a. 82. Insertion of Parentheses. To enclose terms in parentheses, we take the converse of the rules of § 81. Any number of terms may be enclosed in parentheses preceded by a -\- sign, without changing their signs. Any number of terms may be enclosed in parentheses preceded by a — sign, if the sign of each term be changed, from + to — , or from — to -\-. Ex. Enclose the last three terms of a — b + c — d + e in parentheses preceded by a — sign. Result, a — b — (—c-{-d — e).- EXERCISE I 1. Add5(« + 6), -4(a;-y), -6(a + &), S{x-y), -7(x-y), and 8(a + b). 2. Add 7 w2 - 2p - 8 w3, 5 n^ - m^ and 3 xy - 4 m'^ + 2 nK 3. Add a - 9 - 8 aM- 16 a^, 5 + 15 a3 - 12 a - 2 a% and 6 a2 - 10 a^ -f- 11 a - 13. 4. AddU{x + y) - 17 (y + z), i(y + z)- Id^z + x), and -1{z + x) -Six + y). 5. Add^x-^y-^z, -lx + y + ^z,^ud~'lx-^y + lz. 6. Subtract the sum of 8(w + n) and - 15(m + n) from - 19 (w + n). 7. Subtract Sb-6d- 10 c + 7a from 4 (Z + 12 a - 13 c - 9 6. 8. Subtract 41 a:^ - 2 x^ + 13 from 15 a;^ + x - 18. 9. Subtract -p--m-- n from -m--p--n. 3^ 2 4 5 7^ 3 10. Subtract the sum of 4 x'^ - 9 ^/^ + 6 ^2 and 2 a;^ + 8 y2 _ n ^2 from 7 X2 - 3 ?/2 - 5 5!2. 11. From the sum of 2a + 36-4c and 3& + 4c-5(? subtract the sum of 5c-6d — 7a and — 7d! + 8a + 9 6. PARENTHESES 31 Simplify the following : 12. 9m- (3 n + {4 m - [w - 6 ?7i]} - [w + 7 w]). 13. 2a + (-6 6-{3c + (-4&-6c + a)}]. 14. 7 X - {- 6 X - {- 6 X ~ [- 4 X - S X - 2]}). 15. 6n-[Sn-(Sn + 6)-{-6n-\-7n- 5}]. 16. 4a- [a-{-7a-(8a-5a + 3)-(-6a-2a-9)}]. 17. x-{-12y-l2x-\-(-4y-{-lx-6y}-6x-9y)-8x-\-y^}. 18. Enclose the last three terms of a ■}- b — c + d — e in parentheses preceded by a — sign, and in the result enclose the last two terms in parentheses in brackets preceded by a — sign. 32 ADVANCED COURSE IN ALGEBRA IV. MULTIPLICATION OF ALGEBRAIC EXPRESSIONS 83. The Rule of Signs. The results (8), (9), and (12), § 40, hold when a and b are any positive numbers. From these results we may state what is called the Rule of Signs in multiplication, as follows : The product of two terms of like sign is positive; the product of two terms of unlike sign is negative. 84. We have by § 40, (12), (— a) X (— 6) X (— c) = (ab) X (— c) = — abc ; (1) (- a) X (- 6) X (- c) x{-d) = (- abc) x (- d), by (1), = abcd; etc. That is, the product of three negative terms is negative ; the product of four negative terms is positive ; and so on. In general, the product of any number of terms is positive or negative according as the number of negative terms is even or odd. 85. The Index Law. Let it be required to multiply a^ by a\ By § 60, a^ =a xaxa, and a^ = ax a. Whence, a^xa^=axaxaxaxa — a^. We will now consider the general case. Let it be required to multiply a™ by a'\ where m and n are any positive integers. We have a™ = a x a x • • • to m factors, and, a" = a X a X • • • to n factors. Then, a'" x c^" = a X a X ••' to m-\-n factors = a*""^**- MULTIPLICATION 33 Hence, the exponent of a letter in the product is equal to its exponent in the multiplicand plus its exponent in the multiplier. This is called the Index Law for Multiplication. A similar result holds for the product of three or more powers of the same letter. Thus, a^xa^xa^ = a^+^+' = a^\ 86. Multiplication of Monomials. 1. Let it be required to multiply 7 a by — 2 &. By §72, _2& = (-2)x6. Then, 7a X (-26) = 7ax (-2) x h. Then by the Commutative Law for Multiplication (§ 14), 7ax (-2 6) = 7 X (-2) xax 6= -14a& (§83). 2. Required the product of — 2 a^h^, 6 ab^, and — 7 a^c. {-2a^W)X 6ab'x(-7a*c) = (- 2) a'b^ x6ab'x (- 7) a*c = (-2)x6x(-7)xa^Xaxb^xb'xc = 84a36% by §§84and85. We then have the following rule for the product of any number of monomials : To the product of the mcmerical coefficients (§§ 72, 84, 85) annex the letters; giving to each an exponent equal to the sum of its exponents in the factors. 3. Multiply - 5 a^ft by - 8 ah\ (_ 5 a36) X (- 8 ab^) = 40 a^^^b^"^^ = 40 a^b\ 4. Find the product of 4 n^, — 3 n^, and 2 n\ 4^2 X (- 3n'') X 2n^ = - 24ri2+«+'« = - 24 n". 5. Multiply — flj"* by 7 a;^, m being a positive integer. (— a;"') X 7 a;^ = — 7 of-^^ 6. Multiply Q(m-[- n)< by 7 (m + nf. 6 {m + n)* X 7 (m + nf = 42 (m + n)\. 34 ADVANCED COURSE IN ALGEBRA 87. Multiplication of Polynomials by Monomials. By §§ 14, III, and 24, (7), we have the following rule for the product of a polynomial by a monomial : Multiply each term of the multiplicand by the multiplier, and add the partial products. Ex. Multiply 2x^-5x-\-7hj -Sa^. {2a^-5x-^7) X (-8a^) = (2aj2)x (-8a^) + (-5aj)x (-Sa?) + (7) x (-8a^) = _16a^ + 40a;^-56aj3. The student should endeavor to put down the final result in one operation. 88. Multiplication of Polynomials by Polynomials. By the Distributive Law for Multiplication (§ 14), (a + 6) X (c + d) = (a + 6) X c + (a + 6) X c? = ac -{- be + ad -\-bd ', and a similar result holds whatever the number of terms in the multiplicand or multiplier. We then have the following rule : Multiply each term of the multiplicand by each term of the multiplier, and add the partial products. 1. Multiply 3a-4& by 2a-5&. In accordance with the rule, we multiply 3 a — 4 & by 2 a, and then by —5 b, and add the partial products. A convenient arrangement of the work is shown below, similar terms being in the same vertical column. 3a -46 2a -5b 6a'- Sab -15ab + 20 b^ 6 a:'- 23 ab + 20 b\ MULTIPLICATION 35 The work may be verified by performing the example with the multi- plicand and multiplier interchanged. 2. Multiply 4 aa^ + a^ - 8 a^ - 2 a^a; by 2 a; + a. It is convenient to arrange the multiplicand and multiplier in the same order of powers of some common letter (§ 76), and write the partial products in the same order. Arranging the expressions according to the descending powers of a. we have a3 _ 2 a^a; + 4 ay? - 8 a;^ a +2x a* — 2 a^a; + 4 a V — 8 ax^ 2 a^aj - 4 aV + 8 aar^ - 16 aj^ a^ - 16 x\ If the multiplicand and multiplier are arranged in order of powers of a certain letter, with literal coefficients, the operation may sometimes be abridged by the use of parentheses. 3. Multiply aj2 — ax ~hx — ab by x — a. By § 87, — ax — hx can be written — (a + h)x. x^ — (a + l))x — ah y? — (a + V)x^ — ahx — ax^ 4- (5), the product will also be homogeneous, and its degree equal to the sum of the degrees of the multiplicand and multiplier. For if each term of the multiplicand is of the mth degree (§ 64), and each term of the multiplier of the nth degree, each term of the product will be of the (m + n)th degree (§ 85). The examples in § 88 are instances of the above law ; thus in Ex. 2, the multiplicand, multiplier, and product are homogeneous, and of the third, first, and fourth degrees, respectively. The student should always, when possible, apply the prin- ciples of homogeneity to test the accuracy of algebraic work. Thus, if two homogeneous expressions be multiplied together, and the product obtained is not homogeneous, it is evident that the work is not correct. 90. Multiplication by Detached Coefficients. In finding the product of two expressions which are arranged according to the same order of powers of some common letter, the operation may be abridged by writing only the numerical coefficients of the terms. 1. Multiply 3x24-5a;-4by2a;2-7a; + l. 3+5-4 2- 7+ 1 6 + 10- 8 _ 21 - 35 + 28 3+ 5-4 6-11-40 + 33-4. We know that the exponent of x in the first term is 4. Then, the product is 6 a:* - 11 a^ - 40 a;^ + 33 a; - 4. If the term involving any power be wanting, it may be supplied with the coefficient 0. 2. Multiply 4 a^ + 6 aa;2 - 7 a:^ by 2 a^ - 3 a^l In this case the term involving a^x in the multiplicand and the term involving ax in the multiplier are wanting. MULTIPLICATION 37 4 + 0+ 6- 7 2 + 0- 3 8 + + 12-14 _ 12 + . - 18 + 21 8 + 0+ 0-14-18 + 21. We know that the product is homogeneous (§ 89), and that the exponent of a in the first term is 5. Then, the product is 8 a^ - 14 aV - 18 ax^ + 21 x^. 3. Find the value of (2 a; - 3) (3 a; + 5) (6 a; - 1). 2-3 3+5 6 - 9 + 10-15 6 6 + 1-15 - 1 36 + 6-90 -6-1 + 15 Kesult, 36 ar^ -91aj+15. 91. By § 83, (+ ci) X (+ 6) = + ab, (+ a) X (- 6) = - ab, (- a) X (- 6) = + a6, (- a) x (+ 6) = - ab. Hence, in the indicated product of two monomial expressions, the signs of both expressions may be changed without altering the product; but if the sign of either one be changed, the sign of the product will be changed. The above is true for the product of a monomial and a poly- nomial, or of two polynomials. If either expression is a polynomial, care must be taken, on changing its sign, to change the sign of each of its terms. For by § 81, - (a - 6 + c) = - o + 6 - c. Thus, (a — b)(c — d) may be written in the forms (6 — a)(d — c), —(b — a)(c — d), or — (a — 6) (d — c). 38 ADVANCED COURSE IN ALGEBRA In like manner it may be shown that, in the indicated product of more than two expressions, the signs of any even number of them may he changed without altering the product; but if the signs of any odd number of them be changed, the sign of the product will be changed (§ 84). Thus, (a — b)(c — d) (e — /) may be written in the forms (a-b)(d-c){f-e), (b-a){c-d)(f-e), — (6 — a) (d — c) (/— e), etc. EXERCISE 2 Multiply the following : 1. x3 - 6 ic2 + 12 x - 8 and a:2 + 4 a; + 4. 2. w - 5 ^2 + 2 + w3 and 5 n + w^ - 10. 3. 3(a + 6)2 _ (a + &) + 2 and 4(a + 6)2 _ (« + 6) _ 5. 4. 4 x^wi+Si/M-s 4- 5 ajw+SySH-z and 3 x^"^~^y^ — 7 x^y^^+K 5. x2 — (m — n) ic — wiw and a; — p. 6. x^ + ax -bx — ah and a; + 6. 7. a^ + 3 + 5 c^4 _ 6 a _ 2 a2 and 6 + 2 a2 - a. 8. ma; + m?/ — nx — ny and wa; — my ■\- nx — ny. 9. 2 X - 3 y, 3 a; + 2 y, 2 X + 3 y, and 3 X - 2 y. 10. X + a, X + 6, and x — c. 11. 2 2_3 ;^ g^^^ lwi2 + lwi-i 3 4 2 3 9 12. x2 - (a + 6) X + a6 and x2 - (c — d) x — cd 13. a + 6 + c, a — 6 + c, and a + 6 — c. Simplify the following : 14. [3x-(5?/ + 20)][3x-(5?/-2;2)]. 15. [(m + 2n)-(2w- w)][(2wi + n)-(m-2w)]. 16. [2 x2 + (3x - l)(4x + 5)] [5x2 -(4x + S)(x - 2)]. 17. (a - 6) (a8 + &») [a (a + 6) + ft^]. DIVISION - 39 V. DIVISION OP ALGEBRAIC EXPRESSIONS In the present chapter, we shall consider those cases only in which the Dividend, Divisor, and Quotient are rational and integral (§ 63). In such cases, the division is said to be exacts and the dividend is said to be divisible by the divisor. 92. The Reciprocal of a number is 1 divided by that number. Thus, the reciprocal of a is -• a 93. We have a x 1 = a. Regarding a as the quotient, 1 as the divisor, and a as the dividend, we have l = a. . 94. By § 30, (1), ^ = ^xj = axi(§ 93). 1 Hence, to divide by a number is the same thing as to multiply by its reciprocal (§ 92). 95. The Commutative Law for Division. By § 94, every operation in Division can be expressed as an operation in Multiplication. Thus, if a is to be divided by b, c, •••, in succession, the result is ^ ^ ax-X-X— . b c It follows from this, by § 14, I, that if a number is to be divided by any number of numbers in succession, the divisions can be performed in any order. This is the Commutative Law for Division. It may be expressed as follows : (a-j-6) -!-C'" = (a-i-c)-^b-", etc. 96. By § 14, I, II, b(ac) = a(bc). 40 ADVANCED COURSE IN ALGEBRA Then by §31, ^ = ^. (1) That is, a factor common to the dividend and divisor can he removed, or cancelled. Putting 6 = 1, in (1), we have "B^^^a (§93). C JL That is, if a number he hotli multijMed and divided hy another, the value of the former will not be changed. 97. The Rule of Signs. From the results (1), (2), and (3), § 48, we may state the Rule of Signs in Division, as follows : TJie quotient of two terms of like sign is positive; the quotient of two terms of unlike sign is negative. 98. The Index Law for Division. Let it be required to divide a^ by a^. By § 60, a" a X a Cancelling the common factor axa (^ 96), we have a^ We will now consider the general case : Let it be required to divide a"* by a**, where m and n are any positive integers such that m is > n. We have — = a X^ X a X •♦• to m factors ' a" ax ax ax •••tow factors Cancelling the common factor axaxax--- to n factors, ~ = a X axax •••torn — n factors a" = a**"". Then, the exponent of a letter in the quotient is equal to its exponent in the dividend, minus its exponent in the divisor. This is called the Index Law for Division. DIVISION 41 99. Division of Monomials. 1. Let it be required to divide — 14 a^h by 7 a^. Bv § 86 — 14 a^h _ (—2)x7xa^xb ^ ' 7 a' 1X0? ' Cancelling the common factors 7 and a?, we have z:ii^ = (-2)x6=-26. We then have the following rule for the quotient of two monomials : To the quotient of the numerical coefficients annex the letters, giving to each an exponent equal to its exponent in the dividend minus its exponent in the divisor, and omitting any letter having the same exponent in the dividend and divisor. 2. Divide 54 a^6V by - 9 a'h\ 54 a'hh'' -^a'h^ 3. Divide — 2 x^'^y'^z'' by — x'^y'^z^ ; m, n, and r being positive integers, and r > 5. — x^'y^'z^ 4. Divide 35 (a - by by 7 (a - b)*. l{a-by ^ ^ 100. Division of Polynomials by Monomials. By § 94, a a = 5 .- + c .- - (^ .-, by §§ 14, III, and 24, (7), a a a = - + ---, by §94. a a a Hence, to divide a. polynomial by a monomial, we divide each term of the dividend by the divisor, and add the results. This is called the DistributiveLgwforJDivision. 42 ADVANCED COURSE IN ALGEBRA 1. Divide 9 c^W -^o^c-^ 12 a%(? by - 3 al = -3b' + 2a'c-4.ahc\ The student should endeavor to put down the result in one operation. 2. Divide 35 (x + yY - 20 (« + 2/)^ by 5 ( a; + yf. 35(x-{-yy-20(x + yy ^ / , -,2 ^/ . n 101. Division of Polynomials by Polynomials. Ex. Let it be required to divide 12 + 10 or' — 11 x — 21 a^ by 2a^-4.-3x. Arranging the expressions according to the descending powers of X (§ 76), we are to find an expression which, when rnultiplied by 2rc2 - 3a; - 4, will produce 10 a^ - 21 a^ - 11 a; + 12. It is evident that the term containing the highest power of X in the product is the product of the terms containing the highest powers of x in the multiplicand and multiplier. Therefore, 10 x^ is the product of 2 a:^ and the term containing the highest power of x in the quotient. Whence, the term containing the highest power of x in the quotient is 10 a^ divided by 2 x^, or 5 x. Multiplying the divisor by 5x, we have the product 10 af^ — 15 a^ — 20 a; ; which, when subtracted from the dividend, leaves the remainder — 6 a;^ + 9 a; + 12. This remainder must be the product of the divisor by the rest of the quotient ; therefore, to obtain the next term of the quotient, we regard — 6 a;^ + 9 a; + 12 as a new dividend. Dividing the term containing the highest power of a;, — 6 a^, by the term containing the highest power of x in the divisor, 2 x^, we obtain — 3 as the second term of the quotient. Multiplying the divisor by — 3, we have the product —601^ + 9 a; + 12 ; which, when subtracted from the second dividend, leaves no remainder. Hence, 5 x — 3 is the required quotient. DIVISION 48 It is customary to arrange the work as follows : 10a^-15x'-20x 2 ic^ — 3 a; — 4, Divisor. 5 ic — 3, Quotient. - 6^2+ 9a; + 12 - 6a^+ 9 a; + 12 The example might have been solved by arranging the dividend and divisor according to ascending powers of x. 102. From § 101, we derive the following rule : Arrange the dividend and divisor in the same order of powers of some common letter. Divide the first term of the dividend by the first term of the divisor J and write the result as the first term of the quotient. Multiply the whole divisor by the first term of the quotient, and subtract the product from the dividend. If there be a remainder, regard it as a new dividend, and proceed as before; arranging the remainder in the same order of powers as the dividend and divisor. 1. Divide 9a62 + a3-963-5a26 by 3&2-fci2_2a6. Arranging according to the descending powers of a, a^-5a'b-{-9 ab'^ -9 b' a'-2ab-\-3b' a'-2a'b-\-3 ab'^ -Sa'b-\-6ab' -Sa'b-^6ab'-9b' a-3b In the above example, the last term of the second dividend is omitted, as it is merely a repetition of the term directly above. The work may be verified by multiplying the quotient by the divisor, which should of course give the dividend. 2. Divide 4 + 9 a;^-28a;2 by _3a^ + 2 + 4a;. Arranging according to the ascending powers of x, 4_28a;2+ 9a;^ 44 8a; - 6a^ 2 + 4a;-3a;2 2_4aj-3a:2 8a; -22a;2_^ 9^4 8a; -16a;2 + 12a;3 - 6a;2-12ar^ + 9a;^ - 6x'-12o^i-9x* 44 ADVANCED COURSE IN ALGEBRA 3. Divide x^ -\- {a + b — c) a^ -{- (ab — be — cd) x — abe by a; + a. a^ -{- (a + b — e)x^ + (ab - -be — ca)x - - abe X -\-a ar^+ ax" x^ + {b — e)x—bc {b-e)x' (b-e)x^-{-(ab — ca)x — bex — bex — abc 103. It is evident from § 89 that, if the dividend and divisor are homogeneous, the quotient will also be homogeneous, and its degree equal to the degree of the dividend minus the degree of the divisor. 104. Division by Detached Coefficients. In finding the quotient of two expressions which are arranged according to the same order of powers of some common letter, the operation may be abridged by writing only the numerical coefficients of the terms. If the term involving any power is wanting, it may be sup- plied with the coefficient 0. Ex. Divide ea^-\-2a^-9x*-\-5x'-i-lSx~80hy3a^+x^-6. 3+1+0-6 6 + 2-9+ + 5 + 18- 6 + 2 + 0-12 -9 + 12 _9- 3 + + 18 -30 15 + 5 15 + 5+ 0- -30 2+0-3+5 Then the quotient is 2 a^ — 3 a; + 5. 105. By § 37, (6), if a is any number, X a = 0. Regarding as the quotient, a as the divisor, and as the dividend, we have a - = 0. a DIVISION 45 EXERCISE 3 Divide the following : 1. x5 + 37 x2 - 70 ic + 50 by a;2 - 2 a; + 10. 2. 6(x~yy-7(x-y)-20 hj S{x-y) + i. 3. a^+^b^ + a&39+2 i^y Qtp+152 ^ ab^+^ ; p and ^ being positive integers. 4. a& - &5 _ 5 a^b ^^ab^ + 10 a^S^ - 10 a^b^ hy a^-b^-S a'^b + 3 ab^. 5. 6 n^ + 25 # - 7 n3 _ 81 7i2 _ 3 ri + 28 by 2 n^ + 5 n2 - 8 n - 7. 6. 23x2 - 5 x4 - 12 + 12 a;5 + 8 X - 14 a:3 by x - 2 + 3 x2. 7. 16(a + ft)4 - 81 by 2(« + 6) - 3. 8. 8 x6 - 4 x5 - 2 X* + 15 x3 + 3 x2 - 5 X - 15 by 4 x» - a; - 3. 9. a^ + b^ - c^ + S abc hy a + b - c. 10. im4-2m« + -m2--L by ?m2-^wi--. 9 4 16 "^ 3 2 4 11. 52 x3 + 64 + 18 X* - 200 x2 -f x5 by 6 x2 - 8 + x^ - 12 x. 12. a^-6 o4,j2 ^ 9 Qj2„4 _ 4 ^6 by cfS _ 2 a2^ _ an2 + 2 #. 13. x^ - (a — 6 + c)x2 — (ab ~ ac -^ bc)x + a6c by x^ — (a — b)x — a6. 14. x3 + (a - 6 4- c)x2 + (_ a6 + ac - &c)x — abc by x + c. 15. x3 4- (4 a + 2 6 4- 3 c)x2 4- (8 a6 + 12 ac + 6 bc)x 4- 24 abc by x2 + (4 a 4- 3 c)x 4- 12 ac. 16. x3 - (a + 3 & + 2 c)x2 4- (3 a6 4- 2 ac 4- 6 bc)x ~ 6 abc hy x-Sb. 17. (2 w2 4- 10 mn) x^ -\- (S m^ - 9 mn - 15 w2)x - (12 mn-9 n^) by 2 mx — 3 n. 18. x*-(4a4-3)x3 + (12a-5&4-2)x2-(8a-15 6)x- 106 by x2 - 3 X 4- 2. 46 ADVANCED COURSE IN ALGEBRA VI. INTEGRAL LINEAR EQUATIONS 106. The First Member of an equation is the expression to the left of the sign of equality, and the Seco7id Member the expression to the right of that sign. Thus, in 2 a? — 3 = 3 a; + 5, the first member is 2 oj — 3 and the second member 3 x + 5. Any term of either member of an equation is called a term of the equation. The sides of an equation are its two members. 107. An Identical Equation, or Identity, is an equation whose members are the same, or become the same after all the indi- cated operations have been performed ; as, 5 = 5, or (a -{- b) (a — b) — a? — W. The sign =, read "is identically equal to,'''' is frequently usedrin place of the sign of equality in an identity. It is evident that, in an identity involving letters, the mem- bers are equal whatever values are given to the letters, provided the same value is given to the same letter wherever it occurs. All equations considered up to the present time have been identical equations. 108. An equation is said to be satisfied by a set of values of certain letters involved in it when, on substituting the value of each letter in place of the letter wherever it occurs, the equa- tion becomes identical. Thus, the equation x — y = 5 is satisfied by the set of values x = S, y = S; for, on substituting 8 for x, and 3 for y, the equa- tion becomes 8 — 3 = 5, or 5 = 5; which is identical. 109. An Equation of Condition is an equation involving one or more letters, called Unknown Numbers, which is satisfied only by particular values, or sets of values, of these letters. INTEGRAL LINEAR EQUATIONS 47 Thus, the equation x-{-2 = 5 is satisfied by the value x = 3; but not by the value x = 5. Again, the equation x-{-y = 7 is satisfied by the set of values a; = 4, 2/ = 3 ; but not by the set of values x = 6, y = 9. An equation of condition is usually called an equation. Any letter in an equation of condition may represent an unknown number ; but it is usual to represent unknown num- bers by the last letters of the alphabet. 110. Any letter, or set of letters, which satisfies an equation is called a Solution of the equation. If the equation contains but one unknown number, its solu- tions are called Roots. A solution is verified when, on substituting the values of the unknown numbers, and performing the operations indicated, the equation becomes identical. To solve an equation, or a system of equations, is to find its solutions. 111. A Numerical Equation is one in which all the known numbers are represented by Arabic numerals ; as, 2x-7 = x + 6. A Literal Equation is one in which some or all of the known numbers are represented by letters ; as, Sx-\-a = 5x — 2b. An Integral Equation is one each of whose members is a rational and integral expression (§ 63); as, 4ic — 5 = -v + l. 112. If a rational and integral monomial (§ 63) involves a certain letter, its degree with respect to it is denoted by its exponent. If it involves two letters, its degree with respect to them is denoted by the sum of their exponents ; etc. Thus, 2 ab^x^y^ is of the second degree with respect to cc, and of the fifth with respect to x and y. 48 ADVANCED COURSE IN ALGEBRA 113. If an integral equation (§ 111) contains one or more unknown numbers, the degree of the equation is the degree of its term of highest degree. Thus, if X and y represent unknown numbers, ax — hy= c, is an equation of t\iQ first degree ; x^ -\-4,x= — 2, an equation of the second degree ; 2 ic" — 3 xf = 5, an equation of the third degree ; etc. A Linear Equation is an equation of the first degree. 114. Two equations, each involving one or more unknown numbers, are said to be Equivalent when every solution of the first is a solution of the second, and every solution of the second a solution of the first. 116. The following are of use in solving equations : 1. If the same number (or equal numbers) be added to equal ^lumbers, the sums will be equal. 2. If the same number (or equal numbers) be subtracted from equal numbers, the remainders will be equal. These follow from § 22. 3. If equal numbers be multiplied by the same number (or equal numbers), the products will be equal (§ 29). 4. If equal numbers be divided by the same number (or equal numbers), the quotients will be equal, provided the divisor is not 0. This follows from § 29 ; compare § 43. PRINCIPLES USED IN SOLVING INTEGRAL EQUATIONS 116. Addition. If the same expression be added to both members of an equation, the resulting equation will be equivalent to the first. Consider, for example, the equation A = B. (1) To prove that the equation A^-C = B+G, (2) where C is any expression, is equivalent to (1). INTEGRAL LINEAR EQUATIONS 49 Any solution of (1), when substituted for the unknown num- bers, makes A identically equal to B (§ 108). It then makes A-\- G identically equal to J5 -f- (7 (§ 115, 1). Then it is a solution of (2). Again, any solution of (2), when substituted for the unknown numbers, makes A-\- C identically equal to B -\- G. It then makes A identically equal to B (§ 115, 2). Then it is a solution of (1). Therefore, (1) and (2) are equivalent. The above demonstration proves that if the same expression be sub- tracted from both members of an equation, the resulting equation will be equivalent to the first. 117. Transposing Terms. Consider the equation x-\-a — b = c. Adding — a and + & to both members (§ 116), we have the equivalent equation x-{-a — b—a-\-b = c — a-\-b. Or, x = c — a-\-b. In this case, the terms a and — b are said to be transposed from the first member to the second. Hence, if any term be transposed from one member of an equor Hon to the other by changing its sign, the resulting equation will be equivalent to the first. If the same term appears in both members of an equation affected with the same sign, it may be cancelled. 118. Consider the equation a — x = b — c. (1) Transposing each term, we have the equivalent equation — b-\-c= — a-f-ic, or a; — a = c — ^; which is the same as (1) with the sign of every term changed. Hence, if the signs of all the terms of an equation be changed, the resulting equation will be equivalent to the first. 50 ADVANCED COURSE IN ALGEBRA 119. Multiplication. If the members of an equation he multiplied by the same expres- sion, which is not zero, and does not involve the unknown numbers, the resulting equation will be equivalent to the first. Consider the equation A=zB, (1) To prove that the equation AxO=BxO, (2) where C is not zero, and does not involve the unknown num- bers, is equivalent to (1). Any solution of (1), when substituted for the unknown num- bers, makes A identically equal to B. It then makes AxC identically equal to 5 x O (§ 115, 3). Then it is a solution of (2). Again, any solution of (2), when substituted for the unknown numbers, makes AxC identically equal to B x G. It then makes A identically equal to B (§ 115, 4). Then it is a solution of (1). Therefore, (1) and (2) are equivalent. The reason why the above does not hold for the multiplier zero is, that the principle of § 115, 4, is restricted to cases where the divisor is not zero. 120. The necessity for limiting the principle of § 119 to cases where the multiplier does not involve the unknown num- bers is that, if C contains the unknown numbers, the equation AxO=BxO is satisfied by certain values of the unknown numbers which make (7=0. But these values do not, in general, satisfy A=B. Consider, for example, the equation x-\-2 = 3x-4:. (1) Now the equation (x + 2)(x-l) = (Sx-4:)(x-l), (2) which is obtained from (1) by multiplying both members by a? — 1, is satisfied by the value x = l, which does not satisfy (1). Then (1) and (2) are not equivalent. INTEGRAL LINEAR EQUATIONS 51 It follows from this that it is never allowable to multiply- both members of an integral equation by an expression which involves the unknown numbers j for in this way additional solutions are introduced. 121. Clearing of Fractions. Consider the equation 2 5 5 9 -X = -X . 3 4 6 8 Multiplying each term by 24, the lowest common multiple of the denominators (§ 119), we have the equivalent equation 16x-30 = 20a;-27, where the denominators have been removed. Removing the fractions from an equation by multiplication is called '•''Clearing the equation of fractions.'''' 122. Division. If the members of an equation he divided by the same expres- siouj which is not zero,. and does not involve the unknown num- bers, the resulting equation will be equivalent to the first. Consider the equation A = B. (!) To prove that the equation A-^C=B-^C, (2) where C is not zero, and does not involve the unknown num- bers, is equivalent to (1). Any solution of (1), when substituted for the unknown num- bers, makes A identically equal to B. It then makes A-~-C identically equal to B-i- C (% 115, 4). Then it is a solution of (2). Again, any solution of (2), when substituted for tjie unknown numbers, makes A-i-C identically equal to B -i- C. It then makes A identically equal to B. Then it is a solution of (1). Therefore, (1) and (2) are equivalent. 52 ADVANCED COURSE IN ALGEBRA 123. The necessity for limiting the principle of § 122 to cases where the divisor does not involve the unknown num- bers is that, if C contains the unknown numbers, the solution oi A = B contains certain numbers which do not, in general, satisfy ^ ^ (7= 5 -^ a Consider, for example, the equation {x + 2){x-l) = {3x-^){x-.l). (1) Also the equation a? + 2 = 3 a; — 4, (2) which is obtained from (1) by dividing both members by x — 1. Now equation (1) is satisfied by the value ic = 1, which does not- satisfy (2). Then (1) and (2) are not equivalent. It follows from this that it is never allowable to divide both members of an integral equation by an expression which in- volves the unknown numbers; for in this way solutions are lost. SOLUTION OF INTEGRAL LINEAR EQUATIONS 124. To solve an equation containing one unknown number, we put it into a succession of forms, which lead finally to a knowledge of the root or roots. This process is called transforming the equation. If every transformation is effected by means of the principles of §§ 116 to 122, the successive equations will have the same roots as the given equation, and no solutions will be introduced nor lost. EXAMPLES 125. 1. Solve the equation 5x-l = ?>x-\-l. Transposing 3 a; to the first member, and — 7 to the second (§117),wehave 5^_3^^7 + l. Uniting similar terms, 2 a; = 8. INTEGRAL LINEAR EQUATIONS 53 Dividing both members by 2 (§ 122), X = A. To verify the result, put x = 4 in the given equation. Thus, 20 - 7 = 12 + 1 ; which is identical. 2. Solve the equation 7 5_3 1 Clearing of fractions by multiplying each term by 60, the L. C. M. of 6, 3, 5, and 4, we have 70x-100 = 36a:-15. Transposing 36 x to the first member, and — 100 to the second, and uniting similar terms, 34 a; = 85. Dividing by 34, ^ = 1 = 1 3. Solve the equation (5 - 3 ic) (3 + 4 r») = 62 - (7 - 3 a;)(l - 4 x). Expanding, 15 + 11 ic-12 a^ = 62 - (7 - 31 x + 12 x") = 62 - 7 + 31 a; - 12 a;2. Cancelling the — 12 x^ terms (§ 117), and transposing, 11 a; - 31 a; = 62 - 7 - 15. Uniting terms, — 20 a; = 40. Dividing by - 20, x = - 2. To expand an algebraic expression is to perform the operations indicated. 4. Solve the equation .2 a; -f .001 - .03 a; = .113 x - .0161. Transposing, .2 aj - .03 a; - .113 x = - .0161 - .001. Uniting terms, .057 x = — .0171. Dividing by .057, x = — .3. From the above examples, we have the following rule for solving an integral linear equation with one ujiknown number : 54 ADVANCED COURSE IN ALGEBRA Clear the equation of fractions, if any, by multiplying each term by the L. C. M. of the denominators of the fractional coefficients. Remove the parentheses, if any. Transpose the unknown terms to the first member, and the known to the second; cancelling any term which has the same coefficient in both members. Unite similar terms, and divide both members by the coefficient of the unknown number. The student should endeavor to apply more than one principle at one operation. He will also find it excellent practice to verify his solutions. « , , . „ EXERCISE 4 Solve the following : 5 3 9 15 2. .05 jc- 1.82- .7 a: = .008 a; -.504. 3. 4(a; + 14) - 4(3 a: - 32) = 6(a; + 12) - 7 (x - 12). .577 1 55 4. -X x = - X X 6 8 9 18 48 5. (5-3a;)(3 +4a;)-(7 + 3a;)(l-4a;)=-l. 6. .07(8 a: - 5.7) = .8(5 x + .86) + 1.321. 7. (1 + 3 xY - (5 - xy - 4(1 - a;) (3 - 2 X) = 0. 8. 6(a;-4)2 = 5-(3 - 2 x)2 - 5(2 + a;)(7 - 2 a:). 9. (3 a; - 2)3 - 9 a; (a; - 1) (3 a; - 8) = 45 3^2 _ 33. 10. (x + 4)8-(a;-4)8 = 2(3a;-2)(4a; + l). 11. |(4 + a:)-^(l-5x) = |(l + 2a;)-A(2-3x). 12. |[.-l(5. + l)] = |[.-|(3. + 4)] + I. PROBLEMS INVOLVING INTEGRAL LINEAR EQUATIONS WITH ONE UNKNOWN NUMBER 126. For the solution of a problem by algebraic methods no general rule can be given. The following suggestions will be found of service : INTEGRAL LINEAR EQUATIONS 55 1. Represent the unknown number, or one of the unknown numbers if there are several, by x. 2. Every problem contains, explicitly or implicitly, just as many distinct statements as there are unknown numbers involved. Use all but one of these to express the other unknown num- bers in terms of x. 3. Use the remaining statement to form an equation. ILLUSTRATIVE PROBLEMS 127. 1. Divide 45 into two parts such that the less part shall be one-fourth the greater. Here there are two unknown numbers ; the greater part and the less. In accordance with the first suggestion of § 126, we represent the greater part by x. The first statement of the problem is, implicitly : The sum of the greater part and the less is 45. The second statement is : The less part is one-fourth the greater. In accordance with the second suggestion of § 126, we use the second statement to express the less part in terms of x. Thus, the less part is represented by -x. 4 We now, in accordance with the third suggestion, use the first state- ment to form an equation. Thus, x-^-x = i6. 4 Clearing of fractions, ix-{-x = 180. Uniting terms, 5x= 180. Dividing by 5, x = 36, the greater part. Then, i x = 9, the less part. 4 2. A is 3 times as old as B, and 8 years ago he was 7 times as old. Required their ages at present. Let X = number of years in B's age. Then, 3x = number of years in A's age. Also, X — S = number of years in B's age 8 years ago, and 3x — 8 = number of years in A's age 8 years ago. 56 ADVANCED COURSE IN ALGEBRA But A's age 8 years ago was 7 times B's age 8 years ago. Whence, 3x-8 = 7(x-8). Expanding, 3x — 8 = 7x— 56. Transposing, — 4x= — 48. Dividing by — 4, x = 12, the number of years in B's age. Then, 3x = 36, the number of years in A's age. It must be carefully borne in mind that x can only represent an abstract number. Thus, in Ex. 2, we do not say "let x represent B's a^e," but "let x represent the number of years in B's age." 3. A sum of money, amounting to f 4.32, consists of 108 coins, all dimes and cents ; how many are there of each kind ? Let X = the number of dimes. Then, 108 — x = the number of cents. Also, the X dimes are worth 10 x cents. But the entire sum amounts to 432 cents. Whence, 10 ic + 108 - x = 432. Transposing, 9x = 324. Whence, x — 36, the number of dimes ; and 108 — x = 72, the number of cents. 4. At what time between 3 and 4 o'clock are the hands of a watch opposite to each other? Let X = the number of minute-spaces passed over by the minute-hand from 3 o'clock to the required time. Then since the hour-hand is 15 minute-spaces in advance of the minute- hand at 3 o'clock, X — 15 — 30, or x — 45 will represent the number of minute-spaces passed over by the hour-hand. But the minute-hand moveS' 12 times as fast as the hour-hand. Then, x = 12 (x - 46) = 12 X - 540. Transposing, _ 11 x = — 540. Whence, x = 49j\. Then, the required time is i9j\ minutes after 3 o'clock. INTEGRAL LINEAR EQUATIONS 57 5. Two persons, A and B, 63 miles apart, start at the same time, and travel towards each other. A travels at the rate of 4 miles an hour, and B at the rate of 3 miles an hour. How far will each have travelled when they meet ? Let ^x = number of miles that A travels. Then, l(}^ = number of miles that B travels. By the conditions^ x-i-j^x = 6^^ Then, 4 x = 36, number of miles that A travels, and 3 a; = 27, number of miles that B travels. It is often advantageous, as in Ex. 5, to represent the unknown num- ber by some multiple of x, instead of by x itself; EXERCISE 5 1. Divide 66 into two parts such that - the greater shall exceed - the less by 21. '^ ^ 5 S 2. In 9 years, B will be - as old as A ; and 12 years ago he was - as old. What are their ages ? (Let x represent number of years in A's age 12 years ago.) 3. Divide 197 into two parts such that the smaller shall be contained in the greater 5 times, with a remainder 23. 4. After A has travelled 7 hours at the rate of 10 miles in 3 hours, B sets out to overtake him, travelling at the rate of 9 miles in 2 hours. How far will each have travelled when B overtakes A ? 5. At what time between 8 and 9 o'clock are the hands of a watch together ? 6. Find four consecutive odd numbers such that the product of the first and third shall be less than the product of the second and fourth by 86. 7. A sum of money, amounting to $19.30, consists of .^2 bills, 25-cent pieces, and 5-cent pieces. There are 13 more 5-cent pieces than $2 bills, 7 and - as many 5-cent pieces as 25-cent pieces. How many are there of each ? 8. At what times between 4 and 5 o'clock are the hands of a watch at right angles to each other ? 58 ADVANCED COURSE IN ALGEBRA 9. A woman sells half an egg more than half her eggs. She then sells half an egg more than half her remaining eggs. A third time she does the same, and now has 3 eggs left. How many had she at first ? 10. A train leaves A for B, 210 miles distant, travelling at the rate of 28 miles an hour. After it has been gone 1 hour and 15 minutes, another train starts from B for A, travelling at the rate of 22 miles an hour. How many miles from B will they meet ? 11. A man puts a certain sum in a savings bank paying 4% interest. At the end of a year he deposits the interest, receiving interest on the entire amount. At the end of a second year and a third year he does the same, and now has $2812. 16 in the bank. What was his original deposit ? 12. A fox is pursued by a hound, and has a start of 77 of her own leaps. The fox makes 5 leaps while the hound makes 4 ; but the hound in 5 leaps goes as far as the fox in 9. How many leaps does each make before the hound catches the fox ? 13. A clock has an hour-hand, a minute-hand, and a second-hand, all turning on the same centre. At 12 o'clock alt the hands point at 12. How many seconds will it be before the hour-hand is between the other two hands and equally distant from them ? . (^ 14. A freight train travels from ^ to ^ at the rate of 12 miles an hour. After it has been gone 3^ hours, an express train leaves A for B, travel- ling at the rate of 45 miles an hour, and reaches B 1 hour and 5 minutes ahead of the freight. Find the distance from Ato B and the time taken by the express train. 15. A merchant increases his capital each year by one-third, and at the end of each year sets aside $ 1350 for expenses. At the end of three 145 years, after setting aside his expenses, he finds that he has — - of his original capital. What was his original capital ? i SPECIAL METHODS 59 VII. SPECIAL METHODS IN MULTIPLICATION AND DIVISION 128. Any Power of a Power. Required the value of (a"*)% where m and n are any positive integers. We have, (a'")"=: a'" x a"* X ••• to n factors (§ 60) Qm + TO + • • • to n terms ^_. n^n 129. Any Power of a Product. Required the value of (abc ---y, where n is any positive integer. We have, (abc-'-y = {abc •••) X (abc •••) x ••• to n factors = (ax ax '"ton factors) (b xb x -"to n factors) ••• 130. Any Power of a Monomial. 1. Required the value of (5 a-by. We have, (5 a'bf = 5a'bx5a'bx5 a'b = 125 a'b\ 2. Required the value of (— my. We have, (— my = (— m) x (— m) x (— m) x (— m) = m\ 3. Required the value of (— 3 n^y. We have, (- 3 n^y = (-3n') x {-Sn') x {-Sn')= -27 n\ From §§ 128 and 129, and the above examples, we have the following rule for raising a rational and integral monomial (§ 63) to any power whose exponent is a positive integer : Raise the absolute value of the numerical coefficient to the required power, and multiply the exponent of each letter by the exponent of the required power. Give to every poiver of a positive term, and to every even power of a negative term, the positive sign ; and to every odd power of a negative term the negative sign. 60 ADVANCED COURSE IN ALGEBRA 131. Square of a Binomial. We find by actual multiplication, (a + by = {a + h) X {a + b) =0? + 2 ah -\- h\ (1) {a-hf={a-h) x {a-h) = o? -2 ab-\-V\ (2) That is, The square of tJie Bum of two numbers equals the square of the first, plus twice the product of the fii^st by the second, plus the square of the second. The square of the difference of two numbers equals the square of the first, minus twice the product of the first by the second, plus the square of the second. In the remainder of the book, we shall, for the sake of brevity, use the expression "the difference of a and &" to denote the remainder obtained by subtracting b from a. 1. Square 3 0.2 -2 6. By (2), (3 a^ - 2 bf =^ (3 a'f - 2 (3 a') (2 b) + (2 bf = 9 a'- 12 a'b-^- 4. b' (§ 130). If the first term of the binomial is negative, it should be enclosed, negative sign and all, in parentheses, before applying the rule. 2. Square - 4 a^ + 9. (- 4 ar^ + 9)- = [(- 4 x') + 9]^ =.(_4ar^y+2(-4a^)(9) + 9^by (1) = 16a^-72(^ + Sl. 132. Product of the Sum and Difference of Two Numbers. We find by actual multiplication, (a-\-b){a-b) = a'-b\ That is, the product of the sum and difference of two numbers equals the difference of their squares. 1. Multiply Qa + ^b"" by Q>a-5b\ By the rule, (6 a + 5 W) (6a-5b') = (6 a)' - (5 6^)- = 36 a' - 25 b'\ SPECIAL METHODS 61 2. Multiply - ic^ + 4 by - ic^ -^ 4. (-x'2 + 4)(-a^-4) = [(-.r^)4-4][(-:i^)-4] = (_a;2)2_42=;:a^-,16. 3. Expand (a + b — c) (a — b -{- e). To expand an algebraic expression is to perform the operations indi- cated. By § 82, (a + & - c) (a - 6 + c) = [a + (^> - c)] [a - (6 - c)] = ct^ — (b — c)'^, by the rule, = a2 ^ (Z)2 - 2 6c + c2) =.a2-^62^2 5c-cl 4. Expand (x -{- y -i- z) (x — y + 2;). (a; -f- 2/ + 2) (^ - 2/ + 2^) = [(a^ + 2;) + 2/] C(^ + 2;) - 2/] = (x + 2)^-2/^ = x2 + 2 a^2 + ^2 _ ^2^ 133. Product of Two Binomials having the Same First Term. We find by actual multiplication (x -\- a) (x -\^ b) = x^ -{-{a -\-b)x + ab. That is. The product of tivo binomials having the sayne first term equals the square of the first term, plus the algebraic sum of the second terms multiplied by the first term, plus the product of the second terms. 1. Multiply £c — 5 by a; + 3. By the above rule, the coefficient of x is the sum of — 5 and + o, or — 2, and the last term is the product of — 5 and + 3, or - 15. Whence, (a; - 5) (a; + 3) = a^ - 2 a; ^ 15. 2. Multiply x — ^ by a; — 3. The coefficient of x is the sum of — 5 and — 3, or — 8, and the last term is the product of — 5 and — 3, or 15. Whence, (a; - 5) (a; - 3) = a;^ - 8 ic + 15. 62 ADVANCED COURSE IN ALGEBRA 3. Multiply ab — 4: by ab-{-7. By the rule, (ab - 4) (ab + 7) = a%^ + 3 a& - 28. 4. Multiply m + n -\- 6 hj m -i- n -{- S. (m + 71 + 6) (?>i + ii + 8) 3= [(m + n) + 6] [(m + 7i) + 8] = (m + ny + 14 (m + n) + 48. 134r. Square of a Polynomial. By § 131, , (1), (a, + aa)^ = a,' + a/ + 2 a^a^. (1) We also have, (tti + ^2 + «3)^ =[(«! + ^2) + «3]^ = («! +a2)2 + 2 («! + as) X as + ag^ = a/ + 2 aitta + «2^ + 2 ajag + 2 agOEg + a^^ = ai + ai -\-a^ -[-2 a^a^ + 2 a^a^ + 2 a2<^3- (2) The results (1) and (2) are in accordance with the following law: The square of a polynomial equals the sum of the sqiiares of its terms, plus twice the product of each term by each 'of the following terms. We will now prove that this law holds for the square of any polynomial. Assume that the law holds for the square of a polynomial of m terms, where m is any positive integer ; that is, (% + tto + «3 + ••• + a«-i + ^mf = a^^ J^ ai -{-'" + aj + 2 a^ (a2+. .. + «„.) + 2 a2(a3+ - +a.) + ••• + 2 a,,_ia^. (3) Then, (a^ + ag + % H h ot^ + «m+i)^ = [(ai 4- a, 4- . . . + a J + a^^{\\ = (ai + «2H h a^y + 2 (ai + a, H h a J a^+i + a^+i^, by (1) = af + a^^ + • •• + a^' + a^+i' + 2ai(a2H h «« + a^+i) + 2 (I2 («3 + ••• + a,. + f^m+i) + ••• + 2 a^a^+i, by (3). SPECIAL METHODS 63 This result is in accordance with the above law. Hence, if the law holds for the square of a polynomial of m terms, where m is any positive integer, it also holds for the square of a polynomial of m -f- 1 terms. But we know that the law holds for the square of a polyno- mial of three terms, and therefore it holds for the square of a polynomial of four terms ; and since it holds fc^i- the square of a polynomial of four terms, it also holds for the square of a polynomial of five terms ; and so on. Hence, the law holds for the square of any polynomial. The above method of proof is known as Mathematical Induction. Ex. Expand (2x'-Sx- 5)1 In accordance with the law, we have (2 aj2 - 3 a; - 5)2 = (2x^' + {-^xf+{-5y + 2 (2 x") (- 3 aj) + 2 (2 a?) (_ 5) + 2 (- 3 x) (- 5) = 4 a;* + 9 a^ + 25 - 12 a^ - 20 x-^ + 30 a; = 4.x* -12 ^ -11 ix? -{- 2,0 X + 2^. 135. Cube of a Binomial. We find by actual multiplication, (a + 6)3 = a^ -I- 3 a'b + 3 aft^ + W, (1) (a _ &)3 = a^ _ 3 a?h -f 3 aft^ _ 53^ (2) That is. The cube of the sum of two numbers equals the cube of the first, plus three times the square of the first times the seco7id, plus three times the first times the square of the second, plus the cube of the second. The cube of the difference of two numbers equals the cube of the first, minus three times the square of the first times the second, plus three times the first times the square of the second, minus the cube of the second. 64 ADVANCED COURSE IN ALGEBRA 1. Find the cube of a-h^ b. By (1), (a + 2 bf = a« + 3 aX2 &) + 3 a (2 by + (2 bf = a^ + 6a'b + 12 ab' + 8 b\ 2. Find the cube of 2 a?^ — d /. By (2), (2a;^-5 2/^)« = (2 a^.)« - 3 (2 a!«)X5 /) + 3 (2 a^) (5 /)2 - (5 y^ = 8 a;^ - 60 xY + 150 xY - 125 /. 136. Cube of a Polynomial. By § 135, (1), (ai -f ^s)' = ^i' + ^2' + 3 a^% + 3 a^a^l (1) We also have, (ai + aa + ^3)^ = [(ai + a2) + a3]3 = (% + ttg)^ + 3 («! + a2)^«3 + 3 (oi + a2)ot3^ + «3^ = ai^ + 3 ai\«2 + 3 aia^^ + cxg^ + 3 aj^aa + 6 aia2a3 ■ + 3 a2% + 3 ciiag^ + 3 a2«3^ + «3^ = tti^ + ag^ + aa^ + 3 ai% 4- 3 ai% + 3 a2^«i + 3 a2% + 3 ttg^iofi 4- 3 a/ag + 6 ai«2«3- (2) The results (1) and (2) are in accordance with the following law: The cube of a polynomial equals the sum of the cubes of its terms, plus three times the product of the square of each term by each of the other terms, plus six times the product of every three different terms. We will now prove by Mathematical Induction (see § 134), that this law holds for the cube of any polynomial. Assume that th^ law bolds for the cube of a polynomial of m terms, where m is any positive integer ; that is, («! + «2 + ag + • • . + a^_2 + a^-i + «m)^ z=:za^ + ai-\ VaJ 4- 3 a^ia.^ + ag + ••• 4- «^) + 3 a^Xa^ 4- «3 H h « J 4- ••• 4- 3 aj(a^ + ag 4- ••• + a^^-x) 4- 6 ttjaattg 4- • • • 4-6 a^_2«m-i<^w (3) SPECIAL METHODS Qd Then, (a^^ a^ + a^-h •- + a^_i + a,, + a«+i)^ = [(tti + as + ag + . . . + a^_i + a J + a^+^f = («i + aa + a3 + •.. 4- of,„_i + a^f + 3 (ai + 0!2 + % H + «^_i 4- a J^«^+i 4-3(ai + a2+«3+ ••• +an,_, + a^)a,,J+a^^,' (^ 135). Then, by (3) and § 134, («! + as + «3 + ••• + cim-i + a^ + a^+i)8 = ai3 + a2^H \-aJ + 3ai2(a2 + «3+ ••• 4-a^) + 3a2^(ai + a3+ ... +aj+ ... 4- 3 a J(ai + ag + . • • + a^_i) + 6 a^a^a^ + ... +6 a^.ga^.^a^ + 3 a,„+i(ai2 + as^ + ... + aj + 2 aiag H \-2 a^a^ + 2 a^ttg H +2 a^a^ + ... + 2 a„_iaj + 3 a^+i^(ai + «2 + «3 + ••• + ttm) + a^+i or, ai« + a2«+ ... +aj + a.+i' + 3 ai2(a2 + • • • H- «^ + a^+i) 4- 3 a22(ai + ^3 + ••• -f ^m+i) + •- + 3 a^+i\ai + as 4- ••' + «m) 4- 6 aittsag 4- . . . + 6 a^_ia^a,„+i. This result is in accordance with the above law. Hence, if the law holds for the cube of a polynomial of m terms, where m is any positive integer, it also holds for the cube of a polynomial of m + 1 terms. But we know that the law holds for the cube of a polynomial of three terms, and therefore it holds for the cube of a poly- nomial of four terms ; and since it holds for the cube of, a polynomial of four terms, it also holds for the cube of a poly- nomial of five terms ; and so on. Hence, the law holds for the cube of any polynomial. Ex. Expand (2 ar^ - ar' 4- 2 x - 3)^ 66 ADVANCED COURSE IN ALGEBRA In accordance with the above law, we have (2 a^3 _ ^ _^ 2 « - Sy. = {2a^y-]-(-xy-^(2xy-{-(-3f + 3(2a:^)X~x' + 2x-3) -f- 3 ( - ^2)2(2 :ij3 + 2 a; - 3) + 3 (2 xy(2 a^-x'- 3) -i-S(-3y(2a^-x''-\-2x) + 6(2 a^) (- x") (2 x) + 6(2 a^) (- x') (- 3) + 6(2 0^) (2 oj) (- 3) + 6(- x') (2 a;) (- 3) = Sx'-x^ + Sx'-2T-12 x^ + 24 .^^-36 x^+Q, x' + (S x^-^ x' + 24 0^ - 12 x^ - 36 i»2 _^ 54 ar^ - 27 x^ + 54 a^ - 24 x^ + 36 a^ - 72 a;^ + 36 of = S x^ -12 0^ + 30 x' -61 x" + 66 a^ - 93 x* + 98 a:^ - 63 a^ + 54.X-27. EXERCISE 6 Write by inspection the values of the following : 1. (6a3aj2)3. 2. (- i ab^c^)"^. 3. (-Sx^yz^y. 4. (3 + 7x2)2. 7. (-Ox?/ -11x0)2. 5. (2a3_5^2c)2. 8. (8 x? -9x^)2; ^j and g being 6. (— m'^n'^ + 4^9^)2. positive integers. Write by inspection the values of the following : 9. (5a2 + i2 63c)(5a2_ I2&3c). 10. ( - 10 m% + 13 x5) (- 10 m^n - 13 x^). 11. (a2m 4- x^'>')(a^'»' — x^") ; w and n being positive integers. Expand the following : 12. (a - & + c) (a - 6 - c). 14. (x2 -\-xy + y^) (x2 - xy + y^). 13. (x + ?/ + 3)(x-?/-3). 15. (a2 + 6a-4)(a2_ 5a + 4). 16. (4x2 + 3x + 7)(4x2 + 3x-7). 17. (m* + 5 to2w2 + 2 «4) (m* - 5 ?7i2^2 _ 2 w^). Write by inspection the values of the following : 18. (x + 2)(x + 10). 22. (wn + ll)(mn + 2). 19. (x-5)(x + 7). 23. (a26 4.3c3)(a2;,_8c3). 20. (x-2 - 4) (x2 - 14), 24. (a-h -5)(a-b-9). 21. (X4- 7a)(x- 15 a). 25. (x + y - 6 02)(a: + y + 12 5;2). SPECIAL METHODS 67 Expand the following : 26. (3 ic2 + 5 X - 4)'-^. 28. (2 x"^ - 3 a:^ + x - 2)2. 27. {a-h-c + ay. 29. (a* + 3 a^ _ 4 ^^2 _ 2 « + 1)2. Write by inspection the values of the following : 30. (a + 3 6)3. 32. (3 oT-h + 2 c3)3. 31. (7x*-x3)3. 33. (5 wx2 - 4 w2/3)8. Expand the following : 34. (a2 + a _ 2)3. 36. (a - & + c - (Z)3. 35. (2 x2 - 4 X + 3)3. 37. (3 x3 - 4 x2 - 2 x + 1)3. Simplify the following : 38. (3 a2 + 5 &)2(3 ^2 _ 5 5)2, 4I. (^ + i)3(^^ _ 1)3. 39. (x + 5)(x-2)(x-5)(x + 2). 42. (x + y - zY(x - y + zy. 40. (2 - x)(2 + X) (4 + x2). 43. (a + 6 + c)3(a + 6 - g)3. 44. (x + y + 0)2 + (y + - x)2 4- (0 + X - ?/)2 + (X + ?/ - z)\ 45. (a + 6 + c) (6 + c - a) (c + « - &) (a + & - c). 46. (m + ny - (m - ny - 3(m + w)2(?7i - n) + 3(m + n){m - tiy. 137. We find by actual division, tj=L^ = a-h. (1) t^l^ = a + h. (2) That is, ^' i/" the difference of the squares of two nurtibers he divided by the su7n of the numbers, the quotient is the difference of the numbers. If the difference of the squares oftivo numbers be divided by the difference of the numbers, the quotient is the sum of the numbers. Ex. Divide 25 ?/V - 9 by 5 yz" -}- 3. By § 130, 25 ?/V is the square of 5 yzK Then, by (1), ^^ •^'^' ~ ^ = 5yz'- 3. 138. We find by actual division, t±J^=::a:'-ab + b\ (1) t:Z-^ = a^^ab-^b\ (2) a+b a—b 68 ADVANCED COURSE IN ALGEBRA That is, If the sum of the cubes of two numbers be divided by the sum of the numbers, the quotient is the square of the first number, minus the product of the first by the second, plus the square of the second number. If the difference of the cubes of tivo numbers be divided by the difference of the numbers, the quotient is the square of the first number, plus the 2oroduct of the first by the second, plus the square of the second number. Ex. Divide 27 a^ -Whj 3a- b. By § 130, 27 a^ is the cube of 3 a. Then, by (2), ^^ ^^' ~ ^' == 9 a^ + 3 ab + b\ 3 a— b 139. The Remainder Theorem. Let it be required to divide 2 x^ — 7 x^ -\- 10 x — 3 hj x — 2. x-2 2a^- - 7 a.'2 + 10 a; - -3 2a^- -3 of -3x' + 6^ 4:X » 4tx- -8 2x^-3x-{-^ 5 The division is not exact, and there is a final remainder 5. Now if we substitute 2 for x in the dividend, we have 2 X 2^ - 7 X 2^ + 10 X 2 - 3, which equals 5. This exemplifies the following law : If any rational integral polynomial, involving x, be not divisible by X — a, the remainder of the division equals the result obtained by substituting a for x in the given polynomial. The above is called The Remainder Theorem. To prove the theorem, let D be any rational integral poly- nomial, involving x, not divisible hj x — a. SPECIAL METHODS 69 Let the division be carried out until a remainder is obtained which does not contain x. Let Q denote the quotient, and R the remainder. Since the dividend equals the product of the divisor and quotient, plus the remainder, we have Q{x-a)+R==D. Substitute in this equation a for x. The term Q{x — a) becomes zero ; and since R does not con- tain X, it is not changed, whatever value is given to x. Then, R must equal the result obtained by substituting a for X in D. 140. The Factor Theorem. If any rational integral polynomial, involving x, becomes zero when X is put equal to a, the polynomial has si; — a as a factor. For by § 139, the remainder obtained by dividing the poly- nomial by a; — a is zero. 141. It follows from § 140 that. If any rational integral polynomial, involving x, becomes zero when X is put equal to — a, the polynomial has x -i-a as a factor. 142. We will now prove that, if n is any positive integer, I. a" — b"" is always divisible by a— b: II. d^ — 5" is divisible bya-{-bifn is even. III. a" + b"" is divisible bya-^bifn is odd. IV. a" 4- &" is divisible by neither a-^b nw a — b ifn is even. Proof of 1. If b be\ substituted for a in a*" — ?>", the result is 6" — 6", or 0. Then by § 140, a" — 6" has a — 6 as a factor. Proof of 11. If — 6 be substituted for a in a** — 6", the result is ( — 6)"— 6" ; or, since n is even, 6" — 6", or 0. Then by § 141, a'^—b"" has a + & as a factor. 70 ADVANCED COURSE IN ALGEBKA Proof of 111. If — 6 be substituted for a in a"" -\- b"", the result is (— 6)"+ 6"; or, since n is odd, — 6" + b"", or 0. Then, a^ + 6" has a + 6 as a factor. Proof of lY. If — 6 or + 6 be substituted for a in a" + b"", the results are {—by-\- b"" or b"" + 6", respectively. Since n is even, neither of these is zero. Then, neither a-\-b nor a — & is a factor of a" + 6". 143. We find by actual division = a^- a'b 4- «&' - &^ = a^ + «'^ + ab' + &', = a^ _ a^ft + a'b^ - aW + b\ = a^ + a^6 + a'^^ + aW + ^' ; etc. a — b In these results, we observe the following laws : I. The exponent of a in the first term of the quotient is less by 1 than its exponent in the dividend, and decreases by 1 in each succeediyig term. II. The exponent of b in the secoyid term of the quotient is 1, and increases by 1 in each succeeding term. III. If the divisor is a—b, all the terms of the quotient are positive ; 'if the divisor is a-\-b, the terms of the quotient are alternately positive and negative. 144. We will now prove, by Mathematical Induction, that the laws of § 143 hold universally. Assume the laws to hold for , where n is any positive iiitesrer. ^ ~ a' -b' a + b a' -b' a -b a' + b' a + b a' -¥ r SPECIAL METHODS 71 Then, ^^Ll^ = a--i 4. a'^-'-b + a^-'h' + • • • + 6""'. (1) a — b Now, = ■ a — b a — b a — b = a" + ^(a""^ + a^'-^b + a^-^^^ _| ^ ^n-i^^ by (^^^^ = a'* + a^'-'^b + «""■&" H h &"• This result is in accordance with the laws of § 143. Hence, if the laws hold for the quotient of the difference of two like powers of a and b divided hy a — b, they also hold for the quotient of the difference of the next higher powers of a and b divided hy a — b. . , . •^ qO TyO But we know that they hold for , and therefore they a — b hold for ^LJZ^. and since they hold for 9^^^ they hold for a—b a—b a' -V 1 ; and so on. a — b f^n 7^n Hence, the laws hold for , where n is any positive . , a — b integer. Putting — b for b in (1), we have a-_( 6)n ^ ^^_^ _^ _ ^^ _^ ,,, _^ ^_ ^y.,^ a-(-b) If n is even, (- by = b% and (- by-^ = - b^-^ (§ 130). Whence, "^^^^^ = a^'^ - a^-'b + a-'%' b--\ (2) a + b If n is odd, (- 6)" = - 6^ and (- 6)"-^ = + &""'. Whence, ^!±_^ = a^-^ - a"-'6 + a^-^ft^ ^ ^n-i^ (3) Equations (2) and (3) are in accordance with the laws of § 143. 72 ADVANCED COURSE IN ALGEBRA op, tm Hence, the laws hold for , where n is any even posi- a-j- 5 ^w _j_ ^n tive integer, and for — — — , where n is any odd positive . . a + 6 integer. 145. 1. Divide a^ - 6^ by a - 6. Bv § 143, tjul- = a« + a^6 + ^452 ^ ^3^3 j^ ^254 _^ ^^5 _|_ ^^^ a — h 2. Divide 16 0^4-81 by 2 aj + 3. By §130, Ux'={2x)\ Then, l|^Il|l=(2a^)3_(2aj)2.3 + (2a;).32-33 = 8 a^ - 12 a?^ _^ 18 a; - 27. The absolute value of any term after the first, in equations (1), (2), and (3), of § 144, may be obtained by dividing the absolute value of the preceding term by a, and multiplying the result by h. This would be the shortest method if the numbers involved were large. EXERCISE 7 Write by inspection the values of the following : 36 a^ - 49 5. R 216 mhi^ + 343 p^ 6 TO«2 + 7 p3 7. a* -64 a + 6 8, m^ — n^ m ~ n 9. 1-x 0. a^ + x^ 11. 6a + 7 6 TO«2 + 7 p3 n - X 2 121 x^ - 64 2/4^2 ^ q4 _ 54 ^^ gS - 64 56 11x3- 8 2/20 ' ' a + 6 * ' a-26 * w3 - 1 o m^ - n^ 10 ^^25 m4 - 256 n — 1 m ~ n 5m — 4 8 + m6 « 1 - x6 j^ 256 a8 _ a;8 2 + m2 1-x 2a + x 125 gg - 27 x8 j^j «L±^. 15 243 x^ + 1024 y^ ^ 5a2_3a; 'oj^a; * 3x + 4y 146. Symmetry. An expression containing two or more letters is said to be symmetrical with respect to any two of them, when they can be interchanged without altering the value of the expression. Thus, a + & + c is symmetrical with respect to a and h ; for, on inter- changing these letters, the expression becomes & + a + c. SPECIAL METHODS 73 An expression containing three or more letters is said to be symmetrical with respect to them when it is symmetrical with respect to any two of them. Thus, ab -\- bc + ca is symmetrical with respect to the letters a, 6, and c ; for if a and b be interchanged, the expression becomes ba -{- ac-}- cb^ which is equal to a6 + 6c + ca. And, in like manner, ab + be ■}■ ca is symmetrical with respect to b and c, and with respect to c and a. 147. Cyclo-symmetry. An expression containing n letters, a, h, c, •••, m, n, is said to be cy do-symmetrical with respect to them when, if a is sub- stituted for &, 6 for c, •••, m for n, and n for a, the value of the expression is not changed. The above is called a cyclical interchange of letters. Thus, (a — b)(b — c)(c — a) is cyclo-symmetrical with respect to a, 6, and c ; for if a is substituted for b, b for c, and c for a, it becomes (c — a)(a — 6) (b — c), which by the Commutative Law for Multiplication is equal to (a — 6)(6 — c)(c — a). 148. Every expression which is symmetrical with respect to a set of letters is also cyclo-symmetrical with respect to them. For since any two letters can be interchanged without alter- ing the value of the expression, the condition for cyclo-sym- metry will be satisfied. But it is not necessarily true that an expression which is cyclo-symmetrical with respect to a set of letters is also sym- metrical with respect to them ; for it does not follow that any two letters can be interchanged without altering the value of the expression. 149. It follows from §§ 146 and 147 that, if two expressions are symmetrical or cyclo-symmetrical, the results obtained by adding, subtracting, multiplying, or dividing them are, respec- tively, symmetrical or cyclo-symmetrical. 150. Applications. The principle of symmetry is often useful in abridging alge- braic operations. 74 ADVANCED COURSE m ALGEBRA 1. Expand (a + 6 + c)l We have, (a + ^ + c)^ = (a + ^ + c) (a + & + c) (a + 6 + c). This expression is symmetrical with respect to a, b, and c (§ 146), and of the third degree. There are three possible types of terms of the third degree in a, b, and c; terms like a^, terms like a^b, and terms like abc. It is evident that a^ has the coefficient 1 ; and so, by sym- metry, b^ and c^ have the coefficient 1. It IS evident that a^b has the coefficient 3 ; and so, by sym- metry, have b-a, b~c, c^b, c^a, and a^c. Let m denote the coefficient of abc. Then, (a + b + cf = a^ 4_ ^3 + c^ + 3 (a26 -f b"-a + bh + c~b + (?a^ ah) + mabc. To determine m, we observe that the above equation holds for all values of a, 6, and c. We may therefore let a = 6 = c = 1. Then, 27 = 3 + 18 + m ; and m = Q. Whence, (a + 6 + c)^ = a^ + ?>3 _^ c^ _^ 3 (a-'ft + &'a + bh + 0^6 + c^a -f- ah) + 6 a&c. The above result may be written in a more compact form by represent- ing the sum of terms of the same type by the symbol S ; read sigma. Thus, (2a)5 = Sa3 + 3 ^a% + 6 abc. 2. Expand {x —y — z)^ -\-{ij — z — x)"" + {z — x — yf. This expression is symmetrical with respect to x, y, and z, and of the second degree. The possible types of terms of the second degree in x, y, and z are terms like x^, and terms like xy. It is evident, by the law of § 134, that x^ has the coefficient 3 ; and so, by symmetry, have y"^ and z-. Let m denote the coefficient of xy. Then, {x-y -zf -\- {y -z-xf -^ {z- x- yf = 3 (a^ -f 2/^ + 2^) + m{xy -{- yz -f- zx). To determine m, put x = y = z = l. Then, 3 = 9 -f 3 m, or m = - 2. SPECIAL METHODS 75 Whence, (x — y — zY-[-{y — z — xf-\-{z—x — yy = 3(x'-^y'-^z')-2(xy + yz + zx). 3. Expand (a H- 6 + c)3 + (a + 6 - c)3 + (^ + c - a)3 + (c + a - b).^ The expression is symmetrical with respect to a, b, and c, and of the third degree. The possible types of terms are terms like a^, terms like a-6, and terms like abc. It is evident, by the law of § 136, that a? has the coefficient 2 ; and so, by symmetry, have W and cl Also, by § 136, o?b has the coefficient 3 + 3 + 3-3, or 6; and so, by symmetry, have Wa, bh, c^b, c\ and ah. Again, abc has the coefficient % — Q — Q> — Q, or — 12. Whence, (a + 6 + c)3 + (a + 6 - c)3 + (6 + c - a)3 + (c + a - bf = 2 (a' + 63 .^ c^) ^ 6 ((^25 _^ ^2^^ _^ ^2^ _l_ ^2^ ^ ^2^ ^ ^2^^) _ 12 a6c. EXERCISE 8 1. In the expansion of an expression which is symmetrical with re- spect to a, &, and c, what are the possible types of terms of the fourth degree? of the fifth degree? of the sixth degree? 2. If one term of an expression which is symmetrical with respect to a, b, and c, is (2 a — & — c) (2b — c — a), what are the others ? 3. Is the expression a(b — c)^ + 6 (c — a)^ + c(a — b)'^ symmetrical with respect to a, &, and c ? 4. Is the expression (x^ — y^y + (y^ — z^y -{- (^2 _ x^y symmetrical with respect to x, y, and z ? Expand the following by the symmetrical method : 6. (a + 6 + c)2. 6. (a + & + c + dy. 7. (x-hy-zy+(y + z-xy+(z-\-x- yy. 8. (2 a - 3 & - 4 c)-^ + (2 & - 3 c - 4 a)2 + (2 c - 3 a - 4 &)2. 9. (a + 6 + c - d)2 + (6 + c + cZ - a)2 + (c + cZ + a - 6)2 + (d + a + 6 - c)2. 10. (a + 6 + c + d)3. 11. (a + 6 + c)3 + (a - 6 - c)3 + (& - c - a)3 + (c - a - by. 12. (X + ?/ - ^) (y + ;3 - X) (5! + cc - I/). 13. [x-^ -^ y^- + z'2 ^ 2 (xy + yz -\- zx)y. 14. (« + ?; + c) (a + 6 - c) (?) + €•-«) (« + c - 6). 76 ADVANCED COURSE IN ALGEBRA VIII. FACTORING 151. To Factor an algebraic expression is to find two or more expressions which, when multiplied together, shall pro- duce the given expression. 152. In the present chapter we consider only the separation of rational and integral expressions (§ 63), with integral nu- merical coefficients, into factors of the same form. 153. A Common Factor of two or more expressions is an expression which will exactly divide each of them. FACTORING 154. It is not always possible to factor an expression ; there are, however, certain forms which can always be factored; these will be considered in the present treatise. 155. Case I. When the terms of the expression ho>ve a com- mon factor. 1. Factor 14 0,6*- 35 a^^l Each term contains the monomial factor 7 aV^. Dividing the expression by 7 aW, we have 2W — 6a\ Then, 14 ab^ - 35 a^h'' = 7 ab' {2 b' -5 a'). 2. Factor (2 m + 3) a;^ + (2 m + 3) if. The terms have the common binomial factor 2 m -f 3. Dividing the expression by 2 m + 3, we have x' -f- y^. Then, (2m-f 3)a;2 + (2m + 3)2/'=(2m+3) (x'-\-f). 3. Factor (a — b)m+{b—a) n. By § 81, 5 _ a = - (a - b). Then, (a — b)m-]-(b — a)n=(a — b)m — (a — b)n = (a — b) (m — n). 4. Factor 5 a (x—y) —3a(x-\-y). FACTORING 77 5a(x-y)-3a(x-{-y) = al5(x-y)-3(x-\-y)'] = a (5 X — 5 y — 3 X —3 y) = a{2x-8y) = 2 a (a; — 4 y). We may also solve Ex. 3 by writing the first term in the form — (6 — a) m. Thus, (a — 6) TO + (6 — a) w = (6 — a) n — (h — a)m = (h — a)(n — m). This agrees with § 91 ; for, by § 91, the signs of tv^o factors of a product may be changed without altering the value of the expression. We may thus have more than one form for the factors of an expression. 156. The terms of a polynomial may sometimes be so ar- ranged as to show a common polynomial factor; and the expression can then be factored as in § 155. 1. Factor ah — ay-\-bx — xy. By § 155, ab — ay-j-bx — xy = a(b — y)-\- x(b — y). The terms now have the common factor b — y. Whence, ab — ay-\-bx — xy= (a -i-x) (b — y). 2. Factor a3 + 2a2-3a-6. The third term being negative, it is convenient to enclose the last two terms in parentheses preceded by a — sign. Thus, a?-\-2a?-3a~Q={a?+2a?)-{3a + e>) = a2(a + 2)-3(a + 2) = (a2-3)(a + 2). EXERCISE 9 Factor the following : 1. (3x + 5)m + (3a; + 5). 3. x\by -2z) - x\2y -\- z). 2. (rii — n)x -{■ {n — m) {y -{■ z) . 4. 4tx{a—h—c) — by(h + c — d), 5. (« - h) {mP- + xz) - (a - b) (m^ - yz). 6. (to - w)* - 2 to (to - n)3 + to2(w - n)2. 7. Sxy + 12 ay + 10 bx + lb ab. 9. 6 - 10 a + 27 a2 _ 45 «». 8. TO* + 6 to3 - 7 TO - 42. 10. 26ab -2S ad -5 be + 7 cd. 11. ax — ay + az — bx + by — hz. 12. 3 am — au + 4 hm — ^hn + cm — 2 en. 13. ax -I ay — az — bx — by + bz + ex + ey — ez. 78 ADVANCED COURSE IN ALGEBRA 157. If an expression when raised to the nth. power (n being a positive integer) is equal to another expression, the first expression is said to be an nth Root of the second. Thus, if a^ =b, a is an nth root of b. 158. The Radical Sign, V, when written before an expres- sion, indicates some root of the expression. Thus, Va'' indicates a second, or square root of a^ ; V «^ indicates a third, or cube root of a^ ; -\/«^" indicates an nth root of a^" ; etc. The index of a root is the number written over the radical sign to indicate what root of the expression is taken. If no index is expressed, the index 2 is understood. An even root is one whose index is an even number ; an odd root is one whose index is an odd number. 159. A rational and integral expression is said to be a perfect square, a perfect cube, or, in general, a perfect nth power, when it has, respectively, a rational and integral square, cube, or nth root. 160. Since {2 a'bf = ^ a^b^ (§ 130), a cube root of ^ a%^ is 2 a-b. Again, since (m^y = m^, a fourth root of m^ is m?. It is evident from this that every positive term, which is a perfect ?ith power, has a positive nth root. We shall call this its principal nth root. • We also have (— m^)* = m^ ; so that another fourth root of m^ is — 7n^. It is evident from the above that every positive term which is a perfect nth power, has, if n is even, in addition to its positive wth root, a negative 71th. root of the same absolute value. In the present chapter, only the principal nth root will be considered. 161. Since (- 3 a^)^ = - 27 «^ (§ 130), a cube root of - 27 x^ is - 3 a^. It is evident from this that every negative term, which is a perfect 7ith power, has a yiegative nth root. We shall call this its princijml nth root. FACTORING 79 162. It will be shown (§ 756) that a number has two differ- ent square roots, three different cube roots, and, in general, n different nth roots. It will be understood throughout the remainder of the work, unless the contrary is specified, that when we speak of the nth root of a term, we mean the principal nth root. 163. Let n be a positive integer, and a and h two equal perfect nth. powers ; then, by Ax. 5, § 66, That is, if two perfect nth powers are equal, their principal nth roots are equal. 164. Any Root of a Power. Eequired the value of a/o^, where m and n are any positive integers. By § 128, {a'^Y = a"•^ Then, by § 157, v^a^ = a*". 165. Any Root of a Product. Let n be a positive integer, and a, 6, c, •••, numbers which are perfect nth powers. By § 157, {VabG^'y = ahC'". Also, (Va X V^ X Vc X ...)" = (Va)" X (V6)« X {2_n2. 8. m2 - 2 9nn + w2 - a;2 + 2 xy - y^. » 9. 16a2- 8a6 + 62- c2_ I0cd-25d2. 10. 28 x?/ - 36 ^2 ^. 49 2^2 _|. 60 0-25 + 4x2. ., 172. Case V. When the expression is in the form X* + axY + y*- Certain trinomials of the above form may be factored by expressing them as the difference of two perfect squares, and then employing § 171. 1. Factor a* + a'-^^^ _^ 5^ 84 ADVANCED COURSE IN ALGEBRA By § 167, a trinomial is a perfect square if its first and last terms are perfect squares and positive, and its second term plus or minus twice the product of their square roots. The given expression can be made a perfect square by adding a^h^ to its second term ; and this can be done if we subtract a^h'^ from the result. Thus, a' + a^h'' + &* = (a^ + 2 o?h\+ h') - o?h^ = {a" 4- &' + ah){a'' + &' - ah), by § 171, = {o? + a6 + h^ia? -ab-\- b'). 2. Factor 9 x^ - 37 a;2 _|. 4 The expression will be a perfect square if its second term is - 12 x'. Thus, 9 o;^ - 37 x2 + 4 = (9 a:^ - 12 aj2 + 4) - 25 a^ . = (3x'- 2y - (5 xf = {Sx' + 5x-2){3x'-5x-2). The expression may also be factored as follows : 9 ic* - 37 xM- 4 = (9 x^ + 12 a:2 + 4) - 49 x^ = (3 a;2 + 2)2 - (7 x)2 = (3 a;2 + 7 a; + 2) (3 x2 - 7 a; + 2). Several expressions in the following set may be factored in two different ways. The factoring of trinomials of the form x* -f ax^y^ + ?/^, when the factors.involve surds, will be considered in § 459. EXERCISE 13 Factor the following : 1. x* + 5ic2 + 9. 5. 9 cc* + 6 a;2^2 + 49 y4. 2. a* - 21 a262 + 36 64. 6. 16 a^ - 81 a^ + 16. 3. 4-33a:2 + 4x4. 7. 04 + 64^2 + 25 771*. 4. 25 m* + 6 »n2n2 + n*. 8. 49 a* - 127 a'^x^ + 81 xc*. Factor each of the following in two different ways : 9. a;4 - 17 x2 + 16. 11. 16 m* - 104 m'^x'^ + 25 x^. 10. 9 - 148 a2 + 64 a*. 12. S6 a^ - 97 a'^m'^ + S6 m\ FACTORING 85 173. Case VI. When the expression is in the form oc^ -{- ax -\- b. By § 133, £c- 4- (m + n)x -\- mn = (x -\- m)(x -\- 7i). If, then, a trinomial is in the form x^ -\- ax -{- b, and a and b are, respectively, the sum and product of two numbers, the factors are x plus one number and x plus the other. The numbers may be found by inspection. 1. Factor x^ + 14:X + 4:5. We find two numbers whose sum is 14 and product 45. By inspection, we determine that these numbers are 9 and 5. Whence, a^ + 14 a; + 45 = (a; + 9)(a;4- 5), 2. Factor x"^ -5x-\-4:. We find two numbers whose sum is — 5 and product 4. Since the sum is negative, and the product positive, the numbers must both be negative. By inspection, we determine that the numbers are —4 and — 1. Whence, a;^ — 5ic + 4 = (a; — 4)(a; — 1). 3. Factor a;2 + 6a;-16. We find two numbers whose sum is 6 and product — 16. Since the sum is positive, and the product negative, the numbers must be of opposite sign ; and the positive number must have the greater absolute value. By inspection, we determine that the numbers are + 8 and —2. Whence, a^ ^Qx -16 = {x-\- S)(x - 2). 4. Factor x' - abx" - 42 a-b\ We find two numbers whose sum is — 1 and product — 42. The numbers must be of opposite sign, and the negative number must have the greater absolute value. By inspection, we determine that the numbers are —7 and +6. Whence, x^ - abx" - 42 a-b- = (0^-7 ab){x' -\- 6 ab). 86 ADVANCED COURSE IN ALGEBRA 5. Factor l + 2a-99al We find two numbers whose sum is + 2 and product — 99. By inspection, we determine that the numbers are + 11 and — 9. Whence, 1 + 2a -99 ^2= (1 + 11 a)(l -9a). If the x^ term is negative, the entire expression should be enclosed in parentheses preceded by a — sign. 6. Factor 24 + 5 a; — ar^. We have, 24 + 5 a; - a^ = - (a;^ _ 5 a; - 24) = -(x-8)(a7H-3) = (8-aj)(3 + a;). In case the numbers are large, we may proceed as follows : Required the numbers whose sum is — 26 and product — 192. One of the numbers must be + , and the otlier — . Taking in order, beginning with the factors +1 x — 192, all possible pairs of factors of —192, of which one is + and the other — , we have : + 1 X - 192. + 2 X - 96. + 3 X - 64. + 4x- 48. + 6 X - 32. Since the sum of + 6 and — 32 is — 26, they are the numbers required. EXERCISE 14 - Factor the following : 1. x2 + 18a; + 56. 2. aj2 + 16 a; - 67. 3. a2-10a-76. 4. ?/4 - 21 2/2 + 104. 5. 77-4x-x2. 6. 84 + 5 n - w2. 7. 1 + 17 ?n + 70 m2. 8. 1 + 5 a& - 14 ^252. 9. (x-?/)2- Voi^x-y)- 10. (m- w)2+21(w-?2) 11. (a + x)2-28(a + a;) + 192. 12. 95_14x4-a;8. 13. 105 + 8 w8 - m^. 14. l + 36a^2/2 + 68a;2y4. 15. a;6 - 17 x^yz'^ + 72 y'^z^. 16. a2 - 6 a6 - 91 62. 17. a2 + 32amn + 112w2w2. 18. CCV + 7 yfiy^z _ 170 Z'^. 16. 19. a:2-(2w + 3w)x + 6win. 180. 20. a;2— (a — h)x — ah. FACTORING 87 174. Ca SE VII. When the expression is in the form ax^ -\-hx-\-c. If a is a perfect square, and h is divisible by Va, we may factor the expression directly by the method of § 173. 1. Factor 9 ic2- 18 a; + 5. We have, 9aj2-18a; + 5 = (3 a;)2-6(3a;) + 5. We find two numbers whose sum is — 6, and product 5. The numbers are — 5 and — 1. Then, ^ x^-l^x-\-b={Zx-b){^x-l). If h is not divisible by Va, or if a is not a perfect square, we multiply and divide the expression by a, which, by § 96, does not change its value. 2. Factor6aj2 + 5a;-4. Multiplying and dividing the expression by 6, we have 6^ I 5^ ^_ 36r>^ + 30x-.24 _ (6a;y + 5(6x)-24 6 6 The numbers are 8 and — 3. Then, 6^ + 5x-4 = ii^±|M^:^. ^ X o Dividing the first factor by 2, and the second by 3, we have 6a^ + 5aj-4=(3a; + 4)(2a;-l). In certain cases, th"e coefficient of ar^ may be made a perfect square by multiplying by a number less than itself. 3. ractor8a^ + 26a;2/ + 15 2/^ Multiplying and dividing by 2, we have 8 ic^ 4- 26 i»?/ + 15 y^ = o ^ (4a;)^ + 13y(4a^) + 30/ 2 ^ (4a; + 10y)(4a; + 3y) 2 = (2ic + 52/)(4a; + 32/). 88 ADVANCED COURSE IN ALGEBRA 4. Factor 2 + 5 a;- 3 aj2. ^ (3xY-5(3x)-Q -3 ^ (3x-6){3x-\-l) -3 = (2-x)(l-\-3x). EXERCISE 15 Factor the following : 1. 4a;2 + 28a: + 45. 8. 72 + 7 a: - 49 x^. 2. 6 x2 + X - 2. ^ 9. 6 - ic - 15 x2. 3. 25 x'^ - 25 mx - 6 m^. 10. 5 + 9 n^ - 18 w*. 4. 10 x2 - 39 X + 14. . 11. 21 x2 + 23 x?/ + 6 ?/2. 5. 12x2 + llx + 2. 12. 18 x2 - 27 a6x - 35 «262. 6. 20 ^2x2 _ 23 ax + 6. 13. 7(a - 6)2 - 30(a - &)+ 8. 7. 36x2 + 12x-35. 14. 12(x + ?/)2 + 17(x + 2/)- 7. 15. 14(m- w)2 + 39a(w- w)+10a2. 16. acx'^ —(ad + &c)x + 6(Z. 175. It is not possible to factor every expression of the form a^ + ax + bhj the method of § 173. Thus, let it be required to factor x^-\-lSx-{- 35. We have to find two numbers whose sum is ^8, and product 35. The only pairs of positive integral factors of 35 are 7 and 5, and 35 and 1 ; and in neither case is the sum 18. In Chap. XIX will be given a general method for factoring any expression of the form am? + 6ic + c. 176. Case VIII. When the expression is the cube of a hinomial. Ex. Factor 8 a^ - 36 a^h^ + 54 a6* - 27 h\ We must show that the expression is in the form of the cube of a binomial, as obtained by the rule of § 135, and find its cube root. P^ACTORING 89 We can write the expression as follows : (2 af - 3(2 ay {3 6^) +3(2 a) (3 by - (3 b'y. This shows that it is a perfect cube, and that its cube root is 2 a - 3 6^ Then, 8 a^ - 36 a'b^ + 54 ab' - 27 6« = (2 a - 3 b^. EXERCISE 16 Factor the following : 1. x^ + 3x^ + Sx + l. 2. 8-12a + 6a2_a3. 3. 1 + 9 m + 27 m2 + 27 m\ 4. 64 n3 - 48 w2 + 12 w - 1. 5. 8 a3 + 36 a^6 + 54 a62 + 27 &3. 6. 27 a^b^ - 108 a'^b^c + 144 abc^ - 64 c^. 7. 125 x^ - 600 a;2?/ + 960 xy^ - 512 yK 8. 216 m6 + 756 m^x^ + 882 m2a;6 + 343 x^. 177. Case IX. When the expression is the sum or difference of two perfect cubes. By § 138, the sum or difference of two perfect cubes is divisible by the sum or difference, respectively, of their cube roots. In either case the quotient may be obtained by the rules of §138. 1. Factor x^-21fz\ By § 166, the cube root of x^ is a?, and of 27 2/V is 3fz. Then one factor is a;^ — 3 'ifz. Dividing x^ — 27 2/V by a.*^ — 3 ifz, the quotient is a;^4-3ary2 + 9?/V (§ 138). Then, x^ - 27 2/V = (a;^ - 3 fz) {d" + 3 ;x^fz + 9 y^z"). 2. Factor a^-\-b^. One factor is a^ + 6^ Dividing a^ + W by a? + 6^ the quotient is a* — a^ft^ _|. ^4^ Then, a« + 6« = (a^ + ft^s^ {a' - a'b' + &*). 90 ADVANCED COURSE IN ALGEBRA 3. Factor (x + af —{x — of: {x + of — {x — ay = \_{x + a)-{x- a)] [(a; + af -\- {x + a) (x - a) -\- {x - a)^] = {x -\- a — X -\- a){x^ -\-2 ax -\- o? -\- x^ — 0? -[■ x^ — 2 ax -\- a-) = 2a{3x'-\-a'). 178. Case X. Wheii the expression is the sum or difference of two equal odd powers of two numbers. By § 142, the sum or difference of two equal odd powers of two numbers is divisible by the sum or difference, respectively, of the numbers. The quotient may be obtained by laws of § 143. Ex. Factor a^ + 32 6^ By §130, 32h' = {2hf. Then, by § 142, one factor is a + 2 &. Dividing a^ + 32 6^ by a + 2 h, the quotient is a^ - a^ (2 h) + a^ (2 hf - a (2 hf + (2 by (§ 143). Whence, a* + 32 6^=(a + 2&)(a^-2a3& + 4a262_8a&-^ + 16 60. Factor the following : EXERCISE 17 1. 8w3-/i3. 3. a^ + 64. 5. 729 a%^ +612 c^^. 2. x^y'' + 125 z^. 4. 216 a^m^ _ 343 ,^9. e. TO3-(m + w)3. 7. {x-\-yy+{x-yY. 9. (2 a + x)3 - (a + 2 a;)3. 8. 27(a-6)3-8 63. 10. .(5a;-2?/)3+(3x-4y)3. 11. x^ + y^. 14. 1 + a;7. 17. 32 a^ -■ b^ 12. a^-\. 15. m^ + n^ 18, 2iSx^ + y^ 13. a^-h\ 16. a^-1. 19. mi4 + 128n7. 20. 1024 a565 _ 243 cw. 179. By application of the rules already given, an expression may often be resolved into more than two factors. If the terms of the expression have a common factor, the method of § 155 should always be applied first. FACTORING 91 1. Factor 2 ax^y^ — 8 axy*. By § 155, 2 ax^y- - 8 axy' = 2 axf {y? -^y") = 2axf{x-\-2y)(x-2y),hj^lll. 2. Factor a« - h\ By §171, a^-h^={a^Jrh^)(p?-'b^- Whence, by § 177, a^-h^={a + h) (a? -ab + b') (a - b)(a' + ab-{- 6^. 3. Factor af-y\ By § 171, x^-f= {x' + y') {x' - y') = (x' + y^){x^J^y^(x^-y^ = {x*-\-y')(x' + y')(x + y){x-y). 4. Factor 3(m^ny-2(m'-n^). 3(m-{-ny-2 (m^ - n^) = 3 (m -^ nf - 2 (m -{- 7i) (m - n) = (m + n)[3 (m + n) — 2 (m — n)] = (m + n)(Sm-hSn — 2m + 2n) = (m -\- n) (m -{- 5 n). 5. Factor a (a- 1)- 6 (6-1). a(a-l) - b{b -1) = a' - a-b' -{-b = a'-b'-a+b = (a-\-b)(a-b)-{a-b) = (a-b){a + b-l). EXERCISE 18 Factor the following : 1. X^- 625. 10. (16 m2 + n^y - 64 mH\ 2. ai2-l. 11. 2a'^x-Sa^x^ + 2a^z^-Sax\ 3. mi6 _ 1 . 12. 9 a2c2 _ 16 a2^2 _ 36 ?,2c2 + 64 62(^2. 4. x^ - 26 a;3 - 27. 13. x^* - 2 x^ + 1. 5. (a-^ + 4 a6 + 62)2 _ (^2 + ^2)2, 14. 729 - n^ 6. 12 ic6 - 18 xs _ 6 X* + 9 x3. 15. a^ft^ + a'^yS _ 63a;2 _ 3.2^^3. 7. 81 TO* - 256 ?i8. 16. 48 x^y - 52 x^y^ - 140 xy^. 8. ai4 _ a;i4. 17. 16 a? _ 72 ae + 108 ^5 _ 54 a*. ' 9. x6 - 16 x3i/3 + 64 ?/5, 18. (rn + ny-2(m + ny+(m-\-ny. 92 • ADVANCED COURSE IN ALGEBRA 19. Resolve a^ + 512 into three factors by the method of § 177. 20. a2 _ ,,^-2 ^a + m. 22. ii^^ - 1024. 21. (x2 + 4 a;)2 _ 37 (^2 + 4 ic) + 160. 23. m^ + w + x^ + x. 24. a2c2 _ 4 52^2 _ 9 ^2^2 + 35 52^2. ^ 25. (m - n) (x2 - 2/2) + (x -f y) Qm"^ - n^). 26. (X - 1)3 4- 6(x - 1)2 + 9(x - 1). 27. a2_452_^_2&. 28. (m + n) (m2 - x2) - (m + x) (m2 - n^). 29. (x2 + 4 y2 _ ;22)2 _ 16 x22/2. 31. a^b^ + 27 a^y^ - 8 bH^ -216 x^yK 30. (x2-9x)2 + 4(x2-9x)-140. 32. (2 x2 - 3)2 - x2. 33. (m2 + m)2 + 2(w2 + m) (w + 1) + (w + l)^. 34. 64a3x3 + 8a3_8x3- 1. 36. (x + 2 i/)^ - x(x2 - 4?/2). 35. (4 a2 _ 62 _ 9)2 _ 35 52. 37. (i + ^s^ + (i + x)^ 38. (a2 + 6 a + 8)2 - 14(^2 + 6 a + 8) - 15. 39. a4-9 + 2a(a2 + 3). 43. m^ - m^ + 32 m^ - 32. 40. (ix^-\-y^)-xy{x + y). 44. a(a - c)- 6(& - c). 41. (a5_8m3)-«(a-2w)2. 45. m%m + p)+ n'^(n -p). 42. 18 a^d + 22 a^b^ + 8 aft^. 46. x^ + 8 x^ + x^ + 8. 47. (27 m3 - x3) + (3 w + x) (9 m^ - 12 wx + x2). 48. (4 a2 + 9)2 - 24 a(4 a2 + 9) 4. 144 ^2. 49. m^ + m^ - 64 m^ - 64. 50. (X2 + ?/2)3 _ 4 x22/2(x2 + ?/2). 51. a5 4. (^46 + a^b-^ + a263 _^ ^54 + 55. 52. (8w3-27) + (2w-3)(4w2 + 4w-6). 180. Factoring by Substitution. By § 140, if the expression becomes O'wlieii x is put equal to a, then ic — a is a factor. The positive and negative integral factors of 6 are 1, 2, 3, 6, -1,-2,-3, and - 6. It is best to try the numbers in their order of absolute mag- nitude. If ic = 1, the expression becomes 1 — 7 -f 10 -|- 6. If ic = — 1, the expression becomes — 1 — 7 — 10 -}- 6. FACTORING 93 li x = 2, the expression becomes 8 — 28 + 20 + 6. If ic = — 2, the expression becomes —8 — 28 — 20 + 6. If ic = 3, the expression becomes 27 — 63 + 30 + 6, or 0. This shows that re — 3 is a factor. Dividing the expression by a; — 3, the quotient is a^ — 4 a; — 2. Then, x^ - 7a^ + 10 a; + 6 = (a; - 3) (a;^ - 4a; - 2). 2. Prove that a is a factor of (a-hb-\-c) (ab + 6c + cd) - (a + 6) (6 + c) (c + a). Putting a = 0, the expression becomes (6 + c) 6c - 6 (& + c) c, orO. Then, by § 140, a is a factor of the expression. 3. Prove that m + n is a factor of m* — 4 m^n + 2 m^n^ + 5 mn^ — 2 n*. Putting m= —rij we have 7^4 ^ 4^4 + 271^ - 5n* - 2n*, or 0. Then, m + w is a factor. EXERCISE 19 Factor the following : 1. a;3 + 4xa + 7x-12. 4. «» - 9a;2+ 15a; + 9. 2. x4-x3 + 6x2 + 14a; + 6. 5. a;8-18a; + 8. 3. x3 - a;2 - 11 X - 10. 6. cc^ - 5x2 - 8x + 4& ; 7. a;4 + 83c3+13x2-13x-4. 8. 2x4-7a;3 + i0a;2-14a; + 12. Find, without actual division, 9. Whether a; - 3 is a factor of a;^ - 6 x^ + 13 x - 12. 10. Whether x + 2 is a factor of x^ + 7 x2 _ 6. 11. Whether x is a factor of x (y + zy + y {z + xy -{- z (x -{■ yy. 12. Whether a is a factor of a^ (b - cy -^ b^ (c - ay + c^ (a - by. 13. Whether x - y is a factor of (x - yy + (y — zy + {z — xy. 14. Whether m + w is a factor of w (m + 2n)8 — n (2 w + ny. 15. Whether a + & + c is a factor of a (6 + c) + 6 (c 4- a) + c (a + 6) + a2 + 62 _f. c2. 94 ADVANCED COURSE IN ALGEBRA 181. Factoring of Symmetrical Expressions. The method of § 180 is advantageous in the factoring of sym- metrical expressions. (§§ 146, 147.) 1. Factor a{b + cY + h{G->ray + c{a->thy-a\h + c)-h\c + a)-c\a + h). The expression is symmetrical with respect to a, 6, and c. Being of the third degree, the only literal factors which it can have are three of the type a ; three of the type a + 6 ; or a + h -\- c, and a factor of the second degree. Putting a = 0, the expression becomes hd'-^ch^-Wc-c'h, orO. Then, by § 140, a is a factor ; and, by symmetry, h and c are factors. The expression, being of the third degree, can have no other literal factor ; but it may have a numerical factor. Let the given expression = mdbc. To determine m, let a = 6 = c = 1. Then, 4 + 4 +4-2-2- 2 = m, or m = 6. Whence, the given expression = 6 dbc. 2. Factor x\y -\- z) -{- y"^ (z -\- x) + z^ {x -\- y) + ^ xyz. The expression is symmetrical with respect to x, y, and z. The only literal factors which it can have are three of the type X ; three of the type a; + ?/ j ot x -\- y -{- z, and a factor of the second degree. It is evident that neither x, y, nor 2; is a factor. Putting X equal to — y, the expression becomes y%y-\-z)-{-y\z-y)-Sy% which is not 0. Then, x-{-y is not a factor ; and, by symmetry, neither y-\-z nor 2; + a; is a factor. Putting X equal to —y — z, the expression becomes (y + ^)\y ■^z)-f-z^-3 yz{y + z) = y^ -{-^ y'z + 3 yz^ -{- z^ - f - z^ -S fz -Syz^ = 0. Then, a; + 2/ + 2; is a factor. FACTORING , 95 The other factor must be of the second degree ; and, as it is symmetrical with respect to x, y, and z, it must be of the form m {x^ + 2/^ + 2!^), or ii (xy -{- yz + zx). The first of these cannot be a factor ; for, if it were, there woukl be terms involving x^, y^, and z^ in the given expression. Then, the given expression = 7i(x -{-y -{- z)(xy -{-yz-\- zx). To determine n, let x = 1, y = 1, and z = 0. Then, 1 + 1 = 2n, and n = l. Then, the given expression =(x-\- y -^z) (xy -\-yz-{- zx). 3. Factor ab (a — 6) + be (6 — c) + ca (c — a). The expression is cyclo-symmetrical with respect to a, 6, and c. It is evident that neither a, b, nor c is a factor. The expression becomes when a is put equal to b. Then, a — b is a factor; and, by symmetry, b — c and c — a are factors. The expression can have no other literal factor, but may have a numerical one. Let the given expression =m(a— b)(b — c)(c — a). To determine m, let a = 2, 6 = 1, and c = 0. Then, 2 = — 2 m, and m = — 1. Then, the given expression = —(a — b) (b — c) (c — a). EXERCISE 20 Factor the following : 1. m3 + 2 TO% + 2 wn2 + n^ 2. (ab + &c + ca)(a + 6 + c) - a^ (6 + c) - b^ (c + a) - c^ (a + 6). 3. x2 (?/ + ^) + yi (^z-i-x) + z^(x + y) + 2 xyz. 4. a (ft + c)2 + 6 (c + a)2 + c (a + &)2 - 4 a6c. 5. aP- (6 - c) + &2 (c _ a) + c2 (a - 6). 6. {x-\-y + z) (xy + yz + zx) - (x + y) (y + z) (z + «). 7. ab (a + &) + 6c (& + c) + ca (c + a) + 2 a6c. 8. (X + y + 0)3 - (x3 + y^ + z^). 9. (x + y + 0) (xy + yz +zx) - xyz. 10. (x - yy + (y- zy + (z- x)3. 11. a3 (6 - c) + 63 (c - a) + c^ (a - 6). 96 ADVANCED COURSE IN ALGEBRA SOLUTION OF EQUATIONS BY FACTORING 182. Consider the equation AxBxGx-" = 0; (1) where Aj B, C, •••, are integral expressions which involve the unknown numbers. By § 49, if Ax B xG X '" =0, some one of the factors A, B, C, •••, must equal 0. We obtain in this way a series of equations A=0,B = 0,G = 0,'". (2) We will now show that these are equivalent (§ 114) to'(l). Any solution of (1) makes Ax B x G X • • • identically equal toO. It then makes at least one of the factors A, B, G, •••, identi- cally equal to ; and hence satisfies at least one of the equa- tions (2). Again, any solution of any of the equations (2) makes A X B X G X ••' identically equal to 0; and hence satisfies (1). Then, (1) and (2) are equivalent. It follows from the above that the equation AxBxGX"'=(i may be solved by placing the factors of the first member sepa- rately equal to zero, and solving the resulting equations. 1. Solve the equation 2x^ — x = 0. Factoring the first member, the equation becomes aj(2a;-l) = (§155). Placing the factors separately equal to 0, « = 0; and 2 aj — 1 = 0, or a; = -• 2. Solve the equation ar^-f-4x- — ic — 4 = 0. Factoring the first member (§§ 156, 171), (a; + 4)(a^-l) = 0, or (a; + 4) (a; -f 1) (a; - 1) = 0. FACTORING 97 Then, a; + 4 = 0, orx=— 4; aj + 1 = 0, ov x= — 1 ; and a; — 1 = 0, or X = 1. 3. Solve the equation (2x- 3f =(x- ly -\-{x- 2y. Factoring the second member, we have [{x-l) + {x-2)^l{x-iy-(x-l)(x-2) + (x-2y] = (2x-S)(x'-2x-\-l-x'-{-Sx-2 + x'-4:X-\-4:) = (2x-3)(x'-3x + 3). Then the given equation can be written (2a; -3) [(2a;- 3)2- (a^ -3a; + 3)] = 0. Or, (2a;-3)(4a;2-12a;4-9-a;2-f-3a;-3) = 0. Or, (2a;-3)(3a;2-9a; + 6) = 0. Dividing both members by 3, (2a;-3)(a;2-3a; + 2) = 0, or (2 a;- 3) (a;- 1) (a;- 2) = 0. Then, 2a;-3 = 0, or a; = ?; a; — 1 = 0, ora; = l; and a; — 2 = 0, or a; = 2. The above examples illustrate the principle (§ 715) that the degree (§ 113) of an equation involving one unknown number indicates the num- ber of its roots ; thus, an equation of the third degree has three roots ; of the fourth degree, four roots; etc. It should be observed that the roots are not necessarily unequal ; thus, the equation x^ — 2x + l = may be written (x — 1) (a; — 1) = 0, and therefore the two roots are 1 and 1. EXERCISE 21 Solve the following equations : 1. 5a;4 + 35a;3 = 0. 8. ic* - 18 a;8 + 32 ^2 = 0. 2. 3x3- 108a; = 0. 9. 6x2 + 7a; + 2 = 0. 3. (4x-3)(4x2-25)=0. 10. 10x2 - 7x- 12 = 0. 4. x2 + 23 x + 102 = 0. 11. 15x2 + x -2 = 0. 5. x2 + 4x-96 = 0. 12. 12x3-29x2 + 15x = 0. 6. x2 - 17 X- 110 = 0. 13. x2-ax + &x-a6 = 0. 7. (5x + l)(x2 + 22x.+ 121)=0. 14. x2 + mx + 7ix + mn = 0. 98 ADVANCED COURSE IN ALGEBRA 15. x^-2cx-Sx-\-lQc = 0. 16. x^ + Sm'^x-bin^x- I5m^=0. 17. (4x2-28x+49)(ic2-3x-10)(8ic2+ 14X-15) =0. 18. 27x3 + 18x'-3x-2 =0. 19. a;3 + 6x2-ic-30zr:0 (§180). 20. x4 + 2x3-13x2-14x+24 = 0. 21. (x-2)2-4(x-2)+3 = 0. 22. (x - 2)3 + 8x3= (8x- 2)3. 23. x2-4-(x-2)(3x2 + 4x-4)=0. 24. (x2-l)(x2-9)= -3(x-l)(x + 3). 25. (2x+l)3- (x+2)3=(x-l)3. 26. (x2-l)(x3-8) -19(x + l)(x2-3x + 2) =0. • 183. It follows from §§ 140 and 182 that If the first member of the equation A = is a rational and integral polynomial involving the unknown number x, and divisible by x — a, the equation has a as a root. For by § 140, the first member has x — a as a factor. If A is divisible by ax — 6, the equation has - as a root. a 2 Ex. Find whether is a root of the equation o 3x'-{-Sa^-^13x' + 9x-i-2 = 0. Dividing the first member by 3 .x -f- 2, the quotient is aP-\-2x'^ 2 -f 3 a? -f- 1 ; then, is a root. o EXERCISE 22 Find whether : 1. 4 is a root of x3 - x2 - 19 x + 28 = 0. 2. - I is a root of 6x3+ 13x2 + 5x + 25 = 0. 3. - is a root of 4 x3 - 11 x2 - 14 x - 15 = 0. 4 4. - 5 is a root of 4 x* + 22 x^ + 9 x2 - 8 x - 15 = 0. 5. - is a root of 15 x* - 17x3 + 7x2- 19x+ 6 = 0. 6. - 3 is a root of 9x* + 26x3 - 8x2 - 11 x - 3 = 0. HIGHEST COMMON FACTOR 99 IX. HIGHEST COMMON FACTOR. LOWEST COMMON MULTIPLE HIGHEST COMMON FACTOR In the present chapter, we consider only rational and integral expres- sions (§ 63), with integral numerical coefficients. 184. The Highest Common Factor (H. C. F.) of two or more expressions is their common factor of highest degree (§ 64) ; or if several common factors are of equally high degree, it is the one having the numerical coefficient of greatest absolute value in its term of highest degree. There are always two forms of the highest common factor, one of which is the negative of the other. Thus, in the expressions a^ — ah and 6^ _ q,^^ either a — h or h — a will exactly divide each expression. 185. Two expressions are said to be prime to each otKer when unity is their highest common factor. In determining the highest common factor of expressions, it is convenient to distinguish two cases. 186. Case I. When the expressions are monomials, or poly- nomials ivhich can be readily fax^tored by inspection. 1. Required the H. C. F. of 42 a^H', 70 a^bc, and 98 a'bH\ The H. C. F. of 42, 70, and 98^s 14. It is evident 1)y inspection that the expression of highest degree which will exactly divide a^b'\ d^bc, ^nd a*b^d^, is a^b. Then, the H. C. F. of the given expressions is 14 a^b. It will be observed, in the above result, that the exponent of each letter is the loiuest exponent with which it occurs in any of the given expressions. 2. Required the H. C. F. of 5 x'y - 45 x'y and 10 xh/ -f- 40 xhf - 210 xy\ 100 ADVAN^CED COURSE IN ALGEBRA By §§ 155, 171, and 173, 5 x'^y — 45 x^y z=5a^y (x^ — 9) = 5x'y{x-]-S)(x-3), and 10 x^y^ + ^Ox^y^ - 210xy^ = 10 xy^ (x" -{- 4. x - 21) , =10xy'(x-{-7)(x-S). The H. C. F. of the numerical coefficients 5 and 10 is 5. It is evident by inspection that the H. C. F. of the literal portions of the expressions is xy (x — 3). Then, the H. C. F. of the given expressions is 5xy(x — 3). EXERCISE 23 . Find the highest common factor of : 1. 64 x3 + 27?/3, 16 ic2 - 9 y^, and 16 x^-\-2ixy + 9 y'^. 2. 2x^- 12 x2 + 16 x, Sx^-Sx^- 36 a;2, and 6x^ + bx'^- 100 x\ 3. 125 w3 - 8, 10 w2 + m - 2, and 25 m2 - 20 m + 4. 4. a* - 3 a2 _ 28, a* - 16, and a^ + a'-j ^- 4 a + 4. 5. 2 ic3 ^ ic2 _ 6 X - 3, 6 a;2 + 19 a + 8, and 8 x^ + 12 x2 + 6 a; + 1. 6. 27 x3 - ?/8, 243 x^ - y^, and 12 x2 - 25 a;?/ + 7 ?/2. 7. a2 + &2 + c2 - 2 a6 4- 2 ac - 2 6c, a2 + &2 _ c2 _ 2 ah, and flj2 _ 52 _ c2 _|- 2 6c, 8. 27 a3 + 135 a26 + 225 ah^ + 125 6^, 3 a2 + 2 a6 - 5 62, and 3 ac - 6 ad + 5 6c - 10 hd, 9. 4x* + 11 a;2 + 25, 2 x^ - 9 a;2 + 14 x - 15, and 2 x^ + x2 - x + 10. 187. Case II. When the expressions are polynomials which cannot he readily factored hy inspectiori. Let A and B be two polynomials, arranged according to the descending powers of some common letter, and let the exponent of that letter in the first term of A be not lower than its ex- ponent in the first term of B. Suppose that, when A is divided by B, the quotient is p, and the remainder C. We will prove that the H. C. F. of B and C is the same as the H. C. F. of A and B. The operation of division is shown as follows : HIGHEST COMMON FACTOR 101 B)A(p pB ~d~ We will first prove that every common factor of B and G is a common factor of A and B. Let F be any common factor of B and C ; and let B=bF, and C=cF. (1) Since the dividend is equal to the product of the quotient and the divisor, plus the remainder, we have A=pB + a (2) Substituting in (2) the values of B and C from (1), A=pbF-{-cF=F(pb-\-c). (3) It is evident from (1) and (3) that i^ is a common factor of A and B. We will next prove that every common factor of A and B is a common factor of jB and C. Let F' be any common factor of A and B ; and let A = mF', and B = nF', (4) From (2), C=A-pB — mjri — pnF' = F' (m— pn). (5) . From (4) and (5), F' is a common factor of B and C. It follows from the above that the H. C. F. of 5 and C is the same as the H. C. F. of A and B. 188. Let A, B, and C have the same meanings as in § 187. Suppose that when B is divided by (7, the quotient is q, and the remainder D ; that when C is divided by D, the quotient is r, and the remainder E, and so on ; and that we finally arrive at a remainder H, which exactly divides the preceding divi- sor G. By § 187, the H. C. F. of C and D is the same as the H. C. F. of B and C; the H. C. F. of D and E is the same as the H. C. F. of C and D ; and so on. 102 ADVANCED COURSE IN ALGEBRA Hence the H. C. F. of G and H is the same as the H. C. F. of A and B. But since ^exactly divides O, His itself the H. C. F. of G and H. Therefore, H is the H. C. F. of A and B. We derive from the above the following rule for the H. C. F. of two polynomials, A and B, arranged according to the de- scending powers of some common letter, the exponent of that letter in the first term of A being not lower than its exponent in the first term of B : Divide A by B. If there be a remainder, divide the divisor by it; and continue thus to make the remaiyider the divisor, and the preceding divisor the divideyid, until there is no remainder. The last divisor is the H. C. F. required. Note 1. It is important to keep the work throughout in descending powers of some common letter ; and each division should be continued until the exponent of this letter in the first term of the remainder is less than its exponent in the first term of the divisor. Note 2. If the terms of one of the expressions have a common fac- tor which is not a common factor of the terms of the other, it may be removed ; for it can evidently form no part of the highest common factor. In like manner, we may divide any remainder by a factor which is not a factor of the preceding divisor. Note 3. If the first term of the dividend, or of any remainder, is not divisible by the first term of the divisor, it may be made so by multiply- ing the dividend or remainder by any term which is not a factor of the divisor. Note 4. If the first term of any remainder is negative, the sign of each term of the remainder may be changed. (See note to § 184.) Note 5. If the given expressions have a common factor which can be seen by inspection, remove it, and find the H. C. F. of the resulting expressions. The result, multiplied by the common factor, will be the H. C. F. of the given expressions. Note 6. The operation of division may usually be abridged by thej use of detached coefficients (§ 104). HIGHEST COMMON FACTOR 103 1. Kequired the H.C.F. of 6a^-{-7a^b-Sab^ and 4.a^b + S d'b^ -Sa¥-^b\ We remove the factor a from the first expression, and the factor b from the second (Note 2). We then find the H. C. F. of 6a^ + 7ab-3b^ and 4:a^-\-S a'b-S ab^-9 b^ Since 4 a^ is not divisible by 6 a^, we multiply the second expression by 3 (Note 3). 4.a^ + Sa'b-3ab^-9b^ 3 6 a^ + 7 ab -3 b^)12 a^ + 24. a^b -9 ab^ -27 b\2 a 12a^-{-Ua'b-6ab^ 10 a'b- 3 ab^-27b^ Since 10 a^b is not divisible by 6 a^, we multiply this remain- der by 3. 60.^+7 ab-3b^30a'b- 9aW-Slb\6b 30 o?b + 35 ab^ - 15 W ~ r44a62-66 6« Dividing the remainder by — 22 b^ (Notes 2, 4), 2 a + 3 6)6 a2 -f- 7 a6 - 362(3a - 6 -2a6' -2ab-3b^ Then, 2 a + 3 5 is the H. C. F. required. 2. Kequired the H. C.F. of 2x^-3^ -x^ + x and 6a;* - a^ + 3ar^ - 2a;. E-emoving the common factor x (Note 5), and using De- tached Coefficients, 2-3-1 + 1)6- 1+ 3- 2(3* 6-. 9- 3+ 3 8+ 6- 5 104 ADVANCED COURSE IN ALGEBRA 2- 3- 1+ 1 4 8 + 6- _ 5)8 -12- 4+ 4(1 8+ 6- 5 -18+ 1+ 4 4 -72+ 4 + 16(-9 -72-54 + 45 29)58 - 29 2- 1 2-1)8+ 6- 5(4 + 5 8- 4 10-5 10-5 The last divisor is 2x — l. Multiplying this by x, the H. C. F. of the given expressions is ic(2aj — 1). 189. The H. C. F. of three expressions may be found as follows : Let A, B, and C be the expressions. Let G be the H. C. F. of A and JB ; then, every common factor of G and O is a common factor of A, B, and C. But since every common factor of two expressions exactly divides their H. C. F. (§ 188), every common factor of A, B, and C is also a common factor of G and C. Whence, the H. C. F. of G and C is the H. C. F. of A, B, and C. Hence, to find the H. C.F. of three exp7*essions, find the H. C. F. of two of them, and then of this result and the third expression. We proceed in a similar manner to find the H. C. F. of any number of expressions. EXERCISE 24 Find the highest common factor of ; 1. 4 a;2 + 4 x - 3 and 6 x^ + 11 a;2 - a; - 6. LOWEST COMMON MULTIPLE 105 2. 6a^- lla^b -7ab^ + 4 fes and 12 a^ - 13 a'^b + 21 ab^-6b^ 3. 9a;4-21x3 + 48a;2_24x and ISx* - 25a;3 + 25x2 - 55 a; + 30. 4. 6 a* + a8 + 5 a2 + 7 a _ 3 and 8 a'^ - 6 a^ + 7 a^ - 9. 5. 6x5 + a:4 + 3a;3-6x2-4a: and 12x5 +8a;4 - 3x3 - lOx^ -4 a;. 6. 8x2-6x-35, 10x3-27 X2-X + 15, and 6x3-13x2-13x4- 20. 7. 6 a3 _ 19 a% + a62 + 6 63, 8 a3 - 18 a'^b - 17 a&2 _ 3 63, and 6 a3 + 23 a^b - 6 ab'^ - S b^ LOWEST COMMON MULTIPLE 190. A Common Multiple of two or more expressions is an expression which, is exactly divisible by each of them. 191. The Lowest Common Multiple (L. C. M.) of two or more expressions is their common multiple of lowest degree; or if several common multiples are of equally low degree, it is the one having the' numerical coefficient of least absolute value in its term of highest degree. There are always two forms of the lowest common multiple, one of which is the negative of the other; thus, in the expressions a^ — ab and 62 _ ^55, either ab(a — 6) or a6(6 — a) is exactly divisible by each expression. In determining the lowest common multiple of expressions, it is convenient to distinguish two cases. 192. Case I. When the expressions are monomials^ or poly- nomials ivhich can be readily factored by inspection. 1. Required the L. C. M. of 36 A*, 60 ay, and 84ca;^ The L. C. M. of 36, 60, and 84 is 1260. It is evident by inspection that the expression of lowest degree which is exactly divisible by a^x, a^y-, and ca^, is a^cx^y^. Then, the L. C. M. of the given expressions is 1260 a^cx^y^. It will be observed, in the above result, that the exponent of each letter is the highest exponent with ivhich it occurs in any of the given expressions. 2. Required the L. C. M. of a;^ -f a; — 6, a^ — 4 a; -|- 4, and a^ — 9 a;. 106 ADVANCED COURSE IN ALGEBRA By § 173, x^ + x-6 = (x + 'S){x-2). By § 169, x'-4.x + 4. = (x-2y. By § 171, ix^-9x = x{x + 3){x-S). It is evident by inspection that the L. C. M. of these expres- sions is x(x — 2y (x + 3) (a; — 3). EXERCISE 25 Find the lowest common multiple of : 1. x^-16x + b0,x^-{-2x- 35, and x2 - 3 a; - 70. 2. 27 a* + 64 a, 18 a* - 32 a"-, and 3 a2 + 7 a + 4. 3. 6 ic2 + 7 a; - 5, 10 x2 - 9 ic + 2, and 8x^ - 12x'^ -{- 6x -1. 4. Sac+ ad-6bc-2bd,ac-4ad-2bc + S bd, and 3 c2 - 11 cd-Ad^. 6. a2 + 4 62 _ 9 c2 _ 4 a&, a2 _ 4 52 _ 9 c2 + 12 be, and ^2 _{. 4 52 _|_ 9 c2 _ 4 Q5& - 6 ac + 12 6c. 6. 8 TO» - w3, 4 m2 - 4 mw + w2, and 16 m* + 4 w2?i2 + n*. 7. a;2 + 5 X + 6, x3 - 19 X - 30, and x^-7x^ + 2x + 40. 193. Case II. Wheyi the expressions are polynomials which cannot be readily factored by inspection. Let A and B be any two expressions. Let F be their H. C. F., and M their L. C. M. ; and suppose that A = aF, and B = bF. Then, AxB = abF\ (1) Since F is the H. C. F. of A and 5, a and & have no common factors ; whence, the L. C. M. of aF and bF is abF. That is, M= abF. Multiplying each of these equals by F, we have FxM=abF\ (2) From (1) and (2), A x B = F x M. (Ax. 4, § 66) That is, the product of two expressions is equal to the product of their H. C. F. and L. C. M. 194. It follows from § 193 that, to find the L. C. M. of two expressions. Divide their product by their highest common factor. LOWEST COMMON MULTIPLE 107 Or, divide one of the expressions by their highest common factor, and multiply the quotient by the other expression. Ex, Required the L. C. M. of x^ -Sxy -^7 y^ Sind a^ -9 afy + 23 xy^ - 15 f. By the rule of § 188, the H. C. F. of the given expressions is x-y. Dividing x^ — S xy -{- 7 y^ hj x — y, the quotient is x—7 y. Then, the L. C. M. of the given expressions is (x-7y){a^-9a^y-h2S xy^ -15f). 195. It follows from § 193 that, if two expressions are prime to each other (§ 185), their product is their L. C. M. 196. The L. G. M. of three expressions may be found as follows : Let A, B, and C be the expressions. Let M be the L. C. M. of A and B ; then every common mul- tiple of M and (7 is a common multiple of A, B, and C. But since every common multiple of two expressions is ex- actly divisible by their L. C. M., every common multiple of A, B, and C is also a common multiple of M and C. Whence, the L. C. M. of M and C is the L. C. M. of A, B, and C. Hence, to find the L. C. M. of three expressions, find the L. C. M. of two of them, and then of this result and the third expression. We proceed in a similar manner to find the L. C. M. of any number of expressions. EXERCISE 26 Find the lowest common multiple of : 1. 8 ic2 - 6 X - 9 and 6 ic8 - 7 ic2 - 7 X + 6. 2. 6 a8 + a^h - 11 a&2 - 6 68 and 6 «» - 5 a^b -^ab- + Z b\ 3. 8 x8 - 22 a;2 _ 6 X and 8 a:6 ^ 6 a;6 _ 11 x* - 23 x8 - 5 xK 4. X* - 2 x8 - 2 x2 + 7 X - 6 and x* - 4 x^ + x^ + 7 x - 2. 5. 4 a8 + 4 a2 _ 43 )-(2& + a) ^ 3a + 36-25-a ^ 2a + 5 4. Simplify ^ : ^ ^ {x-y){x-z) {y-x)(y-z) {z-x){z-y) By § 202, we change the sign of the factor y — x in the second denominator, at the same time changing the sign before the fraction ; and we change the signs of both factors of the third denominator. FRACTIONS 115 The expression then becomes 1 + 1 {x-y){x-z) (x-y){y-z) (x-z)(y-z) The L. C. D. is now (x — y)(x — z) {y — z)-, then the result (y — z) -{- (x — z) — (x — y) ^ y — z + X — z — X -\- y (x-y)(x-z){y-z) (x-y)(x-z)(y-z) 2y-2z ^ 2(y-z) . ^ 2 (x -y){x- z) (y -z) {x- y) {x -z)(y — z) {x -y)(x- z) 3x-5 5. Simplify 2 a; - 3 x + l 2^ 3 3x-5 ^ (2x-3)(x-{-l)-(3x-5) x-\-l x + l ^ 2x^-x-S-Sx + 5 ^ 2x^-4.x-\-2 X-^1 iC + l 6. Simplify-J- + -^ + ,-^ + ,-^- 1 — x l-{-x l + or l + ic* We first add the first two fractions ; to the result we add the third fraction, and to this result the fourth fraction. 1 1 ^ 1+,^ + !-^ ^ ^ 2 1-a; l + x (l-hx)(l-x) l-x"' 2 2 ^ 2(l+x') + 2(l-a^ ^ 4 l-x^'^l + x' {l-i-x^)(l-af) 1-x'' 4 4 ^ 4(l + a;0+4(l-a;^) ^ 8 1-a^'^l-^x' (l-\-x')(l-x*) 1-a^' 7. Simplify -i- L^+ ^ ^ a — 1 a + l a — 2 aH-2 We first combine the first two fractions, then the last two, and then add these results. _1 1 _ a + l-(«-l) _ 2 a-1 a + 1 (a4-l)(ct-l) a^-l 1 1 ^ a^2-(a-2) ^ 4 a- 2 a + 2 {a-j-2){a-2) 0^-4:' 116 ADVANCED COURSE IN ALGEBRA 2 4 _2a2-8 + 4a2-4_ 6a2-12 4 (a2-l)(a2-4) a^-5a2 + 4 • MULTIPLICATION OF FRACTIONS 210. By § 30, (1), we have the following rule for the product of two fractions : Multiply the numerators together for the numerator of the product, and the denominators for its denominator. 211. By § 210, ^ X c = ^ X J = y. (1) Dividing both numerator and denominator by c (§ 200), fxc = -^. (2) -T- c From (1) and (2), we have the following rule for multiplying a fraction by a rational and integral expression : If possible, divide the denominator of the fraction by the expression; otherwise, midtiply the mimerator by the expressioyi. 212. Common factors in the numerators and denominators should be cancelled before performing the multiplication. Mixed expressions should be expressed in a fractional form (§ 209) before applying the rules. ^^ ... 1 lOa^y . 3M ¥ 10 a^y 3 6V ^ 2 X 5 x 3 x a^b^x'^y ^5b^x dbJ" ^ 4:ay~ 32 X 22 X a^ba^y' 6y ' The factors cancelled are 2, 3, a^, b, x^, and y. 2. Multiply together f + ^^ 2-^^, and ^^. x^-{-x — 6 \ x — Sj x^ — 4: _ y4-2a; 2 x-Ci-x + 4r x'^-9 X ,, X ic2_l_a;_6 x-3 X'-4: x(x-i-2) .. a;- 2 (a; + 3)(a;~3) ^ x (x-\-S)(x-2) ic-3^ lx + 2){x-2) x-2 FRACTIONS 117 The factors cancelled are x-\-2, x — 2, x-\-S, and x — 3. Dividing the denominator by a — 6, we have 3. Multiply V4^ by «-^- «^ + ^^(a-5) = -^ + 4. Multiply by m + 71. m — n Multiplying the numerator by m-\-n, we have m . , X m^ 4- win X (m + n) m—n m—n DIVISION OF FRACTIONS 213. By § 30, (2), we have the following rule : To divide one fraction by another, multiply the dividend by the divisor inverted. 214. By§213, ^^c = ^x-==^- (1) b b c be Dividing both numerator and denominator by c (§ 200), From (1) and (2), we have the following rule for dividing a fraction by a rational and integral expression : If possible, divide the numerator of the fraction by the expres- sion ; otherwise, midtiply the numerator by the expression. 215 1 D" *de ^ ^^^ \) ^ ^^^^ . 5 ^y^ 10 a^y Wehave 6 a^6 . 9 a^&^ ^ 6 a^6 10 a^/ ^ 4 y« ' 5 xh/ ' 10 a^/ 5 0^2/' ^ a^W 3 b^x Mixed expressions should be expressed in a fractional form (§ 209) before applying the rules. 2. Divide 2-^^^ by 3- ^^-)^ . 118 ADVANCED COURSE IN ALGEBRA 2a;-3\ [^ '^x' x-^1 f) a;-|-2-2a; + 3 . 3a^-3-3a^ + 13 a^ - 1 ^ 5(a; + 1) (g; - 1) ^ a; - 1 oj + l 10 2x5x(a; + l) ,3 3. Divide '\ by m Dividing the numerator by m — w, we have -^-— -f- (m - n) ^ — ^^ ; — 4. Divide tA^ by a + &. a-h ^ Multiplying the denominator by a + 6, we have a-h ^ ^ a^-h^ If the numerator and denominator of the divisor are exactly contained in the numerator and denominator, respectively, of the dividend, it follows from § 210 that tlie numerator of the quotient may he ohtained hy dividing the numerator of the' dividend by the numerator of the divisor ; and the denominator of the quotient hy dividing the denominator of the dividend by the denominator of the divisor. 5. Divide 9^-Y by ^l+ll. or —y^ x — y We have, l^ziAt^Sx±2y^Sx-2y^ x^-y^ x-y x-^y 216. By §213, 1 = 1 x- = -- ci a a b Hence, the reciprocal of a fraction is the fraction inverted. COMPLEX FRACTIONS 217. A Complex Fraction is a fraction having one or more fractions in either or both of its terms. FRACTIONS 119 It is simply a case in division of fractions ; its numerator being the dividend, and its denominator the divisor. 218. 1. Simplify ^ . « X K;r^Si 213) = h —- ^^ ~ ^ bd — c^ ^ bd — c d d It is often advantageous to simplify a complex fraction by multiplying its numerator and denominator by the L. C. M. of their denominators (§ 199). a a 2. Simplify ^^^ ^ + ^ + a — b a -\-b The L. C. M. of a + & and a - 6 is (a + b)(a - b). Multiplying both terms by (a -\-b)(a — b), we have g (g 4- 6) — g (g — 6) _ a^ + g6 — a^ + O'b _ 2ab b(a-\-b)-^ala-b)~ab + b'-{-a^-ab~d'-{- b^' 1 3. Simplify 1 + X in examples like the above, it is best to begin by simplifying the lowest complex fraction ; thus, 1 _ 1 _ x-^1 _ x-\-l -j . 1 ~-. X ~ x + l ■i-x~ 2x-^l X EXERCISE 27 Reduce to their lowest terms by factoring : 8x3-125 3 a;2_ -9y2_z2^eyz 2 a;3 + x'^ - 15 X ' x^ - - 9 2/2 + ^2 _ 2 X2; , (x2-49)(x2- 16X + 63) ^ 21 _ x - 10 x2 (x2 - 14 X + 49)(x2 - 2 X - 63) 15 x?/ - 20 x - 21 ?/ + 28 120 ADVANCED COURSE IN ALGEBRA 5 ojg + 28 aW + 27 b^ g 2x^-bx^-x + Q a* + 9 ^262 + 81 ^>6 * ' 27 - 30 x + 54 x'-^ - 8 x^* Reduce to their lowest terms by finding the H. C. F. : ^ 10 a;2 - 7 X - 6 g ex^-x^- llx + 6 * 4x3-4x2-5x- 1* ■ 9x8- 18x2 + 11 X -2' 8 4 gs 4 13 a26 _ 4 q&2 - 6 &3 ^^ 2 x^ - 9 x^y - 2 xy^ _ 15 y^ 8 a2 + 14 <26 _ 15 62 * ' 2 x^ - 7 x'^y - 16 xy2 + 5 y^' Reduce each of the following to a mixed expression : jj 9x^-2x3-20 j2 m^ + n^ ^3 12x3-3x2-22x4-8 12 x^ m — n ' 3 x2 — 5 Simplify the following : "• 4^^x(2^ + j,). 19. we have x -(x^ -\) ^\, ov x - x'^ ^ \ = \, ov x - x'^ = Q. 126 ADVANCED COURSE IN ALGEBRA Factoring the first member, x(l — x) =0 ; and cc = or 1. Now the value x = 1 satisfies the equation x'^ — 1 = ; then it must be rejected, and the only solution is x = 0. 4. Solve the equation --\ - = — + x + S x + 9 a; + 10 x i-6 Adding the fractions in each member, we have 7 a; + 58 ^ 7 a? + 58 (a; + 8)(a^ + 9)~(a; + 10)(x + 6)' Since 7 a; + 58 is a factor of each member, we may place it equal to zero (§ 182). Then, 7 x + 58 =:= 0, or a; = - — • The remaining root is given by 1 1 (a; + 8) (a; + 9) (a; + 10)(a; + 6) or, (x + 8) (aj + 9) = (a? + 10)(aj + 6), or, a^ + 17 a; + 72 = x^ + 1 6 a; + 60. Whence, a; = — 12. 58 Neither nor — 12 satisfies the equation (X + 8) (X + 9) (x + 10) (x + 6) = 0. 5. Solve the equation 1 — = 2. 2 a; — 3 a;^ + 4 Dividing each numerator by its corresponding denominator, we can write the equation in the form l + -^ + l-^ + i = 2, or -1 ^ = 0. 2a;-3 a^ + 4 ' 2a;-3 a;^ + 4 Clearing of fractions, 2a:2^8_2a^_5a; + 12 = 0; whence, a; = 4. EXERCISE 28 Solve the following : J 8x-ll 7x4-4 3x-8 _Q g 12x-5 3x + 4 _ 4x-5 9 12 8 ■ ■219X + 3 7* FRACTIONAL AND LITERAL EQUATIONS 127 3 2x 4a;^-( _2 3 5X + 42ic-l 1 ) 5 4x2-9 *2x + 3 3x-2~ 6 *• x-b 5x + 4_ o 6 ^^ 7x _12(l-a;2) x+8 "' 'l-x 3 + x X2 4-2X-3 7. 2x-15x + 6_o 23x2- 10 3x + 5 7-2x " 6x2-llx-35 8. 3 5 2 4 a;2 _ 9 x2 + 7 X + 12 x2 - 16 x'^ - 7 x + 12 9. X2 4-3 2x-l 1 _Q 2x8-16 3x2 + 6x + 12 6x-12 * 10. x-3x+4_ 8x+2 ^ x^\ x-'2. x^-x-2 11. 2 1 7 1 2x-l 3x + 2 6x2 + x-2 2 12. 8 3 10 5 x-3 x+7 x-9 x+2 13. 5 1 10 4 2x-l 6x + 5 3x-4 4x + l 14. x-fl x + 4_x + 2 x + 3 x-\ x-4 X- 2 x-3 15. x+2 x-3 x+4_3 x-3 x+4 x+2 16. 7 X + 10 x2 - 3 x2 + 2 _ o x2 + X - 6 x2 - 6 X + 8 x2 - X - 12 ~ 17. x2-2x + 5x2 + 3x-7_;, .x2-2x-3 x2 + 3x+l 18. X — 1 1 x-5 x-7 |X — 7_^ x-4 224. Problems involving Fractional Equations, leading to Linear Equations with one Unknown Number. 1. A can do a piece of work in 8 days, which B can perform in 10 days. In how many days can it be done by both working together ? Let X — number of days required. Then, - = the part both can do in one day. 128 ADVANCED COURSE IN ALGEBRA Also, - = the part A can do in one day, 8 and — = the part B can do in one day. By the conditions, - H = - • 8 10 a; 5 X + 4 5c = 40. 9 a; - 40. Whence, x — 4|, number Of days required. 2. The second digit of a number exceeds the first by 2 ; and if the number be divided by the sum of its digits, the quotient is 4^. Find the number. Let X — the first digit. Then, x + 2 = the second digit, and 2 a; + 2 = the sum of the digits. The number itself equals 10 times the first digit, plus the second ; then, 10 a: + (x + 2), or 11 * -^ 2 i= the number. By the conditions, li^±2 ^ 34 ^ ^ 2a; + 2 7 77 x+ 14 = 68x4-68. 9 X = 54. Whence, x = 6. Then, the number is 68. EXERCISE 29 1. The denominator of a fraction exceeds twice the numerator by 4 ; and if the numerator be increased by l4, and the denominator decreased by 9, the value of the fraction is - . Find the fraction. o 2. A can do a piece of work in 3^ hours, B in 3| hours, and C in 3| hours. In how many hours can it be done by all working together ? 3. The second digit of a nnmber of two figures exceeds the first by 5 ; and if the number, increased by 1, be divided by the sum of the digits increased by 2, the quotient is 3. Find the number. 4. The numerator of a fraction exceeds the denominator by 5. If the numerator be decreased by 9, and the denominator increased by 6, the sum of the resulting fraction and the given fraction is 2, Find the fraction. FRACTIONAL AND LITERAL EQUATIONS 129 6. A tank has three taps. By the first it can be filled in 3 hours 10 minutes, by the second it can be filled In 4 hours 45 minutes, and by the third it can be emptied in 3 hours 48 minutes. How many hours will it take to fill it if all the taps are open ? 6. A freight train runs 6 miles an hour less than a passenger train. It runs 80 miles in the same time tliat the passenger train runs 112 miles. Find the rate of each train. 7. The digits of a number are three consecutive numbers, of which the middle digit is the greatest, and the first digit the least. If the number be 220 divided by the sum of its digits, the quotient is — '-. Find the number. 8. A man walks 13J miles, and returns in an hour less time by a carriage, whose* rate is 1| times as great as his rate of walking. Find his rate of walking. 9. A vessel runs at the rate of llf miles an hour. It takes just as long to run 23 miles up stream as 47 miles down stream. Find the rate of the stream. 10. A can do a piece of work in two-thirds as many days as B, and B can do it in four-fifths as many days as C. Together they can do the work in 3/j- days. In how many days can each alone do the work ? 11. The first digit of a number of three figures is three-fourths the second, and exceeds the third digit by 2. If the number be divided by the sum of its digits, the quotient is 38. Find the number. 12. A and B together can do a piece of work in 5} days, B and C together in 6f days, and C and A together in 5| days. In how many days can it be done by each working alone ? 225. Literal Equations, leading to Linear Equations with one Unknown Number. Ex. Solve the equation -^ =^+1* = «' + *'■ X — a x-i-a y? — o? Multiplying each term by a:^ — a^, £c (a? -f a) - (a; -f 2 6) (ic - a) = a^ + 5^, o?-\-ax — x^— 2 6a; 4- aa; + 2 a6 = a^ + W, 2ax-2bx^a^-^2ab-i-b\ Factoring both members, 2 a; (a — 6) = (a — 6)1 Dividing by 2 (a - 6), x = iSll^ = ^Lul. ■ ^ ^ ^. ^\ 2{a-b) 2 130 ADVANCED COURSE IN ALGEBRA Since ^ — — does not satisfy the equation x^ — a^ = 0, it is the correct solution. EXERCISE 30 Solve the following : 1. (x-2a-&)2-(x + a + 2&)2 = 0. o hx d^ + &2 _ g^ x{a - h) a a^ ~ b^ b ' g 5 _ 2 ^ 6m 5 X + 2 m 4 X — 3 TO 20 x"^ — 7 wix — 6 w^ . 4 X + 3 71 4 X - 5 71 _ 10 n^ X + 2 n Sn — X x^ — nx — ii n^ 5. (x + a - 6)3 - (X - a + 6)3 = 2(a - 6)(3 x2 + x). 6. ^^-^' + ^^-^' ^a + 6. X — 6 X — a - X — g X — 6 X — c _ ?)c(x + 6)— «6^ — q'^c 6 c a abc o a b a — b x — a X — b X — c g X L = 1 _ ^^ - 2 « - 4 a2 ' x2 - 4 a^ 2a~ x2 - 4 a2 10 ^ — <^ x4-a _ 2ax4-l^a^ ■, X + 2 a X — 3 a x2 — ax — 6 ^2 117 3 11. 12. x-2a 6x + a 3x-8a 2x-3a 4x X 4x X X— 4?i X + n X -}- 4n x + 3w 13. (x + a)3+(x + by -\- (X + cy = 3(x + a)(x + 6)(x + c). , ^ m + ?i 2 m m — 71 _ q x + wi — w X — m + n X — w — n jg a b a — b X — 26 X — 2a X — a — 6 ^g x2 - 2 ax + 5 a2 3 x2 + 3 ax - 2 g^ _ ^ x2 - 2 ax - 3 a2 x2 + ax + 2 a2 J- x + q I x+6 ■ X — g — 6 _g X — a x — b x + a4-6 ,Q x+w X — 2w_x + 6n x + 3w X — w x + 2?i x — 6 n x — 3 n 19. (x + 2a)3+(x+ 6)3:=(2x+ 2a 4-6)3. FRACTIONAL AND LITERAL EQUATIONS 131 226. Problems involving Literal Equations, leading to Linear Equations with one Unknown Number. Prob. Divide a into two x^arts such that m times the first shall exceed n times the second by h. Let X = the first part. Then, a — x = the second part. By the conditions, mx = n{a — x) + h. mx = an — nx +b. mx + nx = an -\- b. x{in + n) = an -i- b. Whence, x = ^^1+-^, the first part. (1) m + w Also, . a-x = - an + b_am-}-an-an-b m -\- n m -\- n = TILnA^ the other part. (2) m -i- n The results can be used as formuloe for solving any problem of the above form. Thus, let it be required to divide 25 into two parts such that 4 times the first shall exceed 3 times the second by 37. Here, a = 25, m = 4, n = 3, and b = 37. Substituting these values in (1) and (2), the first part ^25x3 + 37^7^+37^112^ 7 7 7 and the second part = ^A^iA^IL = M^lIL = ^1 = 9. 7 7 7 EXERCISE 31 1. Divide a into two parts whose quotient shall be m. 2. Two men, A and B, a miles apart, set out at the same time, and travel towards each other. A travels at the rate of m miles an hour, and B at the rate of n miles an hour. How far will each have travelled when they meet ? 3. Divide a into three parts such that the first shall be one-with the second, and one- nth the third. 4. If A can do a piece of work in a hours, B in 6 hours, C in c hours, and D in d hours, how many hours will it take to do the work if all work together ? 132 ADVANCED COURSE IN ALGEBRA 5. What principal at r per cent interest will amount to a dollars in t years ? 6. In how many years will p dollars amount to a dollars at r per cent interest ? 7. Divide a into two parts such that one shall be m times as much above h as the other lacks of c. 8. A grocer mixes a pounds of coffee worth m cents a pound, h pounds worth n cents a pound, and c pounds worth p cents a pound. Find the cost per pound of the mixture. 9. A was m times as old as B a years ago, and will be n times as old as B in 6 years. Find their ages at present. 10. If A and B can do a piece of work in a days, B and C in 6 days, and A and C in c days, how many days will it take each working alone ? 227. A linear equation containing hut one unknown number cannot have more than one root. Every linear equation containing but one unknown number, can be reduced to the form ic = a. If possible, let this equation have two different roots, ri and rg. Then, by §110, r^ = a, and ra = a. Whence, r^ = r^ (Ax. 4, § 66). But this is impossible, since by hypothesis, r^ and r^ are dif- ferent j hence, a linear equation containing but one unknown number cannot have more than one root. SIMULTANEOUS LINEAR EQUATIONS 133 XII. SIMULTANEOUS LINEAR EQUATIONS CONTAINING TWO OR MORE UNKNOWN NUMBERS 228. An equation containing two or more unknown numbers is satisfied by an ijidefinitely great number of sets of values of these numbers. Consider, for example, the equation a; + .y = 5. Putting x = l, ^Ye have 1 -{- y = 5, or y = 4:. Putting a; = 2, we have 2 + 2/ = 5, ori/ = 3; etc. Thus the equation is satisfied by the sets of values a; = 1, 2/ = 4, and a; = 2, y = 3 3 and this could be extended indefinitely, for we may give to x any numerical value whatever. An equation which has an indefinitely great number of solu- tions is called an Indeterminate Equation. 229. Consider the equations r x + y = 5, (1) I 2a; + 22/ = 10. (2) By § 119, equation (2) is equivalent to (1). Hence, every solution of (1) is a solution of (2), and every solution of (2) is a solution of (1). Again, consider the equations (x-\-y = 5, (3) U - 2/ = 3. (4) Equation (3) is satisfied by the set of values » = 4, ?/ = 1 ; as also is equation (4). But (4) is not satisfied by every solution of (3), nor is (3) satisfied by every solution of (4). Thus, (3) and (4) are not equivalent. If two equations, containing two or more unknown numbers, are not equivalent, they are called Independent 134 ADVANCED COURSE IN ALGEBRA 230. Consider the equations (x-^y = 5, (1) U + 2/ = 6. (2) It is evidently impossible to find a set of values of x and y which shall satisfy both (1) and (2). Two equations which express incompatible relations between the unknown numbers involved are called Inconsistent. If they express possible relations between the unknown numbers, they are called Consistent. 231. A system of equations is called Simultaneous when each contains two or more unknown numbers, and every equation of the system is satisfied by the same set or sets of values of the unknown numbers. A Solution of a system of simultaneous equations is a set of values of the unknown numbers which satisfies every equation of the system. 232. Two systems of equations, involving two or more un- known numbers, are said to be equivalent when every solution of either system is a solution of the other. PRINCIPLES USED IN SOLVING SIMULTANEOUS EQUATIONS 233. If for any equation of a system an equivalent equation be put, the resulting system is equivalent to the first. ^'' |c = A (1) be equations involving two or more unknown numbers; and E = F an equation equivalent to (1). To prove the system of equations jA = B, [E=F, (2) equivalent to the first system. Since every solution of (1) is also a solution of (2), every solution of the first system is a solution of the second. SIMULTANEOUS LINEAR EQUATIONS 135 And since every solution of (2) is a solution of (1), every solution of the second system is a solution of the first. Therefore, the two systems are equivalent (§ 232). In like manner, the theorem may be proved for a system of any number of equations. 234. If for either equation, in a system oftivo, tve put an equor tion luhose first and second members are the sums, or differences, respectively, of the given first and second members, the resulting system will be equivalent to the first. • Let \C = D, be equations involving two or more unknown numbers. To prove the system of equations equivalent to the first system. Any solution of the first system, when substituted for the unknown numbers, makes A equal to B, and C equal to D. It then makes A+ C equal to 5 + ^ (§ 115, 1). Then it is a solution of the second system. Again, any solution of the second system makes A equal to B, and A -\- G equal io B -\- D. It then makes C equal to D (§ 115, 2). Then it is a solution of the first system. Therefore, the two systems are equivalent. In like manner, the first system is equivalent to the system A = B, A-C = B-D, 235. It may be proved, as in § 234, that if the equations A = B, C=D,E = F,etG., (1) involve two or more unknown numbers, and any equation be replaced by an equation whose first member is the sum of any of the given first members, and second member the sum of the corresponding second members, the resulting system will be equivalent to the first. im ADVANCED COURSE IN ALGEBRA The above is also the case if the signs of both members, in any of the equations (1), be changed from + to — . For example, any equation may be replaced by the equation A-\-C-E = B-i-D-F. 236. If any equation of a system be solved for one of the unknown numbers, and the value found be substituted for this unknown number in each of the other equations, the resulting sys- tem will be equivalent to the first. ., ' [A = B, (1) ^^^ \C=D, (2) be equations involving two unknown numbers, x and y. Let E be the value of x obtained by solving (1) ; and let F=GhQ the equation obtained by substituting E for x in (2). To prove the system of equations x = E, (3) equivalent to the first system. We know that (3) is equivalent to (1) ; hence, any solution of the first system is a solution of (3) (§ 233). Again, any solution of the first system makes E identically equal to x ; and also makes G equal to D. Then it must make the expression obtained by putting E for X, in (7, equal to the expression obtained by putting E for x, in D ; that is, it makes F equal to O. Then, any solution of the first system is also a solution of the second. Again, since (3) is equivalent to (1), any solution of the second system is a solution of (1). Also, any solution of the second system makes x identically equal to E, and also makes F equal to G. Then it must make the expression obtained by putting x for E, in F, equal to the expression obtained by putting x for E, in G ; that is, it makes C equal to D. Then, any solution of the second system is also a solution of the first. SIMULTANEOUS LINEAR EQUATIONS 137 Therefore, the two systems are equivalent. In like manner, the theorem may be proved for a system of any number of equations, involving any number of unknown numbers. 237. Any equation involving two unknown numbers, x and y, can be reduced to the form ax -{-by = c. If we have two independent simultaneous equations of the form ax-\- by = c, they may be combined in such a way as to form a single equation involving but one unknown number. This^ operation is called Elimination. There are three principal methods of elimination. I. ELIMINATION BY ADDITION OR SUBTRACTION. 238. 1. Solve the equations Multiplying (1) by 4, Multiplying (2) by 3, Adding (3) and (4), Whence, Substituting a; = 2 in (1), Whence, - 3?/ = 9, or y = - 3. (8) The above is an example of elimination by addition. (The principles demonstrated in §§ 116 to 119, inclusive, and § 122, hold for equations with more than one unknown number.) By § 119, equation (1) is equivalent to (3), and (2) to (4). Then, by two applications of § 233, the given system is equivalent to the system (3) and (4) . By § 234, the system (3) and (4) is equivalent to the system (3) and (5). By § 122, equation (6) is equivalent to (5) ; and then by § 233, the system (3) and (5) is equivalent to the system (3) and (6), or to the sys- tem (1) and (6). By § 236, the system (1) and (6) is equivalent to the system (6) and (7) ; which by § 233 is equivalent to the system (6) and (8). Thus, the given system is equivalent to the system (6) and (8) ; and since no solutions are introduced nor lost, (6) and (8) form the correct solution. 5x-Sy = 19. (1) ,7x + 4y= 2. (2) 20a;-12y = 76. (3) 21x + 12y= 6. (4) 41 a; = 82. (5) X:= 2. (6) 10 -31/ = 19. (7) 138 ADVANCED COURSE IN ALGEBRA 15x-\-Sy= 1. (1) 2. Solve the equations ' 10a;- 72/= -24. (2) Multiplying (1) by 2, S0x-{-16y= 2. (3) Multiplying (2) by 3, 30a;-21y= -72. (4) Subtracting (4) from (3), S7 y = 74, and y = 2. Substituting y = 2m (1), 15 a; + 16 = 1. Whence, 15x= — 15, and x= —1. The above is an example of elimination by subtraction. We speak of adding a system of equations when we mean placing the sum of the first members equal to the sum of the second members. Abbreviations of this kind are frequent in Algebra ; thus we speak of multiplying an equation when we mean multiplying each of its terms. From the above examples, we have the following rule : If necessary, multiply the given equations by such numbers as will make the coefficients of 07ie of the unknowyi numbers in the res tilting equations of equal absolute value. Add or subtract the resulting equations according as the coeffi- cients of equcd absolute vcdue are of unlike or like sign. If the coefficients which are to be made of equal absolute value are prime to each other, each may be used as the multiplier for the other equation ; but if they are not prime to each other, such multipliers should be used as will produce their lowest common multiple. Thus, in Ex. 1, to make the coefficients of y of equal absolute value, we multiply (1) by 4 and (2) by 3; but in Ex. 2, to make the coefficients of X of equal absolute value, since the L.C.M. of 10 and 15 is 30, we mul- tiply (1) by 2 and (2) by 3. II. ELIMINATION BY SUBSTITUTION «o^ -r. cs ^ .^ .' f7a;-92/= 15. (1) 239. Ex. Solve the equations \ [82/-5a^=-17. (2) Transposing — 5 a; in (2), 8 ?/ = 5 a; — 17. Whence, y = — ~^ (3) 8 Substituting in (1), 7 a; - 9 f ^^~^'^ ^ = 15. (4) 56a;-9(5a;-17) = 120. 56 a; -45 03 + 153 = 120. 11 03= -33. 03= -3. (5) 3),,--lS-l^=-4. (6) SIMULTANEOUS LINEAR EQUATIONS 139 Clearing of fractions, (5'r, Uniting terms, Whence, Substituting oj = — 3 o By § 236, the given system of equations is equivalent to the system (3) and (4) ; or, since (4) is equivalent to (5), to the system (3) and (5). By § 236, the system (3) and (5) is equivalent to the system (5) and (6); whence, the given system is equivalent to the system (5) and (6). From the above example, we have the following rule : From one of the given equations find the value of one of the unknoivn numbers in terms of the other, and substitute this value in place of that number in the other equation. III. ELIMINATION BY COMPARISON 240. Ex. Solve the equations 2o: .3o; -52/ = -16. + 72/= 5. (1) (2) Transposing — 5 ?/ in (1), 2o3 = 5?/-16. Whence, 03-^2/-16. 2 (3) Transposing 7 ?/ in (2), 3 03=5-72/. Whence, (4) Equating values of x, — ^-- = — — -^' (5) Clearing of fractions, 15 2/ — 48 = 10 — 14 y. Transposing, 29 2/ = ^^' Whence, y = % (6) Substituting 2/ = 2 in (3), 03 = ^-^ = - 3. (7) By § 233, the given system of equations is equivalent to the system (3) and (4) ; or, since, by Ax, 4, § 66, (4) is equivalent to (5), to the system (3) and (5). 140 ADVANCED COURSE IN ALGEBRA But (5) is equivalent to (6) ; so that the given system is equivalent to the system (8) and (6). By § 236, the system (3) and (6) is equivalent to the system (6) and (7). Prom the above example, we have the following rule : From each of the given equations, find the value of the same unknown number in terms of the other, and place these values equal to each other. 241. If the given equations are not in the form ax-^by = c, they should first be reduced to this form, when they may be solved by either method of elimination. In solving fractional simultaneous equations, we are liable to get results which do not satisfy the given equations. By § 221, we must reject any solution which satisfies the equation obtained by equating to zero the L. C. M. of the given denominators. ■ 242. 1. Solve the equations { 7 3 , -^- =0. (1) [x(y-2)-y(x-5)=-13. (2) Multiplying each term of (1) by (x + 3)(2/ + 4), 7y + 28-3x-9 = 0, ot 7y-Sx = -19, (3) From (2), xy-2x-xy-{-5y = -13, or 5y-2x==-13. (4) Multiplying (3) by 2, Uy-6x = - 38. (5) Multiplying (4) by 3, Wy-6x=-39. (6) Subtracting (5) from (6), y = — 1. Substituting in (4), — 5 - 2 a; = - 13. Whence, —2x = — 8, or a;=4. Since x = 4 and y = — 1 do not satisfy the equation (a; + 3) (2/ + 4) = 0, the solution x = 4, y = —lis correct (§241). {2x-\-3y =13. (1) 2. Solve the equations ] 1 , 1 _ /^ /o^ SIMULTANEOUS LINEAR EQUATIONS 141 Multiplying each term of (2) by (x — 2) (y — 3), y-3-\-x-2 = 0, or y = - x + 5. (3) Substituting in (1), 2 x — 3 a; + 15 = 13, or a; = 2. Substituting in (3), 2/=-2 + 5 = 3. Since x = 2 and y = 3 satisfy the equation (x — 2) (y — 3) = 0, the sohition must be rejected. The above solution satisfies the first given equation, but not the second ; it is impossible to find a solution which will satisfy both given equations. In solving literal simultaneous linear equations, the method of elimination by addition or subtraction is usually to be preferred. 3. Solve the equations Multiplying (1) by &', Multiplying (2) by 6, Subtracting, Whence, Multiplying (1) by a', Multiplying (2) by a, Subtracting (3) from (4), Whence, ^ ab' - a'b Certain equations in which the unknown numbers occur in the denominators of fractions may be readily solved without previously clearing of fractions. ' ax -\-by = c. (1) . a'x + b'y = c'. (2) ab'x + bb'y = b'c. a'bx 4- bb'y = be'. (ab'-a'b)x=b'G- -bc\ ""-ab'- -be' -a'b aa'x + a'by = ca'. (3) aa'x -{- ab'y = c'a. (4) (ab'-a'b)y = c'a- -ca'. ^_c'a- -ca' 4. Solve the equations Multiplying (1) by 5, Multiplying (2) by 3, 10 X 9_ y 8. \h 15 _ y ~ -1. 50 X 45 _ y ~ 40. 1- 45 y ~ -3. (1) (2) 142 ADVANCED COURSE IX ALGEBRA Adding, — = 37. Whence, 74 = 37 x, and x = 2. 9 Substituting in (1), 5 = 8. 9 Whence, = 3, and yz=: — S. y EXERCISE 32 Solve the following : r.08x+ .9?/ = .048. f x + y ^ 1 _ ^' I .3x-.35?/=:.478. 5 J^~^ ^^* 5 7 _ _ 29 ' I 3x + 8 ^ 6^-1 ^ i- + ^ = |- r3x-4y=-ll. ^" '^ ' 6. 2 5 , ' ] 5 X + 2 ?/ , 7 ?/ - 3 X _ 39 ■ r 5 7 J rmrx + 2/) + jK^-?/) = 2. • I a I m- (x + ?/) — rfiix — y) = m — n. [ y Ua + b)x-{-(a-h)y = 2 (a2 + 52). 8. ^ fe ^ a [x — a — b y — a + h { m , n . f 3 24 \x-a y -\- b ^ \2x + y x-4y j n ^ m ^^ I __7 16_^_3^ [x-a y -\- b ' i2x + y x-4:y 2 xy + 3 _ 4y + 5 ^ g ^ a;_2 x + 3 ^* ' j 2 X + 3 y - 1 _ 3 X - 8 ?/ ^ 25j^ [ 2x + ?/ Sx-lly (2x + y)(Sx-lly)' \ {a -{■ b)x -{- {a - b)y = 2 a^ -2b-\ 12. < y X 4 (tT) I a — /> a + & a^ — b'^ SIMULTANEOUS LINEAR EQUATIONS 143 SIMULTANEOUS LINEAR EQUATIONS CONTAINING MORE THAN TWO UNKNOWN NUMBERS 243. If we have three independent simultaneous equations, containing three unknown numbers, we may combine any two of them by one of the methods of elimination explained in §§ 238 to 240, so as to obtain a* single equation containing only two unknown numbers. We may then combine the remaining equation with either of the other two, and obtain another equation containing the same two unknown numbers. By solving the two equations containing two unknown num- bers, we may obtain their values; and substituting them in either of the given equations, the value of the remaining unknown number may be found. We proceed in a similar manner when the number of equa- tions and of unknown numbers is greater than three. The method of elimination by addition or subtraction is usu- ally the most convenient. If any equation is fractional, we should reject any solution which satis- fies the equation obtained by equating to zero the L. C. M. of the given denominators (§ 241). r Qx- 4?/- lz= 17. (1) 1. Solve the equations \ 9x- 1y-lQz= 29. (2) [lOx- 52/- 32= 23. (3) Multiplying (1) by 3, lSx-12y-21z== 51. (4) Multiplying (2) by 2, lSx-Uy-32z= 58. (5) Subtracting, 2y-\-llz=- 7. (6) Multiplying (1) by 5, S0x-20y-S5z= 85. (7) Multiplying (3) by 3, 3Qa;-15y- 9z= 69. ' • (8) Subtracting (7) from (8), 5 y + 26 ^ = - 16. (9) Multiplying (6) by 5, 10 y + 55 z = - 35. (10) Multiplying (9) by 2, 10 ?/H- 52:^^-32. (11) Subtracting, 32;=— 3, or 2 = — 1. (12) Substituting in (6), 2y-ll = -7, or y= 2. (13) Substituting in (1), 6 .r - 8 + 7 = 17, or x = 3, (14) 144 ADVANCED COURSE IN ALGEBRA By §§ 119 and 233, the given system of equations is equivalent to the system (3), (4), and (5) ; which, by § 235, is equivalent to the system (3), (4), and (6), or to the system (1), (3), and (6). But (1) is equivalent to (7), and (3) to (8) ; so that the given system is equivalent to the system (0), (7), and (8) ; this, by § 235, is equiva- lent to the system (6), (7), and (9), or to the system (1), (6), and (9). But the system (1), (6), and (9) is equivalent to the system (1), (10), and (11) ; which, by § 235, is equivalent to the system (1), (10), and (12). The system (1), (10), and (12) is equivalent to the system (1), (G), and (12) ; which, by two applications of § 230, is equivalent to the sys- tem (12), (13), and (14). In certain cases the solution may be abridged by aid of the artifice which is employed in the following example. 2. Solve the equations 2/+ z + u=^ z -\-u-\-x = Adding, 3i* + 3aj + 3?/ + 3 ;s = 30. Whence, u-\-x-\-y -{- z = 10. Subtracting (2) from (5), ^^ = 3. Subtracting (3) from (5), x= 2. Subtracting (4) from (5), 2/ = 1- Subtracting (1) from (5), 2;= 4. 6. (1) 7. (2) 8. (3) 9. (4) (5) EXERCISE Solve the following : 33 '5x+ ?/-4;S=— 5. ' ax -\- by = {a -\- h)c. 1. 3x-5i/-6^=:-20. 8. by -\- cz = (c + a)b. .x-Zy^^z =-27. ■ cz -{- ax = (6 + c)a. {x-y y ~z 7 3 43' 2 a; - 5 ?/ = - 26. 2. y-z z + x_ 13 3 5 15 4. 7x + 6z= -3S. 3 4 z + X X — y _^Z J/-4 ^ + 2 L 2 5 "lO' SIMULTANEOUS LIKEAR tiQUATlONS 145 6. \ 7. ^ u-\-3x-2y - z= -3. 2u-x-y-^Sz = 2S. u + x + Sy -2z z=z -12. Su-2x + y + z = 22. 1,1,1 -+- + - = «. X y z Ui + i = 6. y z u 1,1,1 - + - + - = c. Z U X 1,1,1 ^ - + - + -=(?. X ?/ ;S r a; + y + = 0. 9. j (6 + c)x + (c + a)y + (« + 6);3 = 0. I 6cx + cay + ahz = 1. Sx~y _ 4g - 5 y _ 19 5 ' 2 ~ 2* 2a;-3g x-iy _7 6 4 ~4 4x + g 3y+5g _49 3 2 ~ 3 * r_l-+-l-=2. x + ?/ X — 6 5 x + 4 5 = 1. = 2. y 10. X y + b a+ b c y g + c 6 + c a _s__x_+_a_Q c + a 6 0. PROBLEMS INVOLVING SIMULTANEOUS LINEAR EQUA- TIONS WITH TWO OR MORE UNKNOWN NUMBERS 244. In solving problems where two or more letters are used to represent unknown numbers, we must obtain from the conditions of the problem as many independent equations (§ 229) as there are unknown numbers to be determined. 1. If 8 be added to both numerator and denominator of a fraction, its value is J; and if 2 be subtracted from both numerator and denominator, its value is \ ; find the fraction. Let OJ = the numerator, and y = the denominator. By the conditions, ?^-±-^ = -, 2/4-3 3' a;-2 ^1 2" and y-2 Solving these equations, x = 7, y = 12 ; whence, ttie fraction is 12 146 ADVANCED COURSE IN ALGEBRA 2. A crew can row 10 miles in 50 minutes down stream, and 12 miles in an hour and a half against the stream. Find the rate in miles per hour of the current, and of the crew in still water. Let X = number of miles an hour of the crew in still water, and y = number of miles an hour of the current. Then, x + y = number of miles an hour of the crew down stream, and X — y = number of miles an hour of the crew up stream. The number of miles an hour rowed by the crew is equal to the dis- tance in miles divided by the time in hours. Then, x + y = 10^^ = 12, 6 and x-y = 12-^-= 8. 2 Solving these equations, x = 10, y = 2. 3. The sum of the three digits of a number is 13. If the number, decreased by 8, be divided by the sum of its second and third digits, the quotient is 25 ; and if 99 be added to the number, the digits will be inverted. Find the number. Let X = the first digit, y = the second, and z = the third. Then, lOOx + lOy + z = the number, and 100 z -{- lOy -{■ X = the number with its digits inverted. By the conditions of the problem, x + y + z = lS, loox-f lOy + g-8 _gg and lOOx + lOy + z + 99 = l00z-\-10y + x. Solving these equations, x = 2, ?/ = 8, = 3; and the number is 283. EXERCISE 34 1. If 3 be added to the numerator of a certain fraction, and 7 sub- tracted from the denominator, its value is -; and if 1 be subtracted from 2 the numerator, and 7 added to the denominator, its value is - . Find the 5 fraction. 2. Find two numbers such that one shall be n times as much greater than a as the other is less than a : and the quotient- of their sum by their difference equal to b. SIMULTANEOUS LINEAR EQUATIONS 147 3. If the greater of two numbers be divided by the less, the quotient is 1, and the remainder 6. And if the greater, increased by 14, be divided by the less, diminished by 4, the quotient is 5, and the remainder 4. Find the numbers. 4. A sum of money at simple interest amounted to $ 1868.40 in 7 years, and to $2174.40 in 12 years. Find the principal and the rate. 6. A certain number of two digits exceeds three times the sum of its digits by 4. If the digits be inverted, the sum of the resulting number and the given number exceeds three times the given number by 2. Find the number. 6. A man invests a certain sum of money at a certain rate of interest. If the principal had been $1200 greater, and the rate one per cent greater, his income would have been increased by $ 118. If the principal had been $3200 greater, and the rate two per cent greater, his income would have been increased by $312. What sum did he invest, and at what rate ? 7. The middle digit of a number of three figures is one-half the sum of the other two digits. If the number be divided by the sum of its digits, the quotient is 20, and the remainder 9 ; and if 594 be added to the num- ber, the digits will be inverted. Find the number. 8. A crew row 16| miles up stream and 18 miles down stream in 9 hours. They then row 21 miles up stream and 19^ miles down stream in 1 1 hours. Find the rate in miles an hour of the stream, and of the crew in still water. 9. A and B can do a piece of work in — hours, A and C in — hours, 21 10 ^2 ^ A and D in — hours, and B and C in — hours. How many hours will it 10 7 take each alone to do the work ? 10. A and B run a race of 280 feet. The first heat, A gives B a start of 70 feet, and neither wins the race. The second heat, A gives B a start of 35 feet, and beats him by 6f seconds. How many feet can each run in a second ? 11. A, B, C, and D play at cards. After B has won one-half of A's money, C one-third of B's, D one-fourth of C's, and A one-fifth of D's, tliey have each $10, except B who has $16. How much had each at first ? 12. The fore-wheel of a carriage makes a revolutions more than the hind-wheel in travelling h feet. If the circumference of the fore-wheel were increased by one-jnth, and the circumference of the hind-wheel by one-nth, the fore-wheel would make c revolutions more than the hind- wheel in travelling d feet. Find the circumference of each wheel. 148 ADVANCED COURSE IN ALGEBRA 13. The sum of the four digits of a number is 14. The sum of the last three digits exceeds twice the first by 2. Twice the sum of the second and third digits exceeds three times the sum of the first and fourth by 3. And if 2727 be subtracted from the number, the digits will be inverted. Find the number. 14. A and B run a race from P to Q and back ; the distance from P to Q being 108 yards. The first heat, A reaches Q first, and meets B on his return at a point 12 yards from Q, The second heat, A increases his speed by 2 yards a second, and B by 1 yard a second ; and now A meets B 18 yards from Q. How many yards can each run in a second ? 15. A train running from A to B, meets with an accident which de- lays it a hours. It then proceeds at a rate one-nth less than its former rate, and arrives at B & hours late. Had the accident occurred c miles nearer B, the train would have been d hours late. Find the rate of the train before the accident, and the distance to B from the point of detention. 16. A man buys 60 shares of stock, part paying dividends at the rate of 3| per cent, and the remainder at the rate of 4| per cent. If the first part had paid dividends at the rate of 4i per cent, and the other at the rate of 3| per cent, the total annual income would have been $12 less. How many shares of each kind did he buy ? DISCUSSION OF LINEAR EQUATIONS 149 XIII. DISCUSSION OF LINEAR EQUATIONS VARIABLES AND LIMITS 245. A variable number, or simply a variable, is a number which may assume, under the conditions imposed upon it, an indefinitely great number of different values. A constant is a number which remains unchanged throughout the same discussion. A limit of a variable is a constant number, the difference between which and the variable may be made less than any assigned number^ however small, without ever becoming zero. In other words, a limit of a variable is a fixed number which the variable approaches indefinitely near, but never actually reaches. 246. It is evident that the difference between a variable and its limit is a variable which approaches the limit zero. 247. Interpretation of -• Consider the series of fractions a a ^ a where each denominator after the first is one-tenth of the pre- ceding denominator. It is evident that, by sufficiently continuing the series, the denominator may be made less than any assigned number, how- ever small, and the value of the fraction greater than any assigned number, however great. In other words, if the numerator of a fraction remains constant, while the denominator apinoaches the limit 0, the value of the fraction ino^eases ivithout limit. It is customary to express this principle as follows : a 5 = '»- 150 ADVANCED COURSE IN ALGEBRA The symbol co is called Infinity ; it simply stands for that which is greater than any number, however great. 248. Interpretation of — . Consider the series of fractions a a a a 3' 30' 300' 3000' ' where each denominator after the first is ten times the preced- ing denominator. It is evident that, by sufficiently continuing the series, the denominator may be made greater than any assigned number, however great, and the value of the fraction less than any assigned number, however small. In other words, If the numerator of a fraction remains constant^ while the denominator increases without limit , the value of the fraction ap2)7'oaches the limit 0. It is customary to express this principle as follows : ii = o. 00 It must be clearly understood that no literal meaning can be attached to such results as ^ ^=00, and -^ = 0; ' CO ' for there can be no such thing as division, unless the divisor is a finite number. If such forms occur in mathematical investigations, they must be in- terpreted as in §§ 247 and 248. 249. Interpretation of 5 By § 44, - signifies a number which, when multiplied by 0, gives 0. But by §§37 and 40, if any number be multiplied by 0, the result is 0. Hence, -r may be any number whatever. DISCUSSION OF LINEAR EQUATIONS 151 For this reason the fraction - is called Indeterminate. 250. A Function of a number is any expression which con- tains the number. Thus, the expression 2 x^ — 3ax-\-5a^ is a function of x. 251. Function Notation. A function of x is often represented by the symbol /(a;) ; read "y-f unction of x/' or simply "/-aj." If, in any investigation, /(aj) stands for a certain function of X, then, whatever value a may have, /(a) represents the result obtained by substituting a for x in the given function. Thus, if f(x) =x^-j-3x-2, then /(3)=32 + 3. 3-2=16; /(_3) = (-3)^4-3(-3)-2=-2; etc. Functions of x are also represented by the symbols F(x), (cc), etc. THE THEOREM OF LIMITS 252. If two functions of the same variable are so related that, as the variable changes its value, they are equal for every value which the variable can assume, and each approaches a certain limit, then the two limits are equal. Let y and z be functions of a certain variable, x ; and let them be equal for every value which the variable x can assume, and approach the limits y' and z', respectively. To prove that y' = z'. Let y^ — y — m, and z^ — z = n. Then, m and n are variables which can be made less than any assigned number, however small (§ 246). Then, m — n is either zero, or else a variable which can be made less than any assigned number, however small. But m — n — y^ — y—z'-\-z = y^ — z^', for, by hypothesis, y and z are equal. Since y' — z' is not a variable, m — n is not a variable. Then, m — n is ; and hence its equal, y* — »' is 0, or y' = «'. 152 ADVANCED COURSE IN ALGEBRA PROPOSITIONS IN REGARD TO LIMITS 253. The limit of the sum of a constant ayid a variable is the sum of. the constant and the limit of the variable. Let a be a constant, and x a variable whose limit is x'. Then, x' — x can be made less than any assigned number, however small (§ 245). Whence, (x' + a) — (oj + a), which equals x' —x, can be made less than any assigned number, however small. Then, x' -}- a is the limit of x + a. 254. The limit of the sum of any finite number of variables is the sum of their limits. Let X, y, 2, •••, be variables whose limits are .t', y\ z\ •••, respectively. Then, x^ —x, y' — y, z' — z, •••, can be made less than any assigned number, however small. Whence, (x' — x) -\- (j/ — y) + {z' — z) -\ , or, (x' -\- y' -\- z' -\ ) — (x-{-y + z + '•'), can be made less than any assigned number, however small. Then, x' -\- y' -\- z' -\ is the limit of x-\-j/-{-z-\ . (1) 255. Any two corresponding positive signs, in (1), § 254, may be changed to negative. Thus, x' — y' -^z' -\ is the limit oi x — y + z-] . 256. The limit of the product of a constant and a variable is the constant multiplied by the limit of the variable. Let a, X, and a?' have the same meaning as in § 253. Then, a{x^ — x), or ax^ — ax, can be made less than any assigned number, however small. Whence, ax' is the limit of ax. 257. The limit of the product of any number of variables is the product of their limits. Let X, y, z, •••, be variables having the limits ic', y', z'^ •••, respectively ; and let x' — x = I, y' — y = 7n, z' — z = n, •••. Then, /, m, n, •••, can be made less than any assigned number, however small. DISCUSSION OF LINEAR EQUATIONS 153 Now, x'y'z' "' = (x + ^)(y-\-7n)(z + 7^)"• = xyz h terms involving I, m, n, •••. Then, x'y'z' ••- —xyz--- = terms involving I, m, n, •••. (1) The second member of (1) can be made less than any assigned number, however small. Then, x^y'z' "- is the limit of xyz •••. 258. Let n be a positive integer, and x a variable having the limit a;'; then, limit of a;" = limit oi xxxxxx -" to n factors = «' X cc' X x' X '" to n factors, by § 257, = {xy = (limit of xy. 259. Tlie limit of the quotient of two variables is the quotient of their limits, if the divisor be not zero. Let x and y be variables having the limits x' and y', respec- tively ; and suppose that y' is not zero. Let x' — x = l, and y' — y = m; then, x = x' ~l and y = y' — m. Now, x' X _x' x' — I _x'y' —mx' — (x'y' —ly') _ly' — mx' ,.. y' y~ y' y' — m~ y'{y' — m) ~y''' — my'' Since I and m can be made less than any assigned number, however small (§ 245), the numerator of this fraction can be made less than any assigned number, however small. Also, the denominator can be made to differ from y'^ by less than any assigned number, however small. Then, the fraction (1) can be made less than any assigned number, however small. Whence, — , is the limit of -. 2/' y INTERPRETATION OF NEGATIVE RESULTS 260. A problem is said to be Impossible when its conditions cannot be satisfied. It is said to be Indeterminate when the number of its solu-. tions is indefinitely great. 154 ADVANCED COURSP: IN ALGEBRA 261. 1. The length of a field is 10 rods, and its breadth 8 rods ; how many rods must be added to the breadth so that the area may be 60 square rods ? Let X = number of rods to be added. By the conditions, 10 (8 + x) = 60. Then, 80 + 10x = 60, or a:= - 2. By § 78, adding — 2 rods is the same thing as subtracting 2 rods. Hence, 2 rods must be subtracted from the breadth in "order that the area may be 60 square rods. The above problem is impossible arithmetically ; for the area of the field is at present 80 square rods ; and it is impossible to make it 60 square rods by adding anything to the breadth. If we should modify the problem so as to read : " The length of a field is 10 rods, and its breadth 8 rods ; how many rods must be subtracted from the breadth so that the area may be 60 square rods ? " and let x denote the number of rods to be subtracted, we should find x = 2. Also, if we had solved the given problem by letting x denote the num- ber of rods to be subtracted, we should have found x = 2. 2. A is 35 years of age, and B 20 ; it is required to deter- mine the epoch at which A's age is twice as great as B's. Let us suppose that the required epoch is x years after the present date. By the conditions, 35 + ic = 2 (20 + x). Then, 35 + a; = 40 + 2 a:, or a; = - 5. By § 56, — 5 years after is the same thing as 5 years before the present date. Therefore, the required epoch is 5 years before the present date. If we had supposed the required epoch to be x years before the present date, we should have found x = 5. From the discussion of the above problems, we infer that a negative result may be obtained: 1. In consequence of the fact that the problem is arithmeti- cally impossible. 2. In consequence of a wrong choice between two possible hypotheses as to the nature of the unknown number. DISCUSSION OF LINEAR EQUATIONS 155 In the first case, it is usually possible to form an analogous problem, whose conditions are satisfied by the absolute value of the negative result, by attributing to the unknown number a quality the opposite of that which had been attributed to it. In either case, a positive result may be obtained by attribut- ing to the unknown number a quality the opposite of that which had been attributed to it; and the equations answering to the new conditions may be derived from the old equations by changing the sign of the unknown number wherever it occurs. Similar considerations hold in problems involving two or more unknown numbers. A negative result sometimes indicates that the problem is impossible. 3. If 11 times the number of persons in a certain house, increased by 18, be divided by 4, the result equals twice the number increased by 3 ; find the number. Let X = the number. By the conditions, ^^^"^^^ = 2 x + 3. 4 Whence, 11 x + 18 = 8x+ 12, and a; = -2. The negative result shows that the problem is impossible, 262. A problem may also be impossible when the solution is fractional, or zero. 1. A man has two kinds of money; dimes and cents. The total number of coins is 23, and their value 37 cents. How many has he of each ? Let X = number of dimes. Then, 23 — x = number of cents. The X dimes are worth 10 x cents ; then, by the conditions, 14 10 X + 23 - X = 37 ; and x = — . The fractional result shows that the problem is impossible. 2. The denominator of a fraction exceeds the numerator by 6 ; and if 2 be added to the numerator, the value of the fraction is \. Find the fraction. 156 ADVANCED COURSE IN ALGEBRA Let X — the numerator. Then," ic + 6 = the denominator. X A- 2 1 By the conditions, , = - • Whence, Sx + 6=x + 6; and x = 0. The result shows that the problem is impossible. THE PROBLEM OF THE COURIERS 263. The discussion of the following problem serves to further illustrate the interpretation of negative and zero results, besides furnishing an interpretation of infinite and indetermi- nate results. The ProbTfem of the Couriers. Two couriers, A and B, are travelling along the same road in the same direction, RR', at the rates of m and n miles an hour, respectively. .If at any time, say 12 o'clock, A is at P, and B is a miles beyond him at Q, after how many hours, and how many miles beyond P, are they together ? R P Q R' L I I I Let A and B meet x hours after 12 o'clock, and y miles beyond P. They will then meet y — a miles beyond Q. Since A travels mx miles, and B 7ix miles, in x hours, we l^^v® I y = mx, 1 2/ — (X = nx. Solving these equations, we obtain a -i am X = , and y = m — n m — n We will now discuss these results under different hypotheses. 1. m > n. In this case, the values of x and y are positive. This means that the couriers meet at some time after 12, at some point to the 7nght of P. DISCUSSION OF LINEAR EQUATIONS 157 This agrees with the hypothesis made ; for if m is greater than iiy A is travelling faster than B ; and he must overtake him at some point beyond their positions at 12 o'clock. 2. m<.n. In this case, the values of x and y are negative. This means that the couriers met at some time before 12, at some point to the left of P. This agrees with the hypothesis made ; for if m is less than n, A is travelling more slowly than B; and they must have been together before 12 o'clock, and before they could have advanced as far as P. 3. a = 0, and m > n or m < n. In this case, a; = and y = 0. . This means that the travellers are together at 12 o'clock, at the point P. This agrees with the hypothesis made ; for if a = 0, and m and n are unequal, the couriers are together at 12 o'clock, and are travelling at unequal rates ; and they could not have been together before 12, and will not be together afterwards. 4. m = n, and a not equal to 0. In this case, the values of x and y take the form -, and are injlnite (§ 247). ^ No definite values can be assigned to x and y^ and the prob- lem is impossible. This agrees with the hypothesis made ; for if m = n, and a is not zero, the couriers are a miles apart at 12 o'clock, and are travelling at the same rate ; and they never could have been, and never will be together. Therefore, an infinite result indicates that the problem is im- possible. In this case, as m — n approaches the limit 0, the values of x and y increase without limit. That is, as the difference of the rates of the couriers approaches the limit 0, both the number of hours after 12 o'clock, and the number of miles beyond P, when A and B are together, increase without limit. 158 ADVANCED COURSE IN ALGEBRA 5. m = n, and a = 0. In this case, the values of cc and ?/ take the form -, and are indeterminate (§ 249). This means that any value of x whatever, with the corre- sponding value of y, is a solution of the problem. This agrees with the hypothesis made ; for if m = n, and a = 0, the couriers are together at 12 o'clock, and travelling at the same rate; and they always have been, and always will be together. Thus, an indeterminate result indicates that the number of solu- tions is indefinitely great. EXERCISE 35 Interpret the negative results, and modify the enunciation accordingly, in the following : 1. If the length of a field is 12 rods, and its width 9 rods, how many rods must be subtracted from the width so that the area may be 144 square rods ? 2. A is 44 years of age, and B 12 years ; how many years ago was A three times as old as B ? 3. A's assets are double those of B. When A has gained $250, and B $ 170, A's assets are five times those of B. Find the assets of each. 4. A cistern has two pipes. When both are open, it is filled in 7| hours ; and the first pipe alone can fill it in 3 hours. How many hours does the second pipe take to fill it ? 5. A and B are travelling due east at the rates of 4| and 3^ miles an hour, respectively. At noon, A is 5 miles due east of B. How many miles to the east of A's position at noon will he overtake B ? 6. A has $720, and B .$300. After A has gained a certain sum, and B has gained two-thirds this sum, A has three times as much money as B. How much did each gain ? In each of the following, interpret the solution ; 7. The number of apple and pear trees in an orchard is 23 ; and seven times the number of apple trees plus twice the number of pear trees equals 82. How many are there of each kind ? 8. The number of silver coins in a purse exceeds the number of gold coins by 3. And five times the number of silver coins exceeds three times the number of gold coins by 3. How many are there of each kind ? DISCUSSION OF LINEAR EQUATIONS 159 9. The numerator of a fraction is four times the denominator ; and if the numerator be diminished by 9, and the denominator by 15, the value 3 of the fraction is - • Find the fraction. 5 10. A is a years old, and B h years. After how many years will A be n times as old as B ? Discuss the solution in the cases when n = \ and a is not equal to 6, and when n = \ and a=h. 11. What number must be added to both terms of the fraction - to c ^ make it equal - ? Discuss the solution in the cases when - = - and c is not equal to d. b d when c = d and - is not equal to -, and when c = d and - = -. b ^ d b d 12. Two couriers, A and B, are travelling along the same road, in the in the same direction, at the rates of m' and m" miles an hour, respec- tively. B passes a certain point n hours after A. How many hours after B passes this point will he overtake A ? Determine for what values of the letters the solution is positive, nega- tive, zero, impossible, and indeterminate, and discuss the solution in each case. 13. The circumference of the fore-wheel of a carriage is a feet, and of the hind -wheel b feet. How far will the carriage have travelled when the fore-wheel has made n revolutions more than the hind-wheel ? Discuss the solution in the following cases : 1. n=0, a and b unequal. 2. a = b, n not equal to zero. 3. a = 6, w=0. INDETERMINATE FORMS 264. The indeterminate form - does not always represent a fraction which may have any value whatever. Take, for example, the fraction . or —ax If ic = a, the fraction takes the form -. Dividing both numerator and denominator by a; — a, x^ — a? _ x-{-a m? — ax X ' which holds so long as x does not equal a (§ 115, 4). (1) 160 ADVANCED COURSE IN ALGEBRA The second member of (1) approaches the limit 2 when x approaches the limit a. 2 2 Then, ^ ~^ approaches the limit 2 when x approaches the T ., x^—ax limit a. /w2 ((2 We call the limit approached hy the fraction — , when x ap- x^ — ax proaches the limit a, the value of the fraction when x = a. In any similar case, we divide both terms of the fraction by the ex- pression which makes each term vanish, and find the limit approached by the result. In the problem of § 264, the result - was obtained in consequence of two independent hypotheses, one causing the numerator to vanish, and the other the denominator ; and in any similar case we should find the result - susceptible of the same interpretation. But in the above example, the result - is obtained in consequence of the same hypothesis causing both numerator and denominator to vanish. 265. The Indeterminate Forms — , x oo, and go — oo. 00 1. To find the limit approached by the fraction "^ , when X is indefinitely increased. Dividing each term of the fraction by Xj we have i + 2 ■ lim l + 2a; ^ lim ^ = 2±J ('8 248^ =-• x^oo3 + 5x a; = ao3_^g 0.+ 5^ ^ 5* X We use the notation ^^ for the words "the limit when x is in- x = cc ,. definitely increased of," and the notation ** for the words "the X — a limit as x approaches a of." In the above example, the fraction takes the indeterminate form — when X is indefinitely increased. In any similar case, we divide both numerator and denominator of the fraction by the highest power of x. DISCUSSION OF LINEAR EQUATIONS 161 2. Find the value of the expression (a^ + 8)[ 1 H -J, when The expression takes the indeterminate form x oo when x = -2. We have J':,[(^ + 8)(l + ^)]=J-^(.' + 8 + »^-2x + 4) ^ hm (^3^_^aj2_2a;_^12) = _8 + 4 + 44-12 = 12. X — — Ji In any similar case, we simplify the expression as much as possible before finding the limit. 1 2 X 3. Find the value of ■ when a; = 1. 1 — x 1 — x- The expression takes the indeterminate form oo — oo when a; = l. ■j^Q^ lim / 1 2x \_ Urn l-^x — 2x a; = l\^l— a; 1 — x'v x = l 1 — x^ __ lim 1 — x _ lim 1 _1. «=ll-a;2 a: = ll4-a; 2* ^ EXERCISE 36 Find the values of the following : 1. ^'-^^ whenx = 4. a;2 _ 2 ic - 8 2. 4 + ox - 3:k ^j^gj^ ^ .g in(jefinitely increased. 7 - x + 4 a:^ 1 + -^ 3 8x^-2x-3 ^henx = §. e. -^^ when x = - 1. 12x2-25x + 12 4 . I 1 X2-1 xB + 9x^ + 27x + 27 ^henx=:-3. 7. -i 12_ ^j^^^ ^ ^ 2. x4 -18x2 + 81 x-^ x3-8 5 4xi+2x^+2x + l ^j^^^^^_l. 8 x3-3x^ + 3x-2 ^i,enx = 2. 8x3 + 1 2 x3-7x + 6 9. (2 x2 - 6 X - 3) ^2 + —^-^ when x = 3. 162 ADVANCED COURSE IN ALGEBRA DISCUSSION OF THE SOLUTION OF A SYSTEM OF SIMULTANEOUS LINEAR EQUATIONS 266. A system of simultaneous equations is said to be Inde- terminate when the number of solutions is indefinitely great. 267. Any system of two simultaneous linear equations, in- volving two unknown numbers, can be reduced to the form ■ a^x-\-\y = c^, (1) . a^x -[- h^y = C2\ (2) where «], 6i, Cj, ofg? ^2? ^ii^ ^s? ii^^y have any numerical values whatever, except that %, hi, a^, and 63 cannot be zero. By Ex. 3, § 242, the solution of the above system is » = M:::M, and y = 2^h:zMi. aj&2 — «2^1 tli^a — <^2^1 These fractions have definite values as long as a-fi-i does not equal a2hi. We will now discuss the values of x and y when ttiftg = Gi^i- (A) (1) If h^Ci — biC2 is not zero, x is infinite (§ 247). From (A), a, = ^' (B) Then, c,ai - c^a^ = c^a^ - ^^ = f {\c, - h^c^). (C) Since neither a^, b^, nor 61C2 — 62^1 is zero, Cgai — Cia2 is not zero. Whence, y is also infinite. By aid of equation (B), the given equation (2) can be written ^x-\-b2y = C2, or a^x -{-b^y = -^', bi bz by multiplying each term by — • &2 Thus, the given equations are inconsistent ; for, by (1), we have a^x -^bjy = Cj. DISCUSSION OF LINEAR EQUATIONS 163 Hence, infinite results show that the given equations are incon- sistent. (2) If 62^1 — &1C2 is zero, x takes the form - , and may have any value whatever (§ 249) ; for in this case we have two inde- pendent hypotheses, one causing the numerator to become zero, and the other the denominator. (Compare § 264.) In this case, by (C), C2a-^ — c^az is also zero ; so that y is indeterminate. From the equatij^ns 62^1 — &1C2 = and C2ai — c^a2 = 0, we have b2 = ^, and ^2 = — • Then the given equation (2) can be written ^^x-\--^y = C2f or a^x -{- biy = Ci] which is the same as (1). We thus have a single equation to determine two unknown numbers. Hence, indeterminate results show that the given equations are not independent. Similar considerations hold for any system of simultaneous linear equations, involving more than two unknown numbers. 268. We will illustrate the principles of § 267 by an example. Consider the system of equations f a^x + h^y + c^z = (^, I \ a2X 4- bzy -\- C2Z = dzf [ a^x + bsy-\- c^z = d^. Solving, we find ^ _ d^b^c^ — t^i^gCg + ^^2^3^! — d^b-fi^ + d^iC2 — d^b^i , ,^. ctAcs — a^b^Cz + as^s^i — «2&iC3 + «3^A — «3^zCi ' with results of similar form for y and z. We will now use equation (1) as a formxda for finding the value of X in the following system of equations. 164 ADVANCED COURSE IN ALGEBRA i2x + 3y-2z = -l, [ X — 52/4-32= 6. Here, ai=:8, 5i = — 2, Ci = 1, di == 5, ag = 2, &2 = 3, Cg = — 2, (^2 = — 1, ttg = 1, &3 = — 5, C3 = 3, dg = 6. Substituting these values in (1), we have ^^ 45 -50 + 5-6 + 24 -18 ^0. 27-30-10 + 12 + 4-3 O' and the same result will be found for y and z. The indeterminate results show that the given equations are not independent (§ 267); this may be seen by observing that the first equation is the sum of the second and third. In this case, x may have any value whatever. We will now apply formula (1) to the following system: 2x-{-5y-Sz= 8, a._42/ + 22!= 3, 3x+ y- z = -2. Here, ai = 2, &i = 5, Ci = — 3, c?i = 8, a2=l, 62 = — 4, C2 = 2, (^2 = 3, a3 = 3, 63 = 1, C3 = -l, d^ = -2. Substituting these values in (1), we have ^^ 32-16-9 + 15-20 + 24 ^26^ 8-4-3 + 5 + 30-36 "^ ' and the same result will be found for y and z. The infinite results show that the given equations are incon- sistent (§ 267) ; this may be seen by observing that the sum of the first two equations gives 3x-\-y — z = ll, while the third requires that 3x-{-y — z should equal — 2. 269. Number of Solutions of a System of Simultaneous Linear Equations. If we have a system of m independent and consistent simul- taneous linear equations, involving m unknown numbers, we may eliminate m — 1 of the unknown numbers, and obtain a single linear equation involving one unknown number. DISCUSSION OF LINEAR EQUATIONS 165 By § 227, the latter has one solution. Whence, the given system- of equations has one solution. And, in general, a system of independent and consistent linear equations has a single solution when the number of equations is the same as the number of unknown numbers. If we have a system of m independent linear equations, involving m -\-n unknown numbers, we may eliminate m — 1 of the unknown numbers, and obtain a single linear equation involving the remaining n + 1 unknown numbers. By § 228, the latter has an indefinitely great number of solu- tions; and hence the given system has an indefinitely great number of solutions. And, in general, a system of independent linear equations has an indefinitely great number of solutions when the number of equa- ti(Jns is less than the number of unknown numbers. If we have a system oi m-\-n independent linear equations, involving m unknown numbers, we may find a set of values of the unknown numbers which will satisfy any m of the equations. But this set of values will not satisfy the remaining n equa- tions ; and hence the given system has no solution. And, in general, a system of independent linear equations has no solution when the number of equations is greater than the number of unknown numbers. 166 ADVANCED COURSE IN ALGEBRA XIV. GRAPHICAL REPRESENTATION 270. Rectangular Co-ordinates of a Point. 1 \(-l>,a) P, (P,.a a a ■c' h b •y ^ N M 'JL a a I M-bra) Let XX' and TY' be straight lines, intersecting at right angles at ; OY being above, and Y' below, XX ', when OX is horizontal and extends to the right, and OX' to the left, of 0. Let Pj be any point in the plane of XX ' and YY', and draw line PjJf perpendicular to XX'. Then, OM and MP^ are called the rectangular co-ordinates of Pi ; OM is called the abscissa, and MPi the ordinate. The lines XX' and YY' are called the axis of X and axis of Y, respectively; and the origin. We express the fact that the abscissa of a point is 6, and its ordinate a, by saying that, for the point in question, x= b and y = a; or, more concisely, we speak of the point as the point (6, a) ; where the first term in parentheses is. understood to be the abscissa, and the second term the ordinate. 271. Let M and N be points on OX aud OX', respectively, such that 0M= OJSf= b ; and draw lines P1P4 and P2P3 through M and JV, respectively, perpendicular to XX ', making MPi=:>MP, = NP2 == i^Pa = a. GRAPHICAL REPRESENTATION 167 Then, each of the points P^ P^, Pg, and P4 will have its abscissa equal to h, and its ordinate equal to a. To avoid this ambiguity, abscissas measured to the right of O are considered -|-, and to the left, — ; and ordinates measured above XX' are considered -f, and below,—. Then the co-ordinates of the points will be as follows : A, (b, a) ; P2, (- 6, a) ; P3, (- 6, - a) ; P„ (b, - a). It is understood, in the above convention respecting signs, that the figure is so placed that OX is horizontal, and extends to the right of 0. If a point lies upon XX, its ordinate is zero ; and if it lies upon YY', its abscissa is zero. The co-ordinates of the origin are (0, 0). 272. Plotting Points. To plot a point when its co-ordinates are given, lay off the abscissa to the right or left of 0, according as it is -f- or — , and then draw a perpendicular, equal in length to the ordinate, above or below XX according as the ordinate is + or — . Thus, to plot the point (— 3, 2), lay off 3 units to the left of upon XX\ and then erect a perpendicular 2 units in length above XX\ GRAPH OF A LINEAR EQUATION INVOLVING TWO UNKNOWN NUMBERS 273. Consider the equation y = x + 2. If we give any numerical value to x, we may, by aid of the relation y = x-{-2, calculate a corresponding value for y. If a; = 0, 2/ = 2. {A) If a; = 1, y = s. (B) If oj = 2, 2, = 4. (C) If a; = 3, 2/ = 5. (D) If aj = -l. y = l. (E) •If a; = - 2, 2/ = 0. (F) If aj = -3, 2/ = -l; etc. (GF) 168 ADVAN^CED COURSE IN ALGEBRA Now let these be regarded as the co-ordinates of points ; and let the points be plotted, as explained in § 272. They will be found to lie on a certain line, OD, which is called the Graph of the given equation. By assuming fractional values for a;, we may obtain intermediate points of the graph. 274. We shall always find that a linear equation, involving two unknown numbers, has a straight line for a graph ; this may be proved as follows : Every such equation can be put in the form y = ax-\-h. We will first show that the graph of 2/ = a^ is a straight line. The equation y = ax is satisfied if « = and 2/ = ^ 5 hence, the graph of y = ax passes through 0. Let A and B be any other two points on the graph ; draw lines OA and OB ; also, lines AC and BD perpendicular to OX. Since A and B are on the graph, AC=axOC, and BD axOD. AC BD since each of these equals a. Then, ' OC OD' Then, triangles OAC and OBD are similar ; and OB coincides with OA. Therefore, all points in the graph of y = ax are in a straight line passing through 0. Now the graph of y = ax-\-b can be obtained from that of y = axhy increasing the ordinate of each point of the graph of y = axhj b. Hence, the graph of y — ax-\-h is a straight line parallel to the graph oi y = ax. It follows from the above that the graphs oi y = ax -{- b and y = ax + c are parallel. , 275. A straight line is determined by any two of its points. GRAPHICAL REPRESENTATION 169 Then, it is sufficient, when finding the graph of a linear equa- tion involving two unknown numbers, to find two of its points, and draw a straight line through them. The points most easily determined are those in which the graph intersects the axes. For all points on OX, ?/ = ; hence, to find where the graph cuts OX, put y = 0, and calculate the value of x. To find where the graph cuts Y, put x = 0, and calculate the value of y. Let it be required, for example, to plot the graph of 2x + Sy = -7. 7 2* Put 2/=0; then 2 x=—7, and x- Then plot A on OX', - units to the left of 0. ^ Put a:=0; then 3y= — T, and y= Then plot B on OY', ^ units below O. o Draw the straight line AB ; this is the required graph. The above method cannot, of course, be used for a straight line passing through the origin, nor for the equations of § 276. 276. Consider the equation y = 5. This means that every point in the graph has its ordinate equal to 5. Then the graph is the straight line AB, parallel to XX', and 5 units above it. In like manner, the graph of x = — 3 is the straight line CD, parallel to YY', and 3 units to the left of it. The graph oiy = Ois the axis of X, and the graph of x = is the axis of Y. EXERCISE 37 O Y JL y' D r Plot the graphs of the following : 1. 3x + 2y = 6. 3. a; = 2. 5. 2 a; = 3 y. 2. aj - 4 2/ = 4. 4. y = - 4. 6. a; + y = 0. 7. 16a;-27y = -72. 8. 8 a; + 16 y = - 6. 170 ADVANCED COURSE IN ALGEBRA INTERSECTIONS OF GRAPHS 277. Consider the equations 3 a; - ?/ = - 9. (AB) x-{-2y= 4. (CD) Let AB be the graph of 3 a; — ?/ = — 9, and CD the graph of x-\-2y = 4:. Let AB and CD intersect at E. Since E lies on each graph, its co- ordinates must satisfy both given equa- tions ; hence, to find the co-ordinates of E, we solve the given equations. In this case, the solution is x = — 2, y = 3; and it may be verified in the figure that these are the co-ordinates of E. Hence, if the graphs of any tivo linear equations, ivith two unknown numbers, intersect, the co-ordinates of the point of inter- section form a solution of the system of equations represented by the graphs. EXERCISE 38 Verify the principle of § 277 in the following systems : r 5 X - 4 y = 0. y = -o. *l7x + 6y=:-29. 3x4-7^ = 5. 8 a; + 3 2/ = - 18. r 4 X + 5 y = 24. \sx-2y=-6. 4. r 9 X + 14 y = - 25. 1 3 X - 4 V = 22. ' 278. Graphs of Inconsistent and Indeterminate Linear Equa- tions, with two Unknown Numbers. Consider the equations (3x-2y= 5. (AB) [6x-4:y=-7. (CD) The equations can be written, 3 5 -, 3,7 V = - X , and y = -x-{-- ; 2 2 2 4 and it was shown in § 274 that the graphs of two equations of this form are parallel. GRAPHICAL REPRESENTATION 171 The given equations are inconsistent ; and we shall always find that two inconsistent equations, with two unknown num- bers, are represented by parallel graphs. Again, consider the equations 3x-2y=: 5. \6x — 4:y = 10. In this case, the graphs coincide. The given equations are not independent; and in any similar case, we should find that the graphs were coincident. 279. Graphical Representation of Linear Expressions involving one Unknown Number. Consider the expression 3 x + 5. Put 2/ = 3 a; 4- 5 ; and let the graph of this equation be found as in § 275. r : / 1 B i ■ Y' Putting 2/ = 0, x= ; then the graph cuts XX' ~ units to the left of 0. 3 Putting a; = 0, ?/ = ^ 5 then the graph cuts TY' 5 units above 0. The graph is the straight line AB. It was shown in § 274 that the graph of a linear equation, with two unknown numbers, is a straight line. 280. Graphical Representation of Roots (§ 110). Consider the equation ax -\- h — Q. (1) To find the graph of the first member, put y = ax-\-h. (2) The abscissa of the point in which this graph intersects OX, must, when substituted for x in (2), make 2/ = 0. Then, it makes the first member of (1) equal to zero, and is therefore a root of the given equation. Hence, the abscissa of the point in which the graph of the first member of any linear equation, with one unknown number, inter- sects XX', is the root of the equation. 172 ADVANCED COURSE IN ALGEBRA Consider, for example, the equation 3 x + 5 = 0. The graph of the first member was found in § 279. This intersects XX' at the point -4, whose abscissa is — ; and the root of the equation is — — EXERCISE 39 Verify the principles of § 278 in the four following systems : r3x + 4 l3x + 4 2/= 16. (2x- 7y = U. y= -16. ■ l4x-14?/ = 28. (2x- 6y= 0. ( 5x+ 6y=15. lex -15?/ = 30. ■ ll5x+18?/ = 45. Plot the graphs of the first members of the following equations, and in each case verify the principle of § 280 : 6. 2x + 7 = 0. 6. 5x-4 = 0. INVOLUTION AND EVOLUTION 173 XV. INVOLUTION AND EVOLUTION INVOLUTION 281. Involution is the process of raising an expression to any power whose exponent is a positive integer. We have already given (§ 130) a rule for raising a rational and integral monomial to any power whose exponent is a positive integer. 282. Any Power of a Fraction. We have f - Y* = ? x ? X • • • to n factors bj b b _ axax • " to n factors _ ct" b X b X •" to n factors 6" Then, a fraction may be raised to any power whose exponent is a positive integer by raising both numerator and denominator to the required poioer, and dividing the first result by the second. THE BINOMIAL THEOREM 283. A Series is a succession of terms. A Finite Series is one having a limited number of terms. An Infinite Series is one having an unlimited number of terms. 284. In §§ 131 and 135, we gave rules for finding the square or cube of any binomial. The Binomial Theorem is a formula by means of which any power of a binomial may be expanded into a series. In the present chapter, we shall consider those cases only in which the exponent is a positive integer. 285. Proof of the Theorem for a Positive Integral Exponent. The following are obtained by actual multiplication. 174 ADVANCED COURSE IN ALGEBRA (a -\- xy = a^ -{- 2 ax -\- x^ ; (a-{-xY = a^-\-3a'x-\-3ax' + x^; (a -{- xy = a^ -\- 4: a?x -)- 6 a-x^ + 4 aa^ + ic^ ; etc. In these results, we observe the following laws : 1. The number of terms is greater by 1 than the exponent of the binomial. 2. The exponent of a in the first term is the same as the exponent of the binomial, and decreases by 1 in each succeed- ing term. 3. The exponent of x in the second term is 1, and increases by 1 in each succeeding term. 4. The coefficient of the first term is 1, and the coefficient of the second term is the exponent of the binomial. 5. If the coefficient of any term be multiplied by the exponent of a in that term, and the result divided by the exponent of x in the term increased by 1, the quotient will be the coefficient of the next following term. We will now prove by Mathematical Induction (Note, § 134) that these laws hold for any positive integral power of a-\-x. Assume the laws to hold for (a + x)", where n is any positive integer. Then, (a + xy = a" + wa^-^o; + ^^^^ ~ ^^ a^-^x^ + • • •• (1) Let P, Q, and R denote the coefficients of the terms involv- ing a"~'"a;'", a""''~V+^, and a^'^^'V^^, respectively, in the second member of (1) ; thus, {a -\- xy = or- -\- noT-'^x -\- ♦•• -f- Par-'^x^ -t- Qa"-'-V+^ -f Ita''-'-^x'+^ + .... (2) Multiplying both members by a + cc, we have (a -f x) "+^ = a"+^ -f na^x -\ \- Qa"-''aj''+i + -Ra"-''-^x'+2 ^ -f a"a; H h Pof"~'x!'+^ + Qa"-'-V+2 -j = «"+' + (ji + 1) a'^a; H + (P + Q)a^-''x'+' +(Q + i?)a"-'- V+2 -f .... (3) INVOLUTION AND EVOLUTION 175 This result is in accordance with the second^ third, and fourth laws. Since the fifth law is assumed to hold with respect to the second member of (2), we shall have r + 1 r + 2 Therefore, Q(n-r-l) Q(n + 1) Q-\-B _ r + 2 _ r + 2 _ n-7 P+Q~ Q(r + 1) ^ ~ Q(n + l) ~r + 2' n — r n — r n — r Whence, Q + i2 = (P + Q) r + 2 But n — r is the exponent of a in that term of (3) whose coeffi- cient is P-\-Qi and r + 2 is the exponent of x increased by 1. Therefore, \hQ fifth law holds with respect to (3). Hence, if the laws hold for any power of a + a? whose expo- nent is a positive integer, they also hold for a power whose exponent is greater by 1. But the laws have been shown to hold for (a + xy, and hence they hold for (a + xf ; and since they hold for (a + xy, they hold for (a + xy ; and so on. Hence, they hold when the exponent is any positive integer. By aid of the fifth law, the coefficients of the successive terms after the second, in the second member of (2), may be readily found ; thus, (a + xf = a** + na^'-'^x + ^^!^ ""-*•) a"" V 1 • i& + "^"7^iV^^ «'-V+-. (4) This result is called the Binomial TJieorem. In place of the denominators 1 • 2, 1 • 2 • 3, etc., it is customary to write [2, [3, etc. The symbol [w, read '^factorial n," signifies the product of the natural numbers from 1 to n inclusive. 176 ADVANCED COURSE IN ALGEBRA 286. In expanding expressions by the Binomial Theorem, it is convenient to obtain the exponents and coefficients of the terms by aid of the laws of § 285, which have been proved to hold for any positive integral exponent. 1. Expand {a + xy. The exponent of a in the first term is 5, and decreases by 1 in each succeeding term. The exponent of x in the second term is 1, and increases by 1 in each succeeding term. The coefficient of the first term is 1 ; of the second, 5. Multiplying 5, the coefficient of the second term, by 4, the exponent of a in that term, and dividing the result by the exponent of x increased by 1, or 2, we have 10 as the coefficient of the third term ; and so on. Then, (a + ic)^ = a^ + 5 a*x + 10 aV + 10 aV + 5 ax^-[- x^. It will be observed that the coefficients of terms equally distant from the ends of the expansion are equal ; this law will be proved in § 288. Thus the coefficients of the latter half of an expansion may be written out from the first half. If the second term of the binomial is negative, it should be enclosed, negative sign and all, in parentheses before applying the laws ; in reducing, care must be taken to apply the princi- ples of § 130. 2. Expand (1 - xf. (l-Xy=:\:i + (-X)f = 16 + 6 . 1^ . (- oj) + 15 . 1^ . (- xy + 20 . 13 . (- xy -h^5 'V ' (-xy + 6 '1 ' (-xy-\-{-xy = l-6x-\-15a^-20a^A-15x^-6a^-^x^ If the first term of the binomial is a number expressed in Arabic numerals, it is convenient to write the exponents at first without reduc- tion ; the result should afterwards be reduced to its simplest form. If either term of the binomial has a coefficient or exponent other than unity, it should be enclosed in parentheses before applying the laws. INVOLUTION AND EVOLUTION 177 3. Expand (3 m^ + 2 ri^)*. (3 m^ + 2 ny = [(3 m') + (2 n«)]^ = (3 m^)^ + 4 (3 m')X2 7i') + 6 (3 m^)^ (2 nS)^ + 4(3m2)(2n3)3+(2n3)4 = 81 m« + 216 mV + 216 mV + 96 mV + 16 w^l A polynomial may be expressed as a binomial, and raised to any positive integral power by successive applications of the Binomial Theorem. But for second or third powers, the methods of §§ 134 and 136 are shorter. 4. Expand (a^ - 2 a; - 2y, (x" -2 x-2y=l(x^ -2 X) -^ (-2)y = (x'-2 xy-Jr 4(aj2 _ 2 xfi - 2) + 6(a^ - 2 xy(- 2y + 4.(x' -2 x)(-2y + (-2y = x^-Sx' + 24:a^-S2af + 16x^ -S(x^-6x' + 12x^-Sa^ + 24(a;*-4a^ + 4a;2)_32(a^_2a;) + 16 = ic8_8fl;7_|_i6a;« + 16aj^-56a^-32a^ + 64a^+64a; + 16. 287. To find the rth or general term in the expansion of (a -1- xy. The following laws hold for any term in the expansion of (a + a;)", in equation (4), § 285 : 1. The exponent of x is less by 1 than the number of the term. 2. The exponent of a is n minus the exponent of x. 3. The last factor of the numerator is greater by 1 than the exponent of a. 4. The last factor of the denominator is the same as the exponent of x. Therefore, in the ?'th term, the exponent of x will be r — 1. The exponent of a will be n — (r — 1), or w — r -f 1. The last factor of the numerator will be n — r + 2. The last factor of the denominator will be r — 1. 178 ADVANCED COURSE IN ALGEBRA Hence, the rth term ^ n(n-l)(n-2)".(n-r-\-2) ^n-r+i^r-i n^ 1.2.3...(r-l) * ^^ In finding any term of an expansion, it is convenient to obtain the coefficient and exponents of the terms by the above laws. Ex. Find the 8th term of (3 a - b'y\ We have, (3 a - bY = [(3 a) + (- 6^)]". The exponent of (— ¥) is 8 — 1, or 7. The exponent of (3 a) is 11 — 7, or 4. The first factor of the numerator is 11, and the last factor 4 + 1, or 5. The last factor of the denominator is 7. Then, the 8th term ^ 11 • IQ • ^ • 8 » 7 ■ 6 » 5 .3 ^yr^^y ' 1.2.3.4.5.6.7 ^ ^ ^ ^ = 330 (81 a^) (- 6^) = - 26730 a'b^. If the second term of the binomial is negative, it should be enclosed, sign and all, in parentheses before applying the laws. If either term of the binomial has a coefficient or exponent other than unity, it should be enclosed in parentheses before applying the laws. 288. Multiplying both terms of the coefficient, in (1), § 287, by the product of the natural numbers from 1 to n — r + 1, inclusive, the coefficient of the rth term becomes n(n - 1) . . . (yi - r + 2) . (n - r + 1) . . . 2 . 1 ^ |j^ I r — 1 xl '2 ...(n — r + 1) I r — 1 \ n — r-\-l To find the coefficient of the rth term from the end, which since the number of terms is ti + 1, is the [n — (r — 2)]th from the beginning, we put in the above formula n — r-{-2 for 7\ Then, the coefficient of the rth term from the end is In \ n I — or I n-r + 2-1 I n-(n-r + 2) + l ' | n-r-f 1 | r-1 Hence, in the expansion of (a + xy, the coefficients of terms equidistant from the ends of the expansion are equal. INVOLUTION AND EVOLUTION 179 289. Properties of Coefficients in the Expansion of (a -f- a;)**. I. Putting in (4), § 285, a = l and a; = 1, we have That is, tJie sum of the coefficients, in the expansion of (a+ic)" is equal to 2**. II. Putting in (4), a = 1 and x = — l, we have Q^l y^ I n(n-l) n(n-l)(n-2) _ [2 . [3 Or, l + !%i) + ... = . + ^^"-|V"-^U -. . That is, the sum of the coefficients of the odd terms is equal to the sum of the coefficients of the even terms. 290. The Greatest Coefficient in the Expansion of (a + a;)". By § 287, the coefficient of the (r + l)th term, in the expan- sion of (a + ^)'N is n(n — l) ••' (n — r-\-l) \r This is obtained by multiplying the coefficient of the rth , n — r-\-l n-\-l . term by ^^, or —I^- L r r The latter expression decreases as r increases. It is evident that the successive coefficients, commencing with the first term, will increase numerically so long as ■ IS > 1. r I. Suppose n even ; and let n = 2m, where m is a positive integer. Then, ^I — IIL- becomes r r If r = m, ^^~'' + ^ becomes '^ + ^ , and is > 1. r m 180 ADVANCED COURSE IN ALGEBRA U r = m + l, '^'^-r-Vl becomes ^ . and is < 1. r m + 1 Then, the greatest coefficient will be when r = m-\- 1. As the number of terms in the expansion is 2 m + 1, it fol- lows that the middle term has the greatest coefficient. II. Suppose n odd ; and let n — 2m-\- 1, where m is a posi- tive integer. Then, becomes ^t_, r r If r = m, 'in — r-j- ]3g(;omes !??lir_, and is > 1. r m If r = m + l, — — ±— becomes ^ "^ ^ , and equals 1. r m-\-l T-p ,o2m — r + 2, m -i-^h It r = m-f 2, ■ — becomes , and is < 1. ' r m + 2' There will then be two terms having the greatest coefficient; those where r = m + 1 and r = m-{-2. EXERCISE 40 Expand the following : 1. f-^J^y. 4. (a + 6)6. 7. (2a2_53)6. Expand the following to five terms : 10. (x + yy. 11. (a6-l)io. 12. (m2-2w3)ii. Expand the following : 13. (2 a;2 + x + 3)8. 16. (2 ^2 + x - 4)*. 14. (4 a2 - 3 a6 - 62)8. I7. (i + 2 a; + a;2)5. 15. (1 -3x+'2a;2)*. 18. (?;'2 - 3 x - 2)6. Find the 19. 8th term of (a + x)". 20. 10th term of (1 - w)". INVOLUTION AND EVOLUTION 181 21. 9tli term of (a2 + i)i5. 23. 7th term of {x^ - 2 y^y\ 22. Sthtermof f^- — Y- 24. Middle term of /'ma + ^V^ 25. Term involving a;i4 in ( a;* + — | • 26. Term involving x^i in (2x2 -) . V 3xV EVOLUTION 291. Evolution is the process of finding any root (§ 157) of an expression. We shall consider in the present chapter those cases only in which both the expression and its root are rational (§ 198). We have already given (§ 166) a rule for finding the principal root of a rational and integral monomial, which is a perfect power of the same degree as the index of the required root. 292. If m and n are positive integers, we have, by § 164, Whence, by § 157, V(a^ = ( Va")™. This method is preferable to that of § 166, if the expression whose root is to be found is a power of a number which is a per- fect power of the same degree as the index of the root. Ex. V(32 a}y = ( V32 a}^ = (2 a?f = 8 a\ 293. Any Root of a Fraction. Let n be a positive integer, and a and h numbers which are perfect ?ith powers. By § 165, ^^ X ^b = y;lf^= Va. Dividing both members by -^b, nja_;Va 182 ADVANCED COURSE IN ALGEBRA Then, to find any root of a fraction, each of whose terms is a perfect power of the same degree as the index of the required root, extract the required root of both numerator and denominator, and divide the first result by the second. Ex i 27 ft-'^^^^ ^ -^27 a'b' ^ 3 ab^ ^' 64 c» -^64c"» 4c3' SQUARE ROOT OF A POLYNOMIAL 294. In § 168, we gave a rule for finding the square root of a trinomial perfect square ; and, in § 170, of an expression of the form ^, _^ b'-\-c'+2 ab-\-2ac + 2 be, which could be seen, by inspection, to be a perfect square. We will now consider the method of finding the square root of any polynomial perfect square. Let A and B be rational and integral expressions (§ 63); and suppose them to be arranged in the same order of powers of some common letter, x. Let the exponent of x in the last term of A be greater, or less, than its exponent in the first term of B, according as A and B are arranged in descending, or ascending, powers of X. By § 131, {A-irBf = A^ + 2AB + B', Whence, {A + By-A' = 2AB+ &. If the expression 2 AB + B^ be arranged in the same order of powers of a; as ^ and B, its first term must be twice the product of the first term of A and the first term of B. Hence, the first term of B may be obtained by dividing the first term of 2 AB -|- B^ by twice the first term of A. By the expression "first term of ^," in the above discussion, we mean the sum of all the terms of A containing the highest, or lowest, power of a;, according as A is arranged in descending, or ascending, powers of as. Thus, if 2I = ax* + 6aj* + cx^, the first terra of A ia (a + 6) a*. INVOLUTION AND EVOLUTION 183 A similar meaning is attached to the expressions "last term of ^," and "lirst term of ^." 295. We will now consider an example. Required the square root of 24 a; - 12 a.-^- 7 a^H- 4 a;* + 16. Arranging the expression according to the descending powers of a?, we are to find an expression which, when squared, will produce 4 a;* - 12 ar^ - 7 a:^ ^ 24 aj + 16. It is evident from § 134 that the first term of the expression is the square of the term containing the highest power of x in the square root. Hence, the term containing the highest power of x in the square root must be the square root of 4 x'^, or 2 a?. Denoting the term of the root already found by A, and the remainder of the root, arranged in descending powers of a;, by B, we have (^-f--B)2-^2^4a;*-12a^-7a;2 + 24a; + 16-(2a^2 = _12a^_7a^ + 24aj + 16. (1) By § 294, the first term of B may be obtained by dividing the first term of (1), — 12 a^, by twice ^, or 4 a^^j that is, the first term of 5 is —3x. Hence, the first two terms of the root are 2 a^ — 3 a;. Denoting this expression by A\ and the remainder of the root, arranged in descending powers of a?, by B\ we have {A + By-A^'- = 4 a;^ - 12 «3 _ 7aj2 + 24 ajH- 16- (2 a^- 3 a;)2 = 4a;^-12aj«-7ar^-f24a; + 16-(4a^-12a^ + 9a!2) = _16a^4-24a; + 16. (2) By § 294, the first term of B' may be obtained by dividing the first term of (2), — 16 x^, by twice the first term of A\ or 4 .T^ ; that is, the first term of B' is — 4. Hence, the first three terms of the root are 2 ar^ — 3 a; — 4. 184 ADVANCED COURSE IN ALGEBRA Denoting this expression by A", and the remainder of the root, arranged in descending powers of x, by B", we have (A"-^B"y-A"' = 4.x'-12a^-Tx^-\-24:X-{-16-(2a^-3x-4:y =4:x'-12x'~7x'+24.x+16-(4:x'-12x'~7x^-j-24.x-{-16) =0. Hence, the required square root is 2 a^ — 3 a; — 4. 296. Let the last term of A' be O. Then, A' = A+C, 2ind A" = A' + 2AC + C\ Therefore, {A' + By - A^^ = {A + Bf - A^ -2 AC - C^ = \_{A^-By - A'^- {2 A+ C)a In like manner, if C" denotes the last term of A['j (^" + B"Y - A'" = [{A' + By - A"] -(2A'-{- G')C' ; and so on. That is, any remainder after the first may be obtained by subtracting from the preceding remainder an expression which is formed by doubling the part of the root already found, adding to it the next term of the root, and multiplying the result by this term. The expressions 2 A, 2 A', etc., are called trial-divisors, and 2 A -{• C, 2A'+ C, etc., complete divisors. 297. It is customary to arrange the work as follows, the com- plete divisors and remainders being formed by the rule of § 296 : 4:x'-12x^- 7 a;^ + 24 a; +16 12x^-3 a;- 4 4a;^ 4a^ 3a; 3a; -12 a;^- 7 a;2 -f 24 a; + 16, 1st Eem. -12a.'3+ 90^ 4ar^ 6x — 4 - 4 - 16 x2 -I- 24 a; + 16, 2d Kern. -16x2 + 24 a; + 16 To avoid needless repetition, the last three terms of the first remainderj and the last two terms of the second, may be omitted. INVOLUTION AND EVOLUTION 185 We then have the following rule for extracting the square root of a polynomial perfect square : Arrange the expression according to the powers of some letter. Extract the square root of the first term{% 294), write the result as the first term of the root, and subtract its square from the given expressioyi, arranging the remainder in the same order of powers as the given expression. Divide the first term of the remainder by twice the first term of the root, and add the quotient to the part of the root already found, and also to the trial-divisor. Multiply the complete divisor by the term of the root last obtained, and subtract the product from the remainder. If other terms remain, proceed as before, doubling the part of the root already found for the next trial-divisor. If the expression had been written 16 + 24 a; - 7 x2 - 12 x3 + 4 a:*, the square root would have been obtained in the form 4 + 3 x — 2 cc^, which is the negative of 2 aj2 _ 3 gj _ 4. 298. With the notation of § 296, 2^' = 2^ + 2a In like manner, 2 A^' = 2 A' -\-2 C ; and so on. That is, any trial-divisor, after the first, is equal to the preced- ing complete divisor with its last term doubled. SQUARE ROOT OF AN ARITHMETICAL NUMBER The term '■^ number,^'' in the following discussion, signifies an integral or decimal perfect square, expressed in Arabic numerals. 299. The square root of 100 is 10 ; of 10000 is 100 ; etc. Hence, the square root of a number between 1 and 100 is between 1 and 10; the square root of a number between 100 and 10000 is between 10 and 100 ; etc. That is, the square root of an integer of one or two digits contains one digit; the square root of an integer of three or four digits contains two digits ; etc. 186 ADVANCED COURSE IN ALGEBRA Hence, if a point he placed over every second digit of an integer, beginning at the units' place, the number of points shows the number of digits in its square root. 300. If a is an integral perfect square, then -— , where n is any positive integer, is also a perfect square. But is a number whose decimal part contains an even 10-" number of digits, and which differs from a only in the position of its decimal point. Hence, if a point be placed over every second digit of any number, beginning at the units' place and extending in either direction, the yiumber of points shows the yiumber of digits in its I square root. 301. Let a, b, and c represent positive integers. We have, (a + 5 + c)^-a^ ^ 2a(& + c)4-(& + c)^ ^^ ^ {b^-c)\ 2a 2a 2a That is, if the remainder obtained by subtracting a^ from (a + 6 + cy be divided by 2 a, the quotient is greater than b. In like manner, if the remainder obtained by subtracting a^ from (a -{-by be divided by 2 a, the quotient is greater than b. 302. We will now consider an example. Eequired the square root of 10719076. Pointing the number in accordance with the rule of § 299, we find that there are four digits in its square root. Since the number is between 9000000 and 16000000, the square root is between 3000 and 4000. That is, the first digit of the root is 3. Let a represent the number 3000 ; b the second digit of the root, multiplied by 100 ; and c the number consisting of the last two digits of the root in their order. Then, a + & -f c represents the root ; now, (a^b-{-cy-a' ^ 10719076 - 9000000 _ 1719076 _ ^^^ ^ 2 a 6000 6000 INVOLUTION AND EVOLUTION 187 By § 301, this is greater than h. Hence, 6 is a multiple of 100 less than 286. + . Assume, then, h = 200. Then, the first two digits of the root would be 32. Let a' represent the number 3200 ; h' the third digit of the root, multiplied by 10 ; and c' the last digit of the root. Then, a^ -\-h^ -{-c' represents the root ; now, (g^ 4- 6^ 4. cy _ a'2 ^ 10719076-10240000 ^ 479076 ^ n>^ . 2 a' 6400 6400 ' ' By § 301, this is greater than 5'. Hence, V is a multiple of 10 less than 74. + . Assume, then, 6' = 70. Then, the first three digits of the root would be 327. Let a" represent the number 3270, and 6" the last digit of the root. Then, a" + 6" represents the root ; now, (gn -4- 6>y - a^'2 ^ 10719076 _ 10692900 ^ 26176 ^ ^ _^ 2 a" 6540 6540 ' ' By § 301, this is greater than 6" ; assume, then, &" = 4. Since (3274)2 = 10719076, the required square root is 3274. 303. We have with the notation of § 302, (g' + 5'_l_c')2_a'2=(a+&+-c)2-(a + 6)2 z= {a-\-h + cf - a? -2 ah -W = [(a + 6 + cf - a^] - (2 a + h)h. Similarly, (g" + 6")2 _ g"2 ^ [-(^r _^ 5f _^ ^y _ ^f2-| _ (2 g' + &')6'. That is, any remainder after the first may be obtained by subtracting from the preceding remainder a number which is formed by doubling the part of the root already obtained, adding to it the next root digit followed by as many ciphers as there are digits in the remainder of the root, and multiplying the result by the latter number. The numbers represented by .2 a, 2 a', etc., are called trial-divisors^ and those represented by 2 a + 6, 2 a' + &', etc., complete divisors. 188 ADVANCED COURSE IN ALGEBRA 304. The work of the example of § 302 may be arranged as follows, the complete divisors and remainders being formed by the rule of § 303 : 16719076 I 3000 + 200 + 70+4 a^- 9000000 1st Comp. Div., 2d Comp. Div., 3d Comp. Div., Omitting the ciphers for the sake of brevity, and condensing the operation, it will stand as follows : 6000 + 200 200 1719076 1240000 6400 + 70 70 479076 452900 6540+4 4 26176 26176 10719076 9 3274 62 171 124 647 4790 4529 6544 [ 26176 26176 We then have the following rule for extracting the square root of an integral perfect square : Separate the number into periods by pointing every secoyid digit, beginning with the units' place. Find the greatest square in the left-hand period, and. write its square root as the first digit of the root; sxibtract the squai'e of the first root-digit from the left-hand period, and to the result annex the next period. Divide this remainder, omitting the last digit, by ttvice the jxirt of the root already found, and annex the quotient to the root, and also to the trial-divisor. Multiply the complete divisor by the root-digit last obtained, and subtract the product from the remainder. INVOLUTION AND EVOLUTION 189 If other periods remain, proceed as before, doiibling the part of the root already found for the next trial-divisor. Note 1. It sometimes happens that, on multiplying a complete divisor by the digit of the root last obtained, the product is greater than the remainder. In such a case, the digit of the root last obtained is too great, and one less must be substituted for it. Note 2. If any root-digit is 0, annex to the trial-divisor, and annex to the remainder the next period. 305. We will now show how to obtain the square root of a number which is not integral. Eequired the square root of 49.449024. Wehave, V4al4902i = ^,'^g = I5| (§ 293) = 7.032. The work may be arranged as follows : 49.449024 17.032 49 1403 4490 4209 14062 28124 28124 Since 14 is not contained in 4, we write as the second root-digit, in the above example ; we then annex to the trial-divisor 14, and annex to the remainder the next period, 90. Hence, if any number be pointed in accordance with the rule of § 300, the rule of § 304 may be applied to the result, and the decimal point inserted in its proper position in the root. 306. After n-\-l digits of the square root of an integral perfect square have been found by the rule of § 304, n more may be obtained by simple division only, supposing 2n-\-Xto be the whole number. For let a represent the integer whose first n-\-l digits are the first w -f- 1 digits of the root in their order, and whose last n digits are ciphers; and let h represent the integer consisting of the last n digits of the root in their order. Then, a-\-h represents the root. 190 ADVANCED COURSE IN ALGEBRA We have, (a + by- -a^ = 2ab-{- h\ Whence, (a + by- 2a -a' '-fe That is, (a -\-by- - a^, divided by 2 a, digits of the root, increased by - — 52 ^ We will now prove that -— is less than - ; so that, by neg- Z a Li lecting the remainder arising from the division, we obtain the part of the root required. By hypothesis, h contains n digits. Then, 6^ cannot contain more than 2 n digits. But a contains 2 n + 1 digits. • 52 6^ 1 Hence, - is less than 1 : and therefore — is less than — a 2a . 2 If, then, the (n + l)th remainder be divided by twice the part of the root already found, the remaining n digits of the root may be obtained. The method applies to the square root of any number. Ex. Required the square root of 638.876176. 638.876i76 [25^ 4 45 238 225 502 1387 1004 50.4)3.836176(.076 3528 3081 We obtain the first three digits of the root by the ordinary method, and the other two by the method of § 306; that is, by dividing the third remainder, 3.836176, by twice the part of the root already obtained, or 50.4. The required root is 25.2 + .076, or 25.276. INVOLUTION AND EVOLUTION 191 CUBE ROOT OF A POLYNOMIAL 307. In § 176, we gave a method for finding the cube root of any expression of the form a^ H- 3 a^fe + 3 ab^ + b% or a^ - 3 a'b -\-Sab^- b\ We will now consider the method of finding the cube root of any polynomial perfect cube. Let A and B have the same meanings as in § 294. By § 135, {A + B)3 = ^^ + 3 A^B + 3 AB' -f B\ Whence, {A + Bf -A^ = ?> A'B + 3 AB' + B\ If the expression 3 A^B + 3 AB~ + B^ be arranged in the same order of powers of x as A and B, its first term must be three times the product of the square of the first term of A and the first term of B. Hence, the first term of B may be obtained by dividing the first term of 3 A^B + 3 AB^ + B^ by three times the square of the first term of A. 308. We will now consider an example. Kequired the cube root of A0a^-6a:^-64. + x^-96x. Arranging the expression according to the descending powers of X, we are to find an expression which, when cubed, will produce x^ _ 6 af + ^0 x" -96 x - 64:. It is evident from § 136 that the first term of the expression is the cube of the term containing the highest power of x in the cube root. Hence, the term containing the highest power of x in the cube root must be the cube root of a?^, or a^. Denoting the term of the root already found by A, and the remainder of the root, arranged in descending powers of x, by B, we have (A + Bf-A^ = af-6 3^ + i0x'-96x-6i-(xJ = _ 6 a^ 4- 40 ar^ - 96 x - 64. (1) 192 ADVANCED COURSE IN ALGEBRA By § 307, the first term of B may be obtained by dividing the first term of (1), — 6 x', by three times the square of A, or 3 x"^ ; that is, the first term of jB is — 2 x. Hence, the first two terms of the root are x^ — 2x. Denoting this expression by A', and the renjainder of the root, arranged in descending powers of x, by B', we have (A' + B'f-A" = a;« - 6 a^ + 40 a^ - 96 a; - 64 - (a^ - 2 ar) 3 = a;« _ 6 a^ + 40 a^ - 96 a; - 64 - (a;*5 - 6 a;^ + 12 a;^ - 8 a;3) = _12aj4_|.4g^_9g^_64 (2) Then, the first term of B' may be obtained by dividing the first term of (2), — 12 a;'*, by three times the square of the first term of A', or 3 aj"* ; that is, the first term of B' is — 4. Hence, the first three terms of the root are x^ — 2 x — 4. Denoting this expression by A", and the remainder of the root, arranged in descending powers of x, by B", we have (A" + By - A"' = a;6 - 6 a^ 4- 40 a^ - 96 a; - 64 - (a;2 - 2 a; - 4)3 = 0. Hence, the required cube root is »^ — 2 a; — 4. 309. Let the last term of A' be C. Then, A' = A -{- C; whence, (A' + By - A'' =(A-\-By-A^-S A'C - 3 AC - C^ = l(A + By -A'-] ~(SA'-\-3AC+ C')G. In like manner, if C" denotes the last term of A'\ and so on. That is, any remainder after the first may be obtained by subtracting from the preceding remainder an expression which is formed by adding together three times the square of the part of the root already found, three times the product of the part of the root already found by the next term of the root, and the square of the next term of the root, and multiplying the sum by the latter term. INVOLUTION AND EVOLUTION 193 The expressions SA^, SA''^, etc., are called trial-divisors, and 3J.2 + SAC+ C^SA'^^ + SA'C + C'-^, etc., complete divisors. 310. We arrange the work as follows, the complete divisors and the remainders being formed by the rule of § 309 : x^-6Q^-\-^0a^-9Qx -64 x' x'-2x-4: 3a;4_6ar^ + 4a^ -6a^-^^0x^-96x -64 -6a^-^12x'- Sa^ Sx'-naf + Ux' -12x''-\-4.Sx^-96x -64 -12ar^ + 24a; + 16 3a;*-12af^ +24a; + 16 | - 12a;^ + 48a^ - 96a; -64 The last three terms of the first remainder, and the last two terms of the second, may be omitted. We then have the following rule for extracting the cube root of a polynomial perfect cube : Arrange the expression according to the powers of some letter. Extract the cube root of the first term, icrite the result as the first term of the root, and subtract its cube from the given expres- sion ; arranging the remainder in the same order of powers as the given expression. Divide the first term of the remainder by three times the square of the first term of the root, and write the residt as the next term of the root. ■ - Add to the trial-divisor three times the product of the term of the root last obtained by the part of the root previously found, and the square of the term of the root last obtained. Multiply the complete divisor by the term of the root last ob- tained, and subtract the result from the remainder. If other terms remain, proceed as before, taking three times the square of the part of the root already found for the next trial- divisor. 311. With the notation of § 309, 3 ^'2 = 3 (^ + C)2 = 3 ^2 + 6 ^(7 + 30^ = 3 ^2 _^ 3 ^(7-hC2+ (3 JIC + 2 02). 194 ADVANCED COURSE IN ALGEBRA In like manner, 3 A'" = 3 A" + 3 A'C'-\- C" + (3 A'CT + 2 C") ; etc. Tliat is, if the last term of the expression which is added to any trial-divisor he doubled, the residt, added to the corresponding complete divisor, will give the next trial-divisor. Thus, in the example of § 310, if we add to the first complete divisor Sx*-Qx^ + 4: x^^the expression -6x^-\-Sx^, the result, 3 x* - 12 x^ + 12 x^, is the next trial-divisor. CUBE ROOT OF AN ARITHMETICAL NUMBER The term '■'■ ymmher^'''' in the following discussion, signifies a positive integral or decimal perfect cube, expressed in Arabic numerals. 312. The cube root of 1000 is 10; of 1000000 is 100; etc. Hence, the cube root of a number between 1 and 1000 is be- tween 1 and 10 ; the cube root of a number between 1000 and 1000000 is between 10 and 100 ; etc. That is, the cube root of an integer of one, two, or three digits contains one digit ; the cube root of an integer of four, five, or six digits contains two digits ; etc. Hence, if a. point he placed over every third digit of an integer, heginning at the units'' jolace, the numher of points shows the num- ber of digits in its cube root. 313. If a is an integral perfect cube, then —j^, where w is any positive integer, is also a perfect cube. But T-— is a number, the number of digits in whose decimal part is divisible by 3, and which differs from a only in the position of its decimal point. Therefore, if a point be placed over every third digit of any numher, heginning at the units'' place and extending in either direction, the number of points shows the number of digits in its cube root. 314. Let a, b, and c represent positive integers. INVOLUTION AND EVOLUTION 195 Then (a + & + cy-ft^ ^ 3a^(6 + c) + 3a(&H-cy + (& + c)» ^^ ^ ^ , 3a(b + cy + (b + cY 3 a- That is, if the remainder obtained by subtracting a^ from (a + 6 + cy be divided by 3 a^, the quotient is greater than b. Similarly, if the remainder obtained by subtracting a^ from (a + by be divided by 3 a^, the quotient is greater than b. 315. We will now consider an example. Required the cube root of 9745491456. Pointing the number in accordance with the rule of § 312, we find that there are four digits in its cube root. Since the number is between 8000000000 and 27000000000, the cube root is between 2000 and 3000. That is, the first digit of the root is 2. Let a represent the number 2000 ; b the second digit of the root, multiplied by 100 ; and c the number consisting of the last two digits of the root in their order. Then, a-{-b + c represents the root ; now, (a^b-\- cy - g-^ ^ 9745491456 - 8000000000 So" 12000000 ^ 1745491456 _-^^g^ 12000000 By § 314, this is greater than b. Hence, 6 is a multiple of 100 less than 145. +. Assume, then, b = 100. Then, the first two digits of the root would be 21. Let a' represent the number 2100; b' the third digit of the root multiplied by 10; and c' the last digit of the root. Then, a' -\-b' -\- c' represents the root ; now, (a' -hb' + c'y - a" ^ 9745491456 - 9261000000 3 a'2 13230000 484491456 ^3g^ 13230000 196 ADVANCED COURSE IN ALGEBRA By § 314, this is greater than b'. Hence, b' is a multiple of 10 less than 36. + . Assume, then, b' =!= 30. Then, the first three digits of the root would be 213. Let a" represent the number 2130, and b" the last digit of the root. Then, a" + b" represents the root; now, (a" + b"y - a"^ ^ 9745491456 - 9663597000 3a"2 13610700 ^ 81894456 ^ ^ _^ 13610700 ■ * By § 314, this is greater than b" ; assume, then, b" = 6. Then, since (2136)^ = 9745491456, the required cube root is 2136. 316. We have with the notation of § 315, (a' + &' + Cy -a'^=(a + b + cf - (a + bf = (a + b-{-cy-a^-Sa'b-3ab^-b^ = [ (a + & + c)3 - a^] - (3 a^ + 3 a6 + &-) 6. Similarly, (a" + by-a"^ = [(a'^+ b' + cy-a'']-(3a"-\-S a'b'-{-b")b'. That is, any remainder after the first may be obtained by subtracting from the preceding remainder a number which is formed by taking three times the square of the part of the root already obtained, adding to it three times the product of the part of the root already obtained by the next root-digit fol- lowed by as many ciphers as there are digits in the remainder of the root, plus the square of the latter number, and multiply- ing the result by the latter number. The numbers represented by^ 3 a^, 3 a'^, etc. , are called trial-divisors, and those represented by Sa^ + 3ab + b% 3 a'^ + 3 a'b' + 6'^, etc., com- plete divisors. 317. The work of the example of § 315 may be arranged as follows, the complete divisors and remainders being formed by the rule of § 316. INVOLUTION AND EVOLUTION 197 9745491456 | 2000 + 100 + 30 + 6 a^ = 8000000000 12000000 1745491456 600000 10000 1st. Comp. Div., 12610000 1261000000 13230000 484491456 189000 900 2d Comp. Div., 13419900 13610700 402597000 81894456 38340 36 3d Comp. Div., 13649076 81894456 Condensing the operation, it will stand as follows 974549i456 | 2136 8 1200 : 60 1 L745 1261 : L261 132300 1890 9 484491 134199 402597 13610700 38340 36 81894456 13649076 81894456 We then have the following rule for extracting the cube root of an integral perfect cube : Separate the number into periods by pointing every third digit y beginning with the units' place. 198 ADVANCED COURSE IN ALGEBRA Find the greatest cube in the left-hand period, and write its cube root as the first digit of the root; subtract the cube of the first root- digit from the left-hand period, and to the result aimex the next period. Divide this remainder by three times the square of the part of the root already found, with two ciphers annexed, and write the quotient as the next digit of the 7'oot. Add to the trial-divisor three times the product of the last root- digit by the part of the root previously found, with one cipher annexed, and the square of the last root-digit. Multiply the complete divisor by the digit of the root last obtained, and subtract the product from the reynainder. If other periods remain, proceed as before, takiiig three times the square of the part of the root already found, with two ciphers annexed, for the next trial-divisor. Note 1, § 304, applies with equal force to the above rule. If any root-digit is 0, annex two ciphers to the trial-divisor, and annex to the remainder the next period. 318. With the notation of § 315, 3 a'2 = 3 (a -f bf = 3 a^ + 6 a& + 3 6^ = 3 a^ -f- 3 a6 + 6M- (3 ab + 2 b^). In like manner, 3 a"2 = 3 a'2 + 3 a'V -f b'^ + (3 a'Z>' -f 2 V') ; etc. That is, if the first number and the double of the second number required to complete any trial-divisor be added to the complete divisor, the residt, tvith tivo ciphers annexed, will give the next trial-divisor. 319. We will now show how to obtain the cube root of a number which is not integral. Kequired the cube root of 1073.741824. We.ave, ^1^21 = 4^^ = 12?i(§ 293) = 10.24. 100 ^ ^ The work may be arranged as follows: INVOLUTION AND EVOLUTION 199 1073.741824 10.24 1 30000 73741 600 4 30604 61208 600 12533824 8 3121200 12240 16 3133456 12533824 Here the second root-digit is ; we then ' annex two ciphers to the trial-divisor 300, and annex to the remainder the next period, 741. The second trial-divisor is formed by the rule of § 318. Adding to the complete divisor 30604 the first number, 600, and twice the second number, 8, required to complete the trial-divisor 30000, we have 31212 ; annexing two ciphers to this, the result is 3121200. Hence, if any number be pointed in accordance with the rule of § 313, the rule of § 317 may be applied to the result, and the decimal point inserted in its proper position in the root. 320. After n -{- 2 digits of the cube root of an integral per- fect cube have been found by the rule of § 317, n more may be obtained by division, supposing 2 n -f 2 to be the whole number. For let a represent the integer whose first w + 2 digits are the first n -J- 2 digits of the root in their order, and whose last n digits are ciphers, and b the integer consisting of the last n digits of the root in their order ; then, a+6 represents the root. We have, (a + by-a^=^3 a'b-\-3 aW + b\ Whence, ^02 = ^^ ^■^"i* 3 a^ a 3a^ That is, (a-\-bf — a^, divided by 3 a^, will give the last n 52 ^3 digits of the root, increased by — + ;;—;• a 3a^ By hypothesis, b contains n digits. Then, b^ cannot contain more than 2 n digits; 200 ADVANCED COURSE IN ALGEBRA But a contains 2 n + 2 digits Again, 52 H and hence — is less than — • a 10 — - = — X TT— ; and since — is less than — - , and - — oa^ a o a a 10 oa 53 1 less than 1, — - is also less than — • 3 ci^ 10 WW 1 ' Therefore, — + -— is less than - . a 3a^ . ^ If, then, the (n + 2)th remainder be divided by three times the square of the part of the root already found, the remaining n digits of the root may be obtained. The method applies to the cube root of any number. Ex. Required the cube root of 1452648.865311064. 1452648.865311064 1 113.2 300 452 30 1 331 331 30 121648 2 36300 990 9 37299 111897 990 9751865 18 3830700 6780 4 383748 4 7674968 6780 8 38442.72)2076.897311064(.054 19221360 15476131 INVOLUTION AND EVOLUTION 201 We obtain the first four digits of the root by the ordinary method, and the other two by § 320; that is, by dividing the fourth remainder, 2076.897311064, by three times the square of the part of the root already found, or 38442.72. The required root is 113.2 + .054, or 113.254. ANY ROOT OF A POLYNOMIAL 321. Let A and B have the same meaning as in § 294. By § 285, if n is any positive integer, (A + Bf = J." + nA^'-'^B H . Whence, (A + By - ^" = nA^'-^B + • • •• If the expression nA'^~'^B-\- •••he arranged in the same order of powers of a; as ^ and B, its first term must be n times the product of the (n — l)th power of the first term of A by the first term of B. Hence, the first term of B may be obtained by dividing the first term of the expression nA'^^^B + -"by w times the (n— l)th power of the first term of A. 322. It follows from § 321, exactly as in §§ 295 and 308, that the nth root of a polynomial, which is a perfect power of the nth degree, may be found by the following rule : Arrange the expression according to the powers of some letter. Extract the nth root of the first teryn, and write the result as the first term of the root; subtract f^om the polynomial its first term, and arrange the remainder in the same order of powers as the given expression. Divide the first term of the remainder by n times the (n — l)th power of the first term of the root, and write the result as the second term of the root. Subtract from the given polynomial the nth power of the part of the root already fouyid, and arrange the remainder in the same order of powers as the given expression. If other terms remain, proceed as before, dividing the first term of the remainder by n times the (n — l)th power of the first term of the root; arid continue in this manner until there is no remainder. 202 ADVANCED COURSE IN ALGEBRA ANY ROOT OF AN ARITHMETICAL NUMBER The term " number^'''' in the following discussion, signifies a positive, integral or decimal, perfect power of the degree denoted by the index of the required root, expressed in Arabic numerals. 323. It may be proved, as in §§ 299, 300, 312, and 313, that : If a point he placed over every 7ith digit of any number, beginning at the units' place and exlendiyig in either direction, the number of poiyits shows the number of digits in its nth root. 324. Let a, b, c, and n represent positive integers. Bv^285 (a + & + c)»-a- ^ [a + (6+c)]--a- ^na^~\b + c) + ^-.^ That is, if the remainder obtained by subtracting- d^ from (aH- 6 4- cy be divided by 7ia^~^, the quotient is greater than b. In like manner, if the remainder obtained by subtracting a** from (a + by be divided by na""^, the quotient is greater than b. 325. It is evident, from §§ 323 and 324, that the nth root of a positive integral perfect ?ith power may be found by a process similar to that employed in §§ 302 and 315. The general rule will be as follows : Point the number in accordance with the rule of § 323, and let the number of digits in the root be m. Find the greatest perfect nth power in the left-hayid period, and write its nth root as the first digit of the root. Raise the part of the root already found, with m — 1 ciphers annexed, to the nth power, and subtract the result from the given number. Raise the part of the root already found, with m — 1 ciphers annexed, to the (n — l)th power, and multiply the result by n. Divide the remainder by this number. If the quotient is a number ichose integral part contains m — 1 digits, write its first digit as the next digit of the root; otherwise, write a.9 the next root-digit. INVOLUTION AND EVOLUTION 203 Raise the part of the root already found, with m — 2 ciphers annexed, to the nth power, and subtract the result from the given number. The above process is to be repeated until there is no remainder; the only change being that, in the successive applications of the rule, m — 2, m — 3, etc., are written in place of m — 1 in the fourth, sixth, and seventh paragraphs. The rule may be used to find the wth root of any number. Ex. Find the cube root of 34550.415593. In this case, n = 3 and m = 4. 34550. 415593 | 32.57 3000^ = 27000000000 3 X 30002 = 2700000 0)7550415593( 200+ 3200^ = 32768000000 3 X 3200^ = 3072000 0)1782415593( 50+ 3250^ = 34328125000 3x3250^= 316875 00)222290593( 7+ 3257^ = 34550415593 Hence, the required root is 32.57. Some of the ciphers may be omitted in practice. It sometimes happens that, on raising the part of the root already found to the nth. power, the result is greater than the given number ; in such a case, the digit of the root last obtained is too great, an(J one less must be substituted for it. 326. Let m and n be positive integers, and a a perfect power of the degree mn. By §157, C^~ar- = a, (1) and (VV«)-=Va. (2) Raising both members of (2) to the nth. power, {^WaY^ = a. (3) From (1) and (3), by § 163, mn /— ^/ n/— "va = ^ va; for each of these expressions is the mnih. root of a. 204 ADVANCED COURSE IN ALGEBRA That is, the mnth root of a (§ 162) equals the mth root of the nth root of a. The above is only true of principal roots. It follows from the above that the fourth root of a perfect power of the fourth degree equals the square root of the square root of the expression. The sixth root of a perfect power of the sixth degree equals the cube root of the square root of the expression ; etc. In like manner, if m, n, and p are positive integers, and a a perfect power of the degree mnp, "Ta=V(VvS); and so on. 327. Let m, n, and r be positive integers, and a a rational number whose mth power is positive if n is even. By § 157, {-^/a^y = a-. Raising both members to the rth power, we have {■y/'ory = a'"'". (1) Also, CV^O" = »""■• (2) From (1) and (2), ( Vo^)"'* = (Va^)"'. Taking the nrth root of both members (§ 163), = Va , and Va = va . « This means that the principal ?ith root of a"* is equal to the principal nrth root of a"**". (The general theorems of evolution, in §§ 163, 164, 165, 292, 293, 326, and 327, were there proved only ioT principal roots. That of § 326 is only true for such roots. The others are true for certain values of the roots which are not prin- cipal roots ; take, for example, the equation Vab=^Va^y/h {% 165). If w = 2, « = 4, & = 9, it becomes \/4 x 9 = Vi x \/9. The last equation is true when the value + 6 is taken for V4 x 9, for Vi, and — 3 for V9 ; also, when the value — 6 is taken for y/4t x 9, + 2 for \/4, and - 3 for V9.) INVOLUTION AND EVOLUTION 205 EXERCISE 41 Find the values of the following : 1. v'(125 a3)2. 2. v^Cie xY^)^- 3. v^(- 243 a5&25cio)8. Find by inspection the values of the following : 343 6« > 812/8 >'32p30 Find the square roots of : 7. 49 a* + 16 64 + 14 a8& - 8 aS^ _ 55 ^252. 8. 16 x^ + 9 y"^ + 25 z^ - 24: xy- 40 XZ + 30 yz. 9 2/8 8 2/6 60?/* 5y2"*"25' 10. 4 x2 - 31 a;4 + 4 - 30 a;5 + 44 x^ + 25 x^ - 16 x. 11. 99729.64. 12. 64.91041489. 13. .0063138916. Find the cube roots of : 14. 27 x^ - 27 x^y - 99 ic*?/^ + 71 x^y^ + 132 x'^y^ - 48 xy^ - 64 y^. 15 8a« 12 g^ 10 a 5,56 6^ 53 16. 12 x* - 6 x8 - 27 X + a:9 + 62 x3 + 9x7 + 27 - 45 x^ + 13 x^ - 45 x^. 17. 201.230056. 18. 8831234.763. 19. .537764475968. 20. Find the fourth root of 16 x8 + 32 x' - 72 x6 - 136 x^ + 145 x* + 204 x^ - 162 x2 - 108 x + 81. 21. Find the sixth root of 1 + 12 X + 54 x2 + 100 x3 + 15 x4 - 168 x^ - 76 x^ + 168 x' + 15 x8 - 100 x9 + 54 xw - 12 x" + xi2. 22. Find the fifth root of 32 xi'> - 80 x9 + 240 x^ - 360 x? + 570 x^ - 561 x^ + 570 x* - 360 x3 + 240 a;2 _ 80 X + 32. 23. Find the fourth root of 888.73149456. 24. Find the sixth root of .009229812275335744. 25. Find the fifth root of 8472886.09443. 206 ADVANCED COURSE IN ALGEBRA XVI. INEQUALITIES 328. An Inequality is a statement that one of two expres- sions is greater or less than another. The First Member of an inequality is the expression to the left of the sign of inequality, and the Second Member is the expression to the right of that sign. Any term of either member of an inequality is called a term of the inequality. Two or more inequalities are said to subsist in the same sense when the first member is the greater or the less in both. Thus, a > 6 and c> d subsist in the same sense. PROPERTIES OF INEQUALITIES 329. An inequality will continue in the same sense after the same number has been added to, or subtracted from, both members. This follows from § 23, which is supposed to hold for all values of the letters involved. 330. It follows from § 329, that a term may be transposed from 07ie member of an inequality to the other by changing its sign. If the same terra appears in both members of an inequality affected with the same sign, it may be cancelled. 331. If the signs of all the terms of an inequality be changed, the sign of inequality must be reversed. For consider the inequality a — b>c — d. Transposing every term, d — c>b — a. (§ 330) That is, b — ab; that is, a > b. Again, let a — 6 be a negative number. Then, (a — b)-{-bb. By § 332, a — 6 is a positive number. Hence, if m is a positive number, each of the numbers m(a — b) and , or, ma — mb and , is positive. mm Therefore, ma >m?>, and — > — m m 334. It follows from §§ 331 and 333 that if both members of an inequality be multiplied or divided by the same negative num- ber, the sign of inequality must be reversed. 335. If any number of inequalities, subsisthig in the same sense, be added member to member, the resulting inequality will also subsist in the same sense. For consider the inequalities a>b, a' >b', a">b", •••. Then each of the numbers, a — b, a' — b', a" — b", •••, is positive. Therefore, their sum a-b-{-a'-b' -ha" -b" +*••', or, a + a' + a" 4- (& + &' + b" + .••), is a positive number. Whence, a-fa' + a" + ••• > b ■{- b' -\- b" -{- •••. 336. If two inequalities, subsisting in the same sense, be subtracted member from member, the resulting inequality does not necessarily subsist in the same sense. Thus, it a>b and a' >b', the numbers a — b and a' — &' are positive. But (a — b) — (a' — b'), or its equal (a — a') ~ (b — b'), may be positive, negative, or zero ; and hence a —a' may be greater than, less than, or equal to 6 — b'. 208 ADVANCED COURSE IN ALGEBRA 337. If a > & and a' >b', and each of the numbers a, a', b, b', is positive, then ^^t ^^, For since a—b, a' — b', a, and b' are positive numbers, each of the numbers ^^^, _ ^,^ ^^^ b'(a-b) is positive. Then, aa'>ab', and ab'>bb'. Then by § 335, aa' + ab' > ab' + bb'. Whence, aa' > bb'. 338. If we have any number of inequalities subsisting in the same sense, as a > 6, a' >b', a" >b", •••, and each of the numbers a, a', a", •••, b, b', b", •••, is positive, then aa'a""'>bb'b"'- - For by §337, aa'>bb'. Also, a" > &". Whence, aa'a" > &6'6" (§ 337). Continuing the process with the remaining inequalities, we obtain finally aa'a" ... > 66'6" ... • 339. It follows from § 338 that, if a is > b, and a and b are positive numbers, and n a positive integer, then a" > b\ 340. If 71 is a positive integer, and a and b perfect nth powers such that a is > 6, then ■Va>^b. For, if "Va were < -\/b, raising both members to the nth power, we should have a < b. (§ 339) And, if -\/a = -\/by a would be equal to b. Both of these conclusions are contrary to the hypothesis that a is > 6. Hence, -\/« > 'Vb, INEQUALITIES ^09 341. Examples. 1. Find tlie limit of x in the inequality Multiplying both members by 3 (§ 333), we have 21a; - 23 < 2a; + 15. Transposing (§ 330), and uniting terms, 19x<38. Dividing both members by 19 (§ 333), a;<2. 2. Find the limits of x and y in the following : 3a; + 22/>37. (1) 2a; + 32/ = 33. (2) 9a; + 6?/ > 111. 4a; + 6 2/= ^^• 5x> 45, and a; > 9. 6 a; + 4 2/ > 74. 6 a; + 9 2/ = 99. Multiply (1) by 3, Multiply (2) by 2, Subtracting (§ 329), Multiply (1) by 2, Multiply (2) by 3, Subtracting, —5y> —25. Divide both members by — 5, 2/ < 5 (§ 334). 3. Between what limiting values of a; is a;^ — 4 a; < 21 ? a;2-4a;is<21 if a;2-4a;-21 is<0. That is, if (x + 3)(a; — 7) is negative. Now (x + 3)(a; — 7) is negative if x is between — 3 and 7. Hence, a;^ — 4 a; is < 21 if oj is > — 3, and < 7. 4. Prove that if a and b are positive numbers, b a Wehave(a-6)2<0; ot, a' -2 ab-^b^-^O. Transposing —2ab, a^ + 6^ < 2 ab. Dividing each term by ab, b a 210 ADVANCED COURSE IN. ALGEBRA 5. Prove that, if a and h are unequal positive numbers, Wehave (a-6)2>0; or, a^ -2 a& -f-&'>0. Transposing — ah, o? — ah + 6'^ > ah. Multiplying both members by the positive number a + 6, a^^h^>a%^-h''a. 6. Prove that, if a,h, and c are unequal positive numbers, 2(a3 + 63 _^ c^) > a^?) + h'^a + 6'c + c^h + c^a + a^c. By Ex. 5, d' + h^>a^h + h''a, ¥ + c'>h'c+c% and c^ -|~ a^ > c^a + a^c. Adding, 2(a3 + 6^ + c=^) > a'b + ft^a + b'c + c^ft 4- c^a + a'c. EXERCISE 42 Fiud the limits of x in the following : 1. (2x-3)3_71>4x(2x-5)(a;-2). 2. (2 - 3x)(3 - a:)+ 4x + 39 >3 +(2 + 3a;)(x + 3). 3. (x - l)(x - 2) (x- 3)<(a; - 5)(x + 6)(x - 7). 4. a^(x- l)<2 62(2x- l)-a&, if a-26 ispositive. ^ X - m , 2 ^ x_+n^ •£ ^ ^j^(j ^ ^j.g positive and w < n. n m Find the limits of x and y in the following : 5x + 6 2/<45. ^' ' 3x-42/ = -ll. J 7x-4?/>41. 8. Find the limits of x when 3 X - 11 < 24 - 11 X, and 5 x + 23 < 20 x + 3. 9. If 6 times a certain positive integer, plus 14, is greater than 13 times the integer, minus 63, and 17 times the integer, minus 23, is greater than 8 times the integer, plus 31, what is the integer ? 10. If 7 times the number of houses in a certain village, plus 33, is less than 12 times the number, minus 82, and 9 times the number, minus 43, is less than 5 times the number, plus 61, how many houses are there ? INEQUALITIES 211 11. A farmer has a number of cows such that 10 times their number, plus 3, is less than 4 times the number, plus 79 ; and 14 times their num- ber, minus 97, is greater than 6 times the number, minus 5. How many cows has he ? 12. Between what limiting values of ic is a;^ 4. 3 x < 4 ? 13. Between what limiting values of x is 2 aj^ + 1.3 x > 24 ? 14. Between what limiting values of ic is 6 x^ < 19 x — 10 ? Prove that, for any values of x, 49 15. 9x2 + 25<30jc. 16. x(x-3)<| 2i lb Prove that, for any values of a and 6, 17. (4a + 36)(4a-36)<66(4a-3&). 18. a4 4. ^4^ 2 ah{d?- - ah ■\- V^\ Prove that, if all the letters represent unequal positive numbers, 19. d^ + a2& + a52 + 53 > 2 a&(a + &). 20. a2 + 62 _f. c2 > «& + 6c + ca. 21. aW + &2c2 -I- c2a2 > d?'hc + V^ca + <^Mh. 22. (a + 6 - c)2 + (& + c - a)2 + (c + a - hy > a& + 6c + ca. 23. a26 + a62 + 62c + 6c2 + c2a + ca2 > 6 a 6c. 24. (a2 + 62 + C2) (X2 + ^2 4. ^2)^ (Qja; + 6?/ + czf. 212 ADVANCED COURSE IN ALGEBRA XVII. SURDS. THEORY OF EXPONENTS 342. Meaning of V2. It is impossible to find a rational number (§ 51) whose square shall equal 2 ; but we can find two rational numbers, which shall differ from each other by less than any assigned number, however small, whose squares shall be less, and greater than 2, respectively. For, writing the squares of the consecutive integers 1, 2, etc., we have 1^ =1, 2^ = 4, etc. Hence, 1 and 2 are two numbers which differ by 1, and whose squares are less and greater than 2, respectively. Again, 1.12 = 1.21, 1.22=1.44, 1.32=1.69,1.42 = 1.96, 1.5' = 2.25, etc. Hence, 1.4 and 1.5 are two numbers which differ by .1, and whose squares are less and greater than 2, respectively. Again, 1.412=1.9881, 1.422 = 2.0164, etc. Hence, 1.41 and 1.42 are two numbers which differ by .01, and whose squares are less and greater than 2, respectively. By suificiently continuing the above process, we can find two numbers which shall differ from each other by less than any assigned number, however small, whose squares shall be less and greater than 2, respectively. 343. The successive numbers,, in the illustration of § 342, whose squares are less than 2, are 1, 1.4, 1.41, etc. ; and the numbers whose squares are greater than 2, are 2, 1.5, 1.42, etc. If each series be continued to r terms, the difference between the rth terms of the two series is 1 which can be made less than any assigned number, however small, by sufficiently increasing r. Therefore, the rth terms of the two series approach the same limit (§ 245), when r is indefinitely increased. SURDS. THEORY OF EXPONENTS 213 This limit is taken as the definition of V2. 344. In general, if n is any positive integer, and a a rational number (§ 51), which is not a perfect power of the nth. degree, and which is positive if n is even, it is impossible to find a number whose nth power shall equal a. We can find, however, two rational numbers which shall dif- fer from each other by less than any assigned number, how- ever small, whose nth powers shall be less and greater than a, respectively. 345. If n and a have the same meaning as in § 344, and a^ tta, a^, etc., is a series of rational numbers whose nth powers are less than a, and a'l, a\, a'g, etc., a series of rational num- bers whose nth powers are greater than a, such that a\ ~ ai = 1, a'2 ~ a^ = .1, a'g ~ cig = .01, etc., we may show, as in § 343, that the rth terms of the two series approach the same limit, when r is indefinitely increased. This limit is taken as the definition of Va. The expression Va is called a Surd. The symbol ~ signifies the difference of the numbers between which it is placed. 346. In the illustration of § 342, we also have ( - 1)^ = 1, (-2)2 = 4, etc. It is therefore possible to find two negative rational numbers, which shall differ from each other by less than any assigned number, however small, whose squares shall be less and greater than 2, respectively. This is also the case with every surd of the form Va, when n is even. 347. If a is positive, and lk 10. ^IT. -f 16. €• THE THEORY OF EXPONENTS 355. In the preceding portions of the work, an exponent has been considered only as a positive integer. Thus, if m is a positive integer, a"^ = a X a X a X "• to m factors. (§ 60) 356. We have proved the following results to hold when m and n represent positive integers, and a any rational number : a'^Xa'' = a'^+'* (§ 85). (1) (a"^)" = a*"" (§ 128). (2) 357. It is necessary to employ exponents which are not posi- tive integers ; and we now proceed to define those forms of ex- ponent which are rational numbers, but not positive integers. In determining what meanings to assign to the new forms, it will be convenient to have them such that the above law for multiplication shall hold with respect to them. We shall therefore assume equation (1), § 356, to hold, what- ever number is represented by a, for all rational values of m and n, including the case where either m or n is zero'; and find what meanings must be attached in consequence to fractional, negative, and zero exponents. 358. Meaning of a Fractional Exponent. p Required the meaning of a', where p and q represent positive integers. If (1), § 356, is to hold for all rational values of m and n, we have - ? „ -+-+••• to ? terms -xq a' X a' X • • • to g factors = a« * = a' = a^. 218 ADVANCED COURSE IN ALGEBRA p That is, (a«)' = a^. p ■ Whence, by § 157, a^= ^/a^. p We shall then define a^ as being the gth root of a^. For example, a^ = -\/a'^j a^ = Va^; a^ = Va; etc. p We shall throughout the remainder of the work regard a^ as being the principal gth root of a^. 359. Meaning of a Zero Exponent. Required the meaning of a°. By § 356, (1), if m is any rational number Whence, a^ = — = 1. We shall then define a^ as being equal to 1. 360. Meaning of a Negative Exponent. Required the meaning of a~% where s represents a positive integer or a positive fraction. By § 356, (1), a-" x a" = a-'+' = a« = 1 (§ 359). Whence, a~' = — a' We shall then define a~^ as being equal to 1 divided by a*. For example, a~'^ = — ; a"^ = — ; 3 x~^y~^ = — - ; etc. 361. It follows from § 360 that Any factor of the numerator of a fraction may he transferred to the denominator, or ayiy factor of the denominator to the ^ numerator, if the sign of its exponent he changed. Thus, — may be written in the forms cd^ W d'hh-' aH-^ . -, etc. a-'cd'' d' ' h-'c- SURDS. THEORY OF EXPONENTS 219 362. We will now prove that, with the definitions of §§ 358 and 360, equation (1), § 356, holds for all rational valnes of m, n, and a, provided that a"" and a" are rational numbers or surds. It will be understood that, in all fractional exponents, the results are limited to principal roots. • I. Let m and n be fractions of the form ^ and -, respectively, where p, q, r, and s represent positive integers. By § 358, a^ Xa' = V~a^ X Va'" = V^xV^ (§ 327) = Va^'xa«'- (§ 165) = Va^''+^'- (§85) ps+qr p T = a '' =a' '. We have now proved that (1), § 356, holds when m and n represent any positive rational numbers. II. Let m be rational and positive, and let n = — g, where q is rational, positive, and less than m. By §§ 85, or 362, I, a"'-^ x a' = a"'-«+^ = oT. Whence, a"^-^ = ^ = a"* X cr^ (§ 360). a* That is, a"* X a'^ = oT''^. In like manner, the law may be proved to hold when n is rational and positive, and m=^—p, where p is rational, positive, and less than n. III. Let m be rational and positive, and let n = q, where q is rational, positive, and greater than m. By § 361, a- x a'' = -J— = -i— (§ 362, II) = a«-'. In like manner, the law may be proved to hold when n is rational and positive, and m = — p, where p is rational, positive, and greater than n. 220 ADVANCED COURSE IN ALGEBRA ly. Let m — —p, and n= — q, where p and q are rational and positive. Then, a'P x a"' = — = — (§§ 85, or 362, I) = a"^"*. Hence, a™ x f^ = »*"+" for all rational values of m and n. For example, a^ x a~^ — a}~^ = a~^; axa^ = o}^^ = a2 ; etc. 363. We have for all rational values of a, m, and n, provided that a"* and a" are rational numbers or surds, ^m-n y^gn^ ^m-n+n (§§ 35^ 3(32) =, ^« Whence, — = a"*-^ a" _3 a Por example, ^ = a ^ ^ = a ^; ^ = a^*' = ai, etc. a - 364. We will now prove that equation (2), § 356, holds for all rational values of a, m, and n, provided that a"* and a*"" are rational numbers or surds. In all fractional exponents, the results are limited to princi- pal roots. I. Let 91 be a positive integer. Then, {ary = a"* x a*" X a"* X ••• to ti factors ^ ^m+m+m+... ton terms (§§ §5^ ^Q^) = 0^"". II. Let n =^, where p and q are positive integers. p mp Then, (a'")' = i/Jary = i/aF^ (§ 364, I) = a^ HI. Let n = — s, where s is rational and positive. Then, (a"*)-^ = -J— = — (§ 364, I or II) = a"*"'. Hence, (a*")" = a'"" for all rational values of m and n. For example, (a^)^ = a^^t = a^; (a^)-i =a^x-i =a-'*; etc. SURDS. THEORY OF EXPONENTS 221 365. To prove (abc ...)" = a^'b^'c'' ••• for all rational values of a, b, c, •••, and n, provided that a% 6", c", •••, are rational 7iumbers or surds. The theorem was proved in § 129 for any positive integral value of n, and in § 165 for any value of n of the form — , where m is a positive integer. '"^ I. Let n=^, where p and q are positive integers. By § 364, l(abc ••.)'> = (abc ...)^ = a^b^c^ ... (§ 129). (1) p p p p p p By § 129, (a'6^c« ...)« = (a')«(6«)«(c^)« • • • = a^b^c^ * • .. (2) From (1) and (2), p p pp \_{abc ' . -yy = {a^¥c^ . • •)«'. p p pp Whence, (abc ...)* = a^¥c^ • . . (§ 163). This means that the principal gth root of {abc .••)p is equal to the product of the principal gth roots of a^, 6^, c^, •••. II. Let n = — s, where s is rational and positive. Then, (a6c ...)-' = -—i-— ^ ^ {abc'-y ^ (§§129, or 365, 1) a'b'(f"' = a-'b-'c--". Hence, {abc ...)" = a^'^c'' ... for all rational values of m and w. 366. Examples. In the following examples, every letter is supposed to represent a rational number such that every expression of the form a« is a rational number or surd. The value of a number affected with a fractional exponent may be found by first, if possible, extracting the root indicated by the denominator, and then raising the result to the power indicated by the numerator. 1. Find the value of (-8)i (-8)^ = V(-8)2=(V-8)2 (§292) = (-2)2 = 4. 222 ADVANCED COURSE IN ALGEBRA 2. Multiply a + 2a^-3a3 by 2-4a~^-6a"^. a + 2j - 3a^ 2-4 tt"^- 6a~^ 2 a + 4 a 3 -4: J 6a^ 8a^ + 12 6a^_12 + 18a"^ 2a 20 a* + 18 a"^. 3. Divide lSxy-'--23-{-x~iy-\-6x-y by 3 ic^2/^^ + ic^ — 2 ic~^2/- 18 xy^ - 23 + a;~^2/ + 6 x-y IS xy-^ + 6 x^jr' -12 3 aj%~^ + a?^ — 2 a? ^,j 6xh 2x~^-3x~^y — 6 x^y~^ — 11 + X ^y-{-6 x~y — 6flj^2/"^— 2-\-4.x~^y - 9-3a;~^2/4-6£c - 9-3a;~^?/ + 6 ic~y It is important to arrange the dividend, divisor, and each remainder in the same order of powers of some common letter. / i o, \5 . 4. Expand ( — — Vm^j by the Binomial Theorem. 1 c»^ = (^-1)5 _^ 5 (m-^)X- m^) + 10 (m-^)\- m^f + 10 (m-^)2(- m^f + 5 (m-^)(- m^)^ = m ^' — 5 m~^ • m^ + 10 m~^ • m^ — 10 m~^ .• m* -{-5m ^ -m^ —m^ — ^-¥ _ 5 ^-| _|_ 10 m^^^ — 10 m^ + 5 rrfi^ — m ^. SURDS. THEORY OF EXPONENTS 223 EXERCISE 44 It will be understood, in the following set of examples, that every letter used as the index of a root represents a positive integer, and every other letter a rational number, such that every expression of the form a" or \^a is a rational number or surd. Express with radical signs : 31 251 4 1 ^ I - 1. a^b^. 2. x^y^z^. 3. 6mJn^. 4. a^x^y*". Express with fractional exponents : Express with positive exponents : p 9. x'^y-^. 10. 4a ^b^. 11. rri^n 2. 12. a-»«6«c 2. In each of the following, transfer all literal factors from the denomi- nator to the numerator : 13. ^- 14. -1^. 15. -^^. 16. ^ y '±y z a^b-^c « In each of the following, transfer all literal factors from the numerator to the denominator : 17. ^. 18. ^l!^. 19. 2 rn~^x^ 20. ^If^. 6 Find the values of the following : 21. (aT^)5. 23. (a;*)"^^. 25. 126^ 27. (- 1024)i 22. (a-3)-9. 24. (a"^)T^. 26. 16^. 28. 729^. Multiply the following : 29. a:2 _ 4 a^f _ 5 _|_ 6 x~^ by 2 cc"^ + r7^ - 3 x"^. 30. m~^ + 2 wi-in-i + 3 m~^w-2 by 2 m~^w-i - 4 m~^w-2 + 6 w^. ft ^ 31. 3 a^&^ + 4 a6^ - a^& by 6 a^6~^ - 8 a"^6"^ - 2 a~i ^>^ ' .^ Divide the following : 32. TO"2n — 5 w~i + 4 m^ji-^ by w^w^ — m'^n — 2 m*^. 33. a;2|/"^ - 10 xxf"^ + 9 by Aj^ ^2x\^ - 3 x%. 34. atfti _ 2 62 + a-lfti by a^b^ -26^ + a"^6^. f** v> 4 1 ', |.L/ 224 ADVANCED COURSE IN ALGEBRA (In the following ten examples, use the rules of Chap. VII.) Find the value of : 35. (2a^ + 3&"*)2. 36. ^6 m-^^ - S mH-^y. 37. (4 x^y'i + 7 z-'^) (4 x V^ - 7 z-^) . 38. {Sx^-4.y~^y. 40. (a^b~^ - 2a^ - a-'^bfy. 39. (a-263 + 2 a35-2)3. 41. (x^ - 3 a;^ + 2 ic~^)3. ^^ 25a-6-49m^ ^3 8a;2 + 27y~^ ^^^ a^ - b~^ 5 a-3 - 7 m^ 2x^ +By~^ a^ + &~* 45. Factor a^ - S 6-9 by the rule of § 177. 46. Factor aJ + ah~^ + b~^ by the rule of § 172. Expand the following by the Binomial Theorem : 47. (a^ + 3 6~^)4. 49. (a~^\/¥ - b'^Vcfiy. 48. (zh-i-\y. BO. p_v5! + .?rV. ■ Find the 51. 6th term of (y/a^ + v^a)". 62. 7th term of (x%^ - ^ Y*. 53. 10th term of f 2 v^w —\ ^^ 54. Find the square root of aJb-^ - 6 a^6"2 + 5 ft-i + 12 a"^ + 4 a~^&. 55. Find the cube root of a;^- _ 6 X* + 2 1 x~^ - 44 jc"'^ + 63 x"^ - 54 x"^ + 27 x~i Simplify the following : 56. [^(xV2)-v^(aJ~V)]^. 60. l^C^V^^9)^^r\ M ^ s. V^ '-V/^ CI X^" - 1 , X2« + 1 68. (^^7«4!V- •-. 62. -^il^i:^x ------ + l. a.2u 4. 1 jc2 \ Q^TO-W Qj i _ 5 i a3 _ ^i 59 x^+y^ a; + y , g3 q^ + 2 &^ 7 a^&^ + 6 fe^ SURDS. THEORY OF EXPONENTS 225 REDUCTION OF SURDS It will be understood, in §§ 367 to 398, inclusive, that every letter used as the index of a root, represents a positive integer, and every other letter a rational number such that every expression of the form Va is a surd. 367. If a surd is in the form b-Va, where a and b are rational expressions (§ 198), b is called the coefficient, and n the index; and the surd is said to be of the nth degree. 368. A quadratic surd is a surd of the second degree. 369. Similar Surds are surds which do not differ at all, or differ only in their coefficients ; as 2-\/ax^ and 3 V«^. Dissimilar Surds are surds which are not similar^ 370. Reduction of a Surd to its Simplest Form. A surd is said to be in its simplest form when the expression under the radical sign is rational and integral (§ 63), is not a perfect power of the degree denoted by any factor of the index of the surd, and has no factor which is a perfect power of the same degree as the surd. 371. Case I. When the expression under the radical sign is a perfect power of the degree denoted by a factor of the index. Ex. Reduce V8 to its simplest form. We have, -^S=/&, Vc, •••, are surds. # 2. Reduce V3 a^b - 12 d'b'^ + 12 ab^ to its simplest form. V3 a% - 12 a'b^ + 12 ab^ = V(a' - 4 a& + 4 6^) 3 «& = Va2-4a6 + 4 62V3a6 = (a-2 6)V3a6. 226 ADVANCED COURSE IN ALGEBRA We then have the following rule : Besolve the expression under the radical sign into two factors, the second of ivhich contains no factor which is a perfect power of the same degree as the surd. Extract the required root of the first factor, arid multiply the result by the indicated root of the second. If the expression under the radical sign has a numerical factor which cannot be readily factored by inspection, it is convenient to resolve it into its prime factors. 3. Keduce v 1944 to its simplest form. -^ 12 + 8V3 6 + 4V3' Multiplying both terms of the latter by 6 — 4 V3, 4_V3- V7 ^ (3-V21)(6-4V3) 4 + V3-V7 6^-(4V3/ ^ 18-6V21-12V3 + 12V7 -12 ^ _9-|-3V21 + 6V3-6V7 ~ 6 The example may also be solved by multiplying^ both terms of the given fraction by 4 — V3 + V7, or by 4 — V3 - V7. 385. Case III. When the denominator is the sum of a rational expression and a surd of the nth degree, or of two surds of the nth degree. 1. Eeduce — ^—z to an equivalent fraction having a rational denominator. We have, (a + b) (a' -ab-h b') = a' + b\ Then, (2 + ^3) [2^ - 2 -^3 + {■^/m = 2^ + (^3/. Then, if we multiply both terms of the fraction by 22-2^5+(-\/3)^ the denominator will become irationaL SURDS. THEORY OF EXPONENTS 233 Thus 1 ^ ^ 2^-2-^3+_(-C/3)^ ^ 4-2-^3 + a/9 ' 2^V^ 2^+ {-y/^f 11 2. Reduce -^ ^ to an equivalent fraction having a V7 — V5 rational denominator. We have, (a - h)(a^ H-a^ft + aW + h^) = a' - b\ Then,^ if we multiply both terms of the fraction by i^r + {'27 ^e 19. 5^294-9 V150 + 18^/5 _ 24 J^. \27 \32 20. VeSo^P 4- VTtSo^ + a62 V63 a'^ft - 210 a^ft^ + 175 ahK 21. (m + n)A/?^L±J^ -(m - n)4'^^^^=^ - {mn - 3 n^) J— i Multiply the following : 22. VIW- and V%h^. 23. i/M, -v^, and ^. '8 '27 24. 5 Vm + % — 8 y/m — n and 6 y/m + n — 7 Vm — «. 26. 7 V8 + 3 V27 - 2 a/20 and 7 V2 - 3 V3 - 4 V6. 238 ADVANCED COURSE IN ALGEBRA In the following, introduce the coefficients under the radical signs 6 62 125 as' 26. 2v^. 27. 4:x^y^\/3x^y^. 28. ba^b^-yJ 29. «zi^jzz«^zm:. 30. 5'^ii^/23SHiz. a + X ^a^ + 2 aic - 3 a:2 x2 - 2 ^ (x - 1)2 Divide the following : 81. v^lii by \/l2. 32. y/U^ by v^T^^p. 33. ^ by ^^. 34. \/2i3 + \/i8 by V3. 35. Vx2 + 2 x - 3 by Vx2 - 6 x + 5. 36. \/a2 _ &2 by \/^2p-qr^p-. Simplify the following : 37. (v/54)3. 39. (^nx^)5. 41. v^(V243a^). 38. C^^^S"^^)*. 40. \/(\/l2). 42. v/(3xv^3^). Expand the following by the rules of Chap. VIl^: 43. (5\/2 + 2V6)2. 47. \/4 + 2 V3 x ^^4 - 2 Va. 44. (6a/5 + 7V3)(6V5-7V3). 48. V3 VS - 2 V? x V3 V5 + 2 ^7. 45. (3\/x + ?/-4\/x-2/)2. 49. ( Va + Vft + Vc) ( Va + Vd - Vc) . 46. (3V5-2>/lO)3. 50. (v^ + v^9)( v^ - v^). 51. (2V3 + 5\/2-V5)(2\/3-5\/2+\/5). 52. (\/lO-4V5 + 5V2)2. 53. (2V2 + V6 -VS)^. Factor the following : 54. V2a + \/3a. 55. x-\/x-20. 56. ac ■{■ ay/d - cVb -Vbd. 57. Factor v^ - v^i by the rule of § 171. 58. Factor a — 6 by the method of § 178, taking Va —y/b for the first factor. Reduce each of the following to an equivalent fraction having a rational denominator : 65. 59. __^«!_. 62. V^^Ty^-y/^^^^^ v^27 ab^c^ ' Vx2 + y^ + Vx'^ - ?/2 ""' A^. 16. \/l20 by V^. 18. Vsi by V^^, 15. \/^41 by V^, 17. V^:^^224 by V^. 19. V^^^lGO by VS. 250 ADVANCED COURSE IN ALGEBRA Expand the following by the rules of Chap. VII : 20. (4 V-~3 + 5a/^^)2. 22. (3V^=T + 2V^^)(3\/^^ - 2V^^). 21. (v^ir6-\/^^)3. 23. (3\/^^ + V^^-2\/^^)2. 24. (V3^-v/^r5 + V'^^)(v'^+V^^-V"^). Reduce each of the following to an equivalent fraction having a real denominator : 25. ^-^^ . 26. 2x/^ + 7x/i:3 . ^^^ 1 34-V-7 4\/-5-3>/^^ 2 + 3\/-2-\/-6 Expand the following by the Binomial Theorem : 28. (V^^ + SV'^2y. 29. iVa-y^^y, Simplify the following : o^ 3 , 6 oi 2\/5 + 3>Ar6 2\/5-3\/^ au. — — — -| • oi. — — —^^ • 5_4V-5 7 + 2V^ 2V5-3a/^ 2V5 + 3V-6 COMPLEX NUMBERS It will be understood, throughout the remainder of the present chapter, that every letter represents a real number. 415. The expression a -i-bi (§ 407), where a and b are any- real numbers whatever, is called a Complex Number. In operations with complex numbers, we shall assume the laws of §§ 12 and 14 to hold. It follows from this that the statement in the last paragraph of § 408 is equally true of complex numbers. 416. Addition, Subtraction, Multiplication, and Division of Complex Numbers. 1. Add a-{- bi and c + di. Since the laws of § 12 hold for complex numbers, (a + bi) + (c + cZz) = a + c 4- bi + di = a + c + (b + d)i, by § 12, II, and § 14, III. 2. Subtract c + di from a + bi. (a 4- bi) - (c + di) = a -^bi - c - di (^ 81) = a - c + (6 - d)i, by § 12, and § 24, (7). PURE IMAGINARY AND COMPLEX NUMBERS 251 3. Multiply a-\-bihj c + di. By § 88, (a + bi){c + di) = ac-{- adi + bci + Mi^ — ac — bd-{- {ad -f- bG)i, by § 411. 4. Express the quotient of a + bi by c + di as a complex number. Multiplying both numerator and denominator by c — di, we have a + ^<^' _ (g + &t)(c — di) _ ac — adi + bci — 6d^^ c + di (c + di)(c — di) & — dH^ _ac-\-bd-{-(bc—ad)i_ac-{-bdbG — ad. TTd' ~ c'-\-d' c^-\-d^^' 417. It follows from § 416 that the result of any finite num- ber of additions, subtractions, multiplications, and divisions, performed upon complex numbers, is a complex number. 418. By the definition of 0, 1X0 = i{a - rt) = ia - m(§ 24, (7)) = 0. It follows from this that the complex number a -|- bi becomes a real number when 6 is 0. It also becomes a pure imaginary number when a is 0. Hence, a + 6i cannot equal unless both a = 0, and 6 = 0. 419. Ifa-^bi = c-\- di, where a, b, c, and d are real numbers, then a = c, and b==d. Transposing the terms of the given equation, we have a-~c-\-{b — d)i = 0. Then, by §418, a- c = 0, and 6-d = 0. Whence, a = c, and b = d. 420. Square Root of a Complex Number. We will now prove that Va -h bi, where a and b are re^,!, can be expressed in the form -y/x + i V^, where Va; and Vy are real. 252 ADVAJ^CED COURSE IN ALGEBRA Squaring both, members of the equation Va + M = Va? + i V^, (1) we have a-^bi = x-\-2i -Vxy — y. Then by § 419, a^x-y, (2) and 6i = 2 1 w'xy. Subtracting, a — hi = x — 2i -Vxy — y. Extracting the square roots of both members, Va — hi — -Vx — i -Vy. (3) Multiply (1) by (3), -V^Tb' = x + y. ■ (4) Add (2) and (4), y/W+h^ + a^2x,ovx = ^^^^^' + ^ . Subtract (2) from (4), Va^ + 6^ — a = 2 ?/, or ?/ = Va' + &' - a It is evident from this that ■\/x and V^/ are real. 421. It follows from (1) and (3), § 420, that If Va H- hi = Va; + i V?/, w/iere a, 6, a;, and y, cure real num- hers, then Va — hi = Va; — z V?/- 422. The preceding principles may be used to find the square root of a complex number. 1. Find the square root of 7 — 6 V— 2. Assume, V7- 6V-2 =V^- V^V^^. (1) Then by § 421, Vt+W^ = V^^ + V^ V^^. (2) Multiplying (1) by (2), we have V49-36(-2) = a; + 2/, or, x-\-y = ll. (3) Squaring (1), 7 — 6 V^^ = x — 2 V^V— 1 — 2/. Whence by § 419, x — y = 7. (4) Add (3) and (4), 2 a; = 18, or a; = 9. Subtract (4) from (3), 2 3/ = 4, or 2/ = 2. PURE IMAGINARY AND COMPLEX NUMBERS 253 Substitute in (1), V7 - 6-V-2 = V9 - V2 V- 1=3- V^. The example may also be solved by the method of § 392. We have 6 V^ = 2V9 x(-2). We then separate 7 into two parts whose product is 9 x (— 2). The parts are 9 and -2; then, V7-6V^ = V9-\/^=3- \^^. We may also find by the above methods the square root of an expression of the form a + V6, or a — V6, when a is negative. 2. Find the square root of — 35 + 12 V6. We have V_ 35 + 12 V6 = V^l ^35 - 2 V216. Separating 35 into two parts whose product is 216, V_35 + 12V6 = V^^ V(27 - 2 V27^ + 8) = v^ri( V27 - V8) = 3 V^=^- 2 V"=^. EXERCISE 48 Extract the square roots of the following : 1. 13 + V^^l92. 3. - 38- 8 v^-^30. 5. - 52 - 2 V640. 2. 18 - 5V'^^28. 4. - 26 + V480. 6. 2 V- 16. 423. Putting a = and 6 = 1, in § 420, we have x = y = - Substituting these values in (1), we have 424. Cube Root of a Complex Number. We will now prove that if Va -\- bi = c-{- di, where a, &, c, and d are real, then -Va — bi = c — di. Cubing both members of the equation Va -\-bi = c-\- di, we have, by § 411, a + bi = c^ -\- 3 c'di - S cd^ - dH. Then by § 419, a = c^-S cd^ and bi = 3 cHi — dH. Subtracting, a—bi = (? — 3 &d% — 3 cd^ + d?L 254 ADVANCED COURSE IN ALGEBRA Extracting the cube root of both members, V a — bi = c — di. 425. The complex numbers a + hi and a — hi are called Conjugate. We have {a + hi) x (a - hi) = o?- hH\ = a^ + h\ Also, (a + hi) + (a — hi) = 2 a. Hence, the sum and product of two conjugate complex numbers are real. It will be shown in the Appendix (§ 804), that any even root of a negative number, or any root of a pure imaginary or complex number, can be expressed as a complex number. GRAPHICAL REPRESENTATION OF PURE IMAGINARY AND COMPLEX NUMBERS 426. Let XX' be a fixed straight line, and a fixed point in that line. It was shown in §§57 and 406 /^"^ that any positive real number, + a, X ^/ -^ o -ta A ^ could be represented by the dis- tance from to ^, a units to the right of in OX ; and any negative real number, — a, by the distance from to A', a units to the left of in OX'. 427. Since — a is the same as (-f a) x (— 1), it follows from § 426 that the product of -f a by — 1 is represented by turning the line OA which represents the number + a, through two right angles, in a direction opposite to the motion of the hands of a clock. We may then regard — 1, in the product of any real number by — 1, as an operator which turns the line which represents '\Mq first factor through two right angles, in a direction opposite to the motion of the hands of a clock. 428. Graphical Representation of the Imaginary Unit. By the definition of § 407, — 1 = ^ x i. PURE IMAGINARY AND COMPLEX NUMBERS 255 Y ■B C- +ai +i ^ . Y -i 0-^cb A C] -ai ■b' y' being Then, since one multiplication by i, followed by another multiplication by i, turns the line which represents the first factor through two right angles, in a direc- tion opposite to the hands of a clock, we may regard multiplication by i as turning the line through one right angle, in the same direction. Thus, let XX' and YY' be fixed straight lines, intersecting at right angles at 0, arranged as in the figure of § 270. Then, if +a be represented by the line OA, where ^ is a units to the right of in OX, -f ai may be represented by OjB, and — ai by OB'^ where ^ is ci units above, and B' a units below, 0, in YY\ Also, 4- i may be represented by OC, and — i by OC, where C is one unit above, and O one unit below, 0, in YY\ It will be understood throughout the remainder of the chapter that, in any figure where the lines XX' and YY' occur, they are fixed straight lines intersecting at right angles at 0, the letters being arranged as in the figure of § 270 ; that all positive or negative real numbers ar3 repre- sented by lines to the right or left of O, respectively, in XX' ; and all positive or negative pure imaginary numbers by lines above or below 0, respectively, in YY'. 429. Graphical Representation of Complex Numbers. We will now show how to repre- sent the complex number a + hi. Let a be represented by OA, to the right of if a is positive, to the left if a is negative. Let hi be represented by OB, above if 6 is positive, below if h is negative. ^^ Draw^ line AC equal and parallel to OB, on the same side of XX' as OB, and line OG. Then, OG is considered the result of adding hi to a ; that is, OG represents a + hi. B C'^"" y^ \)V a A Y' 256 ADVANCED COURSE IN ALGEBRA This agrees with the methods already given for adding a real or pure imaginary number, if either a or 6 is zero. The figure represents the* case where a and h are both positive ; if a is positive and 6 negative, OC will lie between OX and OF ; if a is nega- tive, 00 will lie between OY and OX' if h is positive, and between OX^ and Oy if 6 is negative. In accordance with § § 57 and 406, — (a + hi) may be repre- sented by line 0C\ where OO is equal in length to OC, and drawn in the opposite direction from 0. 430. The modulus of a real, pure imaginary, or complex number is the length of the line which represents the number. The argument is the angle between the line which represents the number and OX, measured from OX in a direction opposite to the motion of the hands of a clock. If, ^ for example, in the figure of § 429, ZXOC = 30% the argument of the complex number represented by OG is 30°, and the argument of the complex number represented by OC is 210°. The modulus is always taken positive, and the argument may have any value from 0° to 360°. The real numbers + a and — a have the modulus a, and arguments 0° and 180°, respectively ; the pure imaginary numbers + ai and — ai have the modulus a, and arguments 90° and 270°, respectively. 431. In the figure of § 428, 00 = -^ OA^ + AG^ = Vo^+T"; that is, the modulus of the complex number a-{-biis Va^ + 6^; this is also the modulus of the complex numbers a— hi, —a-\-bi, and —a — hi. 432. Graphical Representation of Addition. We will now show how to represent the result of adding h to a, where a and h are any two real, pure imaginary, or com- plex numbers. Let a be represented by OA, and h by OB. / ~~/!^ Draw line AG equal and parallel to OB, / ^y^ I on the same side of 0^1 as OB, and line OG. / y^ / Then, OG is considered the result of add- // / ing h to a\ that is, line OG represents a-\-h. o a ji PURE IMAGINARY AND COMPLEX NUMBERS 257 This agrees with the method of § 429, which is a special case of the above. In like manner, the sum of any number of real, pure imagi- ary, or complex numbers may be represented by a straight line drawn from 0. 433. Graphical Representation of Subtraction. Let a and h be any two real, pure imagi- nary, or complex numbers. Let a be represented by OA^ and h by OB ; and complete parallelogram OB AG. By § 432, OA represents the result of adding the number represented by OB to the number represented by OC. That is, if h be added to the number represented by OC, the sum is equal to a ; hence, a — 6 is represented by line OC. 434. Graphical Representation of Multiplication. Since + ai may be written ( + 1) x {-\-ai), the product of +1 by -\- ai is represented by turning the line OA, which represents the number + 1, through one right angle, in a direction opposite to the motion of the hands of a clock, and multiplying the result by a. And since —ai may be written (+1) x (— ai), the product of +1 by — ai is repre- sented by a line equal in length to that which represents the product of + 1 by + aiVbut drawn in the opposite direction from 0. This suggests the following : The product of any real, pure imaginary, or complex number by + ai may be represented by turning the line which repre- sents the number through one right angle, in a direction oppo- site to the motion of the hands of a clock, and multiplying the result by a. Y B- +1^ -ai ■B' 258 ADVANCED COURSE IN ALGEBRA c'a^ The product of any real, pure imaginary, or complex number by — at may be represented by a line equal in length to the line which represents its product by + at, but drawn in the opposite direction from 0. 435. Since a + bi may be written (+ 1) x (a + 6i), the prod- uct of -f 1 by a + bi is represented by turning the line OA, which represents the number + 1, through an angle equal to the argument of a-\-bi (§ 430), in a direction opposite to the motion of the hands of a clock, and multi- plying the result by the modulus of a -\- bi. And since — (a -f bi) may be written (+1) x (—a— bi), the product of + 1 by — (a -f- bi) is represented by a line equal in length to that which represents the product of +1 by a + bi, but drawn in the opposite direction from 0. This suggests the following : If a and b are any real numbers, the product of any real, pure imaginary, or complex number by a -f bi may be repre- sented by turning the line which represents the number through an angle equal to the amplitude of a -f bi, in a direction oppo- site to the motion of the hands of a clock, and multiplying the result by the modulus of a -h bi. The product of any real, pure imaginary, or complex number by — (a -f- bi) may be represented by a line equal in length to the line which represents its product by a -f- bi, but drawn in the opposite direction from 0. 436. Let a and b be any two real, pure imaginary, or complex numbers, repre- sented by the lines OA and OB, respec- tively. The result of multiplying a by 6 is represented by line OC, where angle XOC is the sum of angles XOA and XOB, and OC = OA x OB (§ 435). PURE IMAGINARY AND COMPLEX NUMBERS 259 Z XOA - Z XOB, and OC ■ That is, ab is represented by OC. In like manner, the product of any number of real, pure imaginary, or complex numbers may be represented by a straight line drawn from 0. 437. Graphical Representation of Division. Let a and b be any two real, pure imaginary, or complex numbers. Let a be represented by OA, and b by OB. Draw line OC, making ZXOC = OA OB Then, Z XOA == Z XOC + Z XOB, and OA=OCx OB. Whence, by § 436, OA represents the product of the number represented by OC by the number represented by OB. Then, OC represents a number which, when multiplied by b, gives a; and hence 0(7 represents -" b Therefore, the quotient of any two real, pure imaginary, or complex numbers can be represented by a straight line drawn from, 0. 438. Graphical Representation of Roots. Let a be any real, pure imaginar}^, or complex number, represented by line OA. Draw line OB making Z XOB = ~Z XOA, n and having its length equal to the ?ith root of the modulus of a. Then, Z XOA = nxZ XOB, and the modulus of a is the 7ith power of the length of OB. Then, by § 436, OA represents the nt\\ power of the number represented by OB. Whence, OB represents Va. ' 260 ADVANCED COURSE IN ALGEBRA 439. It follows from §§ 432, 433, 436, 437, and 438 that any number which is the result of any finite number of the follow- ing operations performed upon one or more real, pure imaginary, or complex numbers, may be represented by a straight line drawn from 1 Addition; Subtraction; Multiplication; Division ^ raising to any power whose exponent is a rational number (§ 51) ; extract- ing any root. This is a graphical representation of the fact that any such number can be expressed in the form a -j- 6^, where a and b are real numbers, either of which may be zero. (Compare §§417 and 804.) We shall limit ourselves in the present work to numbers of the above form. The discussion of complex exponents requires a knowledge of Higher Trigonometry. QUADRATIC EQUATIONS 261 XIX. QUADRATIC EQUATIONS 440. A Quadratic Equation is an equation of the second degree (§ 113), with one or more unknown numbers. In the present chapter we consider only quadratic equations involving one unknown number. The principles demonstrated in §§ 116 to 119, inclusive, and § 122, hold for quadratic equations. 441. By transposing all terms to the first member, any quadratic equation, involving one unknown number, oj, may be reduced to the form „^ + j^ + , ^ 0. If neither b nor c is zero, this is called a Complete Quadratic Equation. A Complete Quadratic Equation is sometimes called an Affected Quad- ratic Equation. If either or both of the numbers b and c are zero, the equa- tion is called an Licomplete Quadratic Equation. An incomplete quadratic equation of the form ax^ -f- c = 0, is called a Pure Quadratic Equation. In § 183, we showed how to solve quadratic equations of the forms aa;2 _^ 5^; = 0, ax^ + c = 0, a;2 + ax + 6 = 0, and ax'^-\-hx + c = 0, when the first members could be resolved into factors. 442. Consider the equation where ^ is a rational and integral expression involving the unknown numbers. We may write the equation, A^-& = 0, or {A + B){A-B) = 0. By § 182, the latter is equivalent to the set of equations ^ + ^ = 0, and ^-jB = 0. 262 ADVANCED COURSE IN ALGEBRA Or, to the equations A=:-\-B, and A = — B, which may be written together in the form A = ±B. The sign ± , called the double sign, when prefixed to a number, indi- cates that it may be either + or — . Thus, the given equation is equivalent to an equation which is obtained by equating the positive square root of the first member to ± the square root of the second. PURE QUADRATIC EQUATIONS 443. A pure quadratic equation may be solved by reducing it, if necessary, to the form x^=a, and then equating ic to ± Va (§ 442). 1. Solve the equation 3x^ -\-7 = — + 35. 4 Clearing of fractions, 12 a;^ + 28 = 5 a;^ + 140. \ Transposing and uniting terms, 7 x' = 112. X Dividing by 7, x- = 16. Equating ic to ± the square root of 16, x=± 4. 2. Solve the equation 7 x'^ — 5 = 5xF — 13. Transposing and uniting terms, 2 a;^ = — 8. Or, a;2 = - 4. Equating a; to ± the square root of —4, a;=±V— 4 = ±2V=T(§409). In this case, both values of x are imaginary (§ 407) ; it is impossible to find a real value of x which will satisfy the given equation. EXERCISE 49 Solve the following : 1. 2(:l x - 5)2 + 3(a; + 10)2 = 434. 2. (a; + 1)8 -(x - 1)8 = 20. 3 §^' + — + — = ^^^- — -5 * « 10 12 12 8 4. (2 X + 7 ) (o .X - 6) - (4 a; - 3) (7 a; + 5) - 24 X + 59 = 0. 5. (x + 2 a) (a; + 3 6) + (X - 2 a) (a; - 3 6) = a;2 + 4 a2 + 9 62. QUADRATIC EQUATIONS 263 6 3 a;2 + 7 5 a;2 + 3 4 cc^ _ ip ^ ^ 7 14 35 ' 7. (X +l){x- 2)(x - 3)-(3c - l){x + 2)(x + 3) = -52. COMPLETE QUADRATIC EQUATIONS 444. First Method of Completing the Square. By transposing the terms involving x to the first member, and all other terms to the second, and then dividing both mem- bers by the coefficient of x^, any complete quadratic equation can be reduced to the form Q? -\- px = q. A trinomial is a perfect square when its first and third terms are perfect squares and positive, and its second term plus or minus twice the product of their square roots (§ 167). Then, the square root of the third term is equal to the second term divided by twice the square root of the first. Hence, the square root of the expression which must be added to x^ -\-px to make it a perfect square, is ^, or ^, Adding to both members the square of ^, we have ^' + P^ + f = ^ + f = ^^^-T-^- , By § 442, this is equivalent to an equation which is obtained by equating the positive square root of the first member to ± the square root of the second. Then, it is equivalent to Adding to both members such an expression as will make the first member a perfect trinomial square, is called Completing the Square. 445. We derive from § 444 the following rule for solving a complete quadratic equation. 264 ADVANCED COURSE IN ALGEBRA Reduce the equation to the form x^ -\-px = q. Complete the square, by adding to both members the square of one-half the coefficient of x. Equate the positive square root of the first member to ± the square root of the second, and solve the linear equations thus formed. 446. 1. Solve the equation 3cc^ — 8a; = — 4. Dividing by 3, a:2_^ = _|. Adding to both members the square of -, we have ^_8£+m^ = »4 16^1 S [SJ 3 9 9 Equating the positive square root of the first member to ± the square root of -, 4. o 3 3 Transposing - 1, ic = | ± | = 2 or |. If the coefiicient of x^ is negative, the sign of each term must be changed. 2. Solve the equation — 9 o;^ — 21 a; = 10. Dividing by - 9, aj2 + Z^ = _^. 7 \ Adding to both members the square of -, \ 3 [fij 9 36 36 Equating the positive square root of the first member to ± the square root of — > / rr o wi, 7,325 Whence, ^ "" ~ 6 "^6^ ~3 °'' ~ 3* QUADRATIC EQUATIONS 265 447. By § 167, aV-f- 6aj will become a perfect trinomial square by adding to it the square of r-^, or - — 2 ax 2 a Hence, we may complete the square by adding to both mem- bers the square of the quotient obtained by dividing the coefficient of X by twice the square root of the coefficient ofx^. This is usually a more convenient rule than that of § 445, when the coefficient of x^ is a perfect square. 1. Solve the equation 9 a^ — 5 ic = 4. 5 Adding to both members the square of 2x3' \^J 36 36 5 13 Extracting square roots, 3 a; — - = ± — -. 6 6 Then, 3:K = |±^ = 3'or-|. Dividing by 3, ic = 1 or .' y If the coefficient of a? is not a^erfect square, it may be made so by multiplication. 2. Solve the equation 8^ — 15 a; = 2. Multiplying each term by 2, 16 a^ — 30 a; = 4. Adding to both members the square of , or — , 16a^-30a. + ^_j=4 + _ = _. 15 17 Extracting square roots, 4 a; = ± — . 4 4 Then, 4aj = l^±lI = 8or-i. 4 4 2 266 ADVANCED COURSE IN ALGEBRA Whence, ic = 2 or — . ' 8 If the coefficient of x^ is negative, the sign of each term must be changed. 448. Second Method of completing the Square. Every complete quadratic equation can be reduced to the form aa^ + 6x + c = 0, or ax- -\-bx= —c (^ 441). Multiplying both members by 4 a, we have 4 aV + 4 abx = — 4 ac. Completing the square by adding to both members the - (§ 447), or b, we obtain I 4 aV + 4 abx + 6^ = ft^ — 4 a/i. square of (§ 447), or b, we obtain 2 X 2 a Extracting square roots, 2 ax-\-b= ± V 6^ — 4 ac. Transposing, 2ax= —b ± V 6^ — 4 ac. -6± V6'-4a.c Whence, x = 2a We derive from the above the following rule for completing the square: Reduce the equation to the form ax^ + 6x + c = 0. Multiply both members by four times the coefficient of x^, and add to each the square of the coefficient of x in the given equation. The advantage of this method over the preceding is in avoiding fractions in completing the square. 449. 1. Solve the equation 2x^ — lx= — 3. Multiplying both members by 4 x 2, or 8, Wx'-mx=-24t. Adding to both members the square of 7, 16 a^- 56 a; + 7^= - 24 + 49 = 25. QUADRATIC EQUATIONS 267 Extracting square roots, 4 a? — 7 = ±5. 4 a; = 7 ± 5 = 12 or 2. Whence, ■ c» = 3 or ^. If the coefficient of x in the given equation is even^ fractions may be avoided, and the rule modified, as follows : Multiply both members by the coefficient ofx^, and add to each the square of half the coefficient of x in the given equation. 2. Solve the equation 15 «2 + 28 a; = 32. Multiplying both members by 15, and adding to each the square of 14, we have 152a;2 + 15(28 x) + 14^ = 480 + 196 = 676. Extracting square roots, 15 a; + 14 = ± 26. 15a;=-14±26=12or -40. Whence, a; = - or — -• The method of completing the square exemplified in the present section is called the Hindoo Method. EXERCISE 50 The following may be solved by either of the preceding methods, preference being given to the one best adapted to the example under consideration. 1. x2-a; = 12. 8. 4 a;2 - 12 x = 23. 2. 3a;2-17a; = -10. 9. 8a:2 + 7x + 2 = 0, 3. 5«2 + i7x = 12. 10. 36x2 + 3x = 6. 4. 10x2 + 27a; + 14 = 0. 11. 28x2 + 29x + 6 = 0. 5. 36x2-24x = 77. 12. 10 - 37a; = - 30x2. 6. 49a;2 + 21a;-4 = 0. 13. a;(5x + 22)+ 36 =(2x + 5)2. 7. 6 + 23x-18x2 = 0. 14. (2x + l)8-(2x+3)3=_386. 16. (3a; + 2)(2x-3) = (4x-l)2-14. 16. (a; + 4)(2x-l) + (2x-l)(3x + 2)-(3x + 2)(4x-l) = -49. 17. (x + 1) (a: + 3) = (x + 7) V2 + 12. 18. (5 + 2V3)x2-(4 + 14V3)x = 9- 14V3.'' 268 ADVANCED COURSE IN ALGEBRA 450. Solution of Complete Quadratic Equations by Formula. It follows from § 448 that, if ax^ -f- &a^ 4- c = 0, then ^^ -6±Vy-4 ac, ^^^ Z CI This result may be used as a formula for the solution of any complete quadratic equation in the form ao? + 6aj + c = 0. 1. Solve the equation 2a;^ + 5 a; — 18 = 0. Here, a = 2, 6 = 5, and c = — 18 ; substituting these values in (1), we have ^^ -5±V25Tl l^ -5±13 ^^^^_9, 4 4 2 I 2. Solve the equation 110 a;^ — 21 a; = — 1. Here, a = 110, 6 = -21, c = 1. Then, ^^21±VSr3440^21±2^1_^^l, ' 220 220 10 11 Dividing both terms of the fraction in equation (1) by 2, - = -^-^ (§377)=^-Ai (2) This is a convenient formula in case the coefficient of x in the given equation is even. 3. Solve the equation — 5 o;^ + 14 a? + 3 = 0. Here, a = — 5, 6 = 14, c = 3 ; substituting in (2), — 5 — 5 5 Particular attention must be paid to the signs of the coefl&cients in making the substitution. The student should now work the examples of Exercise 50 by formula. I QUADRATIC EQUATIONS 269 451. Fractional Equations involving Quadratics. In solving fractional equations which involve quadratics, we reject any solution which satisfies the equation obtained by equating to zero the L. C. M. of the given denominators (§ 222). 3 2 1. Solve the equation = 1. x—6 x—o Multiplying each term by (x — 6)(x — 5), we have 3a;-15-2a; + 12 = a;2_i;I^^_l_3Q Or, a^-12x = -S3. Completing the square, aj^ — 12 a; + 6^ = 3. Extracting square roots, x — 6 = ± V3. Whence, x=6± V3. Since neither 6 + V3 nor 6 — V3 satisfies the equation (x — 6) (cc — 5) = 0, both roots may be retained. 1 1 a^^Sx-6 2. Solve the equation We may write it 2^x 2-x a^-4 1 , 1 a^ + 3a;-6 x-\-2 x-2 a^-4 Multiplying each term by a;^ — 4, x-2 + x + 2 = x^-\-Sx-6. Or, x^ + x==6. Multiplying by 4, and adding V to both members, 4a;2 + 4x + l = 25. Extracting square roots, 2 a; + 1 = ± 5. Then, 2x = -l±5 = 4or -6. Whence, a; = 2 or — 3. Since 2 satisfies the equation a^ — 4 = 0, it must be rejected, and the only solution is x = — 3. 452. Literal Equations involving Quadratics. In solving literal equations which involve complete quad- ratics, the methods of § 449 are usually the most convenient. 270 ADVANCED COURSE IN ALGEBRA Ex. Solve the equation acx^ — adx -\- hex —bd = 0. We write the equation acxr — (ad — bc)x = bd. Multiplying by 4 ac, and adding (ad — bey to both members, 4 a^c^x^ — 4 ac (ad — bc)x-\- (ad — bey = 4 abed + a'd' - 2 abed + 6V = a'd'+ 2 abed + bh\ Extracting square roots, 2 acx — (ad — be) = ± (ad + be). 2 aca; = ad—be± (ad -{-be) =2 ad or — 2 be. Whence, a; = -or c a If several terms contain the same power of x, the coefficient of that power should be enclosed in parentheses, as shown above. The above equation may be solved more easily by the method of § 183 ; thus, by § 156, the equation may be written (ax + b) (ex — d)=0. Then, ax + b = 0, or a: = — - ; a and ex — d = 0, or a; = — c Several equations in Exercise 51 may be solved most easily by the method of § 183. As a general rule, literal quadratics are best solved by formula. 453. Equations leading to Quadratics, having the Unknown Number under Radical Signs. In solving equations of this kind, only solutions which satisfy the given equation should be retained. In verifying solutions, only principal values of the roots^are considered (§ 397). -t n 1. Solve the equation VS -\-x + VS — x = — • Vs — x Clearing of fractions, ■^^25 — x^ + 5 — cc = 12. Transposing, V25 — x- = x-{-7. Squaring, 25 — x^ = x'^ -}- 14: x -\- 49. QUADRATIC EQUATIONS 271 Or, 2a^-l-14a;=:-24. Multiplying by 2, and adding 7^ to both members, 4x2 _^ 28x4-49 = 1. Extracting square roots, 2 x + 7 = ± 1. 2x = -7±l = -6 or -8. Whence, a; = — 3 or — 4. Putting X = — 3, the given first member becomes V2 + V8 = V2 + 2 V2 = 3 \/2. 12 3x2x2 /~ The second member becomes — :: = ' = — = 3 v 2. V8 2\/2 Then, the solution x = — 3 is correct. Putting X = — 4, the given first member becomes 1 + \/9 = 4. 12 The second member becomes — = 4. V9 Then, the solution x = — 4 is correct. 2. Solve the equation Vx — 1 + V3 x + 3 = 4. Transposing -y/x — 1, V3 x + 3 = 4 — Vx — 1. Squaring, S x-^3 = W-SWx-l + x-l. Transposing, 8 Vx — 1 = 12 — 2 x. Or, 4 Vx — 1 = 6 — X. Squaring, ■l6a;-16 = 36-12x + a^. Or, 0^-28 x = -52. Completing the square, ^ ic2-28x + 142 = 144. Extracting square roots, aj — 14 = ± 12. Whence, x = 14 ± 12 = 26 or 2. Putting X = 26, the given first member becomes 5 + 9, or 14. Then, the solution x = 26 is not correct. Putting X = 2, the given first member becomes 1 + 3, or 4. Then, x = 2 is the only correct solution. 272 ADVANCED COURSE IN ALGEBRA EXERCISE 51 Solve the following : 1. 2x_A=:I^_2i. 14. V5x2-3x-41 = 3a;-7. 3 4x 9 4a: 5 13 1 15. 6 - V5 a;2 _ 9 = 12. 6^ 9x2 18 16. V7 a; + 8 - V5 x - 4 = 2. 3. A + i^ = _§^. 17 _A«___^+AL. = o 4x3 18 *a;-6&3a-10& 4. x2 + 2 wa; = 1 - m2. ^g 1 1 _ a;2 - 17 5. a;2-2ax = -6a + 9. ^ + 3 a;-5 a;2-2x-15 6. a;2 + ax + &x = - ab. 19. ^ ^ + ^^"^^ ^ = _-l^ 7. cc2 — (w — l)a: = 7i. a; - 2 24 (x + 2) x^ - 4 x-2 _ x + 4 _ _ 7 8. 6x + 5 _ 4x + 4 ^Q *";x+5 x-3~ 3* 4x-3 X-3 ' o r o n-i X — a x + a_x2 — 5a2 9. x2 — m2wx + mn^x = m^n^. 10. x2 - 4 ax - 10 X = - 40 a. ^^ 10a;2-3 ^ 6x2 + 6 Gx^-l a X — a x2 — .a2 22. V(a + 2 6)x-2a6 = x - 4 6. 21 18 9 9 x2 - 2' 23. 2 X + \/4 x2 - 7 V4 x2 - 7 12. 6 X? + 4 ax - 15 6x = 10 a6. „ „ . 24, ^ " _ *_ 13. aTOx2 + anx + bmx + bn = 0. '2x + a 3x — 4a 3 25. 2 \/3 X + 4 + 3 V3 X + 7 = ^ 26. (a + x)3 + (6-x)3 = (a + 6)3. 32. V3x + 4 1 1 14 x2 _ 3 X x2 + 4 X 15 x2 27. V^'=r^ + V2x + 3a = \/6a. 33. 2x_±i + §A=_2 ^ IZ . 3x-2 2x+l. 4 28. 3Vx^-— i=_ = 4. 34 __J^ = l_l4.1. Vx — 1 a — 6 + x a b x 29. _^ + «±^ = 2(«!±&!l. 35 If-i ^Usf-i 1^ a-\-b X a'^-b'' 3V4x-l 2/ VSx+l 3^ 30 x^-Jl^_4a6_. 3g ^^ ^-^ = 1 + ' X a2-62 aj2-4 3(x + 2) 2-x 31. J^ iziA^ = _5 37. 3V^T^ + 2V^T6^ = -24a. 4-5x 3x 6 3 x2 - 4 4 x2 + 3 9 x2 - 123 x2 + 6 2 x2 - 1 2 X* + 9 x2 - 5 = 1. QUADRATIC EQUATIONS 273 5 . 7 ^ 8 x^ - 13 a; - 64 * 2a; + 3'^3x-4 6x'' + x-12 40. x + 4a Zx — 2a 30 ax — (jp- 2x-Sa 3x + « 6a;2-7ax-3a2 41. X -{- a x+ b — c = 2. 45. x + 4 5+ 4x x-1 x2 + x + l x^ 2. 42. V^ + V(m-n)x + mn = 2m. 45 3x^ + x-2 ^3x^ + 4x-l^ 2x^-4x-6 2x2-2x-l 47. aV(l + x)2 - 62<^2(i _ ^y = 0. 2x+l 2n+l 43 28(3x+10) 25 ' 8 x3 - 27 2 x2 - 3 X " 2 X - 3 ?i 3x + n _ 10_ 3x + 7i 2x-3?i 3 1 44. 48 50. Vl + x2 + Vl x X + 4 ^ x + 4 \/x + 1 Vn + 1 1 ^ V2, X2* \/r+x2- VI X 51. V6 - 5 X + V2 - 7 X = V12 + 6 x. 5. ^+_« + ^Zl« = «+_^ + ^Jl^. 63. V¥T^-^/^T^ = a-b. X — a x-\- a a — h a — b 54. (X - 2 a + 3 6)2 + (X + 3 a - 2 6)2 = 25 a^ - 30 a6 + 13 62. 55. VxMn + V^^l _ V^MH - Vg ^ ^ 2 V16^^^. Vx2 + 1 1 \/x2+T 4- Vx2 - 1 56. (a - 6 + 2 c)x2 - (2 a + 6 + c)x = - a - 2 6 + c. 57. Va2 + ax + x2 + Va2 - ax + x2 = a ( V7 + Vl3). x+i x^h^ x + 3^3^ a;_l^a:-2 x-3 59 a: + 5 x-7 a; + 9 _.^^ ■x-4 x+6 x-8 60. Vl + X + x2 + Vl - X + x2 = X (1 + V3). 61. (X 4- a - 2 6)3 - (X - 2 a + 6)3 = 9 (a - 6)3. «« 6 1 2 3 x+3 x-5 x+7 9 63. V3 x2 + 4 X + 10 - V3 x2 + 2 X - 8 = 2. 64. -^ + l + -i- + l = o, x + a a x + 6 6 65. +u 1 0. X 6 — X a a + 6 66. (3 mn + n"^) x^-(Gm^-l iim - 2 n2)x +'2 win - 3 n2 = 0. 67. 1 x+3 x+2 x+4 x+7 x+5 x+8 x+6 274 ADVANCED COURSE IN ALGEBRA 68. (8 a'' -j- 10 ab-S 62) x^ - (16 a"- -^ 6 b"^) x + S a^ -I0ab-Sb^ = 0. 69 ^ 4- ^ — ^ _L 5 . 70 ^ ~ ^ _|_ ^ — ^ _ X — 4: a; — 5 x—a x—b b a 'x—2 x—S x—5 x—6 71. Va:2 + 2 X - Sr - Vx^ - 2 x - 3 = V2 x^ - 6. . 72. \/x2 + 7 ax + 12 a^ + Vx'^ - 7 ax + 12 a'^ = V2 x^ + 24 a^. 73. ^ + ^ + ^ + ^ = 0. x+2a+36 x-2a-3& x+2a-3 6 x-2a+36 ^^ x'^ + 6 X + 1 4 x^ 4- 12 X + 1 _ 2 x^ - 12 X - 1 3 x- - 9 x - 1 . x+6 x+3 x-6 x-3 76. (x-i] (x-?) ^x-^] = (X + 2)(x + 3)(x + 4). 77. (x2 + l)(a2 - &2 _ c2 + 2 &c) = 2 x(a2 + 62 _ 2 6c + c2). THEORY OF QUADRATIC EQUATIONS 454. A quadratic equation cannot have more than two different roots. Every quadratic equation can be reduced to the form ax^ 4- 6ic + c = 0. If possible, let this equation have three different roots, rj, rg, and ^3 ; then, by § 110, ar^' + hr^ + c = 0, (1) ar/-}-6r2 + c = 0, (2) and ar^ + 6? 3 + c = 0. (3) Subtracting (2) from (1), Then, a{r^ + r^ {r^ -r^ + h (n - r^) = 0, or, (ri — rg) {ar^ + arg + 6) = 0. Whence, either r^ — 7-2 = 0, or else ar^ + a?-2 + & = (§ 49). But rj — rs cannot be zero, for, by hypothesis, r^ and rg are different. Hence, ar^ -{- ar^ + h = 0, (4) QUADRATIC EQUATIONS 276 In like manner, by subtracting (3) from (1), we have m\ 4- arg + & = 0. (5) Subtracting (5) from (4), arg — ar^ = 0, or rg — Tg = 0. But this is impossible, for, by hypothesis, rg and r^ are different. Hence, a quadratic equation cannot have more than two different roots. 455. Sum and Product of Roots. Let Ti and 7*2 denote the roots of ax^ -\-bx-\- c = 0. Then, by § 450, 2a ' ' 2a Adding these values, ri + Tz = ^ — = 2a a Multiplying them together, 4ta^ 4a^ a Hence, if a quadratic equation is in the form ax^ -\-bx-\-c = 0, the sum of the roots equals minus the coefficient of x divided by the coefficient of x^, and the ^woduct of the roots equals the independent term divided by the coefficient ofx^. 1. Find by inspection the sum and product of the roots of 3a.'2_7a;-15 = 0. 7 —15 The sum of the roots is -, and their product — -— , or — 5. o o 17 2. One root of the equation 6 a;^ + 31 a; = — 35 is — - ; find the other. The equation can be written 6 or' + 31 a; + 35 = 0. q-J OK Then, the sum of the roots is , and their product — Then, the other root is _ §1 + 1, or — -^ - - : that is, - -• 6 2 6 2 3 276 ADVANCED COURSE IN ALGEBRA EXERCISE 52 Find by inspection the sum and product of the roots of : 1. a:2 4- 7 a; + 6 = 0. 5. 4 x - 12 a;^ = 3. 2. x2 - jc + 12 = 0. 6. 9 X - 21 x2 + 7 = 0. 3. 2c2 + 3 a: - 1 = 0. 7. 6 a;2 + x + 4 = 0. 4. 3a;2-a;-6 = 8. 6 ax'^ + 7 abx = 2Q b\ 9. One root of x2 + 7 x = 98 is 7 ; find the other. 10. One root oi28x'^ - x - 15 = is -- ; find the other. 11. One root of x'^-9x + xVS + 2-7V3 = 0is5 + 2V3; find tlie other. 12. If ri and ^2 are the roots of ax'^ + bx + c = 0, express the following in terms of ri and rz : (a) ri2 + rir2 + r^^. (&) ni±l2! . (c) i^ + i^ . (c?) n^ + r^z, nr^ ri2 ^22 456. Discussion of General Solution. By § 448, the roots oi ax^ -\- bx -\- c = are 2a 2a We will now discuss these results for all possible real values of a, 6, and c. I. 6^ — 4 ac positive. In this case, r^ and rg are rea? and unequal. II. 6^ _4ac = 0. In this case, r^ and r2 are real and egi^a?. III. hi — 4, ac negative. In this case, r^ and rg are imaginary (§ 407), or complex (§ 415). IV. 6 = 0. In this case, the equation takes the form ax^ + c = ; whence, x= ±\\ — -» ^ a If a and c are of unlike sign, the roots are real^ equQil irk absolute valuer and unlike in sign. QUADRATIC EQUATIONS 277 If a and c are of like sign, both roots are imaginary, V. c = 0. In this case, the equation takes the form ax^ + bx = 0: whence, x = 0, or a Hence, the roots are both reol^ one being zero. VI. 6 = 0, a7idc = 0. In this case, the equation takes the form aoiy^ = 0. Hence, both roots equal zero. The roots are both rational, or both irrational, according as 6^ — 4 ac is^ or is not, a perfect square. Ex. Determine by inspection the nature of the roots of 2x2-505-18 = 0. Herea = 2,6 = -5,c = -18; and 6^ - 4 ac = 25 -}- 144 = 169. Since b^ — Aac is positive, the roots are real and unequal. Since 6^ — 4 ac is a perfect square, both roots are rational. EXERCISE 53 Determine by inspection the nature of the roots of : 1. 6 x^ + 7 a; - 5 = 0. 6. 16 x2 - 9 = 0. 2. 4x2 -x = 0. 7. 9x2 = 12x + l. 3. 10a;2 + i7a; + 3 = 0. 8. 25x2 + 30x + 9 = 0. 4. 4x2 -20 a; + 25 = 0. 9. 7 a;2 + 3 a; = 0. 6. a;2 - 21 x + 200 = 0. 10. 41 x- 20x2 = 20. FACTORING 457. Factoring of Quadratic Expressions. A quadratic expression is an expression of the form ax^ + hx-\-c. In § 174, we showed how to factor certain expressions of this form hy inspection; we will now derive a rule for factoring any quadratic expression. 278 ADVANCED COURSE IN ALGEBRA We have, ax^ -\-bx -^ c = afx^ + — -^ -\ \ a aj L a \2aJ 4:0? aJ by §171. (1) But by § 450, the roots oi ax^ -{-hx-\-c = are ^ h ^b^-4:ac ^^^ b Vb^-4:ac 2a 2a 2a 2a ' Hence, to factor a quadratic expression, place it equal to zero, and solve the equation thus formed. Theii the required factors are the coefficient of x^ in the given expression, x minus the first root, and x minus the second. 1. Factor Q>x^ + 1 x-Z. Solving the equation 6 a;^ + 7 a; — 3 = 0, by § 450, -7±V49 + 72 _ -7±ll 1^, 3 ''- 12 --"l2~"==3 °' "2* Then, ^x^j^7x-S = q[x-^(x-\-^ = (3a;-l)(2a; + 3). The example may also be solved by using (1) as 2l formula. 2. Factor 4 + 13 a; - 12 a^. Solving the equation 4 + 13 a; — 12 aj^ = 0, we have ^_ -13±V169 + 192 ^ -13±19 ^ 14 -24 -24 43' QUADRATIC EQUATIONS 279 Whence, 4:-\-lSx-12x'=z-12fx-^^fx-^ -<-i)K-')e-i) = (1 + 4 a;) (4 - 3 x). 3. ¥2iGtov2x^-Sxy-2y^-7x + 4.y + 6. We solve the equation 2x^-x{3y-h7)-2y^ + 4:y-\-6 = 0. By §450, ^^ 3y + 7±V(3y + 7)^ + 16/-32y-48 ^ 3y + 7±V25y^ + 10y + l 4 ^ 3y + 7±(5y + l) 4 = -^ ^ or ^-J— = 2 ?/ -f 2 or — ^-^ — 4 4 -^ 2 Then, 2x''-3xy-2y^-7x + 4.y-{-6 = 2[x-(2y + 2)-]^x-^y^'] = (x-2y-2){2x-\-y-3). 458. If the coefficient of x^ is a perfect square, it is prefer- able to factor the expression by completing the square as in § 447, and then using § 171. 1. Factor 9 a^- 9 a; -4. By § 447, 9 ic^ — 9 a? will become a perfect square by adding 9 3 to it the square of - — -, or -; then, 2x3 2 9a52_9a._4 = 9a^_-9a;+/'|Y-?-4 -i''-2)-r 280 ADVANCED COURSE IN ALGEBRA Then, 9x'-9x-4.==(3x-^ + ^Vsx-^-^\ (§171) = {3x-\-l){3x-i). If the x"^ term is negative, the entire expression should be enclosed in parentheses preceded by a — sign. 2. Factor 3 -12 a;- 4 ccl 3 _ 12 a; - 4 a;2 = - (4 a;2 4- 1 2 a; - 3) = _(4a;2 + l2a; + 9-9-3) = -[(2a; + 3)2-12] = (2a; + 3+Vl2) X (-l)(2iv + 3-VT2) = (2V3 + 3 + 2 a;) (2 V3 - 3 - 2 x). In certain cases, the factors of a quadratic expression involve complex numbers. 3. Factor a;2 - 4 a; + 9. ^ x'-4,x-\-9 = (x'-Ax + 4) + 5 = (x-2f-(-^) _ = (a; _ 2 + V- 5) (a; - 2 - V- 5). EXERCISE 54 Factor the following by the method of § 457 : 1. a;2+14a; + 33. 10. 8 3:24.18 a; -5. 2. a;2 - 13 ic + 40. 11. 6 a;2 + 7 x + 2. 3. x^-x- 42. 12. 12 a;2 + 7 X - 45. 4. 4 a;2 + 3 X - 7. 13. 14 x2 - 23 xa + 3 a^. 5. 3x2-llx-20. 14. 24 x2 - 17 xy + 3 ?/2. 6. 2 x2 + 9 X + 9. 15. 28 x2 - mx - 2 m^. 7. 5x2-36x + 36. 16. 5 - 26 x- 24x2. 8. 9-8x-x2. 17. 8 + 14 X- 15x2. 9. 20 + 19x-6x2. 18. 21x2 + 23x?/2 + 6y*. 19. x^-xp-^6y'^-6x + lSy + 5. QUADKATIC EQUATIONS 281 20. x^-Sxy-iy^ + 6x-iy + S. 21. x^-6xy-\-5y^-2x-2y-S. 22. 2 a2 + 5 a& + 2 &2 + 7 a + 5 6 + 3. 23. Zx^-\-7xy-6y^-10xz-8yz + S z\ 24. 2-7?/-7x + 3z/2 + x^-4x2. Factor the following by the method of § 468 : 25. 4 a;2 - 12 a; - 7. 30. 1 + 2 a; - %\ 26. 32 - 12 a; - 9 x2. 31. 16 ^2 - 16 x + 1. 27. 16 x2 + 56 a: + 33. 32, 25 a;2 - 25 ic + 6. 28. 9 x2 + 24 X - 2. 33. 36 ^2 + 72 x + 29. 29. 4x2 + 20x + 19. 34. 11 + 10 x- 25x2. 459. We will now take up the factoring of expressions of the forms a?* + aa^?/^ + 2/^ or a;* + 2/^ when the factors involve surds. (Compare § 172.) 1. Factor a^ + 2 a^^^ + 25 &^ a* + 2 a^h^ + 25 6^ = {p> + 10 a%^ + 25 6^) - 8 o?}}" = (a2 + 5 62)2_(a5V8)' . = (a^ + 5 62 + a6 V8)(a2 + 5 6^ _ a6V8) = (a2 + 2 a6 V2 + 5 62)(a2 - 2 ab-^2 + 5 62). The above expression may also be expressed as the product of two factors involving complex numbers. 2. Factor x'' ^\. a^ + 1 = (iB^ + 2a^ + 1) - 2 a^ = (aj2 + l)2_(a;V2)2 = (a;2 _j_ 3,^2 4- V){y? - ajV2 + 1). EXERCISE 55 In each of the following, obtain two sets of factors, where this can be done without bringing in imaginary numbers : 1. x4-7x2 + 4. 4. 4a4 + 6a2 + 9. 2. otS + 64. 5. 36 x* - 92 x2 + 49. 3. 9 m*- 11 w2 + 1. 6. 26 m* + 28 m2n2 + 16 w*. 282 ADVANCED COURSE IN ALGEBRA 460. Solution of Equations by Factoring. In § 183, we showed how to solve equations whose first mem- bers could be resolved by inspection into first degree factors, and whose second members were zero. We will now take up equations whose first members can be resolved into factors partly of the first and partly of the second, or entirely of the second degree. 1. Solve the equation ic^ — 1 = 0. Factoring the first member, {x — l){x^ -\- x -{-1) := 0. Then, a; — 1 = 0, or a; = 1 ; and a:- + a; + 1 = ; whence, by § 450, l±Vl-4 -1±V- 2 2 2. Solve the equation a;'' + 1 = 0. By Ex. 2, § 459, a;4 -f 1 = {f -f a; V2 -\-l){x' - x^2 + 1). Solving x^ + a; V2 + 1 = 0, we have -V2±V2^^ -V2±V-2 2 Solving a? — a; V2 + 1 = 0, we have ^^ V2±V2-4 ^ V2±V-2 ^ EXERCISE 56 Solve the following equations : 1. 3 a:3- 2x2 +15 a; -10 = 0. 2. x* + 4a;2 - 32 = 0. 3. x3 - 64 = 0. 5. 27 x3 + 8 = 0. 7. 64 x^ - 125 = 0. 4. 3 a;4 + 24 X = 0. 6. 16 x* - 81 = 0. 8. x^ - 729 a^ = 0. g a;2 + 2x + 4 ^4_ ^^ x\x + 4 a) - 9 g^ x^ - 2 g^ ^ ^ ' x2 - 2 X - 4 x^' * x2(x - 4 a) - 9 a3 a:2 + 2 a^ 11. x4 + 81=0. 13. x* + 2x2 + 25 = 0. 15. x^ - 256 «» = 0. 12. x4 - 5 x2 + 1 = 0. 14. x4 - 18 x2 + 9 = 0. 16. 9 x* - x2 + 4 = 0. r QUADRATIC EQUATIONS 283 461. Maxima and Minima Values of Quadratic Expressions. The greatest or least value of a quadratic expression may sometimes be found by the artifice of completing the square. 1. Find the minimum value oi x^ — 5x-\-7. We have x^ — 5x-j-7 = (x ] -\ — Since (x J is positive for every real value of x, the least value of x — -] + - is when x = 2; ' 4 2 3 Thus, the minimum value of the expression is -• 2. Find the maximum value of 4 — 3 ic — 2 a^. We may write the expression 4^2la^-\-lx^ = 4.-2[fx-{-^^' ^ 2 1 I V 47 16 3 41 of . 3 ¥-T + 4 The greatest value is when x = . 41 Thus, the greatest value of the expression is — • 8 EXERCISE 57 Find the maxima and minima values of the following, and determine which : 1. x'^ + Sx-l. 3. 4x2-8x-5. 5. Sx^ + 5x + 4. 2. 6-Sx-x^. 4. 3 + a;-x2. 6. - 2 - 9a: -9x2. 7. 6x2 -7 a; + 3. 8. -7 + 2x-5x2. PROBLEMS INVOLVING QUADRATIC EQUATIONS WITH ONE UNKNOWN NUMBER 462. In solving problems which involve quadratic equations, there will usually be two values of the unknown number ; only those values should be retained which satisfy the conditions of the problem. The considerations of §§ 261 and 262 hold for equations of any degree. 284 ADVANCED COURSE IN ALGEBRA 1. A man sold a watch for $21, and lost as many per cent as the watch cost dollars. Find the cost of the watch. Let X = number of dollars the watch cost. Then, x = the per cent of loss, X x^ and X X — , or = number of dollars lost. 100' 100 By the conditions, — = a; — 21. 100 Solving, aj = 30 or 70. Then, the cost of the watch was either $30 or $70 ; for either of these answers satisfies the conditions of the problem. 2. A farmer bought some sheep for f 72. If he had bought 6 more for the same money, they would have cost him $1 apiece less. How many did he buy ? X = number bought. ■ . number of dollars paid for one. Let X Then, and 72 X 72 cc + 6 By the conditions. 72 X Solving, X = number of dollars paid for one if there had been 6 more. x + 6 = 18 or - 24. Only the positive value of x is admissible, for the negative value does not satisfy the conditions of the problem. Therefore, the number of sheep was 18. If, in the enunciation of the problem, the words "6 more" had been changed to "6 fewer," and "^1 apiece less" to "$1 apiece more," we should have found the answer 24. (Compare § 261.) EXERCISE 58 1. What number added to its reciprocal gives 2^ ? 2. Divide the number 24 into two parts such that twice the square of the greater shall exceed 5 times the square of the less by 45. 3. Find three consecutive numbers such that the sum of their squares shall be 434. 4. Find two numbers whose difference is 7, and the difference of whose cubes is 1267. QUADRATIC EQUATIONS 285 5. rind five consecutive numbers such that the quotient of the first by the second, added to the quotient of the fifth by the fourth, shall 23 equal — • ^ 12 6. Find four consecutive numbers such that if the sum of the squares of the second and fourth be divided by the sum of the squares of the first 13 and third, the quotient shall be — • 7. The area of a certain square field exceeds that of another square field by 1008 square yards. And the perimeter of the greater exceeds one- half that of the smaller by 120 yards. Find the dimensions of each field. 8. A fast train runs 8 miles an hour faster than a slow train, and takes 3 hours less to travel 288 miles. Find the rates of the trains. 9. The perimeter of a rectangular field is 184 feet, and its area 1920 square feet. Find its dimensions. 10. A merchant sold goods for $ 22.75, and lost as many per cent as the goods cost dollars. What was the cost ? 11. A merchant sold two pieces of cloth of different quality for $ 40.25, the poorer containing 28 yards. He received for the finer as many dollars a yard as there were yards in the piece ; and 7 yards of the poorer sold for as much as 2 yards of the finer. Find the value of each piece. 12. A merchant sold goods for $50.69, and gained as many per cent as the goods cost dollars. What was the cost ? 13. A has five-fourths as much money as B. After giving A $ 6, B's money is equal to A's multiplied by a fraction whose numerator is 15, and whose denominator is the number of dollars A had at first. How much had each at first ? 14. A and B set out at the same time from places 247 miles apart, and travel towards each other. A travels at the rate of 9 miles an hour ; and B's rate in miles an hour is less by 3 than the number of hours at the end of which they meet. Find B's rate. 15. A man buys a certain number of shares of stock, paying for each as many dollars as he buys shares. After the price has advanced as many dimes per share as he has shares, he sells, and gains $ 722.50. How many shares did he buy ? 16. The two digits of a number differ by 1 ; and if the square of the number be added to the square of the given number with its digits reversed, the sum is 585. Find the number. 17. A gives $ 336, in equal amounts, to a certain number of persons. B gives the same sum, in equal amounts, to 18 fewer persons, and gives to each $ 6 more than A. How much does A give to each person ? 286 ADVANCED COURSE IN ALGEBRA 18. The telegraph poles along a certain road are at equal intervals. If the interval between the poles were increased by 22 feet, there would be 8 fewer in a mile. How many are there in a mile ? 19. A merchant bought a cask of wine for 1 45. Having lost 3 gallons by leakage, he sells the remainder at f 1.50 a gallon above cost, and makes a profit of 33^ per cent on his entire outlay. How many gallons did the cask contain ? 20. The men in a regiment can be arranged in a column twice as long as it is wide. If their number were 224 less, they could be arranged in a hollow square 4 deep, having in each outer side of the square as many men as there were in the length of the column. Find the number of men. 21. The denominator of a fraction exceeds twice the numerator by 2, and the difference between the fraction and its reciprocal is — • Eind the fraction. ^"^ 22. A man started to walk 3 miles, intending to arrive at a certain time. After walking a mile, he was detained 10 minutes, and was in consequence obliged to walk the rest of the way a mile an hour faster. What was his original speed ? 23. A regiment, in solid square, has 24 fewer men in front than when in a hollow square 6 deep. How many men are there in the regiment ? 24. A rectangular field is surrounded by a fence 160 feet long. The cost of this fence, at 96 cents a foot, was one-tenth as many dollars as there are square feet in the area of the field. What are the dimensions of the field? 25. A crew can row down stream 18 miles, and back again, in 7| hours. Their rate up stream is 1| miles an hour less than the rate of the stream. Find the rate of the stream, and of the crew in still water. 26. A man put $ 5000 in a savings-bank paying a certain rate of in- terest. At the end of a year he withdrew 1 75, leaving the remainder at interest. At the end of another year, the amount due him was f 5278.50. Find the rate of interest. 27. A square garden has a square plot of grass at its centre, surrounded by a path 4 feet in width. The area of the garden outside the path ex- ceeds by 768 square feet the area of the path ; and the side of the garden is less by 16 feet than three times the side of the plot, Find the dimen- sions of the garden. 28. A merchant has a cask full of wine. He draws out 6 gallons, and fills the cask with water. Again he draws out 6 gallons, and fills the cask with water. There are now 25 gallons of pure wine in the cask. How many gallons does the cask hold ? QUADRATIC EQUATIONS 287 29. A and B sell a quantity of corn for ^ 22, A selling 10 bushels more than B. If A had sold as many bushels as B did, he would have received $ 8 ; while if B had sold as many bushels as A did, he would have received $ 15. How many bushels did each sell, and at what price ? 30. Two men are employed to do a certain piece of work. The first receives $48 ; and the second, who works 6 days less, receives $27. If the second had worked all the time, and the first 6 days less, they would have received equal amounts. How many days did each work, and at what wages ? 31. A and B run around a course, starting from the same point, in opposite directions. A reaches the starting-point 4 minutes, and B 9 minutes, after they have met on the road. If they continue to run at the same rates, in how many minutes will they meet at the starting- point ? 32. A carriage- wheel, 15 feet in circumference, revolves in a certain number of seconds. If it revolved in a time longer by one second, the carriage would travel 14400 feet less in an hour. In how many seconds does it revolve ? DISCUSSION OF PROBLEMS INVOLVING QUADRATIC EQUATIONS WITH ONE UNKNOWN NUMBER 463. Interpretation of Complex Results. Prob. Let it be required to find two real numbers whose sum shall be 10, and product 26. Let X = one number. Then, 10 - x = the other. By the conditions, a; (10 - a;) = 26. Solving, x = 5 ± V— 1. "We conclude that the given conditions cannot be satisfied, and the problem is impossible. Hence, imaginary or complex results show that the problem is impossible. 464. The Problem of the Lights. To find upon the line which joins two lights, A and B, the point equally illuminated by them; it being given that the intensity of a light, at a certain distance, equals its intensity at the distance 1, divided by the square of the given distance. 288 ADVANCED COURSE IN ALGEBRA 1 \ \ 1 [ C" A OB C Let J5 be c units to the right of A. Let 'a and h denote the intensities of A and J5, respectively, at the distance 1. Let the point of equal illumination be x units to the right oiA. Then it will be c — a; units from B. By the conditions of the problem, the intensity of A at the distance x units, is —; and the intensity of B at the distance •*/ c — X units, is Then, (c - xy a ^ (c - xf c Va ^^ c Va Solving this equation, x = — :: or Va -I- Vft Va — V6 Since there are two lights, c must always be positive ; then, neither a, 6, nor c can equal zero. The problem then admits of only three different hypotheses : 1. a > 6. In this case, ^^-^ is < 1, and > -• Va4-V6 2 Then, the first value of aj is < c, and > |- Thus, the first point of equal illumination is at (7, between the lights, and nearer 5, the lesser light. Again, ^ ^ is > 1. Va — V6 Then, the second value of cc is > c. Thus, the second point of equal illumination is at C", in AB produced, to the right of the lesser light. 2. axf-2{x^ + (^x) = Z5. Completing the square, (a^ + 6 0^)2-2(032 + 6 a?) +1 = 36. Extracting square roots, (a^ + 6 .t) — 1 = ± 6. Then, aj2 + 6 a; = 7 or -5. Completing the square, x^ + 6 a; + 9 = 16 or 4. Extracting square roots, a; + 3=±4or ±2. Then, .a; = -3±4 or -3±2 = 1, -7, -1, or -5. In solving equations like the above, the first step is to complete the square with reference to the x^ and x^ terms ; by § 447, the third term of the square is the square of the quotient obtained by dividing the x^ term by twice the square root of the a;* term. 3. Solve the equation x^ — Qx-\- B^x" - 6 a; + 20 = 46. Adding 20 to both members, (aj2 _ 6 a; + 20) + 5 Va;2-6a; + 20 = m. Completing the square, (a;2_6a; + 20)+5Va;2-6a; + 20 + ?^ = 66 + ?^ = ^. ^ ^ 4 4 4 Extracting square roots, Vaj^ — 6a3 + 20 + - = ±— -• Then, Va^-6a; + 20=:6 or -11. EQUATIONS SOLVED LIKE QUADRATICS 295 The only solution is Vx^ — Qx-{-20 = 6. Squaring, a;^- 6 a; + 20 = 36. Completing the square, aj^ — 6 a; -f 9 = 25. Extracting square roots, a; — 3 = ± 5, and a; = 8 or — 2. In solving equations of the above form, add such an expression to both members that the expression vsrithout the radical sign in the first member may be the same as that within, or some multiple of it. 4. Solve the equation 2x^-\-5x — 2 xVx^ -f- 5 a; — 3 = 12. The equation may be written a:2 + 5a;-2a;Va^ + 5a;-3 + a^ = 12.. Subtracting 3 from both members. (a;2 + 5a;-3)-2a;Va^ + 5a;-3 + a;2 = 9. Extracting square roots, VxP + 5a; — 3 — a; = ±3. Or, ^ Va^ + 5a;-3 = a;±3. Squaring, a^ + 5a; — 3 = a52±6a; + 9. Then, -x or lla; = 12, and a; = -12 or i?. Neither value satisfies the given equation. 5. Solve the equation ^:=:^ + ^^::^ = §. X^ — X X — O ^ a^ — 3 Kepresenting by y, the equation becomes ai/ — X y + - = t or 2/ + 2 = 52/. Solving this, ?/ = ^ or 2 ; that is, ^^ = ^ or 2. 2 scr — x 2 Taking first value, 2 a;^ — 6 = ar^ — a;. Or, ar^ + a; = 6. Solving, a; = 2 or —3. Taking second value, a;^ — 3 = 2ar^ — 2a;. Or, -a2^2aj = 3. Solving, a; = 1 ± V— 2. 296 ADVANCED COURSE IN ALGEBRA EXERCISE 60 Solve the following : 1. iK* - 29 a;2 = - 100. 9. 6x-2 = nVx. 2. 0.-6 + 19 ^-s = 216. ^^ ^_V- ^ 244 x'^ = - 243. 3. xUl0xJ + 9 = 0. 11 2x-B-35x-* + 48 = 0. 4. x^-33x^ = -32. 13^ 27x6 + 46x3 = 16. 6. 64 + 63x-t-x- = 0. 13 32 ^f _ 33 ^ _ ,-i. 6. 8x-2 + 14x-i = 5. _6 3 14. 16x8-33x4-243 = 0. 7. 5x^+7x^=-2. 2 18 15. 161x5 + 5 =-32x10. 8. 4x^- — = -21. X* 16. 81x~^-308- - 64 x^ = = 0. 17. (2 x2 + 3 x)2 - 4(2 x2 + 3 x) = 45. 18. x* + 12 x3 + 14 x2 _ 132 X - 135 = 0. 19. 5x + 12-5V5x+ 12 = ~4. 20. x2-3 2x _ 17 2 X x2 - 3 4 21. 3 x2 + X + 5V8 :k2 ^ a; _}_ 6 = 30. J. 8 x2 - 1 + 6 X V8 x2 - 1 = - 8 x2. 23. X* - 2 ax8 - 17 a2x2 + 18 a^x + 72 a* = 0. 24. lsfx--Y + 2llx--\=-b. 25 a;^ + 2 2 X - 5 _ 35 *2x-5 x2 + 2 6* 26. x2 - 6 Vx2 -4x + 11 = 4 X - 19. 27. x2 + 3 X + 4 - Vx2 + 3 X + 4 = 2. 28. (4 x2 + 2 X - 7)2 + 4 x2 + 2 X - 189 = 0. 29. \/x2 - 3 X - 3 = x2 - 3 X - 23. 30. (2 x2 - 3 X - 1)3 - 6(2 x2 - 3 X - 1)^ = 16. 81. 3v^x2 - 12 X - 7 v'x2 - 12 X = - 2. 32. x* - 18 x3 + 109 x2 - 252 X + 180 = 0. 33. V3 x2 - 2 X + 16 + 2 v/3 x2 - 2 X + 16 - 16 = 0. 34. 7(x3 - 28)"^ + 8(x3 - 28)"^ = - 1. 35. (3x + 15)* + 9(3x + 15)* = 22. 36. 5 x2 + 6 X - 166 = 4 x Vx^ + 5 x- 8. EQUATIONS SOLVED LIKE QUADRATICS 297 37. 10(1 - x2) - 12 X - V5ic2 + 6x-2 = 0. 38. x* + 28 a;3 + 190 x2 - 84 x ^ 135 = 0. 39. 9(x + a)^ - 22 b^(x + a)^ + 8 6* = 0. 40. x2 + l+Vx2-8x + 37 = 8(x + 12). 41. 25(x + l)-i - 15(x + 1)"^ = - 2. 42. (3 X - 2 a)2 - 4 x(3 X - 2 a) = 4(x + 4 a)2 + 8 x(x + 4 a). 5c2_5a; + i a;2_2x + 2 8 43 2x + 2 x2-5x + l 3 44. 9 x(7 - x) + 9\/x2 _ 7 X - 5 = - 43. 45 3 x2 + 2 X - 5 4 x2 - 7 X*- 1 ^ 5 •4x2_7x-13x2 + 2x-5 2* 46. 9 X* - 30 x3 - 185 x2 + 350 x + 1176 = 0. 47 / x2 + 3x+i0 / x2-5x + \x2-5x + 2 "^ \x2 + 3x + 48. "^ ■ " ' -^ ^ 10 5* V^^-aI x2 + 3 2 298 ADVANCED COURSE IN ALGEBRA XXI. SIMULTANEOUS QUADRATIC EQUATIONS On the double signs ± and :p. If two or more double signs are used in an equation, it will be under- stood that the equation can be read in two ways ; first, reading all the upper signs ; second, reading all the lower signs. Thus, the equation a ±b = ±c can be read either a -\-b = c, OT a — b = — c. And the equation a±b = =p c can be read either a + 6=— c, or a — 5 = c. The same notation will be used in the case of a system of equations, each involving double signs. Thus, the equations x = ±2, y = ±S, can be read either X = + 2, 2/ = + 3, or a; = - 2, y=-3. And the equations x= ±2, y = =p3, can be read either aj = + 2, y=-S, or X = - 2, y=+S. The principles demonstrated in §§ 233 to 236, inclusive, hold for simul- taneous equations of any degree. 470. The following principle is of frequent use in solving simultaneous equations of higher degree than the first : The system of equations AxB = 0, (1) CxD = 0, (2) where A, B, C, and D are rational and integral expressions which involve the unknown numbers, is equivalent to the systems lo=o, [d=o, \c=o, [d=o. For any solution of (1) and (2) makes Ax B and G X D identically equal to 0. SIMULTANEOUS QUADRATIC EQUATIONS 299 It then makes at least one factor oiAxB and C x D identi- cally equal to 0, and hence satisfies some one of the systems (3). Again, any solution of any one of the systems (3) makes either A or B, and also either C or D, identically equal to ; and hence satisfies (1) and (2). Then, the system (1) and (2) is equivalent to (3). The principle holds for any number of equations, with any number of factors. 471. Two equations of the second degree (§ 113) with two unknown numbers will generally produce, by elimination, an equation of the fourth degree with one unknown number. Consider, for example, the equations V-f2/=a. . (1) .x-^f=b. (2) From (1), y = a — oi^; substituting in (2), a; + a^ — 2 ax^ -f a;* = 6 ; an equation of the fourth degree in x. The methods already given are, therefore, not sufficient for the solution of every system of simultaneous quadratic equa- tions, with two unknown numbers. In certain cases, however, the solution may be effected. 472. Case I. WJien each equation is in the form ax^ + by^ = c. In this case, either x^ or y^ can be eliminated by addition or subtraction. 3^2+ 42/2 = 76. (1) 1. Solve the equations , „ ^ „ ^ '32/2-11x2 = 4. (2) Multiply (1) by 3, 9x' + 12y^ = 228. Multiply (2) by 4, l2/-44a^= 16. Subtracting, 53 a^ = 212. Or, a^= 4. (3) Whence, x =±2, 300 ADVANCED COURSE IN ALGEBRA Substituting a; = ± 2 in (1), 12 + 4 / = 76. Then, / = 16. (4) Whence, 2/ = ± 4. The solution is a; = 2, 2/ = ± 4 j or, a; = — 2, 2/ = ± 4. (5) It follows, precisely as in Ex. 1, § 238, that the given system is equiva- lent to the system (3) and (4). Now (3) and (4) may be'written (x + 2) (cc - 2) = 0, and (y + 4)(?/ - 4) = 0. And by § 470, these are equivalent to which are the same as (5). The method of elimination by addition or subtraction may be used in other examples. Sx'-4.y= ^7. (1) 62/= 33. (2) Multiply (1) by 3, 9 x^ - 12 ?/ = 141. Multiply (2) by 2, ' Ux'^12y= 66. Adding, 23 a:^ ^ 207. Then, o^ ^ 9^ and a; = ± 3. Substituting a;= ±3 in (1), 27-42/ = 47. Then, - 4 2/ = 20, and 2/ = - 5. It is possible ^o eliminate one unknown number, in examples (1) and (2), by substitution (§ 239), or by comparison (§ 240). EXERCISE 61 Solve the following : 3 ic2 + 2 y2 = 66. ^ r 4 a;2 + 9 2/2 _ 13. r3a:2_ 2. Solve the equations \ ^ I7aj2 + ■ 9 x2 + 5 2/2 = 189. I 8 x2 - 27 2/2 = 6. g , 3a;- 52/2 = _ 116. ^ (^xy+ y'^=-75. 7x + 42/2 = 121. '1 a-y - 3 2/2 = - 95. 3 a;2 - 2 X2/ =; 24. j 11 a;2 - 6 2/2 = 84. 4 a;2 - 5 a;2/ = 46. '* L 7 x2 + 15 y^ = 204. 2/2 + 4 a;?/ - 3 ?/ = 42. 2 2/2 _ xy + 6y=- 10. SIMULTANEOUS QUADRATIC EQUATIONS 301 (2x^-xy-Sy'^ = 0. (2x^ + 3y^ = 61 - x. ' \ ^2 + xy + 3 2/2 = 27. ■ '1 ic2 - 2 2/2 = 17. ^ r4x2-2/2=(3a+&)(a+36). ^^ l4 2/2-x2=(3a-6)(a-3&). ( Sx + 2y Sx-2y ^41 11 J 3 X - 2 2/ 3 iK + 2 2/ 20' [ 8 2/2 + 3 x2 = 29. ^g r (3a: + 42/)2- (6a:- 2/) (6 x + 5 ?/) = 57. 1 (5 a: - 2 2/)2 - 4(x - 4 2/) (2 a: + 3 2/) = 225. 473. Case II. When one equation is of the second degree, and the other of the first. Equations of this kind may be solved by finding the value of one of the unknown numbers in terms of the other from the first degree equation, and substituting this value in the other equation. Ex. Solve the equations [2 a^' - a^2/ = 6 y. (1) 1 X +2y=7. . (2) From(^, 2/ = ^- (3) Substituting in (1), 2x'-x f^^) = 6 C^^^) ' (4) (5) Clearing of fractions, 4a ^- -7x + x' = = 42- ■6 a;. Or, ^x'-x^ = 42. Solving, x = = 3 or _I4 Substituting in (3), y = 7- = 2 3 or ^ = = 2 or 49 10* The solution is x = 3, y = 2 -, or, x = — —, y = —-. (6) 5 10 By § 236, the given system is equivalent to (3) and (4) ; or, since (4) is equivalent to (5), to the system (3) and (5). Now, (5) can be written (x — 3)(5x + 14) = 0. Then, by § 470, the system (3) and (5) is equivalent to the systems (3) and X - 3 = 0, and (3) and 5 x + 14 = ; that is, to (6). 302 ADVANCED COURSE IN ALGEBRA EXERCISE 62 Solve the following : \x -2y =9. 9. '-r-f 1 xy = 2 a2 + 3 a - 2. \Sx + 4:y = na-\-2. ^+!=¥- ■ "^ I* (4x + y-Sxy = -6. (2xy + x = ~S6. ixy — Sy = — 6. 10. \x-6y + 2xy = 10. (x^-2y^ + Sx=-S. 11. X ^ — y _ 40 X - w X 21 [.x^-2y^-4y = -2. 2y+3x = -l. (x^-xy + 2y'^ = 8. ' \ Sx + y = 10. 12. ' X ?/ _29 . 2y Sx 24' (Sx^-xy-y^ = -S. 4?/-x=-2. L 2x-Sy = 6. ft;2 + 4 x?/ = 13. . 2 x?/ + 9 y^ = 87. \2x Sy_^ 3^2 13. '* 1- + ^-^- r ^-1 ^-1 = 0. 1 3 2 ^ 2 X Sy 14. . a;2 _ y y^ -X Q I 32 15 ' .^f-^-l 15. ' x'^y^ - 24 x?/ + 95 = 3 X - 2 ?/ = - I X + ?/ = (2 a — &)(a + 7 6). :3( a — &) . 13. 474. Case III. When the given equations are symmetrical with respect to x and y ; that is, when x and y can he inter- changed without changing the equation. Equations of this kind may be solved by combining them in such a way as to obtain the values oi x-\-y and x — y. [x + y = 2. (1) 1 xy = -W. . (2) Squaring (1), o^ + 2xy + f-= 4. (3) Multiplying (2) by 4, 4 xy = - 60. (4) Subtracting, x^ — 2xy-\-y^= 64. (5) Extracting square roots, x — y = ±S. (6) 1. Solve the equations SIMULTANEOUS QUADRATIC EQUATIONS 308 Adding (1) and (6), 2 a; = 2 ± 8 = 10 or - 6. Whence, ic = 5 or — 3. Subtracting (6) from (1), 2 ?/ = 2 t 8 = - 6 or 10. Whence, 2/ = — 3 or 5. The solution is a; = 5, 2/ = — 3 ; or, a; = — 3, ?/ = 5. (7) In subtracting db 8 from 2, we have 2 ^ 8, in accordance with the nota- tion explained on page 298. In operating with double signs, ± is changed to T , and =F to ± , when- ever + should be changed to — . Equation (4) is equivalent to (2) ; but (3) is equivalent to x-\- y = 2 and X -\- y = — 2. If, then, we use only the value +2 tor x+ y, the given system is equivalent to (3) and (4). By § 234, the system (3) and (4) is equivalent to (3) and (5). Then, the given system is equivalent to the positive value oi x -{■ y in (3), and (5) ; that is, to (1) and (5). Now (5) may be written {x — y — 8) (x — y + S) = 0. Then, by § 470, the system (1) and (0) is equivalent to (1) and X — y — S=0, and (1) and x — y + S =0; that is, to (7). The above equations may also be solved by the method of Case II ; but the symmetrical method is shorter and neater. 2. Solve the equations a;2 4- 1/2 = 50. xy = -7. Multiply (2) by 2, 2xy = - 14. Add (1) and (3) x" -\- 2 xy -\- y' = 36. Whence, x-\-y=±6. Subtract (3) from (1), x^-2xy-{-y^ Whence, Add (5) and (7), Whence, Subtract (7) from (5), Whence, The solution is x= ±7, ?/ = =F 1 ; or, a; = ± 1, i/ = T 7. = ±6 = 64. J X -y = -.±S.^ 2x = = 6 ± 8, or -6±8. x = = 7,-1,1, or -7. 2y = ^ 6 T 8, or -6q=8. ?/ = = -1,7,- 7, or 1. 304 ADVANCED COURSE IN ALGEBRA The given system is equivalent to the system (4) and (6). Now (4) and (6) can be written (X + y -{- 6){x + y - 6) = 0, a^nd (x - y + S)(x - y -S)=0. Then, the system (4) and (6) is equivalent to (5) and (7), with every possible combination of signs. We may solve by the method of Case III other systems in which the equations are symmetrical, except in the signs of terms ; as, for example, the system fx-y = a. \ xy = b. We may also solve certain non-symmetrical systems ; as, for example, the system r a2x2 + 62^2 ^ c. [ ax+ by = d. EXERCISE 63 Solve the following : ^ |x2 + 2/2 = 29. 'x^-\-xy + y^=:^ I x + y = -3. 9. 4 lx-y = n. ^' I xy = -2S. x^-xy-^-y-^^^j- ^ ra;2 + 2/2 = i3o. * 1 x-y = -8. 10. X^ 2/2 1-1=12. (x-^y = 2a-l. ' \ xy = a'^- a-2. I X y f 1 1 _ 289 x^ y^ 36 1 _io. .. {'-'•='^- 11. ^ I xy = ab. xy 3* 'I x-y = S. 12. ^ 'a;2 + 9?/2 = 50. , x-Sy = 0. r? + |/ = -12. 13. xy = -16. . 2 X + 2/ = 14. 1. . y X 3 [x-y = l. r 1 (x'^-xy + y^ = a^ + ' [ x-hy = 2a. 3 62. 14. ^ - = 24. xy x + y = llxy. SIMULTANEOUS QUADRATIC EQUATIONS 806 15. 16. 1 + i + l x2 xy y^ 49. 1 + 1 = 8. X y r36x2 + I 6x + -I x^ — xy -{■ y2 3:2^/2 iC?/ _7_ 324* J_ 18* 21. l + i. + l = r2 icy y2 _i_ + l = ^ 8y =11. 25 x2 + 16 y2 = 544. 5x - 4 2/ =32. r 2 x2 - 3 icy + 2 y2 = 92. ' 15x2 + 4 a;?/ + 5^2 -161. 16 x22/2 _ 104 xy = - 105. X — y = — 2. 3 g^ - 3 a262 + 54 a2(a - 6)2 4 _ ^262 + 54 ■■I x?/ y^ a^^a — &)2 475. Case TV. When each equation is of the second degree, and homogeneous / that is, when each term involving the unknown numbers is of the second degree with respect to them. Certain equations of this form may be solved by the method of Case I or Case III. (See Exs. 1, § 472, and 2, § 474.) The method of Case IV should be used only when the example cannot be solved by Cases I or III. Ex. Solve the equations \ „ „~ ' Putting in the given equations y = vx, 5 we have a^ — 2vx^= 5 ; or, aj^ = and 0^ + 1)^0^ = 29', or, 0^ = Equating values of a^, Or, Solving this equation. 29 5 ^ 29 l-2v l + u^' 5'u2+58'y = 24. (1) (2) (3) (4) (5) -y = - or — 12. 5 Substituting these values in (2), we have x" or 1- 1 + 24 25 or 9; then, x = ±5 or ± 5 V6 306 ADVANCED COURSE IN ALGEBRA Substituting the values of v and x in the equation y = vx, |(,5)or-12(.^) ±2 or T 1?. The solution is x= ±5, y 1 12 ± 2 ; or, a; = ± — z, 2/ = T -— • The given system, and (1), are three equations with three unknown numbers ; by § 236, they are equivalent to the system (1), (2), and (3). Then, precisely as in Ex., § 240, the system (1), (2), and (3) is equivalent to (1), (2), and (4), or to (1), (2), and (5). We may write (6) in the form (6v — 2){v + 12) = 0. Then, the system (1), (2), and (5) is equivalent to the systems (1), (2), and5v-2 = 0, and (1), (2),and v+ 12 = 0. Then, the given system is equivalent to the systems and y z=—12x, x"^ 1- 1 + 24 In finding y from the equation y = vx, care must be taken to multiply each value of x by the value of v which was used to obtain it. Solve the following : 25. xy = 4. + 3 xi/ = - 5. EXERCISE 64 1^2+ 2/2 Ix"^ — XV '^1 2xy -y^ = - 24. x^ + xy + y^ = 19. 2x'^ + xy = -2. ( ix^ — xy — y^ L Sxy + y^ 16. 28. ^( A-l = 9. a;2 2/2 ±.-^ = -90, xy 2/2 x2 -f- ojj/ - 5 2/2 = 25, a;2 + 4 2/2 ^ 40. 7. 10. 11. IB. \'^- a;2 - 2 a:2/ - 4 y2 = _ 41. ic2 - 5 ici/ + 8 2/2 = 58. r 2 ic2 + 7 iC2/ + 4 y2 = 2. 1 3 x2 + 8 X2/ - 4 y2 = _ 72. r 11 x2 - xy- 2/2 = 45. I 7 a;2 + 3 £C2/ - 2 2/2 = 20. 5 ic2 + xy-3y^ = 27. 4 «2 _ 4 x2/ + 3 2/2 = 72. x2 xy y^ i-i = -12. xy y^ 4 ic2 - 2 iC2/ - 2/2 = - 16. 4 a;2 + 7 a;2/ + 2 2/2 = 104. xy - 40 x2 = 30 x^y^. Sxy - 72 ic2 = 38 x^y^. 12. SIMULTANEOUS QUADRATIC EQUATIONS 307 476. Solution of Simultaneous Equations of Higher Degree by Factoring. 1. Solve the equations \^ ^n n- j ; . [(x + y){x-Sy + 2)=0. By § 470, the given system is equivalent to the systems x-y = 0, x + y = 0, x-y = 0, [x-Sy-{-2 = 0, (x + 2y-l = 0, ^^^ cx-^2y-l = 0, I xi-y = 0, ^"^ \x-3y + 2 = 0. The solutions of these are x = 0,y = 0', x = l, y = l; x = -l, y = l; x = --, y = o 2. Solve the equations | x'-^xy-2y^ = 0. (1) 2x4-32/ + 6 = 0. We may write (1) in the form {x -]- 2 y)(x — y) = 0. Then the given system is equivalent to the systems r x + 2y = 0, ^^^ I x-y = 0, \2x-{-Sy-{-6 = 0, ^^ [2x + 3y-^6 = 0. /» f* Solving these, x = — 12, y = 6; or, x = — -, y = — -' 5 5 The example can be solved by the method of § 473 ; but the above method is shorter. Sx^+f-=7x. (1) xy-^f=2x. (2) Multiplying (1) by 2, 6 a^+2 /=14 x. Multiplying (2) by 7, 7 xy-\-7y^= 14 x. Subtracting, 6x^-7 xy-5y^=0, or (2x-\-y)(Sx-5y)=0. Then, the given system is equivalent to the systems (Sx'-^y' = 7x, ^^^ (Sx'-h f = 7x, 120? +2/ =0, ^^ l3a;-52/=0. 25 5 Solving these, a; = 0, y = ; a;=l, 2/ = — 2 ; or, ic = — , 2/ = j* 3. Solve the equations \ 308 ADVANCED COURSE IN ALGEBRA 4. Solve the equations j ^ a^= x + y. y^=3y-x. Subtracting, x^ — y^=2x—2y,OT{x—y)(x-\-y—2)=0. a^ = x + y, Then , I ^ = ^ + ^' and j ^ =^ + ^^ Solving these, x = 0, y = 0] x = 2, y = 2) or, x=± V2, y = 2T-^2. EXERCISE 65 Solve the following : (2x-3?/)(a;-4?/ + l) = 0. ^ j (x + 2 ?/)(4ic - 3y) = 0. 5 a; + 4 ?/ = 23. ' L 3 x2 - 5 a:?/ - 4 ?/2 = 18. ^ r(2x+ 2/)(3a:-4?/ + 5) = 0. ^ j ^^^ _ 4 ^2 ^ 3a;. * 1( x-3?/)(2x + 5y-8)=0. ' 1 a:?/ + 8 2/2 = X. :,2_3^y_4y2^0. 10 j3^' + ^2/-22/2 = -4y. o„ r.. .« 1 6x2 -2X2/ + 1/2 = 3?/. r 3 x2 + 3 ?/2 = 10 xy. 1 + 1=.! X v 3 ( «,.i.. "■ hi-' J 2 x2 - xy - 15 2/2 = 0. 1 6x + ^ r 12x2 - 11 x?/ + 2 2/2 = 0. I (a; _ y)(3.2 _ sc?/ - 6 2/2)= 0. • 1 7x-32/ = -5. ^2- t 2x-5v + l = 0. r 2 x3 + 5 xs?/ - 8 X2/2 - 20 ^f = 0. 3x + 42/ = 7. x2 = mx — wy. xy -\-y'- = 44. ly"^ = nx — my. J 2 x2 - xy = 0. r ^- 1x2 + 22/2 = 9. ^^- 1 7 r x2-2xy = 0. ^^ I 2 x2 + xy + 2/2 = 44. 477. Solution by Division. n 'A .1. ,' {AxB = axb, (1) Consider the equations j ^^ , I B = b', {z) where A and B are rational and integral expressions which involve the unknown numbers, and a and b any numbers. B7 § 236, the given system is equivalent to Axb = a xb, ( A = a, (3) _ , or, to j ^ , SIMULTANEOUS QUADRATIC EQUATIONS 309 Here the first member of (3) is obtained by dividing the first member of (1) by the first member of (2), and the second mem- ber by dividing the second member of (1) by the second member of (2). (a^-y^ = 9. (1) Ex. Solve the equations \ ^ [ x-y = S. (2) Divide (1) by (2), a^ + xy-{-f = 3. (3) Squaring (2) , a?~2xy + / = 9. (4) Subtract (4) from (3), 3 x?/ = - 6, or 0*2/= - 2. (5) Add (3) and (5), a^ + 2 a;?/ + / = 1- / Extracting square roots, x-{-y=±l. (6) Add (2) and (6), 2a; = 3±l=4or2. Then, aj = 2orl. Subtract (2) from (6), 22/=-3±l=-2or-4. Then, y=-lov-2. The solution is a? = 2, 2/ = — 1 ; or, a; = 1, 2/ = — 2. The equations (2) and (3), though not symmetrical, are solved as in § 474. 478. Consider the equations \ where A, B, C, and D are rational and integral expressions which involve the unknown numbers. By § 236, the given system is equivalent to (AxB=CxB, ^ (B(A-C) = 0, i or, to 1 B==D; ' 1 B = D. By § 470, the latter is equivalent to the systems (B = 0, , (A-C = 0, {d=d, and B = D. Then, the given system is equivalent to the systems li) = 0, [b=d. 310 Ex. Solve the equations ADVANCED COURSE IN ALGEBRA (1) x-y = 3x-\-2. (2) Dividing (1) by (2), the given system is equivalent to the systems x-{-y = 9x^ — 6x-{-4:j x-y = 0, 3 a; + 2 = 0, and ( x-{-y = 9 or [ x — y = S X + 2. 2 2 The solution of the first system is x= , y= . o o To solve the second, add (2) and (3) ; then, 2x=9o^-Sx-\-6,ov9x~-5x-\-6 = 0, 5 ± V25-216 5 ± V-m. By § 450, By (2), x = 18 18 y=^2x-2=-^^^-^^^-2 ^ 9 ^ -23q:V-191 9 (3) If we try to solve by substituting the value of y from (2) in (1), we shall have an equation of the third degree in x. Solve the following : a;3 - 2/3 _ 26. x-y = 2. x^ + y^ = 280. x^-xy-^y^ = 28. r x + y = S5. I v^a; 4- v^ = 6. = 189. y EXERCISE 66 ra:3_8?/3r=U 'I X - 2 V = 9. flj3 2/8 X y l+± + l = 12. I x^ xy 2/2 1 X 9. 10 + x^2 _ 56. + ?/ = -l. aJ* + «22/2+y4 = 481. x2 - xy + 2/2 = 37. r(x2-16)(9 2/2-4)=-2100. I (x + 4) (3 2/ - 2) = 60. r 25 x2 - 25 2/2 = 24 x22/2. I 5x + 5y = -4x2/. SIMULTANEOUS QUADRATIC EQUATIONS 311 11. 12. 13. r 8 a;3 + 27 y^ = 91. U .x2 _ 6 xy + 9 2/2 = 13. jx^-y^ = ic2?/2 - 1. I X2 — y2 — a;?/ — ] . rx2 i 2/ ■ a; 6* 1 1 + Ul. I X y 6 14. 15. 16. 2/ x 15 ^ a;-2/ = 2. a;2 _ ^2 + 3 a- _ 3 y ^ 12. (x-2/)(3x-52/-4)=-9. x'^y + ?/2x = 42. 1+1=1. X y 6 479. Special Methods for the Solution of Simultaneous Equa- tions of Higher Degree. r Q^ y^ z=z 19. 1. Solve the equations j „ on' [ ocry — xy' = o. Multiply (2) by 3, 3x'y-3xy' = 18. Subtract (3) from (1), a^ -Sx'y +3 xy^ - f = 1, Extracting cube roots, x — y = l. Dividing (2) by (4), xy = 6. Solving equations (4) and (5) by the method of §474, we find a; = 3, 2/ = 2 ; or, x = — 2, ?/ = — 3. (1) (2) (3) (4) (6) « o. , T . (a^ + y^ = 9xy. 2. Solve the equations i , n ^ [x -{-y =6. Putting x — u-\-v and y = u — v, (u 4- vy + (u - vf = 9(u -\.v) (u-v); (1) and (u + v) + (w - v) =6. (2) Reducing (1), 2 1^^ + 6 ^^2 ^ 9(^^2 _ ^2^^ (3^ Reducing (2), 2 w = 6, or w = 3. Putting w = 3 in (3), 54 + 18 v^ = 9(9 - v"). Whence, i;^ = 1, or v = ± 1. Therefore, a; = w + 'V = 3±l = 4or2; and 2/ = w — 'y = 3Tl = 2or4. The artifice of substituting u + v and u — v for x and y is advantageous in any case where the given equations are symmetrical (§ 474) with respect to x and y. See klso Ex. 4. 812 ADVANCED COURSE IN ALGEBRA aj2 + 2/' + 2a; + 22/ = 23. 3. Solve the equations xy = 6. Multiplying (2) by 2, 2 xy = 12. Add (1) and (3), a?-^2xy -\-y'' -\-2x + 2y = Z5. Or, (a;4-2/)2 + 2(aj + 2/) = 35. Completing the square, {x + yY + 2(x + ?/) + 1 = 36. Then, (a; + 2/) + 1 = ± 6 ; and a; + ?/ = 5 or — 7. Squaring (4), a^ -4- 2 a;?/ + 2/^ = 25 or 49. Multiplying (2) by 4, 4 a??/ = 24. Subtracting, y? — 2xy-\-.y^— 1 or 25. Whence, a; — 2/=±lor ±5. Adding (4) and (5), 2 ic = 5 ± 1, or — 7 ± 5. Whence, x = S,2, — 1, or — 6. Subtracting (5) from (4), 2 ?/ = 5 ^F 1, or — 7 q= 5. Whence, 2/ = 2, 3, - 6, or - 1. fx'^ -\-y^ =: 97. aj+2/=-l. Putting x = u-{-v and y = u — v, (u + v)^ + (w - 'y)^ = 97, and (u-\-v) +(u — v) = —1. Reducing (1), 2 ?^^ + 12 u^v'- + 2v^ = 97. Reducing (2), 2w=— 1, orw=— -. Substituting in (3), 1 -f 3 v^ _^ 2 ^* = 97. 8 Solving this, v» = ?^ or -51; and -1;= ± ? or ± ^^nB. £> ^ 4 4 ' 2 2 Therefore, a; = w + i; = - - ± ^, or _! ^^ V-31 ' , ' ^ 2 2' 2 2 = 2,-3,or-^^y-^^. SIMULTANEOUS QUADRATIC EQUATIONS 313 A ^ l-r^ l^V-31 And, y = u-v = -^T-^ov --T—^ — In solving fractional simultaneous equations of higher degree, we must reject any solution which satisfies the equation obtained by equating to zero the L. C. M. of the given denominators (§ 222). Also, in solving simultaneous equations of higher degree having un- known numbers under radical signs, we retain only those solutions which satisfy the given equations, when the principal values of the roots are taken (§ 397). EXERCISE 67 Solve the follow ing : {x'^ -Vy^- 257. r a;2y - a: = - 6. y x — y — b.. I x^y^ _ aj3 _ _ 72. x2 + 2/2 + a; _ y = 32. r X - ?/ + v^ = 11. ^y = 6. •^' 1 ^Wy - Vx^ = 30. J x3 + ?/3 = 2a3 + 24a. . 8. ( xy + yz — zx = b. xy — yz + zx = c. _ - xy + yz -\- zx = a. 3. yz + zx = -S. -2x2 4-2/2-02 = 43. x-Sy + z = 17. [ x + y-Sz = lS. izx-\-xy=- 35. 9. (x + y){x + z) = 14. 4. 6. iM + z)(y + x) = Xz + x){z-\-y) = Sx"^ — xy - xz = bx-2y = 2. 7. :4. = 1. 10. x + y -^ z = l2. xy + yz + zx = 47. X2 + 2/2 - 02 _ 0. 4x4-3;? = :-5. ■ 6 X + 6 2/ = 5 xyz. - 3 2/ + 3 = 2 xyz. L 2 4- 2 X = xyz. [l + l-Uis. xyz 11. 6. ^ 1-1=1. y X x2 4- 2/2 + 02 = 110. 1 + 1 = 0. xy z 12. X 4- y - = 4. X0 + ^0 = 77. 816 ADVANCED COURSE IN ALGEBRA PROBLEMS INVOLVING SIMULTANEOUS EQUATIONS OF A HIGHER DEGREE THAN THE FIRST 481. In solving problems which involve simultaneous equa- tions of a higher degree tfian the first, there will usually be more than one set of values of the unknown numbers ; only those values should be retained which satisfy the conditions of the problem. The considerations of §§ 261, 262, and 463, hold for simul- taneous equations of any degree. EXERCISE 69 1. The product of the sum of two numbers by the smaller is 21, and the product of their difference by the greater is 4. Find the numbers. 2. The difference of the squares of two numbers is 260 : and the sum 13 of the numbers is — their difference. Find the numbers. 5 3. The sum of the squares of two numbers is 61, and the product of their squares is 900. Find the numbers. 4. The difference of the cubes of two numbers is 316 ; and if the prod- uct of the numbers be added to the sum of their squares, the sum is 79. Find the numbers. 5. Two numbers are expressed by the same two digits in reverse order. The sum of the numbers equals the square of the sum of the digits, and the difference of the numbers equals 5 times the square of the smaller digit. Find the numbers. 6. A party at a hotel spent a certain sum. Had there been five more, and each had spent fifty cents less, the bill would have been $24.75. Had there been three less, and each had spent fifty cents more, the bill would have been $9.75. How many were there, and what did each spend ? 7. The square 'of "the sum of two numbers exceeds their product by 84 and the sum of the numbers, plus the square root of their product, equals 14. Find the numbers. 8. The difference of the cubes of two numbers is 728 ; and if the prod- uct of the numbers be multiplied by their difference, the result is 72. Find the numbers. 9. If $ 700 be put at simple interest for a certain number of years, at a certain rate, it amounts to $883.75. If the time were 4 years less, and the rate 1^ per cent more, the amount would be $810.25. Find the time and the rate. SIMULTANEOUS QUADRATIC EQUATIONS 317 10. If the digits of a number of two figures be inverted, the quotient 4 of this number by the given number is -, and their product 2268. Find the number. 11. The square of the smaller of two numbers, added to twice their product, gives 7 times the smaller number. And the square of the greater exceeds the product of the numbers by 6 times the smaller number. Find the numbers. 12. A and B travel from P to Q, 14 miles, at uniform rates, B taking 20 minutes longer than A to perform the journey. On the return, each travels one mile an hour faster, and B now takes 15 minutes longer than A. Find the rates of travelling. 13. A and B run a race around a course two miles long, B winning by two minutes. A now increases his speed by two miles an hour, and B diminishes his by the same amount, and A wins by two minutes. Find their original rates. 14. A man ascends the last half of a mountain at a rate one-half mile an hour less than his rate during the first half, and reaches the top in 3f hours. On the descent, his rate is one mile an hour greater than during the first half of the ascent, and he accomplishes it in 2f hours. Find the distance to the top, and his rate during the first half of the ascent. 15. The square of the second digit of a number of three digits exceeds twice the sum of the first and third by 3. The sum of the first and second digits exceeds 4 times the third by 1. And if 495 be subtracted from the number, the digits will be inverted. Find the number. 16. A rectangular piece of cloth, when wet, shrinks - in its length, and — in its width. If the area is diminished by lOf square feet, and the length of the four sides by 5J feet, what were the original dimensions ? 17. A ship has provisions for 36 days. If the crew were 16 greater, and the daily ration one-half pound less, the provisions would last 30 days. If the crew were 2 less, and the daily ration one pound greater, they would last 24 days. Find the number of men, and the daily ration. 18. The sum of two numbers is 5', and the sum of their fifth powers is 1025. Find the numbers. 19. A man lends $2100 in two amounts, at different rates of interest, and the two sums produce equal returns. If the first portion had been loaned at the second rate, it would have produced .^48 ; and if the second portion had been loaned at the first rate, it would have produced .^27. Find the rates. 318 ADVANCED COURSE IN ALGEBRA 20. A can do a piece of work in two hours less time than B ; and together they can do the work in 1^ hours less time than A alone. How long does each alone take to do the work ? 21. A starts to travel from P to §, and at the same time B starts to travel from Q to P, both travelling at constant rates. A reaches ^ in 8 hours, and B reaches P in 18 hours, after they have met on the road. How many hours does each take to perform the journey ? 22. A and B travel from P to ^ and back. A starts one hour after B, overtakes him at a point two miles from Q, meets him 32 minutes after- wards, and reaches P \\ hours before B. Find the distance from P to §, and the rates of travel of A and B. GRAPHICAL REPRESENTATION OF SIMULTANEOUS QUAD- RATIC EQUATIONS WITH TWO UNKNOWN NUMBERS 482. 1. Consider the equation 2/^ = 4 cc -f 4. We have y = ±2 Vx + 1. If 0^ = 0, y = ±2. (A,B) If x = l, y = ±2V2. (C,D) If x = -l, y = 0. (E) etc. For any positive vahie of x^ or for any- negative value between and —l,y has two values ; the graph extends to an indefi- nitely great distance to the right of 0. For any negative value of x< —1, y is imaginary; then, no part of the graph lies to the left of a perpendicular to XX' at E. 2. Consider the equation x^ ■ x^-l Here, 2/ or y r- 2 \ JA' lfx = ±l,y=.0. {A, A') If X is between 1 and —1, y is imaginary ; then, no part of the graph lies between per- pendiculars to XX' at A and A\ SIMULTANEOUS QUADRATIC EQUATIONS 319 If x = ±2, 2/ = ±^. iB,C,B',a) For any positive value of a; > 1, or any negative value < — 1, y has two values ; then, the graph has two branches, each of which extends to an indefinitely great distance from 0. 3. Consider the equation x^ -\- y^ — 4: x -{■ 2 y =A. In this case, it is convenient to first locate the points where the graph intersects the axes. If y = 0, X^ — 4:X = 4:j and a; = 2 ± V8. {A, B) If ic = 0, / + 22/ = 4, and 2/ = -l±V5. {C, D) We may write the given equation or, {y + iy = (l + x){^-x). If x = -l or 5, 2/ + l = 0, and y = -l. (E, F) If x has any positive value > 5, or any negative value < — 1, (1 -\-x)(5 — x) is negative, and y + 1 imaginary ; then, no part of the graph extends to the right of F, or to the left of E. Again, we may write the given equation ar^_4a; + 4 = 8-22/-2/^ or, (x-2y = (4.-\-y){2-y). If 2/ = -4 or 2, iB-2 = 0, and x = 2. {O, H) If y has any positive value > 2, or any negative value < — 4, (4 + 2/) (2 — y) is negative ; then, no part of the graph extends above H, or below G. It is shown, in works on Analytic Geometry, that the graph of any equa- tion of the second degree, with two unknown numbers, is one of the conic sections^ so-called from being the sections of a cone made by a plane ; either a circle^ a parabola, an ellipse, a hyperbola, or a pair of straight lines. The graph of Ex. 1 is a parabola, as also is the graph of any equation of the form y'^ = ax, or y"^ = ax + b. (The graphs of §§ 465 and 467 are parabolas.) The graph of Ex. 2 is a hyperbola, as also is the graph of any equation of the form aoc^ — by^ = c, it a and b are numbers of like sign. (The hyperbola has two branches. The graph of any equation of the form xy = a is a hyperbola. ) 320 ADVANCED COURSE IN ALGEBRA The graph of Ex. 3 is a circle ; as also is the graph of any equation of the form x^ -}- y'^ = a, or x^ -\- y"^ -{- ax -{■ by -\- c = 0. The graph of any equation of the form ax'^ + by"^ = c, where a, 6, and c are numbers of like sign, and a and b unequal, is an ellipse. (The graph of any equation of the form a^x^ — b^y"^ = is the pair of straight lines whose equations are y ±f.) 1. Consider the equations 483. Graphical Representation of Solutions of Systems of Simul- taneous Quadratic Equations. I 2/2=4 a;. The graph of 2/^=4 a; is the parabola AOB. The graph of Sx — y = 5 is the straight line AB, intersecting the parabola at the points A Sind B, respectively. To find the co-ordinates of A and B, we solve the given equations (§ 277) ; the solu- tion is £c = 1, y = — 2, and x — — , y = — - y o It may be verified in the figure that these are the co-ordinates of A and B, respectively. Hence (compare § 277), if any two graphs intersect, the co-ordi- nates of any point of intersection form a solution of the system of equations represented by the graphs. 2. Consider the equations (x^-\-f = 17. The graph of x^ -\-y^=: 17 is the circle AD, whose centre is at 0, and whose radius is Vl7. The graph of xy = 4: is a hyper- bola, having its branches in the angles XOY and X'OY', respec- tively, and intersecting the circle at the points A and B in angle XOY, and at the points C and D in angle X'OY', ^m SIMULTANEOUS QUADRATIC EQUATIONS 321 The solution of the given equations is a; = 4,^=1 5 x=l, y=4:; x=—l,y = —4:; and a;= — 4, 2/= — 1. It may be verified in the figure that these are the co-ordinates of A, B, Cj and D, respectively. 3. Consider the equations r a^ + 42/' = 4. [2x +Sy =-5. The graph of ic^ + 4 2/^ = 4 is the ellipse AB, intersecting XX' at points 2 to the right and 2 to the left of 0, and YY' at points 1 above and 1 below 0. The graph of2x-{-Sy = — 5 is the straight line CD. Substituting x = -^ "^ in cc^ + 4 2/^ — 4, we obtain the equation 25 2/^ + 30 2/ + 9 = 0, which has equal roots. 9 + 5 Thus, y ■; and x = — 5' 2 5 The equal roots signify that the two points of intersection coincide, and the line is therefore tangent to the ellipse. In general, if the equation obtained by eliminating one of the unknown numbers has equal roots, the graphs are tangent to each other. 4. Consider the equations r9aj2- 2/' = -9. 1 x-2y=-2. The graph of 9 o^ — 2/^ = — 9 is a hyper- bola, having its branches above and below 0, respectively. The graph ofa;— 22/=— 2is the straight line AB. Substituting cc=2 2/— 2 in 9 a;^— 2/^= — 9, we obtain the equation 35 2/^ — 72 2/ + 45 = 0, which has complex roots. 322 ADVANCED COURSE IN ALGEBRA Then, the graphs have no point of intersection. In general, if the equation obtained by eliminating one of the unknown numbers has no real root, the graphs do not intersect each other. EXERCISE 70 Find the graphs of the following : 1. xy =- 6. 3. x2 -f y2 _ 4. 5. 9 x2 - 4 2/2 = 0. 2. x2 = 3 2/. 4. 2/^ = 5x-l. 6. 4x2 + 9 2/2 = 36. 7. 4 x2 - 2/2 = - 4. 8. x2 + y2 + 6 X - 2 2/ = 15. Find the graphs of the following systems, and in each case verify the principles of § 483 : 9. r x2 + 4 2/2 = 4. ix - 2/ =l• 12. ' x2- 2/2 = 9. .5a; _4y ^_9. 15. |x2 + 2/2 = 29. I X2/=10. 10. |x2+2/2=16. I 2/"' = 6 X. 13. ' 2 x2 - 3 2/ = 5. ,5x +6y=-12. 16. r 2x2+5 2/2=53. I 3x2-4 2/2= -24. 11 fy2_Sx=-S. I x + 22/ = -2. 14. ' 9 x2 + 2/2 = 148. X2/ = - 8. 17. 'ic2+2/H3x =22. - 4x2-92/2=0. XX. INDETERMINATE LINEAR EQUATIONS 323 XXII. INDETERMINATE LINEAR EQUATIONS 484. It was shown in § 269 that a system of m independent linear equations containing more than m unknown numbers, has an indefinitely great number of solutions. Such a system is called indeterminate (§ 266). If, however, the unknown numbers are required to satisfy other conditions, the number of solutions may be finite. 485. Solution of Indeterminate Linear Equations in Positive Integers. We shall consider in the present chapter the solution of indeterminate linear equations, in which the unknown num- bers are restricted to positive integral values. 1. Solve 7 x-\-By — 118 in positive integers. Dividing by 5, the smaller of the two coefficients, x-\- — -{-y = 23-\--; or, — ^ — = 2S-x-y. Since, by the conditions of the problem, x and y must be 2x 3 positive integers, must be an integer. 5 Let this integer be represented by p. Then, ^^~^ =p,ot2x-S = 5 p. (1) Dividing (1) by 2, a;-l-i = 2p+|; ov,x-l-2p=^^ Since x and p are integers, x — l — 2p is an integer ; and therefore ^-~— must be an integer. z Let this integer be represented by q. Then, £±l = g. oy^p^2q-\. 324 ADVANCED COURSE IN ALGEBRA Substituting in (1), 2x-3 = 10q-5. Whence, x = 5q — l. (2) Substituting this value in the given equation, 35 q -7 -\-5y = 11S', OT,y = 25-7q. (3) Equations (2) and (3) form the general solution in iyitegers of the given equation. By giving to q the value zero, or any positive or negative integer, we shall obtain sets of integral values of x and y which satisfy the given equation. If q is zero, or any negative integer, x will be negative. If q is any positive integer > S, y will be negative. Hence, the only positive integral values of x and y which satisfy the given equation are those obtained from the values 1, 2, 3 of q. That is, a; = 4, 2/ = 18 ; a; = 9, 2/ = 11 ; and ic = 14, ?/ = 4. 2. Solve ^x — X3y = 100 in positive integers. Dividing by 8, the coefficient of smaller absolute value, x-y-^ = 12 + ~; 0T,x-y-12 = -^^—. Then, ^ J' must be an integer. 8 Multiplying by 5, — ^^ must also be an integer. 8 Then, Sy -\-^-{-2-\-- must be an integer, and hence ^"^ 8 8 8 must be an integer ; let this be represented by p. Then, ^"^ =p, ot y^Sp — 4:. 8 Substituting in the given equation, 8 X - 104p + 52 = 100, or a; = 13p 4- 6. In this case p may be any positive integer. li p = l, x = 19 and ?/ = 4 ; if p == 2, a.* = 32 and 2/ = 12 ; etc. Thus, the number of solutions is indefinitely great. INDETERMINATE LINEAR EQUATIONS 325 The artifice of multiplying ^ "^ by 5 saves much work in Ex. 2. 8 The rule in any case is to multiply the numerator of the fraction by such a number that the coefficient of the unknown quantity shall exceed some multiple of the denominator by unity. If this had not been done, the last part of the solution would have stood as follows : Let ^1^ =p, or by + ^ = 8p. (1) o Divide hj 6, y + - = p -{- -^ ; then ^~ must be an integer. ' 5 5 6 . Let Sp-4: ^ ^^ or 3 j9 - 4 = 5 g. (2) 5 Divide hy3,;?-l-- = g + ^; then ^^"^ must be an integer. Let 2gjfl = r, or 2q + l=3r. (3) 3 Divide by 2, q-\-- = r-\--; then must be an integer. 2 2 2 r — 1 Let = = s, or r = 2s + l. Substituting in (3), 2 g + 1 = 6 s + 3, or g = 3 s + 1. Substituting in (2), 3^) — 4 = 15s-f5, orj? = 5s + 3. Substituting in (1) , 5 ?/ + 4 = 40 s + 24, or ?/ = 8 s + 4. Substituting in the given equation, 8x- 104s -52 = 100, oric = 13a + 19. These values of % and y differ in form from those obtained above ; but it is to be observed that 13 s + 19 and 8 s + 4, for the values 0, 1,2, etc., of s, give rise to the same series of positive integers as \Zp + 6 and 8 j? — 4 for the values 1, 2, 3, etc., of p. We will now show how to solve in positive integers two equations involving three unknown numbers. 3. In how many ways can the sum of $ 14.40 be paid with dollars, half-dollars, and dimes, the number of dimes being equal to the number of dollars and half-dollars together ? Let X — number of dollars, y — number of half-dollars, and » =*« number of dimes. 326 ADVANCED COURSE IN ALGEBRA Then by the conditions, (10x-^5y + z = 144, 1 x + y = z. (1) Adding, llx + 6y-\-z = 144 + z, or, 11 a; + 6 2/ = 144. (2) Dividing by 6, a; + ^ + 2/ = 24. o Then — must be an integer ; or, x must be a multiple of 6. 6 Let x = 6p, where p is an integer. Substitute in (2), 66 p + 6y = 144, or 2/ = 24 - lip. Substitute in (1), 2; = 6 p + 24 - 1 1 j9 = 24 - 5 p. The only positive integral solutions are when p = l or 2. Therefore, the number of ways is two; either 6 dollars, 13 half-dollars, and 19 dimes ; or 12 dollars, 2 half-dollars, and 14 dimes. EXERCISE 71 Solve the following in positive integers : 1. Sx-\-6y = 29. 6. 10x + 7y = 2Q7. 2. 7 a; + 2 2/ = 39. 6. 23 a: + 17 y = 183. 3. 6 a; + 29 2/ = 274. 7. 8 x + 71 2/ = 1933. 4. 4 X + 31 y = 473. 8. 13 a; + 50 ?/ = 1089. Solve the following in least positive integers : 9. 6x-7y = lS. 12. 8 a; -31 2/ = 10. 10. 5 X - 8 ?/ = 31. 13. 15 x - 38 y = - 47. 11. 14 X- 5?/ = 64. 14. 64x-19y = 507. Solve the following in positive integers : ri2x+72/ + 25! = 53. r3x-32/-f-7;2 = 101. I 2x-lly + s=-25. ' l4x-|-2y-3« = 6. 17. In how many different ways can $ 1.65 be paid with quarter-dollars and dimes ? 18. In how many different ways can £2 Is. be paid with half-crowns, worth 2s. 6d. each, and florins, worth 2s. each ? INDETERMINATE LINEAR EQUATIONS 327 19. Find two fractions whose denominators are 5 and 7, respectively, whose numerators are the smallest possible positive integers, and whose 17 difference is — . 35 20. In how many different ways can $ 7.15 be paid with fifty-cent pieces, twenty -five-cent pieces, and twenty- cent pieces, so that twice the number of fifty-cent pieces, plus twice the number of twenty-cent pieces, shall exceed the number of twenty-five-cent pieces by 31 ? 21. A farmer purchased a certain number of pigs, sheep, and calves, for .$138. The pigs cost |4 each, the sheep $7 each, and the calves $9 each ; and the whole number of animals purchased was 23. How many of each did he buy ? 22. In how many different ways can ^ 5.45 be paid with quarter-dol- lars, twenty-cent pieces, and dimes, so that twice the number of quarters, plus 5 times the number of twenty-cent pieces, shall exceed the number of dimes by 36 ? 486. Every linear equation, with two unknown numbers, x and y, can be reduced to one of the forms ax±by=±c, where a, b, and c are positive integers which have no common divisor. The equation ax + by=—c cannot be solved in positive integers ; for, if x, y, a, and b are positive integers, ax -f- by must also be a positive integer. Again, the equations ax±by = c and ax—by = — c cannot be solved in positive integers if a and b have a common divisor. For, if X and y are positive integers, this common divisor must also be a divisor of ax ± by, and consequently of c ; which is contrary to the hypothesis that a, b, and c have no common divisor. 328 ADVANCED COURSE IN ALGEBRA i XXIII. RATIO AND PROPORTION \ 487. The Ratio of one number a to another number & is the quotient of a divided by b. Thus, the ratio of a to 6 is ^ ; it is also expressed a:b. In the ratio a:b, a is called the first term, or antecedent, and b the second term, or consequent. If a is > 6, the ratio a : & is called a ratio of greater inequality; if a is < &, it is called a ratio of less inequality. The ratio of the product of the antecedents of a series of ratios to the product of the consequents, is said to be compounded of the given ratios. Thus, ac : hd is compounded of the ratios a : h and c: d. The ratio a^ : 6^ is called the duplicate ratio., the ratio a^ : 6^ the tripli- cate ratio ^ and the ratio Va : Vft the sub-duplicate ratio., of a : 6. RATIO OF CONCRETE MAGNITUDES 488. In § 487. we considered the ratio of abstract numbers only; it is, however, necessary to consider the ratio of two concrete magnitudes of the same kind. If a concrete magnitude is a times a certain unit, and another b times the same unit, we define the ratio of the first magnitude to the second as being the ratio of a to b. Thus, the ratio of two lines whose lengths are 2\ and 3|- * inches, respectively, is 2\ -^ 3J, or ■^- If the ratio can be expressed as a rational number, as in the above illustration, the magnitudes are said to be Commensurable. If it cannot be expressed as a rational number, they are said to be Incommensurable. rjiCUi'UKTlUJN ' - 489. A Proportion is an equation whose members are equal RATIO AND PROPORTION 329 6 = c : d, or - = -, b d Thus, if a: b and c : d are equal ratios, is a proportion. The symbol : : is sometimes used in place of the sign of equality in a proportion. 490. In the proportion a:b = c:d, a is called the first term, b the second, c the third, and d the fourth. -^ The first and third terms of a proportion are called the ante- cedents, and the second and fourth terms the consequents. _X The first and fourth terms are called the extremes, and the second and third terms the means. vy If the means of a proportion are equal, either mean is called a Mean Proportional between the first and last terms, and the last term is called a Third Proportional to the first and second terms. Thus, in the proportion a:b = b:c, b is a mean proportional between a and c, and c is a third proportional to a and b. A Fourth Proportional to three numbers is the fourth term of a proportion whose first three terms are the three numbers taken in their order. Thus, in the proportion a:b = c:d, d is a fourth proportional to a, b, and c. N A Continued Proportion is a series of equal ratios, in which each consequent is the same as the following antecedent ; as, a:b = b:c = c:d = d:e. PROPERTIES OF PROPORTIONS ^ 491. In any proportion, the product of the extremes is equal to the product of the means. Let the proportion be a:b = c:d. Then by §489, - = -. b d Clearing of fractions, ad = 6c. 330 ADVANCED COURSE IN ALGEBRA 492. From the equation ad = be (§ 491), we obtain be J ad ad -, -, be a = — , = — , c = — , and d = — deb a That is, m any proportion, either extreme equals the produet of the means divided by the other extreme; and either mean equals the produet of the extremes divided by the other mean. ^ 493. (Converse of § 491.) If the produet of two numbers be equal to the produet of two others, one pair may be made the extremes, and the other pair the means, of a proportion. Let ad = be. (1) ' Dividing by bd, - = -• b d Then, a : b = e : d, ov e : d = a : b. In like manner, by dividing the members of (1) by ab, then by cd, and then by ae, we have d:b = e: a, ov c : a = d . b, a: e = b : d, ov b: d = a:c, and d : e = b : a, ov b : a = d: c. \ 494. In any proportion, the terms are in proportion by alter- nation ; that is, the means ean be interehanged. Let the proportion be a: b = c: d. Then by § 491, ad = be., (1) Whence, by § 493, a:e = b:d. We also have from (1), by § 493, d:b = c:a. That is, in any proportion, the extremes can he interchanged. >» 495. In any proportioyi, the terms are in proportion by Inver- sion ; that is, the second term is to the first as the fourth term is to the third. Let the proportion be a : b = e : d. Then by § 491, ad = be. RATIO AND PROPORTION 331 Whence, by § 493, h'.a = d'.c. It follows from the above that, in any proportion^ the means can he written as the extremes^ and the extremes as the means. ~< 496. A mean proportionalJ)etween two numbers is equal to the square root of their product. Let the proportion be a:b — b : c. Then by § 491, b^ = ac, and b = Vac. 1 497. In any proportion, the terms are in proportion by Com- position; that is, the sum of the first two terms is to the first term as the sum of the last two terms is to the third term. Let the proportion be a : b = c : d. Then, ad = be. Adding each member of the equation to ac, ac -j- ad = ac-\- be, or a{c -\- d) = c(a + 5). Then by § 493, a-\-b\ a = c -{- d:c. We may also prove a -\- b :h ■= c ■{■ d: d. 4 498. In any proportion^ the terms are in proportion by Division ; that is, the difference of the first two terms is to the first term as the difference of the last two terms is to the third term. Let the proportion be a:b = c:d. Then, ad — be. Subtracting each member of the equation from ac, ac — ad = ac — be, or a(c — d) = c(a — b). Then, a — b:a = c — d:c. We may also prove a — b:b = c — did. 499. In any proportion, the terms are in proportion by Com- position and Division ; that is, the sum of the first two terms is to their difference as the sum of the la^t two terms is to their difference. Let the proportion be a:b = c:d 832 ADVANCED COURSE IN ALGEBRA Then by §497, a±h^c_±d^ a c ^ And by § 498, ^—^ = ^^. (2) a c ^ Dividing (1) by (2), a + b c-{-d a — b c — d Whence, a-\-b:a — b = c-{-d:G — d. 500. In any proportion, if the first two terms be multiplied by any number, as also the last two, the resulting numbers will be in proportion. Let the proportion be a'.b = c:d. Then, » = «; and hence ??^ = !^. b d mb nd Therefore, ma : mb = nc : nd. We may also prove ^ : A = ^ . ^. m m n n (Either m or n may be unity ; that is, the terms of either ratio may be multiplied or divided without multiplying or dividing tlje terms of the other.) ,^, , _, 501. In any proportion, if the first and third terms be multi- plied by any number, as also the second and fourth terms, the resulting numbers loill be in proportion. Let the proportion be a:b = c:d. Then, ^ = ^; and hence ™! = ™«. b d nb nd Therefore, ma :nb = mG: nd. trr 1 Ct b C d We may also prove _:_ = — : — m n m n (Either w or w may be unity. ) ^ 502. In any number of proportions, the products of the cor- responding terms are in proportion. Let the proportions be a : b = c : d, and e:f=g:h. \ RATIO AND PROPORTION 333 Then, ^ = ' and^ = f. ' b d f h ,r ,,. T . a e c q ae cq Multiplying, _x_^ = -x2,o.-- = X Whence, ae:hf=cg: dh. In like manner, the theorem may be proved for any number of proportions. 503. The quotients of the corresponding terms of two pro- portions are in proportion. Let the proportions be a : 6 = c : c?, and e:f=g:h. Then, -=-, and :£ = -• b d eg TT71, Ci f c^,h abed Whence, -x=^ = -X-; or, --f-- = --f--. b e d g e f g h Then, «:A = f:f. e f g h 504. In any proportion, like powers or like principal roots of the terms are in proportion. Let the proportion be a:b=^ c:d. Then, - = - : and hence ^ = ^. ' b d' b"" d"" Therefore, a** : ^'^ = c" : d\ We may also prove \/a : Vb = Vc : Vd. ^^505. In a series of equal ratios, any antecedent is to its con- sequent as the sum of all the antecedents is to the sum of all the consequents. Let a:b = c:d = e:f Then by § 491, ad = be, and af= be. Also, ab = ba. 334 ADVANCED COURSE IN ALGEBRA Adding, a(b -\- d -i-f) = b(a -\- c -^ e). Whence, a:b = a -{- c -{- e:b -]- d-j-f. (§493) In like manner, the theorem may be proved for any number of equal ratios. Q 506. To prove that if - = - = !=..., ^ b d f 1 then each of these equal ratios equals { ^ ~ ^ — IL JI — \pb^J^qd^-^rr+-- If - = - = -1 = ... = yfc, then a = bk, c = dk, e =fk, etc. b d e Then, pa"" + qc"" + re"" -{-"-= p (bky + q (dky + r (fky + • • • = A;"(p6'* + gd" + rf" + •••). Therefore, k^ = P^" + gc- + re- + ... ^ Or A; = /^ i>«" + gc" + re- + --Y ' \pb'' + qd'' + rf''-{- '"J Q 507. 7/* ^/iree numbers are in continued proportion, the first is to the third as the square of the first is to the square of the second. Let the proportion be a:b = b:c. Then, ? = 5. b c Therefore, ^ x ^ = ^X ?, or « = ^. b c b b c b^ Whence, a:c = a^:W. \J 508. If four numbers are in continued proportion, the first is to the fourth as the cube of the first is to the cube of the second. Let the proportion be a:b = b \ c = c:d. rT^^ a b C Then^ - = - = -• "* bed Therefore, ^ x ^ x ^ = f X ? X « or « = ^'. b G d b b b d b" Whence, a:d = a^: ¥. RATIO AND PROPORTION 335 509. Examples. 1. If x:y = {x-{-z)'^'.{y-\-zf, prove z a mean proportional between x and y. From the given proportion, by § 491, y{x -\- zf = x(y + zy, or ib-y + 2 xyz + yz^ = xy^ + 2 ic?/^! ■+- xz^- Transposing, x^y — xy^ = xz^ — 2/2;l Dividing hj x — y^ xy = z\ Therefore, 2; is a mean proportional between x and y. The theorem of § 499 saves work in the solution of a certain class of fractional equations. 2. Solve the eq uation ^' + a; - 1 ^ ^-2 ^ ar — x — 1 x-\-2 By composition and division, 2a;^-2 _ 2x ^ -\ _ x 2x --4''''' X ~ 2 Clearing of fractions, 2 af' — 2 = — aj^ Then, 3 x^ = 2, and a; = ±-^-. 3. Prove that if - = -, then h d a2 _ 52 : a2 - 3 ao = c2 - ^2 : c2 - 3 cd. Let - = - = Xj then, a = 6a;. 6 d Therefore, c" 1 a'- -b' Wx' -b' x'- -1 d? ^ & -d^ a'- Sab 6V- ■3bh : x'- -3 a; d^ 3 c c^- d -3c(i Then a^-ft^ a'- Sab = c'- d^',& -3cd EXERCISE 72 1. Find the third term of a proportion whose first, second, and fourth terms are - , - , and - , respectively. 4 6 9^ 2. Find a third proportional to — and — y 12 3S6 ADVANCED COURSE IN ALGEBRA 3. Find a mean proportional between l\l and 24|. 4. Find a fourth proportional to 4|, 5|, and If. 5. Find a third proportional to a^ + 27 and a + 3. ^2 /v. 1 2 x^ 9 jC 4- 20 6. Find a mean proportional between and '- — — — . . x-6 ic + 3 Solve the following equations : - 3a;-8 _ 2a;- 5 _ g x^-\-2x-S _ Sx + 2 '3a; + 4~2x + 7' ' x^-2x-S 3ic-2* 9 a;2 + 3x-l _ a;g + 2a;+l x^-Zx+1 x2-2x-l 10. 11. Va;2 + 1 + V a;2 - 1 _ Vgg _ 2 g + 2 + Va^ _ 2a Vic2 + l - Vx2- 1 \/a2 _ 2 a + 2 - Va'^ - 2 a X — y a — b x^ + x + y'^ _a^-\- a + b^ 2 - X + ?/2 a2 - a + 62 12. If a : & = c : c? and e :f = g :h, prove ae + bf-.ae — bf= eg + dh :cg — dh. 13. Find two numbers such that, if 9 be added to the first, and 7 sub- tracted from the second, they will be in the ratio 9:2; while if 9 be sub- tracted from the first, and 7 added to the second, they will be in the ratio 9:11. 14. Find two numbers in the ratio a : 6, such that, if each be increased by c, they shall be in the ratio m:n. 15. Find three numbers in continued proportion whose sum is 28|, 2 such that the quotient of the first by the second shall be -• o 16. Find a number such that, if it be added to each term of the ratio 8 : 5, the result is — of what it would have been if the same number 20 had been subtracted from each term. 17. The second of three numbers is a mean proportional between the other two. The third number exceeds the sum of the other two by 20 ; and the sum of the first and third exceeds three times the second by 4. Find the numbers. 18. If8a — 56:7a — 45 = 86-5c:76 — 4 c, prove c a third propor- tional to a and 6. 19. lia + b + c + d:a + b = a — b + c — d:a — b, prove a:b = C:d. RATIO AND PROPORTION 337 20. lix + y:y + z = y/x^ — y'^ : y/y'^ — 2^, prove y a mean proportional between x and z. 21. A is following B along a certain road, when B turns and walks in the opposite direction; if A and B approach each other five times as fast as 'before, compare their rates. 22. If 4 silver coins and 11 copper coins are worth as much as 2 gold coins, and 5 silver coins and 19 copper coins as much as 3 gold coins, find the ratio of the value of a gold coin, and the value of a silver coin, to the value of a copper coin. 23. Given 2 (a^ + ah)x + (a^ + 2 h'^)y = {a^ - b^)x + (2 a^ + b^)y, find the ratio of x to y. 24. Given ? + ^ = ^ + ^ = ^-f^, fi^d the ratio of x to y, and of x to z. b a c a c b 25. If - = -, prove, b d (a) 3 a2 - 4 a6 : 2 a6 + 7 62 = 3 c2 - 4 cci : 2 ctZ + 7 (^2. lb) a^ + 6 ab'^ : a^b - bb^ = c^ + Q cdC^ -.cH - b #. 26. The sum of four numbers in proportion is 32. The sum of the means exceeds the sum of the extremes by 4 ; and the sum of the conse- quents exceeds the sum of the antecedents by 16. Find the numbers. 27. A passenger observes that a train passes him, moving in the opposite direction, in 3 seconds ; while, if it had been moving in the same direction, it would have passed him in 13 seconds. Compare the rates of the trains. 28. Each of two vessels contains a mixture of wine and water. A mixture consisting of equal measures from the two vessels is composed of wine and water in the ratio 3:4; another mixture consisting of 14 meas- ures from the first and 21 measures from the second, is composed of wine to water in the ratio 2 : 3. Find the ratio of wine to water in each vessel. 29. If ^ = ^ = -^, prove b d f (a) a^ + c"^ + e^ :b^ + d^ + P = ac + ce + ea :bd -h df + fb. (6) a* + c* + c* : 6* + d* +/* =(«« + c2 + e'^y : (b^ + d^ +Py. 30. If a + 6, 6 + c, and c + a are in continued proportion, prove a + b:b + c = c — a:a — b. 31. Find four numbers in proportion such that the sum of the means is 21, and of the extremes 39 ; and twice the last term exceeds three times the sum of the first two terms by 30. 32. If a, 6, c, and d are in continued proportion, prove 3a + 4d:2a-5d = 3a8 + 4 68:2a3_6 63. 338 ADVANCED COURSE IN ALGEBRA XXIV. VARIATION 510. One variable number (§ 245) is said to vary directly as another when the ratio of any two values of the first equals the ratio of the corresponding values of the second. It is usual to omit the word "directly," and simply say that one number varies as another. Thus, if a workman receives a fixed number of dollars per diem, the number of dollars received in m days will be to the number received in n days as m is to w. Then, the ratio of any two numbers of dollars received equals the ratio of the corresponding numbers of days worked. Hence, the number of dollars which the workman receives varies as the number of days during which he works. The symbol cc is used to express variation; thus, ace 6 is read " a varies as 6.'^ 511. One variable is said to vary inversely as another when the first varies directly as the reciprocal of the second. Thus, the number of hours in which a railway train will traverse a fixed route varies inversely as the speed ; if the speed be doubled, the train will traverse its route in oyie-half the number of hours. One variable is said to vary as two others jointly when it varies directly as their product. Thus, the number of dollars received by a workman in a certain number of days varies jointly as the number which he receives in one day, and the number of days during which he works. One variable is said to vary directly as a second and inversely as a third, when it varies jointly as the second. and the recip- rocal of the third. Thus, the attraction of a body varies directly as the amount of matter, and inversely as the square of the distance. VARIATION 339 512. IfxcxLy, then x equals y multiplied by a constant number. Let «' and 2/' denote a j^a;ed pair of corresponding values of X and y, and x and y any other pair. By the definition of § 510, - = -7; or, x = -,y. a;' Denoting the constant ratio — by m, we have X = my. 513. It follows from §§ 511 and 512 that: 1. Ifx varies inversely as y, x = — . 2. Ifx varies jointly as y and z, x — myz. mil 3. Ifx varies directly as y and inversely as 2;, x = ^-- z The converse of each statement of §§ 512 and 513 is also true ; that is, if X equals y multiplied by a constant, xazy, etc. 514. Ifx-^ By § 512, iixS = |[2a + (n-l)d]. >^ 344 ADVANCED COURSE IN ALGEBRA 520. Ex. In the progression 8, 5, 2, —1, —4, ••., to 27 terms, find the last terra and the sum. Here, a = 8, d = 5-8 = -3, n = 27. Substitute in (I), Z = 8 + (27- 1)(-3) =-70. Substitute in (II), S=^~{^-10)=- 837. ; Jj The common . difference may be found by subtracting the first term from the second, or any term from the next following term. EXERCISE 74 In each of the following, find the last term and the suxn : 1. 5, 14, 23, ... to 18 terms. ^ ^^ > '^^ ^^"'fuiL 2. 9, 2, - 5, ... to 23 terms: "^ ^ , ' ^ ^ ^ V Z. —51, —43, —35, ... to 15 terms. 4. --,-—,- 3, ... to 16 terms. 4 8 / 6. ^ 1 _ 1 ... to 28 terms. 6 6 2 6. §, il, ^, ... to 25 terms. 9' 36 18 7. -— , -—; -1, ... to 39 terms. y ^^8. -1, -^, -il, ...to52terms. ^' . ^l t 10' 6' 30' v3)L^ 9. 3a + 46, 8a + 2 6, 13 a, ... to 10 terms. ^^V^ ' ^ v/ 10. ^^, % % ... to 9 terms. ' "'•''' y^ 3 6 3 J A^ 521. If any three of the five elements of an arithmetic pro- gression are given, the other two may be found by substituting the known values in the fundamental formulae (I) and (II), and solving the resulting equations. 5 5 1. Given a = — -, w = 20, 8 = — \-\ find d and I. o o Substituting the given values in (II), -^ = 10(^-^4-^), or -1 = -^ + ?; then, Z = 5_l = |. 3 ^3/' 6 3' ' 362 PROGRESSIONS 846 Substituting the values of a, n, and I in (I), - = \-19d', whence, 19 d = — , and d = -' 2. Given d = -3, l = -39, >S = -264; find a and n. Substituting in (I), _39 = a+(7i-l)(-3), or a = 3n-42. (1) Substituting the values of I, S, and a in (II), -264 = -(37i-42-39), or - 528 = 3^2-81 w, or n2-27?i = -176. wv, 27±V729-704 27 ± 5 -,^^1-, Whence, n = ^-^ = — ~- — 16 or 11. Substituting in (1), a = 48 -42 or 33-42 = 6 or -9. The solution is, a = 6, ?i = 16 ; or, a = — 9, ti = 11. The significance of the two answers is as follows : If a = 6 and n = 16, the progression is 6, 3, 0, - 3, - 6, - 9, - 12, - 15, - 18, - 21, - 24, - 27, - 30, -33, -36, -39. If a = — 9 and 72 = 11, the progression is - 9, - 12, - 15, - 18, - 21, - 24, - 27, - 30, - 33, - 36, - 39. In each of these the sum is — 264. 113 3. Given a = -, d = — — , S — — -\ find I and n. Substituting in (I), Z = | + (« - 1) (- X) = 5^. (1) Substituting the values of a, S^ and I in (II), 3 n /I , 5 — n\ o /9 — ?i\ 2 n oa -2 = 2(3+^' ""' -^ = '\-l2-> "' »'-9" = 36- Solving this, n = 12 or — 3. The value 71 = — 3 must be rejected, for the number of terms in a progression must be a imsitive integer. 5-12 7 Substituting n = 12 in (1), I ■■ 12 12 346 ADVANCED COURSE IN ALGEBRA Any value of n which is not a positive integer must be rejected, together v^'ith all other values dependent on it. From (I) and (II), general formulce for the solution of exam- ples like the above may be readily derived. 4. Given a, d, and S ; derive the formula for n. By §519, 2jS = 7il2a-\-(n-l)d'], or dji"" + (;2a-d)n = 2 S. This is a quadratic in n; solving by formula (1), § 450, _ d-2a± V(2 a ~ d f -^^dS ih — „ • 2d EXERCISE 75 ^ ry^ ^ 1- Given d = S, 1 = 147, w =^19 ; find a and S. 2. Given d = -6, w = 14, S = - 616 ; find a and L 3. Given a =-69, n = 16, I = S6 ; find d and S. 4. Given a = 8, w = 25, S = - 2500 ; find d and l. ^ 5. Given a =-, ? = - — , ^ = - 78 ffind d and n. 4 4 6. Given I =—, w = 25, ^9 = ?^ ; find a and d. 4 4 7. Given a = - -, d = _ A, ^ ^ _ ^ ; find w and I. 5' 10 5 ' 8. Given a = - -, Z = — , (? = - ; find n and S. 3 2 6 9. Given d = . w = 55, S = -166; find a and I 12 10. Given ? = ^, w = 24, ^ = 241 ; find a and d. 11. Given ? = ^, d = -, 8 = — -, find « and n. 6 6 6 12. Given a = --, Z = -— , 8 = -^; find c? and w. 5 10 2 ' 13. Given a = -—, n = 21, /S' = — : find cZ and l. 22 22 14. Given I = ^, d = —, 8 = - — ; find a and ?i. 12 12 3 16. Given «=-— , d = \ 8 = - — ; find ri and I. 6 3 2 ' PROGRESSIONS 347 16. Given a, Z, and n ; derive the formula for d. 17. Given a, n, and S ; derive the formul?e for d and I. 18. Given d, ?z, and /S' ; derive the formulae for a and I. 19. Given a, d, and Z ; derive the formulae for n and S. 20. Given d, Z, and n ; derive the formulae for a and /S^. 21. Given Z, n, and S ; derive the formulae for a and d 22. Given a, (f, and /S ; derive the formula for I. 23. Given a, Z, and 8 ; derive the formulae for d and n. 24. Given cZ, Z, and 8 ; derive the formulae for a and n. 522. Arithmetic Means. To find an arithmetic progression of m + 2 terms, whose first and last terms are two given numbers, a and h, is called inserting m arithmetic means hetiwen a and h. Ex. Insert 5 arithmetic means between 3 and — 5. We find an arithmetic progression of 7 terms, in which a = 3, and Z = — 5 ; substituting n = 7, a = 3, and Z = — 5 in (I), _5 = 3 + 6d, or cZ = --. '3 The progression is 3, -, -, -1, --, -~, ~ 5. 523. Let X denote the arithmetic mean between a and h. Then, x~a = b — x, or 2x = a-^b. Whence, ^ = ^^- That is, tlie arithmetic mean between two numbers equals one- half their sum. EXERCISE 76 1. Insert 7 arithmetic means between 4 and 10. 5 6 7 3 Insert 6 arithmetic means between — - and 6 2 7 3. Insert 9 arithmetic means between — and 6. 33 4. Insert 8 arithmetic means between — 3 and — — • 348 ADVANCED COURSE IN ALGEBRA 6. Insert 5 arithmetic means between - and 9 3 3 29 6. How many arithmetic means are inserted between — - and — , 9 when the sum of the second and last is - ? 5 7. If m arithmetic means are inserted between a and 6, find the first three. Find the arithmetic mean between : 8. 1| and - 2| . 9. (3 m + n)2 and (m - 3 ny. x-l x^-1 524. Problems. 1. The sixth term of an arithmetic progression is -, and the -I p D fifteenth term is — Find the first term. The common difference must be one-ninth the result obtained by subtracting the sixth term from the fifteenth. Then, a = m-'-] = l- 9V3 6/ 2 Again, the first term must equal the sixth term minus five times the common difference. Then, a = = — -• ' 6 2 3 2. Find four numbers in arithmetic progression such that the product of the first and fourth shall be 45, and the product of the second and third 77. Let the numbers be represented by x — 3 ?/, x — y, x + y, and x + Sy, respectively. Then by the conditions, i ^ 1.^2 _ 2/2 = 77. Solving these equations, x = 9, y = ±2; or, x=-9, y = ±2 (§472). Then the nunfbers are 3, 7, 11, 15 ; or, — 3, — 7, — 11, — 15. In problems like the above, it is convenient to represent the unknown numbers by symmetrical expressions. Thus, if five numbers had been required to be found, we should have represented them hy x — 2 y, x — y, x, x -{■ y, and x'+ 2}/. PROGRESSIONS 349 EXERCISE 77 13 1. The seventh term of an arithmetic progression is , and the 7 ^ thirteenth term — - . Eind the twenty-second term. 2. The first term of an arithmetic progression is 1, and the sum of the sixth and tenth tferms is 37. Find the second, third, and fourth terms. Q 3. The first term of an arithmetic progression of eleven terms is - , and the seventh term — 3. Find the sum of the terms. 4. In an arithmetic progression, the sum of the first and last terms is two-ninths the sum of all the terms. Find the number of terms. 5. How many positive integers of three digits are multiples of 13 ? What is their sum ? 6. Find five numbers in arithmetic progression such that their sum shall be 25, and the sum of their squares 135. 7. Find four numbers in arithmetic progression such that the product of the first and third shall be — 21, and the product of the second and fourth 24. 8. If the constant difference of an arithmetic progression equals tvsrice the first term, the quotient of the sum of the terms by the first term is a perfect square. 9. In any arithmetic progression, the sum of the first m terms, less twice the sum of the first w + 1 terms, plus the sum of the first m + 2 terms, equals the common difference. 10. The sum of the first ten terms of an arithmetic progression is to the sum of the first five terms as 13 to 4. Find the ratio of the fii*st term to the common difference. 11. The sum of six numbers in arithmetic progression is 36, and the sum of their squares is 286. Find the numbers. ' 12. A man travels -^— miles. He travels 20 miles the first day, and A increases his speed one-half mile in each succeeding day. How many days does the journey require ? 13. Find three numbers in arithmetic progression, such that the square of the first added to the product of the other two gives 16, and the square of the second added to the product of the other two gives 14. 14. A traveller sets out from a certain place, and goes 3^ miles the first hour, 3f the second hour, 4 the third hour, and so on. After he has been gone 5 hours, another sets out, and travels 8^ miles an hour. After how many hours are the travellers together ? (Interpret the two answers. ) 350 ADVANCED COURSE IN ALGEBKA 15. Find the sum of the terms of an arithmetic progression of 11 terms, in which 121 is the middle terra. 16. A man climbing a mountain, ascends the first hour 1000 feet, the second hour 800 feet, the third hour 600 feet, and so on. After how many hours will he be at the height of 2800 feet ? (Interpret the two answers.) 17. If a person saves $ 120 each year, and puts this sum at simple interest at 3| per cent at the end of each year, to how much will his property amount at the end of 18 years ? iLft-;44^ 18. There are 12 equidistant balls in a straight line. A person starts from a position in line with the balls, and beyond them, his distance from the first ball being the same as the distance between the balls, and picks them up in succession, returning with each to his original position. He finds that he has walked 6460 feet. Find the distance between the balls. 19. A and B travel around the world, the circuit being 23661 miles. A goes east one mile the first day, two miles the second day, three miles the third day, and so on. B goes west at a uniform rate of 20 miles a day. After how many days will they meet ? (Interpret the negative answer.) GEOMETRIC PROGRESSION 525. A Geometric Progression is a series in which each term, after the first, is obtained by multiplying the preceding term by a constant number called the Ratio. Thus, 2, 6, 18, 54, 162, ••• is a geometric progression in which the ratio is 3. 9, 3, 1, -, -, ••• is a geometric progression in which the ratio — 3, 6, —12, 24, —48, ••• is a geometric progression in which the ratio is — 2. A Geometric Progression is also called a Geometric Series. 526. Given the first term, a, the ratio, r, and the number of terms, n, to find the last term, I. The progression is a, ar, ar^, ai^, •••. We observe that the exponent of r in any term is less by 1 than the number of the term. PROGRESSIONS 351 ^ Then, in the nth term the exponent of r will be n — 1. ^^ ~ That is, l=ar''-\ (I) c^ 527. Given the first term, a, the last term, Z, and the ratio, r, to find the sum of the terms, S. 4(9t yd '^ S = a +X^ -\-ar^A h c*?*"~^ + ot^"~^ + ^""^ (1) Multiplying each term by r, r/S = ^ + ar^ + ar' H h ar"-^ + ow:^-i 4. «?''*. (2) Subtracting (1) from (^,/aS - >S = ar^ - a, or /S = ^El^. , But by (I), § 526, "^ ;' r? = ar\ ^--Z^ Therefore, ^^ ^ = !lz:^. v (II) ) r-1 ■ The first term, ratio, number^ terms, last term, and sum of the terms, .are called the elements of the progression. 528. Examples. 1. In the progression 3, 1, -, •••, to 7 terms, find the last term and the sum. Here, a = 3, r = -, n =1 ; substituting in (I) and (II), o ^3; 3^ 243 1 _1__3 _1._3 2186 ^_ 3 243 _ 729 _ 729 _1093 1_1 ~ _? ~ _? "243* 3 3 3 The ratio may be found by dividing the second term by the first, or any term by the next preceding term. 2. In the progression —2, 6, —18, •••, to 8 terms, find the last term and the sum. Here, a = — 2, r = = — 3, n — %; therefore, — z l = -2(-Sy = -2x(- 2187) = 4374. ^ ^ - 3 X 4374 - (- 2) ^ - 13122 + 2 ^ g^g^ 352 ADVANCED COURSE IN ALGEBRA EXERCISE 78 In each of the following, find the last term and the sum of the terms — 2, 4, ... to 10 terms. 3, - 12, - 48, ... to 6 terms. 1. 1, - 2, 4, ... to 10 terms. « 3 . 4 .„ . _,„ ' ' ' 6. —-, 1, —-,... to 7 terms. / 3. _||,_^,...to9ter,„s. 4. A, ||,... to 6 terms. 5. -, *, 2, -.. to 7 terms. 4' ' 3' 16 4 1 , «, y ~3' 3' *" ^^^*^^"^^- 11 A, ...to 5 terms. 529. If any three of the five elements of a geometric pro- gression are given, the other two may be found by substituting the given values in the fundamental formulae (I) and (II), and solving the resulting equations. But in certain cases the operation involves the solution of an equation of a degree higher than the second ; and in others the unknown number appears as an exponent, the solution of which form of equation can usually only be affected by the aid of logarithms (§ 604). In all such cases in the present chapter, the equations may be solved by inspection. 1. Given a = — 2, n = 5, I = — 32; find r and S. Substituting the given values in (I), we have _32 = -2r*; whence, r' = 16, or r = ±2. Substituting in (II), If r= 2, /S = ^^~^P~/~^^ =-64-f 2 = -62. If ,^_2, ^^ (-2)(-32)-(-2) ^64 + 2^_ -2-1 -3 The solution is, r = 2, /S = - 62 ; or, r = - 2, JS = -22. The significance of the two answers is as follows : If r= 2, the progression is —2, —4, —8, —16, —32, whose sum is —62. If r= —2, the progression is —2, 4, —8, 16, —32, whose sum is — 22. 1640 -1'- -3 ~ 4 729 ~ 1 3 -1 Z + 9 = 6560 ■ 729' or, l = - 1 729 PROGRESSIONS 868 2. Given a = S r = -i S = ^^; find n and Z. Substituting in (II), Whence, Substituting the values of a, r, and I in (I), Whence, by inspection, w — 1 = 7, or n = S. From (I) and (II) general formulae may be derived for the solution of cases like the above. If the given elements are n, I, and 8, equations for a and r may be found, but there are no definite formulce for their values. The same is the case v^hen the given elements are a, n, and S. The general formulae for n involve logarithms ; these cases are dis- cussed in §604. EXERCISE 79 1. Given r = 3, w = 8, Z = 2187 ; find a and S. 2. Given a = 6, n = 7, Z = — ; find r and /S. 243 3. Given r = -5, n = 5, S = - 1042 ; find a and I. 4. Given a = - 3, r = --, I = -^: find n and S. ' 2' 128' 5. Given r = -2, n = 10, S = - ^^ ; find a and l. 2 ' 6. Given a = ^, w = 6, Z = - — ; find r and /S. 2' ' 125' 7. Given a =-1 Z = -^, S = - — ; find r and n. 3' 64 ' 192 ' 8. Given a = -,r = -,S = ?^: find I and n. 4' 4' 1024 ' 9. Given I = 384, r = - 4, S = ^^ ; find a and n. 10. Given a = -, Z = 1458, /S = ^^ ; find r and w. 9 9 :) 354 ADVANCED COURSE IN ALGEBRA 11. Given a, r, and S ; derive the formula for I. 12. Given a, I, and S ; derive the formula for r. 13. Given r, I, and S ; derive the formula for a. 14. Given r, «, and I ; derive the formulse for a and S. 15. Given r, n, and &' ; derive the formulse for a and I. 16. Given a, n, and I ; derive the formulse for r and S. 630. Sum of a Geometric Progression to Infinity. The limit (§ 245) to which the sum of the terms of a decreas- ing geometric progression approaches, when the number of terms is indefinitely increased, is called the sum of the series to infinity. Formula (II), § 527, may be written -rl S 1-r It is evident that, by sufficiently continuing a decreasing geometric progression, the absolute value of the last term may be made less than any assigned number, however small. Hence, when the number of terms is indefinitely increased, I, and therefore rl, approaches the limit 0. / Then, the fraction approaches the limit 1 — r ^^ 1 — r /^ Therefore, the sum of a decreasing geometric progression to) (^infinity is given by the formula ^ 1. rind the sum of the series 4, — Here, a = 4, r = — -• Substituting in (III), S = This signifies that, the greater the number of terms taken, the more nearly does their sum approach to — ; but the sum will never exactly equal this value. 1-r (III) ries4 -^ 1*5 . "es 4, -, -, •• to infinity. 4 12 '-1 ' PROGRESSIONS 355 A repeating decimal is a decreasing geometric progression, ^ and its value may be found by formula (III). 2. Find the value of .85151 •••. ■oooS'l^ ,o / -A/ We have, .85151 ... = .8 + .051 + .00051 + .... • ^ ^^' The terms after the first constitute a decreasing geometric progression, in which a = .051, and r = .01. Q K f f V • /TTTN c -051 .051 51 17 Substituting in (III), ^ = ^_^ = _ = _ = _. 8 17 281 Then, the value of the given decimal is tx + ^^j or — -• . 10 oo\) oo\) ^ EXERCISE 80 \ Find the sum of the following to infinity : \ 1. 6, -2, 1, .... 5 8 14 49 5' 15' 90' ^' ^^' ^'4' •"• ®- "10' 25' "125' •••• Q 4 2 1 3 15 _J5^ _ ^- ~ ' "3' "9' **'• 4' 32' 256' ' 4 _25 25 _50 o 5 5 10 • 6 ' 9 ' 27' "" ■ 6' 27' 243' ^ Find the values of the following : 9. .8181 .... 11. .91777 .... 13. .23135135 .... 10. .629629 .... 12. .75959 .... 14. .587474 .... 531. Geometric Means. To find a geometric progression of m + 2 terms, whose first and last terms are two given numbers, a and b, is called insert- ing m geometric means between a and b. Ex. Insert 5 geometric means between 2 and — — . ^ 729 We find a geometric progression of 7 terms, in which a = 2, 1 98 1 28 and I = — ; substituting w = 7, a = 2, and I = — - in (I), i Zv i ZJ II = 2 r« i whence, ,■« = ^, and r = ± |. 356 ADVANCED COURSE IN ALGEBRA The result is 2, ±^J, ±1^,??, ±^,^28^ ' 3' 9' 27' 81' 243' 729 532. Let X denote the geometric mean between a and 6. Then, ^ = ^, or a;2 = a6. a X Whence, x = Va6: That is, the geometric mean between tivo numbers is equal to the square root of their product. EXERCISE 81 o 1. Insert 8 geometric means between - and — 192. 2. Insert 7 geometric means between — 3 and — 19683. 3. Insert 6 geometric means between — and 5120. 16 4. Insert 4 geometric means between — - and — • 6 729 o 5. Insert 5 geometric means between — 48 and — — . 256 6. Insert 3 geometric means between — and — ^ 32 10 7. If w geometric means are inserted between a and 6, what are the last two means ? Find the geometric mean between : 9. t±^^ and ?^^. 10. a2 _ 4 «5 + 4 52 and 4:0? + 4:ah + h\ 533. Problem. Find 3 numbers in geometric progression such that their sum shall be 14, and the sum of their squares 84. Let tlie numbers be represented by a, ar, and ar^. ^, , r a + ar -\- ar"^ = 14. Then, by the conditions, -{ „ U2 + a2^2 + Qj2^ ^ 84. (1) (2) Divide (2) by (1), a - ar + ar^ = 6. (3) Subtract (3) from (1), 2 ar = 8, or r = -• (4) PROGRESSIONS 357 Substituting in (1), a + 4 + — = 14, or a^ - 10 a + 16 = 0. a Solving this equation, a = 8 or 2. Substituting in (4), r = ^ or - = - or 2. 8 2 2 Then, the numbers are 2, 4, and 8. EXERCISE 82 1. What number must be added to each of the numbers a, 6, and c, so that the resulting numbers shall form a geometric progression ? 4 2. The sixth term of a geometric progression is — , and the eleventh 128 ' term Find the third term. 6561 3. Find an arithmetic progression whose first term is 2, and whose first, fourth, and tenth terms form a geometric progression. 4. The product of the first five terms of a geometric progression is 243. Find the third term. 5. Find four numbers in geometric progression such that the sum of the first and fourth is 27, and of the second and third 18. 6. Find six numbers in geometric progression such that the sum of the first, third, and fifth is 147, and of the second, fourth, and sixth 294. 7. The sum of the terms of a geometric progression whose first term is 1, ratio 3, and number of terms 4, equals the sum of the terms of an arithmetic progression whose first term is 4, and common difference 4. Find the number of terms in the arithmetic progression. 8. A man who saved every year five-fourths as much as in the preced- ing year, had saved in four years $9225. How much did he save the first year? 9. The population of a state increases from 100000 to 161051 in five years. What is the rate of increase per year ? 10. The difference between two numbers is 72, and their arithmetic mean exceeds their geometric mean by 8. Find the numbers. 11. The sum of the first eight terms of a decreasing geometric progrea- sion is to the sum to infinity as 16 to 25. Find the ratio. 12. There are three numbers in geometric progression whose sum is Q\. Q fi 7 If the first be multiplied by - , the second by - , and the third by - , the resulting numbers will be in arithmetic progression. What are the numbers ? ^ HARMONIC PROGRESSION 358 ADVANCED COURSE IN ALGEBRA 13. The digits of a number of three figures are in geometric progres- sion, and their sum is 14. If 594 be subtracted from the number, the digits will be reversed. Find the number. 14. The mth term of a geometric progression is p, and the nth. term is q ; prove that the first term is ^'Vp^-^'q'"--^. 15. The sum of three rational numbers in geometric progression is — , and the sum of their reciprocals — Find the numbers. 15 3 16. If the numbers a, b, and c are in geometric progression, prove 1 + 1 + 1 ^ «^ + &^ + c3 a^ b^ c^ a%'^c^ 17. The sum of the first four terms of a geometric progression is 45, and of the first lix terms 189. Find the first term and the ratio. 18. If oj, y, and z are, respectively, the pth, gth, and rth terms of a geometric progression, prove xQ-r X zP-^ = yp-^. U 534. A Harmonic Progression is a series of terms whose reciprocals form an arithmetic progression. Thus, 1, -, -,-,-,••• is a harmonic progression, because the 3 5 7 9 reciprocals of the terms, 1, 3, 5, 7, 9, •••, form an arithmetic progression. A Harmonic Progression is also called a Harmonic Series. Any problem in harmonic progression, which is susceptible of solution, may be solved by taking the reciprocals of the terms, and applying the formulae of the arithmetic progression. There is, however, no general method for finding the sum of the terms of a harmonic series. 2 2 Ex. In the progression 2, -, -, ••• to 36 terms, find the last , 3 5 term. Taking the reciprocals of the terms, we have the arithmetic . 13 5 progression -, -, -, .... Here, a = -, d = 1, n = 36. PROGRESSIONS 359 Substituting in (I), § 518, Z = i+ (36 - 1) x 1 =^- IThen, — is the last term of the given harmonic series. ^'^ ' 71 ^ 535. Harmonic Means. To find a harmonic progression of m + 2 terms, whose first and last terms are two given numbers, a and 6, is called insert- ing m harmonic means between a and h. Ex. Insert 5 harmonic means between 2 and — 3. We have to insert 5 arithmetic means between - and 11 - 2 3 Substituting a = i ? = - ±, « = 7, in (1), § 518, Z o = - 4- 6 a, or a = 3 2' 36 Then the arithmetic series is 1 13 2 1^ _1^ _]_ _1 2' 36' 9' 12' 18' 36' 3* Therefore, the required harmonic series is 2, ||, I 12, -18, -^, -3. 536. Let X denote the harmonic mean between a and h. Then, - is the arithmetic mean between - and — X ah Then, by § 523, 1 = 5LJ = ^+^, and ^ = 1^- ' -^ 'a; 2 2a6' a + 6 EXERCISE 83 In each of the following, find the last term : 1. ^ 12 12 ^Q 22 terms. 6' 43' 71 2. --, --, -—, ...to 19 terms. 8' 6' 11 360 ADVANCED COUKSE m ALGEBKA 3. -3, 2, -, ... to 26 terms. 14 2 4. — , — , — , ••• to 11 terms. 11 39' 17 5. --, --, --,... to 37 terms. 6' 3' 9' 6. Insert 6 harmonic means between 2 and . 9 4 1 7. Insert 8 harmonic means between — and — • 5 5 2 2 8. Insert 7 harmonic means between - and 7 5 Find the harmonic mean between : 9. 2 and -3. 10. ^ + ^ and «^^:l^. 3 4 a-b a^ + b-^ 11. Find the (w — l)th term of the harmonic progression a, b, ••• to n terms. 12. If m harmonic means are inserted between a and 6, what is the third mean ? 13. The first term of a harmonic progression is^j, and the second terra q. Continue the series to three more terms. 14. The arithmetic mean between two numbers is 1, and the harmonic mean — 15. Find the numbers. 15. The fifth term of a harmonic progression is — , and the eleventh term — Find the fifteenth term. 3 16. The geometric mean between two numbers is 4, and the 1 fi ^ harmonic mean — Find the numbers. 5 17. The arithmetic mean between two numbers exceeds the geometric mean by -, and the geometric mean exceeds the harmonic mean by 6 ^ -;- Find the numbers. xo (Represent the sum of the numbers by x, and their product by y.) 18. Prove that, if any three consecutive terms of a harmonic progres- sion be taken, the first is to the third as the first minus the second is to the second minus the third. 19. If «2, 62^ and c^ are in arithmetic progression, prove that & + c, c -\- a, and a + 6 are in harmonic progression. PROGRESSIONS 361 20. If a, 6, and c are in arithmetic progression, 6, c, and d in geomet- ric progression, and c, d, and e in harmonic progression, prove a, c, and e in geometric progression. 537. Let A, 6r, and H denote the arithmetic, geometric, and harmonic means, respectively, between a and 6. Then, by §§ 523, 532, and 536, « + & ri_^/:T. o^.i 7j-_ 2a5 . z But, ^x^^ = a6=(V^)^. a + & Whence, ^ X 11= G\ or G = V^ X H. That is, ^/ie geometric mean between two numbers is also the geometric mean between their arithmetic and harmonic means. 538. Let a and b be two positive real numbers. By § 537, their geometric mean is intermediate in value between their arithmetic and harmonic means. But c^ + ?> 2ab ^ (a-{-by-4:ab ^ (a-b)\ ' 2 a + b 2(a + b) 2(a4-&)' a positive number. Hence, of the three means, the arithmetic is the greatest, the geometric next, and the harmonic the least. 862 ADVANCED COURSE IN ALGEBRA XXVI. OONVERGENOY AND DIVERGENCY OF SERIES 539. An Infinite Series (§ 283) may be developed by Divi- sion, or by Evolution. Let it be required, for example, to divide 1 by 1 — a?. 1-x X x — x^ The quotient is obtained in the form of the infinite series l + x+ic^H . Again, let it be required to find the square root of 1 + ic. l^x 1 ^2 -8^. 2 + X + ^ 2-i-x- x' The result is obtained in the form of the infinite series ^2 8 Infinite series may also be developed by other methods, one of the most important of which will be considered in Chap. XXVII. 540. Convergent and Divergent Series. An infinite series is said to be Convergent when the sum of the first n terms approaches a fixed finite number as a limit (§ 245), when n is indefinitely increased. CONVERGENCY AND DIVERGENCY OF SERIES 363 This limit is called the Value of the Series. A finite series may be regarded as a convergent series. An infinite series is said to be Divergent when the sum of the first n terms can be made numerically greater than any assigned number, however great, by taking n sufficiently great. Consider, for example, the infinite series developed by the fraction (§ 539). I. Suppose X = Xi, where x^ is numerically < 1. In this case, the given series is a decreasing Geometric Pro- gression ; and by § 530, the sum of the first n terms approaches the limit when n is indefinitely increased. J. iCj That is, the sum of the first n terms approaches a fixed finite number as a limit, when n is indefinitely increased. Hence, the series is convergent when x is numerically < 1. Let us take, for example, x = .1. The series now takes the form 1 + .1 + .01 + ... ; while the value of the fraction from which the series was developed is — = — , or — . 1 - .1' 9 In this case, however great the number of terms taken, their -sum never exactly equals — , but approaches this value as a limit (§ 530). y Thus, if an infinite series is convergent, the value of the series (§ 540) equals the value of the expression from which the series was developed. II. Suppose X = fl^i, where x^ is numerically > 1. The sum of the first n terms is now r"- 1 + a^ + a^i^ + - + xr"^ = ^i^ (§ 143), tL^ By taking n sufficiently great, -^ can be made to numer- al— 1 ically exceed any assigned number, however great. Hence, the series is divergent when x is numerically > 1. Let us take, for example, x = 10. The series now takes the form 1 + 10 + 100 +"•••; while the value of the fraction from which the series was developed is , or — -• ^ 1-10' 9 364 ADVANCED COURSE IN ALGEBRA In this case, the greater the number of terms taken, the more does their sum diverge from the value . Thus, if an infinite series is divergent, the greater the number of terms taken, the more does their sum diverge from the value of the expression from which the series was developed. III. Suppose x = l. In this case, each, term of the series equals 1, and the sum of the first n terms equals n; and this sum can be made to exceed any assigned number, however great, by taking n suffi- ciently great. Hence, the series is divergent when x = l, IV. Suppose x = —l. In this case the series takes the form 1-1+1-1 + ...; and the sum of the first n terms is either 1 or according as n is odd or even. If the sum of the first n terms of an infinite series neither approaches a fixed finite limit, nor exceeds any assigned num- ber, however great, when n is indefinitely increased, the series is called an Oscillating Series. Hence, the series is an oscillating series when x== — l. 541. It follows from § 540 that an infiiiite series cannot he used for the purposes of demonstration unless it is convergent. It will be understood, throughout the remainder of the work, that, in every expression involving a convergent infinite series, the value of the series is meant. For example, the product of two convergent infinite series will be understood as signifying the product of their values. THEOREMS ON CONVERGENCY AND DIVERGENCY \ OF SERIES , ; ;. < -' ' '^ 542. If an infinite series is convergent, the last term approaches the limit zero, when the number of terms is indefinitely increased. Let the series be Wi + 1^2 H h ^^« + ^^«+i H • By § 540, ? 1, and divergent when k = l or /c < 1. I. If A; is > 1, the second and third terms are together < — -f— , or — ; the next four terms are together <— ; the next o eight are together < — ; and so on. Then, the series is less than the series 1|2,4,8, i,l,l.i_i_ or l + 2lri + (2^.J+(2^)+-; which was proved, in § 530, to approach a finite limit when the number of terms was indefinitely increased. Therefore, the given series is convergent. II. VLk = \, the series becomes 1 + 1 + 1 + 1 + .... (1) 2 3 4 ^^ 368 ADVANCED COURSE IN ALGEBRA o -t The third and fourth terms are together > - or - ; the next 4 1 4 2 four terms are together > - or - ; and so on. 8 2 Then, by taking a sufficiently great number of terms, their sum may be made greater than any assigned number, however great, and the series is divergent. The series (1) is a harmonic series (§ 534). III. If A; is < 1, each term of the given series, commencing with the second, is greater than the corresponding term of (1), and the series is divergent (§ 547). As examples of the above general theorem, the series 22 32 42 is convergent ; and the series a/2 VS V4 IS divergent. 549. If, commencing with a certain assigned term, each term of a series of positive terms is greater than some assigned finite number, however small, the series is divergent. For the sum of the first n terms can be made to exceed any assigned number, however great, by taking n sufficiently great. Thus, in the series - + - + ? + 7 + ---, 2 3 4 5 each term, commencing with the second, is greater than — Then, the series is divergent. 550. If, in two series of positive terms, the ratio of two corre- sponding terms is ahcays finite, the first series is convergent if the second is convergent, and divergent if it is divergent. Let the series be Ui-\-U2-{-u^-{- "-, and Vi4- ^2 + ^3 + •••• I. Let the second series be convergent. Let A: be a finite number greater than the greatest of the ratios — , — , •••; then, —k, ^>A;, ..., -">fc. Then, Ui > kv^, ii^ > ^^2} -•-, '^n> ^'^n- Whence, (wi + ^2 H h w„) >k(t\ + V2-\ h v^). Now, since the second series is divergent, Vj + VgH- ••• + v„ can be made greater than any assigned number, however great, by sufficiently increasing n. Then, Wi + Wg + hu^ can be made greater than any assigned number, however great, by sufficiently increasing n, and the first series is divergent. 1. Prove the series 1 1 [- ••• convergent. 1x2 2x3 3x4 We compare the series with the series which was shown to be convergent in § 548. The ratio of the nth. term of this to the nth. term of the given series is ^- — , or ^^^ "^ , or 1 H — 1 n^ n n(n + 1) This is always between 1 and 2, and is therefore finite for every value of n. Then the given series is convergent. 370 ADVANCED COURSE IN ALGEBRA 2. Prove the series 1 1 h ••• divergent. 1x2 3x4 5x6 We compare it with the series 1 + 1 + 1 + -, (1) which was shown to be divergent in § 540, III. The ratio of the wth term of the given series to the nth term of (1) is n' ^^ 1 271(2.^-1/'''' 2(^2-1^ This is always between - and -, and is therefore finite for every value of n ; then the given series is divergent. '^ 551. If, commencing with a certain assigned term, the ratio of each term of an infinite series of positive terms to the preceding term is numerically less than a fixed positive number, which is itself less than unity, the series is convergent. Let the series be 2^^ + Wg + ^3 + •*•• • (1) Suppose —Kk, —2, the 2 ^ ratio is < -• o Then, commencing with the fourth term, the ratio of each 2 term to the preceding is < -, and the series is convergent. -^ 552. If, commencing with a certain assigned term, the ratio of each term of an infinite series of positive terms to the preced- ing term is either equal to, or greater than, unity, the series is divergent. For, commencing with the assigned term, each term of the series is either equal to, or greater than, the assigned termj and the series is divergent by § 549. 2 2^ 2^ Ex. Prove the series + 1 1 divergent. 1x2 2x3 3x4 The ratio of the (71 + l)th term to the Tith is 2n+l (n + l) (7. + 2) ^^ 271 2" 71 -f- 2 n (n + 1) If n = 2, the ratio equals 1 ; and for any value of n>2, the ratio is > 1. Then, commencing with the third terra, the ratio of each term to the preceding is equal to, or greater than, 1, and the series is divergent. 553. The method of § 551 does not apply when the ratio of the (n + l)th term to the nth is less than 1, but approaches the limit 1 when n is indefinitely increased. 372 ADVANCED COURSE IN ALGEBRA For in this case the ratio will not always be less than a fixed positive number, which is itself less than 1. In such cases, the convergency or divergency of the series must be determined by other tests. Consider, for example, the series 1 + 1 + 1+1+.... The ratio of the (n + l)th term to the nth is -J— 2-1 2n + l 2n — 1 n 2»+l 2+1 n This is always < 1, but approaches the limit 1 when n is indefinitely increased. The series is divergent, which may be shown as in § 548, II. Q 554. If an infinite series of positive terms is convergent, it will , remain so after the signs of any of its terms have h^en changed. For the sum of the first n terms will still be finite, however great n may be, and the series is convergent by § 545. It follows from the above that the theorems of §§ 546 and 551 hold when any or all of the terms are negative ; and also the first statement of § 550 when all the terms are negative. 555. It follows from §§ 551, 552, and 554 that I. If the ratio of the (n + V)th term of an infinite series to the nth term approaches a limit numerically < 1, when n is indefi- nitely increased, the series is convergent. 2 Thus, in the example of § 551, • approaches the limit 0, when n is indefinitely increased. < Then, the series is convergent. II. If the ratio of the (n -f l)th term of an infinite series to y the nth term approaches a limit numerically >1, ivhen n is indefinitely increased, the series is divergent. / CONVERGENCY AND DIVERGENCY OF SERIES 373 Thus, in the example of § 552, the ratio of the (n + l)th term to the nth term can be written ; and this approaches the limit 2 when n is indefinitely increased. ^ '^ ^ Then, the series is divergent. If the ratio of the (w + l)th term to the nth term approaches the limit 1, when n is indefinitely increased, other tests must be applied to deter- mine whether the series is convergent or divergent (§ 553). 556. If the terms of an infinite series are alternately positive and negative, and each term numerically less than the preceding term, the series is convergent if the nth term approaches the limit when n is indefinitely increased. Let the series be ii-^^ — U2 -[- u^ — Uj^+ • • •. It may be written in the forms {u^ - U2) -f- (us - W4) + (% - -Me) + •", (1) and u-L — {u2 — u^ — (u^ — u^) • (2) Since each, of the expressions Ui — u^, u^ — u^, etc., is posi- tive, it is evident from (1) and (2) that the sum of the first n terms is positive, and < u^. That is, the sum of the first n terms of the given series is finite, however great n may be. The sum of the m terms commencing with the {n + l)th is the upper or lower sign being taken before u^+m according as m is odd or even. The expression in parentheses can be written in the forms and U^+i - K+2 - Un+s) - (^n+i — Wn+s) + ' ' ' ; which shows that it is positive and < w„+i. Now, if n is indefinitely increased, ?<„+i, and therefore expression (3), approaches the limit 0. Thus, the sum of any finite number of terms commencing with the (7i4-l)th approaches the limit when n is indefinitely increased, and the series is convergent by § 543. 374 ADVANCED COURSE IN ALGEBRA As an example of the above theorem, the series 1-1 + ^-1+... 2 3 4^ is convergent. "*i 557. To prove the infinite series convergent when x is numerically < 1. The ratio of the (n + l)th term to the nth. term is 71 71 — 1 n \ ^, which approaches the limit x when n is indefinitely increased. Then, if x is numerically < 1, the series is convergent (§ 555, I). 558. To prove the infinite seyies convergent for every value of x. a?" The ratio of the (n + l)th term to the nth. is \n \n-l which approaches the limit when ^i is indefinitely increased. Then, the series is convergent (§ 555, I). \ 559. To prove the infinite series [2 [3 lohere n is any positive fraction, or negative integer or fraction, convergent when x is nuTnerically < 1. By § 287, the ratio of the (r + l)th term to the rth term is n(n-l)^"{n-r-\-V) ^, _^ y?(n - 1) ■■■(n ~r + 2) ^.^^ \r ' \ r — l CONVERGENCY AND DIVERGENCY OF SERIES 375 n — r-\-l^ fn-\-l That is, ■ — X, or -1^. r If n does not equal — 1, this approaches the limit — x when ' r is indefinitely increased ; and the series is convergent if x is numerically < 1 (§ 555, I)., If n = — 1, the series takes the form 1 — x-{-x' — x^ -\ , . which is convergent when x is numerically < 1 by § 556. 560. Tlie infinite series a + bx-\-cx^-{-dx^-] ^kx^-^ + laf -\- •" is convergent when the numerical value of x is taken sufficiently small, and for ariy numerically smaller value of x, including zero. Ix^ Ix For the ratio of the (n + l)th term to the nth is z— ^, or — • Whatever the values of k and I, x may be taken so small that — shall always be numerically < g, where q is numerically < 1. k Then, for this value of x the series is convergent. And by § 546, it is convergent for any numerically smaller value of X, including 0. 561. If X has any value (not including zero) which makes the series a + bx + cx' -{- ... (1) convergent, a + &a; + caj^ + ..., and therefore bx + cx^ -\ , or x(b-\-cx-^ '"), is finite, for this value of x. Then, since x is not zero, 6 + ccc + ••• is finite for this value of X ; and x(b-{-cx-\ ), or 5ic + ca^ H approaches the limit when x is indefinitely decreased. Since 6 -f ca; -f- • • • is finite for any value of a;, except 0, which makes series (1) convergent, it is itself convergent for this value of X (§ 545). 562. Absolutely and Conditionally Convergent Series. A convergent infinite series, having negative terms, is said to be absolutely convergent when it remains convergent after the signs of the negative terms are changed. 376 ADVANCED COURSE IN ALGEBRA It is said to be conditionally convergent when this is not the case. Thus, the convergent infinite series (§ 556) if the signs of the negative terms are changed, becomes 1 + 1.1-1.... which was proved divergent in § 548. Then, the series (1) is conditionally convergent. In a conditionally convergent series, the sum of the terms approaches a different limit by a different arrangement of the terms. Thus, in series (1), we may write the terms ,0.5. ('-M)-(i-r(i-l)--- which shows that the sum of the terms is <(1~q + q) l^Q may also write the terms M-i)+G-l-i)H^n-j)-- which shows that the sum of the terms is > - • y Then the terms of a series cannot be arranged in any desired \ order unless the series is absolutely convergent. EXERCISE 84 Expand each of the following to four terms : *l + 2ac '3-6x2 + 7x3 1 — 5 x2 7. 8. 9. \/9 a* - 3 62. 2. ■ . . -- . l-5x-2x2 3 ^^ 6. Vl + 3x. \/8x8-l. 6. Vx2 + x2/ + y2. y/a^ + 2 &. 2 + 6 X - x2 CONVERGENCY AND DIVERGENCY OF SERIES 377 Determine whether the following infinite series are convergent or divergent : 10, 1 + 1 + 1 + 1 + .... (oU 17 1 + A + _!!_+... 13. l + 2! + 33 l+2_3 3^ 43 22^32 42^ 2.3 3.4 4.5 15. 1 + L^ + L^5 ._ 22 l__l„4._i !_+.... 33. 63. 6- 9 3 2.32 3.33 4.34^ 16. 1 + 1 + 1+.... 23. ? + -§_ + -^ + -A. + .... 3^6^9^ 1^1.2 2.3 3.4^ Determine for what values of x each of the following infinite series is convergent or divergent : 24. 1 + 22X + 32^:2 + .... 25. 1+-^ + -^ + -^ + .... 26. Prove the infinite series 1 +^JL_ + ,i +... 1 + X 1 + a:2 1 + x3 convergent when a: is > 1, and divergent when x — \ or aj < 1. 27. Prove the infinite series _1_ + -JL- + __^ + ... 1 + x l + a;2^1+x4 convergent for every value of x. In determining the convergency or divergency of a series, it is usually best to commence with the tests of § 655 ; if the limit approached is 1, then other methods may be tried. 378 ADVANCED COURSE IN ALGEBRA XXVII. UNDETERMINED COEFFICIENTS THE THEOREM OF UNDETERMINED COEFFICIENTS 563. An important method for expanding expressions into series is based on the following theorem : Let the members of the equation A + Bx+ C:»? + Dx^ + '" = A + B'x + Ox" + Dy? ^- '" (1) be injinite series which are equal for any value of x which makes them both convergent ; then^ the coefficients of like powers ofx in the series will be equal; that is, A = A', B = B', C=a,etc. By § 560, each series is convergent for a certain finite value of X, and for all smaller values ; and since this is the case, each of the expressions Bx-j-Gx"-^-" 3indB'x+C'x'+'" approaches the limrt when x is indefinitely decreased (§ 561). Then, the given series approach the limits A and A', respec- tively, when X is indefinitely decreased. Now, since the given series are functions of x, which are equal for every value of x which makes them both convergent, their limits when x is indefinitely decreased are equal (§ 252). Therefore, A = A\ and hence Bx-{-Cx^-\ = B'x -f Ox^ -\ . Since x is not 0, we may divide through by x. Then, B+Gx+"' = B^ + ax + ": (2) By § 561, each member of (2) is convergent for the same values of x as the corresponding member of (1). That is, (2) is satisfied by any value of x which makes both members convergent; and letting x be indefinitely decreased, we have j^ ^ ^r UNDETERMINED COEFFICIENTS 379 Proceeding in this way, we may prove C = C\ etc. The above proof holds when either or both of the given series are finite. 564. 1. Expand EXPANSION OF FRACTIONS in ascending powers of x. I_2a; + 3x2 We saw, in § 539, that a fraction could be expanded into a series by dividing the numerator by the denominator. We therefore know that the proposed expansion is possible. Assume then 2 _ 3 a;2 - a;3 = A+Bx+Cx'-{-D3? + Ex'' + (1) l-2i« + 3a;2 where A, B, C, D, E, -", are numbers independent of x. Clearing of fractions, and collecting the terms in the second member involving like powers of x, we have 2-3x'-a^ = A-{- B x^ G x'-{- D a^+ E -2B -2C -2D + 3^ -{-SB + 30 ^+ (2) A vertical line, called a bar, is often used in place of parentheses. Thus, + ^ I a; is equivalent to (^ — 2 A) x. -2a\ Equations (1) and (2) are satisfied when x has any value which makes the second member a convergent series. Then, by § 563, the coefficients of like powers of x in (2) must be equal ; that is, A= 2. B-2A= 0; or, B = 2A =4. (7_2B + 3^ = -3; or, = 2 5-3 ^-3 = -1. i)_2 0+3 5 = -l; or, 2) = 2 0-3J5-l=-15. ^_2D + 30= 0; OT,E=2D-3C =-27; etc. Substituting these values in (1), we have 2-Sx'-x^ 2^4.x-a^-15a^-27x'--., The result may be verified by division. 380 ADVANCED COURSE IN ALGEBRA The series expresses the value of the fraction only for such values of x as make it convergent (§ 540). If the numerator and denominator contain only even powers of x^ the operation may be abridged by assuming a series con- taining only the even powers of x. Thus, if the fraction were 2 + 4a^-a;^ we should assume it equal to ^ + B:^ + Cx'' + Dx^ + Ex^ + .... In like manner, if the numerator contains only odd powers of £17, and the denominator only even powers, we should assume a series containing only the odd powers of x. If every term of the numerator contains ic, we may assume a series commencing with the lowest power of x in the numerator. If every term of the denominator contains a;, we determine by actual division what power of x will occur in the first term of the expansion, and then assume the fraction equal to a series commencing with this power of a?, the exponents of x in the succeeding terms increasing by unity as before. 1 2. Expand 3a^-a^ in ascending powers of x. Dividing 1 by 3 a^, the quotient is "We then assume, Z^-^ ^^^ -(— JL_/a/ ^^ ^ n — j^ •^ T- ^-^ Clearing of fractions. 1=3^+35 0^ + 3(7 x' + SD o^ + 3E - A - B - G - D Equating coefficients of like powers of a;. 3^ = 1 35-J. = 3(7-5 = 0, 3i)-C = ' 3E-1 D = et c. (3) x* + UNDETERMINED COEFFICIENTS ' 881 mence, ^ = |, B = |, C^l^, i> = i, E^^^,etc. Substituting in (3), we have JL «c I "^ III ^^ I 3aj2-aj3 3 ' 9 ' 27 ' 81 ' 243 In Ex. 1, E = 2D — 3C] that is, the coefficient of x* equals twice the coefficient of the preceding term, minus three times the coefficient of the next but one preceding. It is evident that this law holds for the succeeding terms ; thus, the coefficient of a;^ is 2 x (- 27) - 3 x (- 15), or - 9. After the law of coefficients has been found in any expansion, the terms may be found more easily than by long division ; and for this reason the method of § 564 is to be preferred when a large number of terms is required. The law for Ex. 2 is that each coefficient is one-third the preceding. EXERCISE 85 Expand each of the following to five terms : l+4:x-x^ 9 x^-Sx l-x + 3x2* * 3-x-2x3 "" 2x^ + x* -x + Sx^ jQ 2 - 3 x2 j^ 34-5X- 1+2x2 * ■ 3-2x + x3' ' x2-3x3 + x* 1 + 4x2 ^^ 2 + X - 3 x8 jg l-4xg + 6x8 1 + 4 X - x2* ■ 1 - 4 X + 5 x2' * X + 2 x2 - x8 * a; - 7 x^ - 4 x^ ^2 3-4xg ^g l-4x + 2x« 6 x2 "■ 1 _ 5 X - 2 x2 * ' 2 + x2 - 3 x3* * 2 x^ - 3 x» - x^* EXPANSION OF SURDS 1-x l-6x l + 4x 4 + x2 1 - 3 x2 2x 565. Ex. Expand Vl — a; in ascending powers of x. We saw, in § 539, that the square root of an imperfect square could be expanded into a series by Evolution. We therefore know that the proposed expansion is possible. Assume then, VT^^ = A + BX+ Cx" -\- Dx^ -i- Ex^ -{- .... (1) 382 ADVANCED COURSE IN ALGEBRA Squaring both members, we have by § 134, l-x = A' + 2AB x-\- B^ + 2AC x" + 2AD + 2BC Equating coefficients of like powers of x, A'= 1; or, ^ = 1. 2AB = -1; or, B = - B'-{-'2AC=: 0; or, C=- 2AD-^2BG= 0;or, X> = - C' + 2AE-\-2BD= 0; or, ^ = - 2 A Substituting these values in (1), we have 2 8 16 128 The result may be verified by Evolution. xl+ 0' -{-2AE + 2BD »* + 1 2A 1 2* B' 2 A 1 8' BG A 1 16* G' + 2BD 128' etc. The series expresses the value of Vl — a; only for such values of x as make it convergent. EXERCISE 86 Expand each of the following to five terms : (2 vi^vV 1. VI - 4 a;. 2. V1 + 5X. 3. Vl + 2 X - QoK 4. Vl - a; + 3 ic2. 5. VI + 3 a;. 6. Vl - X - 2 x2. PARTIAL FRACTIONS 566. If the denominator of a fraction can be resolved into factors, each of the first degree in x, and the numerator is of a lower degree than the denominator, the Theorem of Undeter- mined Coefficients enables us to express the given fraction as the sum of two or more imrtial fractions, whose denominators are factors of the given denominator, and whose numerators are independent of x. UNDETERMINED COEFFICIENTS 383 567. Case I. No factors of the denominator equal. 1. Separate ' ^"*" into partial fractions. ^ (3a;-l)(5a; + 2) Assume -—- ^ = z r + p — r^' (■•■) where A and B are numbers independent of x. Clearing of fractions, we have l^x-{-l = A(hx + 2) + B(Zx-l) = {^A + ^B)x^2A-B. (2) The second member of (1) must express the value of the given fraction for every value of x. Hence, equation (2) is satisfied by every value of x] and by § 563, the coefficients of like powers of x in the two members are equal. That is, 5^ + 35 = 19, and 2^- J5 = l. • Solving these equations, we obtain A = 2 and B = 3. Substituting in (1), we have 19x + l _ 2 _^ 3 (3a;-l)(5a5 + 2) 3a;-l 5x + 2 The result may be verified by finding the sum of the partial fractions. ' 2. Separate ^-t — — into partial fractions. ^ X — X^ — 3/ The factors of 2x — x^ — x^ are x, 1 — x, and 2 + a;. (a 4.V a; + 4 A , B , C Assume then -^ = - + z h tt-, — I 2x — Qif — x^ X 1 — x 2 + x Clearing of fractions, we have x-\-^ = A(l-x)(2-j-x)-{-Bx(2 + x)-{-Ox(l-x). This equation, being satisfied by every value of x, is satisfied when x = 0. /f IV 384 ADVANCED COURSE IN ALGEBRA Putting ic = 0, we have 4 = 2 A, or A = 2. Again, the equation is satisfied when x = l. Putting a; = 1, we have 5 = 3 B, or B = — The equation is' also satisfied when x = — 2. Putting x = — 2, we have 2 = — 6(7, or C= o 5 1 mt. ^ + 4 2 3 3 Then, o^ ^ ^ = " + q z + a; 3(1-0?) 3(2 + a;) The student should compare the above method of finding A and B with that used in Ex. 1. 568. Case II. All the factors of the denominator equal. ^2 11 £c -I- 26 Let it be required to separate — ^ — • into partial fractions. ^^ ~ ^) Substituting ^ + 3 for x, the fraction becomes (y + 3)^-ll(y + 3) + 26 ^ i/^-5y + 2 ^1 5 ^ 2 f f y y^ t Replacing 2/ by aj — 3, the result takes the form 1 5 2 a;_3 {x-ZY {x-Zy^' This shows that the given fraction can be expressed as the sum of three partial fractions, whose numerators are indepen- dent of X, and whose denominators are the powers of a; — 3 beginning with the first and ending with the third. Similar considerations hold with respect to any example under Case II; the number of partial fractions in any case being the same as the number of equal factors in the denomi- nator of the given fraction. ^x. Separate — — ^ into partial fractions. \0 X -\- Oj UNDETERMINED COEFFICIENTS 385 In accordance with the above principle, we assume the given fraction equal to the sum of two partial fractions, whose de- nominators are the powers of 3 a? + 5 beginning with the first and ending with the second. That IS, — — ^'^— - = -f (3i» + 5)2 3x + b (3a; + 5)2 Clearing of fractions, 6a?-f-5 = ^(3a; + 5)+5 JEquating coefficients of like powers of x, SA = 6, and 5^4-^ = 5. Solving these equations, A = 2 and B = — 5. Whence 6a;4-5 ^ 2 o (3x-j-5y 3a; + 5 (3x-{-5y 569. Case III. Some of the factors of the denominator equal. Ex. Separate — - — — — into partial fractions. x{x-{- ly The method in Case III is a combination of the methods of Cases I and II ; we assume xix^lf X x-\-l (x+Vf {x + 1) Clearing of fractions, 3x + 2 =: A(x -^If + Bx(x^iy+ Cx{x -{-!)+ Dx = (^ + 5)i»3 + (3^ + 2 jB + (7)a^ -^{3A-\-B-\~C+D)x^-A. Equating coefficients of like powers of x, A + B^O, 3^ + 2^4-0 = 0, 3J. + J3+C + 7) = 3, and A = 2. 386 ADVANCED COURSE IN ALGEBRA Solving, we have A = 2, B = -2, C=-2, and D = l. Substituting in (1), 3x + 2 _2 2 2 1 x{x-^iy X x + 1 (a; + 1)2 {x + Vf The following general rule for Case III will be found convenient : A fraction of the form should be assumed equal to (x + a){x+hy.-ix + my... . A B E F T^ + — ^+--+-^^^+ . .. + ••• + x-{- a x-\-h X -\- m {x + my (x + my Single factors like x -\- a and x -i-b having single partial fractions cor- responding, arranged as in Case I ; and repeated factors like (x + my having r partial fractions corresponding, arranged as in Case II. EXERCISE 87 Separate the following into partial fractions : 75 g 5x2 + 4a;-l 1. Sx-6 5 4x2-9 2. 19 a; - 30 5 x2 - 6 a; 6. ' 3. 2 a; -10 (2x-3)2 7. • 4. 2 a;2 + 15 a: + 34 8 {x + 5y ^ 62 a; - 38 a:3-25x (5x + 2)8 9a;2-15a; ^^ 5 ax2 - 2 a'^x -Sa^ (Sx-iy ' x^ + Sax^-4a^x' 43x + 6 J, J 4x2-22x4-63 6x2 + 5x-6 x(x-3)2 8x2 j2 42-27X 8 - 14 X + 5 x2 2x^+ 19x2-17x + 12 _ (2x-l)(12x2-x-6) "" x4-|-4x3 16 a;8 - X - 4 ^g 18x-4V2 ^^ 9x2-9x-18 (2x- -3)3 14. ■ 18 X- -4V2 (x-l)4 9x2-f6x-l (x2 - 2 x) (x2 - 9) 18 14x2+lla;-f29 21. ^x^ - 24 xMi20x--4 19. (3x-l)(2x-h3)2 (3x-2)* x3 + 14x2-1- 9 X oo 2 x3 -f 12 x2 -h 12x+i (x4-2)4 x(x+l)(x-f-2)2 2Q x3-3x2-7x-f4 23 19 x - 32 x2(x-l)2 ' ' (2x-3)(8x2- lOx-3)' 570. If the degree of the numerator is equal to, or greater than, that of the denominator, the preceding methods are inapplicable. UNDETERMINED COEFFICIENTS 387 In such a case, we divide the numerator by the denominator until a remainder is obtained which is of a lower degree than the denominator. Ex. Separate — into an integral expression and partial fractions. ~ We have, by division, — ^ ^^— =x — 2-\ . (1) x^ — x x^—x 2ic — 1 We can now separate into partial fractions by the Ou — X -I Q method of Case I; the result is -• X X— 1 Substituting in (1), — ^ ^— = x — 2-\ . Olf — X X X — 1 Another way to solve the above example is to combine the methods of §§ 664 and 667, and assume the given fraction equal to C . D Ax + B + X X — 1 EXERCISE 88 Separate the following into integral expressions and partial fractions : J 9 a;3 + 21 x2 + 22 a; - 17 ^ 2x^-2 x^ -3 x"^-]-! (a: + 2)(3a;-l) * * x^ - x^ 2 4 a;3 - 26 a;2 + 20 X - 1 _ ^ 6x^ + 6x^-2x^ + 6x + l (X-2)3 * • ic2(^X + l)3 5 x6 + 15 x5 4- 3 x* - 14 x2 - 18 5. X* + 3 x3 571. If the denominator of a fraction can be resolved into factors partly of the first and partly of the second, or all of the second degree, in x, and the numerator is of a lower degree than the denominator, the Theorem of Undetermined Coeffi- cients enables us to express the given fraction as the sum of two or more partial fractions, whose denominators are factors of the given denominator, and whose numerators are inde- pendent of X in the case of fractions corresponding to factors of the first degree, and of the form Ax-\-B in the case of fractions corresponding to factors of the second degree. 388 ADVANCf:D COURSE IN ALGEBRA The only exceptions occur when the factors of the denorai- nator are of the second degree and all equal. Ex. Separate — — - into partial fractions. The factors of the denominator are x-\-l and x' — x-\-l. Assume then ^^— = -^ + ^^ + ^ . (1) Clearing of fractions, we have l = A{x'-x + l) + {Bx + C){x^-l) = {A + B)x' + {-A + B + C)x + A-^a Equating coefficients of like powers of a?, A + B = 0, ■^^ + 5+0 = 0, and ' A-\-G=l. 1 1 2 Solving these equations, ^ = q? -^ = ~q' ^^^ ^~q' o o o Substituting in (1), ~ 4-1 3(a) + l) Six'-x + l) EXERCISE 89 Separate the following into partial fractions 1. 2. 3. 6 x2 + 5 X + 2 5 a;2 + a^ - 14 4. 5. 6. 4:X^~6x^ + ex-\-S x*-l 22X-6 (3x + l)(a:2_x + 3) 3 a:2 - ic + 25 2 a^3 _ 5 a;2 + 4 a; _ 10 8 a;3 - 27 4x^+lSx + S ic^ + 3 a;2 + 4 572. We will now show how to find an expression for the nth term of the expansion of a fraction in ascending powers of Xy when the denominator can be resolved into binomial factors of the first or second degree in x. Ex. Find the nth term in the expansion of the fraction l + 7a; . ,. p :, ~ -—: m ascending powers of x. l-j-2 x^3x^ UNDETERMINED COEFFICIENTS 889 Separating the fraction into partial fractions by the method of § 667, we have l + 7aj 2 1 l + 2a;-3a^ 1-x l + 3a? (1) By division, -^ = 2(l + x-{-a^-{-a^+ -"), (2) J. ~~~ 35 and \ =i_3a; + 3V-3V+-. (3) J- "T" O 3/ The nth term of (2) is 2a;'*~^, and the nth. term of (3) is (_3)»-ia;«-i. Subtracting, the nth term of (1) is 2 x^-i - (- 3)"-ia;«-S or [2 - (- 3)"-i] a;"-^ ITie above may be used as a formula to obtain the successive terms of the expansion. If w = 1, the expression becomes (2 — l)x'', or 1. li n = 2, the expression becomes (2 + 3)aj, or 5 x. If n = 3, the expression becomes (2 — 9)x^, or — 7 x^ j etc. Then, l±l^—=l + 6x-7x^ + ".. 1 + 2 X - 3 x2 EXERCISE 90 In each of the following, find the nth term of the expansion of the fraction in ascending powers of x : 1 3 ^ 5 a; -12 l-x-2a:2 ex2 + 5x-6 2 2x-l g 7 4- 4 x - x2 ' 1 - 9 X + 20 a?* ' 6 + 3 X - 4 x2 _ 2 x3* 8 6 + 11 a; g 1 + 5 x -f 6 x^^ ' 2 + 5x-3x2' • (l_a;)(l+x2)* In Exs. 5 and 6, it should be observed that there are two forms for the nth term according aa n is even or odd. REVERSION OP SERIES 573. To revert a given series y = a-\- baf* -f ca;" + ••• is to ex- press X in the form of a series proceeding in ascending powers of 2/. 390 ADVANCED COURSE IN ALGEBRA Ex. Eevert the series 2/ = 2aj — 3x^-1-40^ — 5x*H — Assume, x = Ay-{- By^ -f Cy^ + Dy^ H . Substituting in this the given value of y, + 5(4a;2 + 9ic*_12a^ + 16a;*4-...) + 0(8 a;3 - 36 o;^ +...)+ J9(16 a;^ +•••) + ■ a;2+ 4^ aj3- 5^ -12 5 + 25 5 + 8(7 -36(7 + 16i) That is, a; = 2^a;-3^ + 45 Equating coefficients of like powers of Xy 2A = 1 -SA+4B=0 4^-125 + 8(7=0 -5^ + 255-36 (7+16 2) = 0; etc. a7^+.. Solving, A = ^, B = j, C: Substituting in (1), x: 16' i> = ^,etc. 1 3 2,5_, 2^ "^8^ '^W 128^ (1) If the even powers of x are wanting in the given series, the operation may be abridged by assuming x equal to a series con- taining only the odd powers of y. Thus, to revert the series 2/ = a; — o^ + a^ — a;^ H , we assume x = Ay + Bf+Cf-hDf + '':. If the odd powers of x are wanting in the given series, we substitute t for x^, and revert the series, expressing t in ascend- ing powers of 2/ ; by taking the square root of this result, x itself may be expressed in ascending powers of y. If the first term of the given series is independent of x, it is impossible to express x in ascending powers of y, though it is possible to express it in the form of a series whose terms are functions of y. UNDETERMINED COEFFICIENTS 391 Thus, let it be required to revert the series ^ [2 |3 The series may be written 2/ — 1 = a? + — + — + •••. [2 [3 We then assume 05 = ^(2/ - 1) + J5(2/ - 1)2 + (7(2/ - 1)3 + Z>(2/ - 1)* + -. Proceeding as before, we find aj = (2^-l)-|(2/-l)2 + |(2/-l)«-J(2/-l)*+.... EXERCISE 91 Revert each of the following to four terms : 1. ^/zzzXTf 3x2 + 5a;3 + 7 a:* + •-. 2. ?/ = x- 2x2 + 3x3-4x4+ •••• />»2 /v<3 />*4 ^ 2 3 4 4. ?/ := 1 + 2 X + 5 x2 + 8 x3 + 11 X* + />• 0*2 'vS />»4 5. ^ = - - — + — --+ .... ^2 4^6 8 « ,, X , x2 , x^ . 7. y = 2x-4a;3 + 6x5-8x'' + ?! + ^_i-?! 13 [6 II a:+.^ + g+5! + 392 ADVANCED COURSE IN ALGEBRA XXVIII. THE BINOMIAL THEOREM ANY RATIONAL EXPONENT 574. It was proved in § 285 that, if ?i is a positive integer, (l + ^)- = l + na^ + ^^^^^^ + "^^-l^,^^-^> x3+ -... (1) In this case, the second member is a finite series of n-\-l terms. If n is a negative integer, or a positive or negative fraction, the series in the second member becomes infinite ; for no one of the expressions n — 1, n — 2, etc., can equal zero. We will now prove that in this case the series gives the value of (1 -f xy, provided it is convergent; this we know to be the case when X is numerically < 1 (§ 559). 575. Proof of the Binomial Theorem for Any Rational Exponent. If m and n are positive integers, (l+x)-^=l + mx-\- ^'(^ ~ ^^ a^ + -, (2) If and (l+xy = l + nx-{- '*^^'\'~^^ x^+ .... (3) But, (1 + x)"^ X (1 + xy = (1 4- a?)'""^". • Then the product of the series in the second members of (2) and (3) must equal the expanded form of (1 -f 0?)"^+" ; that is, = l+(m + n>+(^ + ^>^,^ + ^~^V+ .... (4) We proved the above result on the hypothesis that both m and n were positive integers. THE BINOMIAL THEOREM 898 But the form of the product will evidently be the same what- ever the values of m and n. Therefore, (4) holds for all rational values of m and n, pro- vided X is numerically <1; for in this case each series is convergent. (We assume that the product of two convergent series is convergent.) Now let the symbol /(m) stand for the' series ^ , , m (m — 1) o , for any rational value of m. Then, if m is a positive integer, /(m) = (1 + x)"^, (5) Then, by § 251, (4) may be written ■ /(m)x/(r0=/(m+7i), (6) which holds for all rational values of m and n. Then by (6), if p is also a rational number, f(m) X f(n) X /(p) = [/(m + n)] x /(p) =/(m + n H-p) ; etc. Thus, /(m) x/(n) x/(p) X ••• to r factors =f(m + w +p + • • • to r terms). (7) ISTow let m, ti, p, ••• be each equal to -, where q and r are positive integers; then (7) becomes [/(f)]=/(fx.)=/(,). But since ^ is a positive integer, f(q) =(l-\-xy. Then, [fg)j = (l + a!)'. Taking the rth root of both members, ' /N -f--0 (l + xX=/(?j = l + fx+^l^^a^ + ...; (8) which proves the theorem for a positive fractional exponent. The result is proved only for the case where x is numerically < 1. 394 ADVANCED COURSE IN ALGEBRA Again, in (6), let m = — w, where n is a positive integer or a positive fraction. Then, /(- 71) xf(n) =/(- n + n) =/(0). But/(0) stands for {1 + xf, which equals 1 (§ 359). Therefore, /( - n) = -J- • Now since w is a positive integer or a positive fraction, it follows from (5) and (8) that/(w) = (1 + x)\ Whence, f(—n) = ' ^v ; (l + iC)'* Then, which proves the theorem for a negative integral or negative fractional exponent. The result is proved only in the case where x is numerically < 1. 576. Putting - for x, in (1), § 574, \ aj a \2_ a^ ^ ^ Multiplying each term by a% we have (a + xY == a" + 7ia''-^x + ^^^ ""-*-) a'*- V + . . . . (lO) We know that the second member of (9) is convergent when- - is numerically < 1. a Hence, the second member of (10) is convergent when x is numerically < a. If x is numerically > a, we can expand (a + xy in ascending powers of a ; thus, (a; + ay = aj" + nx^'-^a -f ^' ^^ ~ -^^ a;"- V + • • • ; and this series will be convergent, since a is numerically < x, THE BINOMIAL THEOREM 395 577. Examples. In expanding expressions by the Binomial Theorem when the exponent is fractional or negative, the exponents and coefficients of the terms may be found by the laws of § 285, which hold for all rational values of the exponent. 1. Expand (a + x)^ to five terms. The exponent of a in the first term is -, and decreases by 1 in each succeeding term. The exponent of x in the second term is 1, and increases by 1 in each succeeding term. The coefficient of the first term is 1 ; of the second term, -. 2 1 Multiplying -, the coefficient of the second term, by , o o the exponent of a in that term, and dividing the product by the exponent of x increased by 1, or 2, we have — ^ as the coefficient of the third term ; and so on. Then, (a + x)^ = at -f ?a-^ a; - ^ oT^ x' + ^orix^- ^oT^^ ic^ + •... ^ ^ 3-9 81 243 It follows from § 576 that the series expresses the value of (a + xy only when x is numerically < a. ^ If X is numerically > a, (a + xY is equal to i , x^ + -x~^ a --x' ^ a^ -{■ —x'"^ a^ . 3 9 81 2. Expand (1 + 2a;~^)-2 to five terms. i. Enclosing 2x~^ in parentheses, we have (1+2 a;-^)-2 = [1 + (2 a;-^)]-^ = 1-2- 2 . 1-3 . (2«-^) +3.1-^. {2x-^Y - 4 . 1-^ . (2 x-'^f 4- 5 . l-« • (2 x-^y . = 1 _4 aj-i 4_ 12 x-^ - 32 ic"^ + 80 x-^ -\ . By writing the exponents of 1, in expanding [1 + (2a;~^)]-2, we can make use of the fifth law of § 285. 396 ADVANCED COURSE IN ALGEBRA The series expresses the value of (1+2 a;~2)-2 only when 2 x"^ is numerically < 1, or x~^ numerically < -• 1 2 That is, when x^ is numerically > 2, or when x is positive and > 4. 3. Expand ^ — to four terms. Enclosing a~^ and — 3 tc^ in parentheses, we have 3, ^ = ^^ = [(«-') + (- 3 x*)]-* = (a-')'* - \ («-'r^ (- 3 ^') + 1 (a->)'^ (- 3 x^f -|f(«-T^(- 3. *)» + ••• = a^ + aM + 2 a^a;^ + ^a^x -\ . The result expresses the value of the given fraction only when Zx^ is numerically < a-i ; if 3 x"3 is numerically > a-i, the fraction can be expanded in ascending powers of a-i. EXERCISE 92 Expand each of the following to five terms : 1. {a + xy. 2. (l + ^)-«. 8. (1 - xy. 4. ^a-x. 5 1 (a + 6)8 6. 1 7. {x^-2yy. 12. — ^— i ^. (m^ - 2 n~^Y 8. H+V 9. ia^-2xh-k ^' ""' 13. \h a) 10. 1 14. {x~^-Zyh'^. 15. f— 1 ^^2\l v^n^ 11. ^[(a:-3 + 6?/0)7]. \16\/^ / 678. The formula for the rth term of {a + xf (§ 287) holds for fractional or negative values of n, since it was derived from an expansion which has been proved to hold for all rational values of the exponent. THE BINOMIAL THEOREM 397 Ex. Find the 7th term of (a - 3 x'^y^. Enclosing — 3 x~^ in parentheses, we have (a - 3 x-^y^ = [a 4- (- 3 x-^)Y'^. The exponent of (— 3 x~^) is 7 — 1, or 6. 1 19 The exponent of a is 6, or • o o The first factor of the numerator is — -, and the last factor 19 , -, 16 ^ -^ + l,or--. The last factor of the denominator is 6. Hence, the 7th term _1 _4 _7 _10 _13 _16 3*3*3* 3 * 3 * 3 1.2.3.4.5.6 728 -Jta,Q6 -9N 728 - a-¥(_3a.-f)6 hr EXERCISE 93 Find the 1. 6th term of {a + xy. 7. 7tli term of (a* - sc^) 2. 5th term of (a + 6)""^. 8. 10th term of ^ 3. 12th term of (1 - a;) -5. ^^ ^ "*^^ 4. 7th term of (^-1 + 2 2/^)-2. 9. 8th term of ( m^ - 2 Q -J 5. 9th term of (a + 2 x)^. ^^^ ^^^^ ^^^"^ ^^ ^^^ " ^^'• 1 11. 6th term of {a^ - 4 6-2) ^. 6. 5th term of ^ 1 _2 V(l - a:)5 12. 8th term of (x'^ + 3 y~^y^. 13. Term involving xr^^ in f x Vw^ + -rrz I * _9 1 _2 '9 14. Term involving a -^ in (a^ — 4 & ^)*. 579. Extraction of Roots. The Binomial Theorem may sometimes be used to find the approximate root of a number, which is not a perfect power of the same degree as the index of the root. 398 ADVANCED COURSE IN ALGEBRA Ex. Find V25 approximately to five places of decimals. The nearest perfect cube to 25 is 27. We have ■^/25 = 1, a— = 4 = - = (§ 248). a 00 Whence, by § 580, log„0 = - oo. The above result must be interpreted as follows : If, in any system whose base is greater than unity, a number approaches the limit 0, its logarithm is negative, and increases without limit in abso- lute value. (Compare § 248.) 588. In any system whose base is less than unity, the logarithm of zero is infinity. For if a is <1, a"^ = ; whence, log^O = oo. This means that if, in any system whose base is less than unity, a number approaches the limit 0, its logarithm increases without limit. 589. In any system, the logarithm of a product is equal to the sum of the logarithms of its factors. ^ U C?/' ' /" Assume the equations a'' = m, a^ = n. '^ Then, by § 580, x = log« m, y — log,, n, X' '^ Multiplying the assumed equations, a" X a^ = mn, or a^^^ = mn. Whence, log„ mn = x-\-y = log^ m + log^ n. In like manner, the theorem may be proved for the product of any number of factors. 1^ ADVANCED COURSE IN ALGEBRA By aid of the above theorem, the logarithm of any positive integer may be found when the logarithms of its factors are known. Ex. Given log 2 = .3010, and log 3 = .4771; find log 72. log72 = log(2x2x2x3x3) = log2 + log2 + log2 + log3 + log3 = 3xlog2 + 2xlog3 * = .9030 + .9542 = 1.8572. 590. In any system, the logarithm of a fraction is equal to the logarithm of the numerator minus the logarithm of the denominator. ^ Assume the equations a"" = m, a^ = n. Then, x = log„ m, y = log„ n. a^ m> 771 Dividing the assumed equations, — = — , or a*~^ = — a^ 7i' n Whence, log,, ~-z=x — y = log„ m — log^ n. Ex. Given log 2 = .3010; find log 5. log 5 = log i^ = log 10 - log 2 = 1 - .3010 = .6990. 591. In any system, the logarithm of any power of a quantity is equal to the logarithm of the quantity multiplied by the exponent of the power. Assume the equation a^'^m-, whence, a? = log„m. Raising both members of the assumed equation to the ^th power, (jfPx _ ^^p . -whence, log„ (m^) =px=p log„ m. 592. In any system, the logarithm of any root of a quantity is equal to the logarithm of the quantity divided by the index of the root. For, log« -Vm = log„ (m^-) = - log„ m (§ 591). LOGARITHMS ^^S - 403 593. Examples. 1. Given log 2 = .3010; find \og2i log 2* = ^ X log 2 = I X .3010 = .5017. o o To multiply a logarithm by a fraction, multiply first by the numerator, and divide the result by the denominator. 2. Given log 3 =.4771; find log ■y'S.^ . - J^ -^ ^ log -V3 = 1^ = 4^ = .0596. 8 8 3. Given log 2 = .3010, log 3 = .4771, find log (2^ x 3^). By § 589, log (2^ X 3^) = log 2^ + log 3* = i log 2 + 1 log 3 = .6967. EXERCISE 95 Given log2 = .3010, log 3 = .4771, log 7 = .8451, find the logarithm of: 1. 84. 6. 2^. 11. V105. 16. 18522. 2. 392. 3.f. 7. ^3. 8. 5292. 12. 43I,. 13. 2807. 17. 18. 3^ /18\* 4. 12^f. 9 40^ 14* 14. 75^. 19. ^36 7* 5. 76. 10. 453. 15. \/98. 20. (2^ X 15^). 594. To prove tJie relation log^m log„m ' Assume the equations a" = m, ¥ = m. ^ Then, x = log„m, y = log^m. ^ ^^ ^ From the assumed equations, a' = ¥, or a" = b. i Mr^A "^ ^y^ X X flf"^ Therefore, log„ 6 = - , or 2/ = ,--^- ^^ \ v^^-^ ^ ^ y log«6 '>WA)r^\ } 404 ADVANCED COURSE IN ALGEBRA That is, logjam = -2£«^. By aid of this relation, if the logarithm of a number m to a certain base a is known, its logarithm to any other base b may be found by dividing by the logarithm of b to the base a. 595. To prove the relation logj a X log„ 6 = 1. Putting m = a in the result of § 594, log.a = |-^ = ^ r§586). l0ga& log,, 6 Whence,^ log^ a x log„ 6=1. 596. In the common system, the maiitissm of the logarithms of numbers having the same sequence of figures are equal. Suppose, for example, that log 3.053 = .4847. Then, log 305.3 = log(100 x 3.053) = log 100 + log 3.053 = 2 4- .4847 = 2.4847 ; log .03053 = log (.01 X 3.053) = log .01 + log 3.053 = 8 - 10 + .4847 = 8.4847 - 10 ; etc. In general, if n is any positive or negative integer, log (10" X m) = n log 10 + log m = n + log m. But 10** X m is a number which differs from m only in the position of its decimal point, and n + log m is a number having the same decimal part as log m. Hence, if two numbers have the same sequence of figures, the mantis see of their logarithms are equal. For this reason, only mantissae are given, in a table of Common Logarithms ; for to find the logarithm of any number, •we have only to find the mantissa corresponding to its sequence of figures, and then prefix the characteristic in accordance with the rules of §§ 583 and 584. This property of logarithms only holds for the common system, and constitutes its superiority over other systems for numerical computation. LOGARITHMS 405 Ex. Given log 2 = .3010, log 3 = .4771 ; find log .00432. log 432 = log(2^ X 3«) = 4 log 2 4- 3 log 3 = 2.6353. Then, the mantissa of the result is .6353. Whence by § 584, log .00432 = 7.6353 - 10. EXERCISE 96 Given log 2 = .3010, log 3 = .4771, log 7 = .8451, find the logarithm of : 1. 87.5. 3. 6750. 5. .0324. 7. .784. 2. 2.592. 4. 274.4. 6. .000175. 8. .001875. USE OF THE TABLE 597. The table (pages 406 and 407) gives the mantissae of "the logarithms of all integers from 100 to 1000, calculated to four places of decimals. 598. To find the logarithm of a number of three figures. Look in the column headed " No." for the first two signifi- cant figures of the given number. Then the required mantissa will be found in the corre- sponding horizontal line, in the vertical column headed by the third figure of the number. Finally, prefix the characteristic in accordance with the rules of §§ 583 and 584. For example, log 168 = 2.2253 ; log .344 = 9.5366 - 10 ; etc. - L*^ 7^ t ^^ For a number consisting of one or two significant figures, the column headed may be used. Thus, let it be required to find log 83 and log 9. By § 596, log 83 has the same mantissa as log 830, and log 9 the same mantissa as log 900. Hence, log 83 = 1.9191, and log 9 = 0.9542. 599. To find the logarithm of a number of more than three figures. 406 ADVANCED COURSE IN ALGEBJIA. No. 1 2 3 4 5 6 7 8 9 lO oocto 0043 0086 0128 0170 0212 0253 0294 0334 0374 II 0414 045J" 0492 0531 0569 0607 0645 0682 •0719 0755 12 079^ 0828 0864 0899 0934 0969-- 1004 1038 1072 iio6' 13 "30 1173 1206 1239 1271 1303 1335 .1367 1399 1430 14 1461 1492 1523 1553 1584 1614 1644 1673 i703_ 173^ 15 1 761 1790 1818 1847 1875 1903 1931 1959 1987 2014 i6 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 17 2304 2330 2355 23S0 2405 2430 2455 2480 2504 2529 i8 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 19 2788 2810 28^3 2856 2878 29CX) 2923 2945 2967 2989 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 23 3617 3636 3838 385^ 3692 3711 3729 3747 3766 3784 24 3802 3820 3874 3892 3909 3927 3945 3962 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 26 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 27 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 28 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 32 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 33 5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 34 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 35 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 36 5563 5575 5587 5599 561 1 5623 5635 5647 5658 5670 37 5682 5694 5705 5717 5729 5740 5752 5763 5786 38 5798 5809 5821 5832 5843 5855 5866 llll 5888 5899 39 591 1 5922 5933 5944 5955 5966 5977 5999 6010 40 602 1 6031 6042 6053 6064 6075 6085 6096 6107 6117 41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 44 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 45 6532 6542 6551 6561 6571 6580 6590 6599 6609 66i8 46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 49 6902 691 1 6920 6928 6937 6946 6955 6964 6972 6981 SO 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 51 7076 7084 7093. 7101 7110 7118 7126 7135 7H3 7152 52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 54 7324 7332 7340 7348 7356 7364 7372 7380 7388 7396 1.0, 1 2 3 4 5 6 7 8 LOGARITHMS. 407 No. 1 2 3 4 6 6 7 8 9 55 7404 7412 7419 7427 7435 74+3 7451 7459 7466 7474 56 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 58 7634 7642 7649 7657 7664 7672 7679 .7686 7694 7701 59 7709 7716 7723 7731 773B 7745 7752 7760 7767 7774 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 62 7924 7931 7938 7945 7952 7959 7966 7973 7987 63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 64 8069 8075 8082 8089 8096 8102 8109 8116 8122 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 66 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 72 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 89 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 90 9542 '9547 9552 9557 9562 9566 9571 9576 9581 9586 91 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 93 9685 9694 9699 9703 9708 9713 9717 9722 9727 94 9731 9736 9741 9745 9750 9754 9759 9763 97.68 9773 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 No. 1 2 3 4 5 6 7 8 408 ADVANCED COURSE IN ALGEBRA 1. Kequired the logarithm of 32.76. We find from the table, log 32.7 = 1.5145, log32.8 = 1.5159. That is, an increase of one-tenth of a unit in the number produces an increase jif40014 in the logarithm. Therefore, a^-rricregtse of six-hundredths of a unit in the number wjiiproduce an increase of .6 x .0014 in the logarithm, or .0008/to the nearest fourthdacimaJ place. Hence, log 32.76 = 1.5145 + -0008 A,l-5153. that the differences of eir corresponding num- The above method is based on the assumption logarithms are proportional to the differences of bers ; which, though not strictly accurate, is sufficiently exact for practi- cal purposes. The difference between any mantissa in the table and the mantissa of the next higher number of these figures, is /Called the Tabular Difference. The following rule is derived from the above Find from the table the mantissajj of the first three significant figures, and the tabular diifei^em Multiply the latter by the remaining figures of the number, with a decimal point before them. Add the result to the mantissa of the first three figures, and pre- fix the proper characteristic. In finding the correction to the nearest units' figure, the decimal por- tion should be omitted, provided that if it is .5, or greater than .5, the units' figure is increased by 1 ; thus 13.26 would be taken as 13, 30.5 as 31, and 22.803 as 23. '^ ' i)8. - . ^ 2. Find the logarithm of .0215^ Mantissa 215 = -3324 Tab. diff. =21 r 2 -08 .3326 Correction = 1.68 = 2, nearly. The result is 8.3326 - 10. 600. To find the number corresponding to a logarithm. 1. Required the number whose logarithm is 0.6571. LOGARITHMS 409 Find in the table the mantissa 6571. In the corresponding line, in the column headed "No.," we find 45, the first two figures of the required number, and at the head of the column we find 4, the third figure. Since the characteristic is 0, there must be one place to the left of the decimal point (§ 583). Hence, the number corresponding to 0.6571 is 4.54. 2. Eequired the number whose logarithm is 1.3934. We find in the table the mantissse 3927 and 3945. The numbers corresponding to the logarithms 1.3927 and 1.3945 are 24.7 and 24.8, respectively. That is, an increase of .0018 in the mantissa produces an increase of one-tenth of a unit in the number corresponding. Then, an increase of .0007 in the mantissa will produce an in- crease of y^^ of one-tenth of a unit in the number, or .04, nearly. Hence, the number corresponding is 24.7 -f .04, or 24.74. The following rule is derived from the above : Find from the table the next less mantissa^he tho^ee figures cor- responding, and the tabular difference. -^ a?^'^ 4^ Subtract the next less from the f/iven mWatissu^' Un^l divide the remainder Jjy the tabular difference Annex the quotient to the first three figures of the number, and point off the result. The rules for pointing off are the reverse of tJiose of §§ 583 and 584 : I, 7/* — 10 is not written after the mantissa^ add 1 to the characteris- tic, giving the number of places to the left of the decimal point. II. 7/" — 10 is written after the mantissa, subtract the positive part of the characteristic from 9, giving the number of ciphers to be placed between the decimal point and first significant figure. 3. Find the number whose logarithm is 8.5265 — 10. 5265 Next less mant. = 5263 ; figures corresponding, 336, . 0,^ Tab. diff. 13)2.00 (.15 = .2, nearly. . I.- L3_ • r. - CJ^a^. 70 410 ADVANCED COURSE IN ALGEBRA By the above rule, there will be one cipher to be placed be- tween the decimal point and first significant figure. The result is .03362. The correction can usually be depended upon to only one decimal place ; the division should be carried to two places to determine the last figure accurately. EXERCISE 97 1 Find the logarithm of : \ 1. 70. 4. .0337. 7. .9617. 10. .0064685. 2. .59. 5. 82.95. . 8. .0003788. 11. 4072.6. 3. 98.4. 6. 253.07. 9. 7.803. 12. .013592. Find the number corresponding to : 13. 1.7782. 17. 0.8744. 21. 1.8077. 14. 8.4314-10. 18. 3.3565. 22. 7.6899-10. 15. 0.6522. 19. 6.2998 - 10. 23. -9.9108-10. 16. 9.0128-10. 20. 8.9646-10. 24. 2.5524. APPLICATIONS 601. The approximate value of a number in which the operations indicated involve only multiplication, division, invo- lution, or evolution may be conveniently found by logarithms. The utility of the process consists in the fact that addition takes the place of multiplication, subtraction of division, multiplication of involution, and division of evolution. In computations with four-place logarithms, the result cannot usually be depended upon to more than four significant figures. 1. Find the value of .0631 x 7.208 x .51272. By § 589, log (.0631 x 7.208 x .51272) = log .0631 + log 7.208 + log .51272. log .0631= 8.8000-10 log 7.208= 0.8578 log .51272 = 9.7099 - 10 Adding, log of result = 19.3677 - 20 = 9.3677 - 10 (See Note below.) LOGARITHMS 411 i Number corresponding to 9.3677 - 10 = .2332. If the sum is a negative logarithm, it should be written in such a form that the negative portion of the characteristic may be — 10. Thus, 19.3677 - 20 is written 9.3677 - 10. 336.8 2. Find the value of 7984 By § 590, log ^ = log 336.8 - log 7984. log 336.8 = 12.5273 - 10 (See Note below.) log 7984 = 3.9022 Subtracting, log of result = 8.6251 - 10 Number corresponding = .04218. To subtract a greater logarithm from a less, or a negative logarithm from a positive, increase the characteristic of the minuend by 10, writ- ing — 10 after the mantissa to compensate. Thus, to subtract 3.9022 from 2.5273, write the minuend in the form 12.5273 - 10 ; subtracting 3.9022 from this, the result is 8.6251 - 10. 3. Find the value of (.07396)^ By § 591, log (.07396)^ = 5 x log .07396. log .07396 = 8.8690 - 10 44.3450 - 50 = 4.3450 - 10 = log .000002213. 4. Find the value of ^.035063. ^ W/J^^' By § 592, log a/.035063 = - log .035063. ^ ^_ log .035063 = 8.5449 - 10 3)28.5449 - 30 (See Note below.) 9.5150 - 10 = log .3274. " To divide a negative logarithm, write it in such a form that the nega- tive portion of the characteristic may be exactly divisible by the divisor, with — 10 as the quotient. Thus, to divide 8.5449 — 10 by 3, we write the logarithm in the form 28.5449 - 30. Dividing this by 3, the quotient is 9.5150 - 10. 412 ADVANCED COUllSE IN ALGEBRA 602. Arithmetical Complement. The Arithmetical Complement of the logarithm of a number, or, briefly, the Cologarithm of the number, is the logarithm of the reciprocal of that number. Thus, colog409 = log-^=logl-log409. log 1 = 10. - 10 (See Ex. 2, § 601.) log 409= 2.6117 0-. colog409= 7.3883-10. Again, colog .067 = log— = log 1 — log .067. .067 log 1 = 10. -10 log .067= 8.8261-10 .-. colog .067= 1.1739. It follows from the above that the cologarithm of a number may he found by subtracting its logarithin from 10 — 10. • The cologarithm may be found by subtracting the last significant figure of the logarithm from 10 and each of the others from 9, — 10 being written after the result in the case of a positive logarithm. * .51384 Ex. Find the value of 8.708 X .0946 log -51^-^^ = log^.51384 X -^ X -^^ ^ 8.708 X .0946 ^V ^-^^^ •^946;' = log .51384 + log-i- + log- ^ 8.708 °.0946 = log .51384 + colog 8.708 + colog .0946. log .51384 = 9.7109 - 10 colog 8.708 = 9.0601 -10 colog .0946 = 1.0241 9.7951 -10 = log .6239. It is evident from the above example that, to find the loga- rithm of a fraction whose terms are the product of factors, we add together the logarithms of the factors of the numerator, ayid the cologarithms of the factors of the denomiyiator. y LOGARITHMS" ."^ ^ 'L ^ 418 The value of the fraction may be found without using cologarithms by . ^ > ^ the following formula : '^ I (, *; f log -^^^^^ = log .51384 - (log 8.708 + log .0946) (§8 589, 590). ^8.708 X .0946 '^ ^ =' ^ J vs^ ^ J The advantage in using cologarithms is that the written wor]j: ^^q^-»V 7 / - putation is exhibited in a more compact form. \ / kj'^H ' 603. Examples. 1. Find the value of ?^. 3f \^ i^#^ log ?^ = log 2 + log -v/S + colog 3f (§ 602) = log2 + ilog5 + ?colog3. o,«33^ log 2= .3010 log 5= .6990; ^3= .2330 colog 3 = 9.5229 -10; x ^ = 9.6024-10 .1364 = log 1.369. A negative number has no common logarithm (§ 580). If such numbers occur in computation, they may be treated as if they were positive, and the sign of the result determined irrespective of the logarithmic work. Thus, to find the value of 721.3 x (- 3.0528), we find the value of 721.3 X 3.0528, and put a negative sign before the result. 2. rind the value of *'' ~ '^^^^^ "^ \ 7.962 3 ^ 7.9( 7.962 3296 962 = i (log .03296 - log 7.962). o log .03296 = 8.5180 -10 log 7.962 = 0.9010 , 3)27^6170-30 9.2057 -10 = log .1606. The result is - .1606. 4:U ADVANCED COURSE IN ALGEBRA 1. 2. 5. 6. 11. 13. 14. 15. EXERCISE 98 Find by logarithms the values of the following : 3. ( - 54.375) X ( - .00061488). 18. 31. 33. 34. 35. 7. 2414.7 X .09348. 832.4 X 4.1639. 51.29 6.348' 834.32 2192.4* °' - ( -.009213) x(- 73.36) (-.0832) X 2.8087 .004497 .09769 ' 3.3629 -.75438' 4. .38142 X (-.0053909). 718 X (-.02415) (-.5157) X 1420.6* .87028 X 3.74 12. (2.514)5. (-83.28)3. (.035127)4. 19. VAm. 20. V- .037368. 21. 16. 100*. ^/7 Va i 17. (- .007795) 't^. 23. -^-IM§. V1553. (837.5 X .0094325) ' V3929 X v^6K48 y/T2TM 31 24. (m]i V6937y V.05287 V.374 X V.0078359 (- .0001916)' ^68i: .27556 3801.4 38. 39. 40. 9. 10. .0006589 x(- 42.318) 3.8961 X .6945 x .01382 25 27. 4694 X .00457 v^. 0009657 \/.'0049784 -(.25693)^; (- .8346)^ 28. 30. / 76.1 X .0593 \| V 1.307 j ^r 7.544 31.4 X .415 V5106.5 X .00003109. 36. .83184 X (.2682)3 X (56.1)i 37. .0005616 X V424.65 (6.73)4 X (.03194)* 485.7 X (.7301)7 xv^ 1000 (9. 1273)6 X (.7095)^ 3 /r (- .95048)5 x(8473)h >L (- 2080.9) xvC05^ J v/-. 0030 12 X 1955 V. 04142 x(-.947?^) 41.^ (-.843)8 X ^17959 x(- 560. fi^?" 04813 X (5.6074)^ .002988)^ (10.115)5 x( ^x (.65034)0 S)^ X \/73r27J LOGARITIIxMS 415 EXPONENTIAL AND LOGARITHMIC EQUATIONS 604. An Exponential Equation is an equation in which, the unknown number occurs as an exponent. To solve an equation of this form, take the logarithms of both members ; the result will be an equation which can be solved by ordinary algebraic methods. 1. Given 31" = 23 ; find the value of x. Taking the logarithms of both members, log (31") = log 23 ; or x log 31 = log 23 (§ 591). ^, log 23 1.3617 Q-,on_u Then, x = ^ & = =.9130+. ' log 31 1.4914 2. Solve the equation .2" = 3. Taking the logarithms of both members, x log .2 = log 3. Then, 0.= !^^=— i^Zi— = ^iI^ = -.6825+. log .2 9.3010-10 -.699 An equation of the form a' = b may be solved by inspection if b can be expressed as an exact power of a. 3. Solve the equation 16" = 128. We may write the equation (2*)" = 2^ or 2*" = 21 • . ■ 7 Then, by inspection, 4 a? = 7 ; and ^' = t* If the equation were 16^ = , we could write it (2*)* = — = 2-^ ; then 7 4 X would equal — 7, and x = 4 Certain logarithmic equations are readily solved by aid of the principles of §§ 589 to 591. 4. Given 2 log^ x = m ; find the value of x. By § 591, 2 log« x = log« (^x^) ; whence, log« (a^) = m. Then by § 580, a'^ = x^- whence, x=± Va"* =±a^. 5. Given log (ic + 4) — log x = 3; find the value of x. By § 590, log (x + 4) - log x = log ^^^' 416 ADVANCED COURSE IN ALGEBRA Then, log^+i = 3; and by § 581, W = ?i±i. X X Therefore, 1000 x = x-\-4:, and x 999 Solve the following : EXERCISE 99 1. 13- = 8. 7. .2-+5 = .S---*. 12. fi-V = JL, 2. .06- = .9. 8. .3-4 = 100. \27/ 81 3. 9.347- = .0625. 9. 16- = 32. 13. 52X-1 = 1.. 25 4. .005038-= 816.3. 5. 34—1 = 42^+3. 6. 73-4-2 =.8x. 10. 32-= ^ . 128 14. 15. 16. Given a, r, and I ; derive the formula for n (§ 529). 17. Given a, r, and /S'; derive the formula for n. 18. Given a, Z, and S ; derive the formula for n. 19. Given r, I, and >S' ; derive the formula for n. Solve the following : 20. 3 loga X = 4 loga w. 23. log3 - log (x + 1) = - 1. 21. log(x-5)-log(2x+l)=2. 24. logx + log (4 a: + 3)= 1. 22. log 5 + log (3 X- 2) =3. 25. log2 + 2 logx = log (5 a; - 2). 605. Logarithm of a Number to Any Baseo 1. Find the logarithm of .3 to the base 7. • By § 594, ^' logioT .8451 .8451 Examples of this kind may be solved by inspection, if the number can be expressed as an exact power of the base. 2. Find the logarithm of 128 to the base 16. Let logiel28 = a;; then, by § 580, 16* = 128. Then, as in Ex. 3, § 604, a; = - ; that is, logig 128 = - • 4 4 LOGARITHMS 417 EXERCISE 100 Find the values of the following : 1. log7 59. 3. log.482. 5. log68 2.915. 2. log6.7. 4. log.9 .00453. 6. logoi .06038. Find by inspection the values of the following : 7. log.5l25. 8. log,.(l). 9. log^,^(3). 10. log^^(^)- EXPONENTIAL AND LOGARITHMIC SERIES- 606. Let n be a real number greater than unity. By §364, [{^ + iy]=(i+iy- Expanding both members by the Binomial Theorem, L ^ [2 n' [3 n'^ J n [2 n^ nx(nx— l)(nx — 2) J^ .^. Since, by hypothesis, n is > 1, - is numerically < 1 ; and both members of (1) are convergent series (§ 559). We may write equation (1) in the form [2 [3 ■J....] f 1\ f l\f 2\ x[x x[x a; which holds however great n may be. Now let 71 be indefinitely increased. 12 Then, the limit of each of the terms -, -, etc., is (§ 248). n n 418 ADVANCED COURSE IN ALGEBRA Hence, the limiting value of the first member of (2) is and the limiting value of the second member is ' + -'" + [2 + g + -- By § 252, these limits are equal ; that is, x^ . X }+'+tt The series in the second member is convergent for every value of X (§ 558) ; and the series in brackets is also conver- gent, for it is obtained from the series in the second member by putting 1 in place of ic. . Denoting the series in brackets by e, we have which holds for every value of x. Putting mx for x, in (3), e"'^ = l + ma;+— — + -—- + .... (4) \A Lr Let m = logg a, where a is any positive real number. Then e'" = a (§ 580), and e'^' = a\ Substituting these values in (4), we obtain a' = l+(log.a)x + (log.a)^| + (log.a)'g+...; (5) which holds for all values of x, and all positive real values of a. The result (5) is called the Exponential Series. 607. The system of logarithms which has e for its base is called the Napierian System, from Napier, the inventor of logarithms. Napierian logarithms are also called Natural Logarithms. LOGARITHMS 419 The approximate value of e may be readily calculated from the series of § 606, and will be found to equal 2.7182818.... 608. To expand log^ (1 + x) in ascending powers of x. Substituting in (5), § 606, l-\-x for a, and y for x, (1 + a;)^ = 1 4- [loge (1 + ic)] ?/ 4- terms in y^, /, etc. ; which holds for all values of y, provided x is real, and alge- braically greater than — 1. Expanding the first member by the Binomial Theorem, = 1 + [log, (1 + i»)] 2/ + terms in y% y% etc. (6) The first membet of (6) is convergent when x is numerically less than 1 (§ 559). Hence, (6) holds for all values of y, provided x is real, and numerically less than 1. Then, by the Theorem of Undetermined Coefficients, the coefficients of y in the two series are equal ; that is, l^ l£ 11 Or, log,(l + a^) = a^-f + f-f + f--; (7) which holds for all values of x numerically less than 1. This result is called the Logarithmic Series. It was proved in § 557 that this series was convergent when x was numerically less than 1. It was also shown in § 556 that it was convergent when x = l, and in § 548 that it was divergent when x = — 1. Then, series (7) can be used to calculate Napierian logarithms, pro- vided X is taken equal to, or numerically less than, 1. Unless X is small, it requires the sum of a great many terms to ensure any degree of accuracy. 420 ADVANCED COURSE IN ALGEBRA 609. We will now derive a more convenient series for the calculation of Napierian logarithms. Putting — X for x, in (7), § 608, we have ^ 2 3 4 5 Subtracting (8) from (7), we obtain »log,(l + a.')-log,(l-a^) = 2a^ + ?^ + ?^+.... (8) Or (§ 590), lo^^l±l = 2(x + t + t ^"1-x \ 3 5 (9) Let X = ; m and n being positive, and m > n. m-\-n' m — n This IS a valid substitution, since in this case — — - < 1. m-{-n m — 71 Then, 1 + 1 — x -i m — n 2 7i~ n Substituting these values in (9), we obtain 2 m _ Srti — n^ \f m — n V l/ m — n V "n _m-{-n 6\m-\-nj 5\m-{-nJ But by § 590, log^— = log^ m — log« n ; whence, loge m = loge n + 2 "m — ^* I 1/wi — n\^ . 1/ m m + n S\m + ny 5V m + ^)'-} 610. Let it be required, for example, to calculate the Napierian logarithm of 2 to six places of decimals. Putting m = 2 and w = 1 in the result of § 609, we have log,2 = log,l + 2 hm-w^-] Or since log,l = (§585), . log, 2 = 2(.3333333 + .0123457 + .0008230 + .0000653 + .0000056 -f .0000005 + •••) = 2 X. 3465734 = .6931468. LOGARITHMS 421 Then, log^ 2 = .693147, to the nearest sixth place of decimals. Having found log^2, we may calculate log^3 by putting m=3 and w = 2 in the result of § 609. Proceeding in this way, we shall find log^ 10 = 2.302585 611. To calculate the common logarithm of a number, having given its Napierian logarithm. Putting 6 = 10 and a = e in the result of § 594, ^^^^«" = iS == 2:3485 ^ log.m = .4342945 x log.m. Thus, logio2 = .4342945 x .693147 = .301030. 612. The multiplier by which logarithms of any system are derived from Napierian logarithms, is called the modulus of that system. Thus, .4342945 is the modulus of the common system. Conversely, to find the Napierian logarithm of a number when its common logarithm is given, we may either divide the common logarithm by the modulus .4342945, or multiply it by 2.302585, the reciprocal of .4342945. EXERCISE 101 Using the table of common logarithms, find the Napierian logarithm of each of the following to four significant figures : 1. 1000. 3. 9.93. 6. .04568. 2. .0001. 4. 243.6. 6. .56734. 7. What is the characteristic of logs 758 ? 8. What is the characteristic of logy 500 ? 9. If log 3 = .4771, how many digits are there in 31^ ? 10. If log 8 = .9031, how many digits are there in 828 ? 11. If log 11 = 1.0414, how many digits are there in the integral part of 11"^? 422 ADVANCED COURSE IN ALGEBRA XXX. COMPOUND INTEREST AND ANNUITIES 613. The principles of logarithms may be applied to the solution of problems in Compound Interest. Let P = number of dollars in the principal ; n = number of years ; t = the ratio to one year of the time during which sim- ple interest is calculated; thus, if interest is compounded semi-annually, ^ = ^ 5 M = number of dollars in the amount of one dollar for t years ; A = number of dollars in the amount of P dollars for n years. Since one dollar amounts to JR dollars in t years, P dollars will amount to PE dollars in t years ; that is, the amount at the end of the 1st interval is PR dollars. In like manner, the amount at the end of the 2d interval is PE x E, or PE'^ dollars ; 3d interval is PE^ X E, or PE^ dollars ; etc. Since the whole number of intervals is -, the amount at the end of the last one, in accordance with, the law observed above, will be PE^ dollars. n That is, A = PEl (1) By § § 589, 591, log ^ = log P + 'i log E. (2) 1. What will be the amount of $7326 for 3 years and 9 months at 7 per cent compound interest, interest being com- pounded quarterly ? Here, P = 7326, n = 3|, i = ^ i? = 1.0175, ^ = 15. COMPOUND INTEREST AND ANNUITIES 423 log P = 3.8649 log i2 = 0.0075; X 15 = 0.1125 log ^ = 3.9774 .-.^ = $9493. 2. What sum of money will amount to ^ 1763.50 in 3 years at 5 per cent compound interest, interest being compounded semi-annually ? From (2), log P = log ^ - - log R. Here, n = ^,t = \,R = 1.025, A = 1763.5, - = 6. Z t log A = 3.2464 log J? = 0.0107; X 6 = 0.0642 . log P - 3.1822 .-. P = $ 1521. 3. In how many years will ^300 amount to $398.60 at 6 per cent compound interest, interest being compounded quar- terly ? From (2), ^^ t(log A-log P) ^ log B Here, P= 300, t = j, P = 1.015, ^ = 398.6. ^ log 398.6 - log 300 ^ 2.6006 - 2.4771 ^ .1235 *** *^ ~ 4 log 1.015 4 X .0065 .0260 = 4.75 years. 4. At what rate per cent per annum will $ 500 amount to ^.83 in 6 years and 6 months, interest being compounded semi-annually ? ^C^)> \logi? = _ — o " --0 n ,p= 500, n = 6i, t = t = 1 ^ = 688.83, n 1 = 13. log^ = logP = log 12 = 2.8381 2.6990 13)0.1391 0.0107 .-. R 1.025. 424 ADVANCED COURSE IN ALGEBRA That is, the interest on one dollar for 6 months is $ .025, and the rate is 5 per cent per annum. 614. To find the present worth of A dollars due at the end of n years, interest being comjjounded annually. Putting ^ = 1 in (1), § 613, we have A = PBT; whence, ^ = ^• ANNUITIES 615. An Annuity is a fixed sum of money payable at equal intervals of time. In the present chapter we shall consider those cases only in which the payments are annual ; in finding the present worth of such an annuity, it is customary to compound interest annu- ally. When we speak of the annuity as beginning at a certain epoch, it is understood that the first payment becomes due one year from that time. 616. To find the present worth of an annuity to continue for n successive years, allowing compound interest. Let A = number of dollars in the annuity ; E = number of dollars in the amount of one dollar for one year ; P = number of dollars in the present worth of the annuity. By § 614, the present worth of the 1st payment = —; H 2d payments—; nth. payment = — • I COMPOUND INTEREST AND ANNUITIES 425 Hence, the sum of the present worths of the separate pay- ments, or the present worth of the annuity, is E^ E''-' E' E That is, P=A R- ^ i2«-i ^ ^ E' EJ The expression in brackets is the sum of the terms of a geometric progression, in which a = —-, r = E, and I = — • it" E Then, by II, § 527, P=^-^_^. (3) Ex. Find the. present worth of an annuity of $150 to con- tinue for 20 years, allowing 4 per cent compound interest. Here, A = 150, n = 20, E = 1.04, E-l = .04. Whence, ^=^'[l " (di).} l°g^r^ = 20cologl.04. colog 1.04 = 9.9830 -10 20 9.6600 - 10 = log .4571. Then, P= ^(1 - .4571) = 3750 x .5429. log 3750 = 3.5740 log .5429 = 9.7347 -10 log P= 3.3087 .-. P=$2036. 617. We have from (3), § 616, ^_ P(Ii-l) _ PE^(E-l) , ^- ^_\_ - E^-1 ' (4) E- which is a formula for finding the annuity to continue for n successive years, when the present worth and the amount of one dollar for one year are given. 426 ADVANCED COURSE IN ALGEBRA Formula (4) may also be used to find what fixed annual payment must be made to cancel a note of P dollars due n years hence, B being the number of dollars in the amount of one dollar for one year. 618. If in (3), § 616, n be indefinitely increased, the limit- ing value of the second member is A R-1 (§ 248). That is, the present worth of a perpetual annuity is equal to the amount of the annuity divided by the interest on one dollar for one year. 619. To find the present worth of an annuity to begin after m years, and continue for n years, allowing compound interest. With the notation of § 616, the number of dollars in the value of the annuity one year before the first payment becomes due, is ^ 2Z^ or 4(^!^li}. By § 614, the present worth of this amount, due m years R\R-1) ' Therefore, P = ^^^^/"^^ » ' R'^+XR - 1) 620. By § 618, the present worth of a perpetual annuity to begin after m years, is given by the formula A P = R^{R - 1) EXERCISE 102 1. What will be the amount of $1300 for 16 years at 5 per cent compound interest, the interest being compounded annually ? 2. What sum of money will amount to $981.75 in 8 years and 9 months at 4 per cent compound interest, the interest being compounded quarterly ? COMPOUND INTEREST AND ANNUITIES 427 3. In how many years will $859 amount to $1012.80 at 3 per cent compound interest, the interest being compounded semi-annually ? 4. What is the present worth of a note for $625.34 due 12 years hence, allowing 3| per cent compound interest, the interest being com- pounded annually ? 6. At what rate per cent per annum will $3700 gain $678 in 4 years and 3 months, the interest being compounded quarterly ? 6. In how many years will a sum of money double itself at 6 per cent compound interest, the interest being compounded annually ? 7. In how many years will a sum of money treble itself at 4| per cent compound interest, the interest being compounded semi-annually ? 8. What is the present worth of an annuity of $ 300 to continue 14 years, allowing 4 per cent compound interest ? 9. What is the present worth of a perpetual annuity of $506.70, allow- ing 3^ per cent compound interest ? 10. What is the present worth of an annuity of $2238 to continue 4 years, allowing 6 per cent compound interest ? 11. What is the present worth of an annuity of $2680 to begin after 10 years and continue for 7 years, allowing 5 per cent compound interest ? 12. What fixed annual payment must be made in order to cancel a note for $3500 in 5 years, allowing 4| per cent compound interest? 13. What is the present worth of a perpetual annuity of $297.50, to begin after 8 years, allowing 3| per cent compound interest ? 14. What annuity to continue 12 years can be purchased for $3149, allowing 7 per cent compound interest ? 15. A person borrows $6365 ; how much must he pay in annual instalments in order that the whole debt may be discharged in 10 years, allowing 4^ per cent compound interest ? 428 ADVANCED COURSE IN ALGEBRA XXXI. PERMUTATIONS AND COMBINA- TIONS 621. The different orders in which things can be arranged are called their Permutations. Thus, the permutations of the letters a, b, c, taken two at a time, are ab, ac, bal be, ca, cb ; and their permutations taken three at a time, are abc, acb, bac, bca, cab, cba. 622. The Combinations of things are the different collections which can be formed from them without regard to the order in which they are placed. Thus, the combinations of the letters a, b, c, taken two at a time are ab, be, ca; for though ab and ba are different permu- tations, they form the same combination. 623. To find the number of permutations of n different things taken two at a time. Consider the n letters a, b, c, •••. In making any particular permutation of two letters, the first letter may be any one of the n ; that is, the first place can be filled in n different ways. After the first place has been filled, the second place can be filled with any one of the remaining 7i — 1 letters. Then, the whole number of permutations of the letters taken two at a time is n(7i — l). We will now consider the general case. 624. To find the number of permutations ofn different things taken r at a time. Consider the n letters a, b, c, -". In making any particular permutation of r letters, the first letter may be any one of the n. After the first place has been filled, the second place can be filled with any one of the remaining n — 1 letters. PERMUTATIONS AND COMBINATIONS 429 After the second place has been filled, the third place can be filled in n — 2 different ways. Continuing in this way, the rth place can be filled in n — (r — 1), or n — r-\-l different ways. Then, the whole number of permutations of the letters taken r at a time is given by the formula „P, = 7i(n _ 1) (n - 2) ... (ii - r + 1). (1) The number of permutations of n different things taken r at a time is usually denoted by the symbol nPr- 625. If all the letters are taken together, r = n, and (1) becomes ^P^ = n{n-l)(n-2) -3 -2 'l = \n. (2) Hence, the number of permutations of n different things taken n at a time equals the product of the natural numbers from 1 to n inclusive. 626. To find the number of combinations of n different things takeri r at a time. ^ The number of permutations of n different things taken r at a time, is ^^ (^ - 1) (n - 2) ... (71 - r + 1) (§ 624). .^ , But by § 625, each combination of r different things may have I r permutations. '^'^"'^'^^J^ ^ Hence, the number of combinations of n different things taken r at a time equals the number of permutations divided by \r. p That is, ^o^^ ^(^-l)(^-2)-(n-r + l) . ^^1^^^) The number of combinations of n different things taken r at a tinit^ is usually denoted by the symbol nCr- 627. Multiplying both terms of the fraction (3) by the prod- uct of the natural numbers from 1 to 71 — r inclusive, we have ^ ^ n{n-V)'"(n-r^l)»{n-r)-"2'l ^ J!^ . \rxl'2"-(n-r) \r \n-r ' which is another form of the result. 430 ADVANCED COUKSE IN ALGEBRA 628. Tlie number of combinations of n different things taken r at a time equals the number of combinations taken n — r at a time. For in making a selection of r things out of ri, we leave a selection of n — r things. The theorem may also be proved by using the result of § 627. 629. Examples. 1. How many changes can be rung with 10 bells, taking 7 at a time ? Putting n = 10, r = 7, in (1), § 624, ioPy.= 10 . 9 . 8 . 7 . 6 . 5 . 4 = 604800. 2. How many different combinations can be formed with 16 letters, taking 12 at a time ? By § 628, the number of combinations of 16 different things, taken 12 at a time, equals the number of combinations of 16 different things, taken 4 at a time. Putting n == 16, r=4, in (3), § 626, p 16 . 15 ♦ 14 . 13 .^^^ ''^'^ 1.2.3.4 =^^^^- 3. How many different words, each consisting of 4 consonants and 2 vowels, can be formed from 8 consonants and 4 vowels ? The number of combinations of the 8 consonants, taken 4 at a time, is ^^1^1^, or 70. The number of combinations of the 4 vowels, taken 2 at a time, is q-^, or 6. Any one of the 70 sets of consonants may be associated with any one of the 6 sets of vowels ; hence, there are in all 70 x 6, or 420 sets, each containing 4 consonants and 2 vowels. But each set of 6 letters may have |6, or 720 different permutations (§ 625). Therefore, the whole number of different words is 420 X 720, or 302400. <^ PERMUTATIONS AND COMBINATIONS 431 EXERCISE 103 Find the values of the followhig : 1. 14^6. 3. 17 P7. 5. 17 On. 2. 9P9. 4. \5G^. ^ 6. 29C'24. 7. In a certain play, there are five parts to be taken by a company of twelve persons. In how many different ways can they be assigned ? 8. How many different words, of nine different letters each, can be formed from the letters in the word flowering^ if the vowels retain their places ? 9. How many different numbers, of seven different figures each, can /^ be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, if e5ch number begins with ^ tT I and ends with 9 ? 10. How many even numbers, of five different figures each, can be formed from the digits 4, 5, 6, 7, 8 ? ^ 11. How many different committees, of 8 persons each, can be formed from a corporation of 14 persons ? In how many will any particular ■- — __^ individual be found ? In how many will any particular individual be excluded ? 12. A and B are in a company of 72 men. If the company is divided into squads of 6, in how many of them will A and B be in the same squad ? 13. In how many different ways can six persons be seated at a round table ? 14. There are 15 points in a plane, no three in the same straight line. How many quadrilaterals can be formed, having four of the points for vertices ? 15. If 32 soldiers are drawn up in line 4 deep, in how many different ways can they be arranged so as to have a different set in the front rank ? In how many ways, if the front rank is always to contain 3 particular men? 16. If the number of combinations of 2 n different things taken w — 1 at a time, is to the number of combinations of 2/i — 2 different things taken II at a time as 132 : 35, find the value of n. 17. How many different crews, each consisting of eight oarsmen and a steersman, can be formed from 16 boys, of whom 12 can row but cannot steer, and the others can steer but cannot row ? 18. A person has 22 acquaintances, of whom 14 are males. In how many ways can he invite 17 guests from them so that 10 may be males ? 19. Out of 10 soldiers and 15 sailors, how many different parties can be formed, each consisting of 3 soldiers and 3 sailors ? 432 ADVANCED COURSE IN ALGEBRA 20. From 3 sergeants, 8 corporals, and 16 privates, how many different parties can be formed, each consisting of 1 sergeant, 2 corporals, and 5 privates ? 21. Out of 3 capitals, 6 consonants, and 4 vowels, how many different words of six letters each can be formed, each beginning with a capital, and having 3 consonants and 2 vowels ? 22. How many different words of 8 letters each can be formed from eight letters, if 4 of the letters cannot be separated ? How many if these four can only be in one order ? 23. In how many different ways can ten soldiers be drawn up in double rank, if three particular men are. always in the front rank, and three others always in the rear ? 24. How many different numbers of seven figures each can be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, if the first, fourth, and last digits are odd numbers ? 25. How many different words of six letters each can be formed from the letters in the woid percolating, if each word has a consonant for its first and last letter, and a vowel for its second and fifth ? 26. There are 2 n guests at a dinner-party. If the host and hostess have fixed places opposite to each other, and two specified guests cannot sit next each other, in how many ways can the company be seated ? 630. To find the number of permutations of n things which are not all different, taken all together. Let there be n letters, of which p are a's, q are 6's, and r are c's, the rest being all different. Let JSf denote the number of permutations of these letters taken all together. Suppose that, in any assigned permutation of the n letters, the p a's were replaced by p new letters, differing from each other and also from the remaining n~p letters. Then by simply altering the order of these p letters among themselves, without changing the positions of any of the other letters, we could from the original permutation form [p differ- ent permutations (§ 625). If this were done in the case of each of the N original per- mutations, the whole number of permutations would be Nx[p. PERMUTATIONS AND COMBINATIONS 433 Again, if in any one of the latter the q 6's were replaced by q new letters, differing from each other and from the remaining n — q letters, then by altering the order of these q letters among themselves, we could from the original permutation form \q different permutations ; and if this were done in the case of each of the N x\p original permutations, the whole number of permutations would be ^ x Lp X |^. In like manner, if in each of the latter the r c's were re- placed by r new letters, differing from each, other and from the remaining n — r letters, and these r letters were permuted among themselves, the whole number of permutations would be Nx\jpx\qx\r. But the number of permutations on the hypothesis that the n letters are all different, is \n (§ 625). \n Therefore, JVx |i> X I g X |r = 1^1 : or, iV= — =— * L_ L_ L_ i_- Lp[2[>; Any other case may be treated in a similar manner. Ex. How many permutations can be formed from the letters in the word Tennessee, taken all together ? Here there are 4 e's, 2 n's, 2 s's, and 1 t. Putting in the above formula n = 9, p = 4, g = 2, r = 2, we have \9 _5.6.7.8.9^3,g()^ [4[2[2 2-2 631. To find the number of permutations ofn different things, taken r at a time, vjhen each may occur any number of times from once up to r times, inclusive. Consider the n letters a, b, c, •••. In making any particular permutation of r letters, the first letter may be any one of the ii ; that is, the first place can be filled in n different ways. The second letter can also be any one of the n ; that is, the second place can be filled in n different ways. 434 ADVANCED COURSE IN ALGEBRA Continuing in this way, the rth place can be filled in 71 differ- ent ways. f Then, the number of permutations is 71 xnx "' to 7^ factors, or n"". Ex. How many different words of four letters each can be formed from nine letters, if each letter may occur any number of times from once up to four times, inclusive ? Here, n = 9, r = 4. Then, the number of different words is 9^ or 6561. 632. To find the entire number of combinatio^is of n different things J when each may he taken any number of times from once to n times, inclusive. Consider the n letters a, 6, c, •••. We can take them one at a time in 71 ways. We can take them 2 at a time in '^('^-^) ^^ays (§ 626). We can take them all in one way. Then, the entire number of ways is n + '-^i^^y^ + -+l = 2"-l (§289,1). \A Ex. In how many ways can a selection of one or more vol- umes be made from 5 books ? Here, n = 5 ; then, the entire number of selections is 2^-1, or 31. EXERCISE 104 1. In how many different orders may the letters of the word denomi- nation be written ? 2. There are four white billiard balls exactly alike, and three red balls, also alike. In how many different ways can they be arranged ? PERMUTATIONS AND COMBINATIONS 435 3. In how many ways can six things be given to five persons, if there is no restriction as to the number each may receive ? 4. How many different numbers less than 10000 can be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8 ? 5. In how many different orders may the letters of the word indepen- dence be written ? 6. How many different signals can be made with 7 flags, of which 2 are blue, 3 red, and 2 white, if all are hoisted for each signal ? - 7. A railway signal has m arms, and each can be placed in n positions. How many different signals can be made with it ? 8. A man has eight friends. In how many ways can he invite one or more of them to dinner ? 9. How many different words of eight letters each can be formed from the letters in the word arranged^ if the first, fourth, and seventh letters are always vowels ? 10. A house has nine windows in front. How many different signals can be given by having one or more of the windows open ? 11. In how many ways can 13 books be arranged on a shelf, when* five volumes are alike, and four other volumes are also alike ? 12. How many different numbers greater than 1000000 can be formed from the digits 4, 3, 3, 3, 2, 2, ? 13. In how many ways can two dimes, three quarters, four halves, and five dollars be distributed among 14 persons, so that each may receive a coin ? 14. A bag contains a cent, a half-dime, a dime, a twenty-cent piece, a quarter-dollar, a half-dollar, and a dollar. In how many ways can a sum of money be drawn from the bag ? 436 ADVANCED COURSE IN ALGEBRA XXXII. PROBABILITY 633. Suppose that a bag contains 5 white balls, 4 red balls, and 3 black balls, and that one ball is drawn at random. Any one ball is. as likely to be drawn as any other. The drawing of a hall can occur in 12 different ways ; for any one of the balls may be drawn. The drawing of a white hall can occur in 5 different ways ; for any one of the white balls may be drawn. 5 We may then consider — as the likelihood that,* if a ball is drawn, it is a white ball. The drawing of a white hall can fail to occur in 7 different ways ; for any one of the red or black balls may be drawn. 7 We may then consider — as the likelihood that, if a ball is ^ 12 ' drawn, it is not a white ball. 634. We may take the following definition for the term prohahility : If an event can happen in a different ways, and fail to happen in h different ways, and all these ways are equally likely to occur j the probability of the happening of the event is r, and the h CL + h probability of its failing is . a-{-h "We say the odds are a to 6 in favor of the event, if a is greater than 6, and a to & against the event, if a is less than h. It follows that if the probability of the happening of an event is p, the probability of its failing is 1—p. 635. Examples. 1. A bag contains 5 white, 4 red, and 3 black balls, (a) If 3 balls are drawn, what is the probability that they are all white ? <^^ PROBABILITY 437 The number of combinations of the 5 white balls, taken 2^ 5.4.3 at a time, is - — - — - (§ 626), or 10 ; that is, the drawing of 3 white balls can ha gp^ninl O different ways. The number of combinations of the 12 balls, taken 3 at a time, is , or 220; that is, the drawing of 3 balls can JL • Z • o occur iiL^^ different ways. Then, the probability of drawing 3 white balls is -— , or — . ZZO ZIi (b) If 6 balls are drawn, what is the probability that 2 are white, 3 red, and |L black ? The nuniber of combinations of the 5 white balls, taken 2 at 5*4- a time, is , or 10 ; the number of combinations of the 4 ^•^ 4.3.2 red balls, taken 3 at a time, is , or 4. ' '1.2.3' We may associate together any one of the 10 combinations of white balls, any one of the 4 combinations of red balls, and any one of the 3 black balls ; hence, there are in all 10 X 4 X 3, or 120, different combinations, each consisting of 2 white balls, 3 red balls, and 1 black ball. Also, the number of combinations of the 12 balls, taken 6 at . 12 . 11 . 10 . 9 . 8 . 7 ciOA " '^"'' '' 1.2.3.4.5.6 ' '' ''^- Hence, the required probability is— — , or — . 2. A bag contains 30 tickets numbered 1, 2, 3, •••, 30. (a) If four tickets are drawn, what is the chance that both 1 and 2 will be among them ? The number of combinations of the 28 tickets numbered 28 . 27 3, 4, ..., 30, taken 2 at a time, is ' : that is, there are 28 . 27 ' different ways of drawing four tickets, two of which Jl ' Z are numbered 1 and 2. 438 ADVANCED COURSE IN ALGEBRA The number of combinations of the 30 tickets, taken 4 at a , . . 30 . 29 . 28 . 27 ^""^-^^ 1.2.3.4 ' Hence, the probability that, if four tickets are drawn, two of them will be 1 and 2, is 28 . 27 . 30 • 29 . 28 . 27 ^ 3.4 ^ 2 1.2 * 1.2.3.4 30 . 29 145* (6) If four tickets are drawn, what is the chance that either 1 or 2 will be among them ? Either 1 or 2 will be among the tickets drawn, unless each ticket drawn bears a number from 3 to 30 inclusive. The number of combinations of the 28 tickets numbered 3, 4, ..., 30, taken 4 at a time, is ^^ ' ^'^ ' ^^ ' ^^ . ' ' ' ' 1.2.3.4 The number of combinations of the 30 tickets, taken 4 at ,. . 30.29.28.27 ^^"^'-^^ 1.2.3.4 • Hence, the probability that each of the 4 tickets drawn bears . o o ^ OA • 1 • • 28 . 27 . 26 . 25 65 a number from 3 to 30 inclusive, is , or — ' 30 . 29 . 28 . 27 87 Then, the probability that none of the tickets drawn bears 65 22 a number from 3 to 30 inclusive, is 1 (§ 634), or — . 87 87 This then is the probability that either 1 or 2 will be among the tickets drawn. EXERCISE 105 1. A bag contains 6 white balls, 5 red balls, and 4 black balls ; find the probability of drawing : (a) One black ball. (d) Four white balls. (&) Two white balls. (e) Two balls of each color, (c) Three red balls. (/) Four red and three white balls. {g) Two red, five white, and two black balls. 2. A bag contains 24 tickets numbered 1, 2, 3, ..., 24 ; if three tickets are drawn, find the probability : (a) That they are 1, 2, and 3. (6) That either 1, 2, or 3 is among them. PROBABILITY 439 3. What is the probability of throwing not more than 5 in a single throw with two dice ? 4. What is the probability of throwing at least 5 in a single throw with two dice ? 5. What is the probability of throwing 10 in a single throw with three dice? 6. If six persons seat themselves at random at a round table, what is the probability that two specified persons will sit together ? 7. If four cards are drawn from a pack, what is the probability that they are of the same suit ? 8. There are 8 books, of which 4 are on mathematics and 3 on science. If the books are placed together on a shelf, what is the probability that the mathematical volumes, and also the scientific, will be together ? 636. Mutually Exclusive Events. If an event can happen in more than one way, and if it happens in any of the ways, cannot at the same time happen in any of the other ways, these various ways are said to be mutually exclusive. If an event can happen in more than one way, and these ways are mutually exclusive, the probability of the happening of the event equals the sum of the probabilities of its happening in the separate ways. Suppose that an event can happen in a certain way a times out of b, and in another way a' times out of b ; all these ways being equally likely to occur. Also, suppose that the two ways in which th§ event can happen are mutually exclusive. Since the event happens a + a' times out of b, the probability of its happening is ^^^^ (§ 634), or - + -• b b b But - is the probability that the event happens in the first a' way, and — the probability that it happens in the second way. Hence, the probability that it happens equals the sum of the probabilities of its happening in the separate ways. In like manner, the theorem may be proved when there are more than two ways in which the event can happen. 440 ADVANCED COURSE IN ALGEBRA 637. Examples. 1. Find the probability of throwing 4 in a single throw with two dice. The event can happen in two ways ; either by throwing 3 and 1, or by throwing double-twos ; and these ways are mutu- ally exclusive. Each die can come up in 6 ways ; and hence the pair can be thrown in 6 x 6, or 36 ways. Of these different throws, two will be 3 and 1 ; hence, the 2 probability of throwing' 3 and 1 is — 36 Again, double-twos can be thrown in only one way ; hence, the probability of throwing double-twos is — • 2 11 Therefore, the probability of throwing 4 is — - -f- — -, or -— • 36 36 1^ This example can be solved more easily by the method of § 635 ; the above method is given simply as an illustration of § 636. 2. A bag contains four $10 gold pieces and six silver dol- lars. If a person is entitled to draw two coins at random, what is the value of his expectation ? If a person has a chance of winning a certain sum of money, the product of the sum by the probability of his winning it is called his expectation. The number of combinations of the four gold pieces, taken 2 at a time, is — ^, and the number of combinations of the ten 10 '9 coins, taken 2 at a time, is -i-^', hence the probability of 4'3 2 drawing two gold coins is z-;—r, or --. 10 '9 15 Then the value of the expectation, so far as it depends on 2 -8 the drawing of two gold coins, is — X 20, or - dollars. 6 '5 1 The probability of drawing two silver coins is — -— , or -; the value of the corresponding expectation is - dollars. PROBABILITY 441 Again, the probability of drawing a gold coin and a silver coin is (6 ■ 4) -^ ' or -— ; the value of the corresponding ex- 1 '2 15 8 88 pectation is -— x 11, or -— dollars. 15 15 /8 2 88\ Hence, the value of the expectation is f - + - + — J dollars, or $9.20. ^^ ^ ^^^ EXERCISE 106 1. A bag contains 20 tickets numbered 1, 2, 3, •••, 20 ; if a ticket be drawn, what is the probability that its number is a multiple of 3 or 7 ? 2. Find the probability of throwing at least 9 in a single throw with two dice. 3. A bag contains 4 half-dollars and 6 quarter-dollars. If a person is entitled to draw a single coin, find the value of his expectation, 4. Find the probability of throwing 13 in a single throw with three dice. 5. A bag contains 3 dimes, 4 five-cent pieces, and 2 twenty-cent pieces. If a person is entitled to draw two coins, what is the value of his expectation ? 6. Find the probahility of throwing 7 in a single throw with four dice. 7. A bag contains 7 gold dollars, and 5 five-dollar gold pieces. If a person is entitled to draw four coins, what is the value of his expectation ? 8. A bag contains 6 fifty-cent pieces, and four other coins which have all the same value. If a person's expectation on drawing three coins is 120| cents, find the value of each of the unknown coins. COMPOUND EVENTS 638. Independent Events. If there are two independent events whose respective probabilities are known, the probability that both will happen is the product of their separate probabilities. Two events are said to be independent when the occurrence of one is not affected by the occurrence of the other. Let a be the number of ways in which the first event can happen, and b the number of ways in which it can fail; all these ways being equally likely to occur. 442 ADVANCED COURSE IN ALGEBRA Also, let a' be the number of ways in which the second event can happen, and b' the number of ways in which it can fail ; all these ways being equally likely to occur. We may associate together any one of the a-\-b cases in which the first event happens or fails, and any one of the a' -\-b' cases in which the second happens or fails ; hence there are (a-f6) (a'-\-b') cases, equally likely to occur. In aa' of these cases both events happen. Therefore, the probability that both events happen is aa' (a + 6)(a' + 6')* But — - — is the probability that the first event happens, a + 6 a' and the probability that the second happens. a +6' Hence, the jDrobability that both events happen is the product of their separate probabilities. And in general, if pi, P2,P3, •••, are the respective probabilities of any number of independent events, the probability that all the events happen is piP2P3*"« 639. Examples. 1. Find the probability of throwing an ace in the first only of two successive throws with a single die. The probability of throwing an ace at the first trial is ^* 5 The probability of not throwing one at the second is -• Hence, the probability of throwing an ace in the first only of 15 5 two successive throws is - X -, or -— • 6 6 oD 2. Find the probability of throwing an ace at least once in three throws with a single die. There will be an ace unless there are three failures. 5 The probability of failing at the first trial is -; and this is also the probability of failing at each of the other trials. PKOBABILITY 443 Hence, the probability that there will be three failures is 5^55 125 Then the probability that there will not be three failures is 3. A bag contains 5 red balls, 4 white balls, and 3 black balls. Three balls are drawn in succession, each being replaced before the next is drawn. What is the probability that the balls drawn are one of each color ? The probability that the first ball is red is — ; the proba- A i bility that the second is white is — , or - ; and the probability 3 1 that the third is black is — , or -. ' 12' 4 Hence, the probability of drawing a red ball, a white ball, 5 11 5 and a black ball, in this assigned order, is — x - X -, or . But a red ball, a white ball, and a black ball may be drawn in 1^, or 6 different orders (§ 625) ; and in each case the proba- bility is A. Then by § 636, the probability of drawing a red ball, a white ball, and a black ball, without regard to the order in which 5 5 they are drawn, is - — x 6, or — . 640. Dependent Events. TJie probability of the concurrent happening of two dependent events is the probability of the first, multiplied by the probability that when the first has hapjjened the second will follow. Let a and b have the same meanings as in § 638. Also, suppose that, after the first event has happened, a' represents the number of ways in which the second will follow, and b' the number of ways in which it will not follow; all these ways being equally likely to occur. 444 ADVANCED COURSE IN ALGEBRA Then there are in all (a + b) (a' + b') cases, equally likely to occur, and in aa' of these both events happen. Therefore, the probability that both events happen is (a + 6) (a' + 6') Hence, the probability that both events happen is the proba- bility of the first, multiplied by the probability that when it has happened the second will follow. And in general, if there are any number of dependent events such that pi is the probability of the first, p2 the probability that when the first has happened the second will follow, pg the probability that when the first and second have, happened the third will follow, and so on, then the probability that all the events happen is P1P2P3 ••*• 641. Examples. 1. Solve Ex. 3, § 639, if the balls are not replaced after being drawn. The probability that the first ball is red is — ; the probabil- ity that the second is white is — ; and the probability that the 3 third is black is — . 10 Hence, the probability of drawing a red ball, a white ball, 5 4 3 and a black ball, in this assigned order, is — x — X — • x^ 11 10 But the balls may be drawn in [3, or 6 different orders. Therefore, the probability of drawing a red ball, a white ball, and a black ball, without regard to the order in which they are drawn, is —- x -— x — : X 6, or — • ^ pjj ' 12 11 10 '11 2. An urn contains 5 white balls and 3 black balls ; another contains 4 white balls and 7 black balls. What is the proba- bility of obtaining a white ball by a single drawing from one of the urns taken at random ? PROBABILITY 445 Since the urns are equally likely to be taken, the proba- bility of taking the first urn is -; and the probability of then Li 5 drawing a white ball from it is -• 8 Hence, the probability of obtaining a white ball from the 15 5 first urn is - X -, or — • In like manner, the probability of obtaining a white ball from the second urn is - x — , or — Z 11 11 5 2 87 Hence, the required probability is 1 , or - — • ' ^ ^ -^ 16 11 176 642. Given the probability of the happening of an event in one trial, to find the probability of its happening exactly r times in n trials. Let p be the probability of the happening of the event in one trial. Then 1 — p is the probability of its failing (§ 634). JThe probability that the event will happen in each of the first r trials, and fail in each of the remaining n — r trials, is But the number of ways in which the event may happen exactly r times in n trials is equal to the number of combina- tions of n things taken r at a time, or n(n-l).-(n-r + l) ^^ ^26). \L Hentie, the probability that the event will happen exactly r times in n trials is \l Ex. What is the probability of throwing exactly three aces in five throws with a single die ? 446 ADVANCED COURSE IN ALGEBRA Here, ^~'a* r = 3, n = 5, and n — r -h 1 = 3. Substituting in (1), the required probability is '•^•'xr^Yxr^Yor^^^ 1-2.3 \(oj \6J' 3888 643. It follows from § 642 that, if the probability of the happening of the event in one trial is p, the probability of its failing exactly r times in oi trials is \l 644. Given the probability of the happening of an event in 07ie trial, to find the probability of its happening at least r times in n trials. The event happens at least r times if it happens exactly n times, or fails exactly once^, twice, •••, n — r times. Then the probability ^at it happens at least r times equals the sum of the probabilities of its happening exactly n times, or failing exactly once, twice, •••, n — r times. By §§ 642, 643, the required probability is \n~r Ex. What is the probability of throwing at least three aces in five throws with a single die ? Here, P = -^, r = 3, n = 5', then the required probability is 67 \6J \6 1-2 \6 \6r 648 EXERCISE 107 1. Find the probability of throwing exactly four sixes in six throws with a single die. 2. Find the probability of throwing at least four doublets in six throws with a pair of dice. PROBABILITY 447 3. A purse contains 5 dollars and 7 five-cent pieces, and another 3 dollars and 12 five-cent pieces. Find the probability of obtaining a dollar by drawing a single coin from one of the purses taken at random. 4. If a coin is tossed eight times, what is the probability that the head will turn up at least live times ? 5. A bag contains 5 white and 3 black balls. If 4 balls are drawn and not replaced, what is the probability that the balls drawn are alternately of different colors ? 6. What is the probability of throwing 10 with a pair of dice exactly three times in four trials ? 2 7. The probability of a certain event is - , and of another independent a ' of the first — . Find the probability that one at least of the events will happen. 8. If two coins are tossed up five times, find the probability that there will be five heads and five tails. 9. Each of four persons draws a card from a pack. Find the proba- bility that there will be one of each suit. 10. A, B, C. and D throw a die in succession, in the order named, until one throws an ace. Find their respective chances of throwing an ace at the first trial. » 11. A bag contains three white and six black balls. A person draws three balls, the balls when drawn not being replaced. What is the probability of drawing a white ball ? 12. A person has four tickets in a lottery in which there are three prizes and seven blanks. Find his chance of drawing a prize. 13. A box contains ten counters numbered 1, 2, 3, ..., 10. After one is drawn, it is put back, and the process is repeated indefinitely. Find- the probability that No. 1 will be drawn in four trials. 14. A bag contains six balls. A person takes one out, and replaces it. After he has done this six times, find the probability that he has had in his hand every ball in the bag. 15. In a series of games, the probability that the winner of any game 2 wi^ the next game is -. Find the probability that the winner of the o first game wins three or more of the next four. 16. A bag contains three tickets numbered 1, 2, 3. A ticket is drawn, and replaced. After this has been done four times, what is the probability that the sum of the numbers drawn is even ? 448 ADVANCED COURSE IN ALGEBRA 17. A purse contains a silver dollar and four dimes ; another contains five dimes. Four coins are taken from the former and put in the latter ; and then four coins are taken from the latter and put in the former. Find the probability that the dollar is still in the first purse. 18. A and B, with six others, draw lots for partners, and play four two- handed games, all the players being of equal skill. The four winners draw lots for partners, and play two games, and the winners in these games play a final game. Find the probability that A and B have played together. 19. If four whole numbers taken at random be multiplied together, find the chance that the last digit in the product is 1, 3, 7, or 9. 20. An urn contains 3 white and 3 black balls, and another 4 white and 4 black balls. A ball is taken from one and put in the other. If a ball be drawn from one of the urns chosen at a random, what is the probability that it is white ? CONTINUED FRACTIONS 449 XXXIII. CONTINUED FRACTIONS 645. A Continued Fraction is an expression of the form h h d a-\ 5 — ; or, a-\- c-{- e-\ as it is usually written. We shall consider in the present work only continued fractions of the form 1 1 & + c + ... ' where each numerator is unity, a a positive integer or 0, and each of the numbers &, c, •.., a positive Integer. 646. A terminating continued fraction is one in which the number of denominators is finite ; as, ,111 h-\- c-\- d An iyijinite continued fraction is one in which the number of denominators is indefinitely great. 647. In the continued fraction ,11 1 Ct2 + «3 + <^4 + • • • «! is called the^?'s^ convergent; tti H — the second convergent; ai -\ the third convergent; and so on. If ai = 0, as in the continued fraction 11 1 «2 + «3 + a4 + ••• then is considered the first convergent. 648. Any ordinary fraction in its loivest terms may he con- verted into a terminating continued fraction. 450 ADVANCED COURSE IN ALGEBKA Let the given fraction be -, where a and h are prime to each other. Divide a by h, and let aj denote the quotient and h^ the re- mainder; then, a , hi ,1 h h h_ Divide b by bi, and let a2 denote the quotient and &2 the re- mainder; then, a , 1 ,1 «2 + 7^ ^2+- &2 Again, divide 6i by &2? ^-nd let ag denote the quotient and 63 the remainder ; then, a , 1 ,1 - = a, + -—^ = a, + -—^-. a2-\ ajH T- The process is the same as that of finding the H. C. F. of a and b (§ 188) ; and since a and b are prime to each other, we must eventually obtain a remainder unity, at which point the operation terminates. Hence, any ordinary fraction in its lowest terms can be con- verted into a terminating continued fraction. Ex. Convert — into a continued fraction. 23 23)62(2 = ai 46 16)23(1 = a^ 16 7)16(2 = a3 14 2)7(3 = a4 6 1 CONTINUED FRACTIONS 451 Therefore, ^ = 2 + — -r^-i-^- '23 1+2+34-2 649. A quadratic surd (§ 368) may be converted into an infinite continued fraction. Ex. Convert V6 into a continued fraction. The greatest integer in V6 is 2 ; we then write V6 = 2 + (V6-2). Keducing V6 — 2 to an equivalent fraction with a rational numerator (§ 387), we have V6+2 V6+2 2 The greatest integer in ^ is 2 ; we then write Z V6 + 2 ^^, I V^-^ ^2 I (V^-2)(V6 + 2) ^^ J 1 2 2 . . 2(V6 + 2) V6 + 2* Substituting in (1), V6 = 2 H ^ (2) 2 + -J— V6 + 2 The greatest integer in V6 + 2 is 4 ; we then write V6 + 2 = 4 + (V6-2) = 4 + (^^^^:iMV6_±2} V6 + 2 = 4+-^ = 4+ ^ Substituting in (2), we have\ V6 = 2 + Lj ' + 4- / V6 + 2 452 ADVANCED COURSE IN ALGEBRA The steps now recur, and we have V6=V' 1111 2 + 4^-2 + 4 + ... An infinite continued fraction in which the denominators recur is called a periodi c continued fraction. 650. A periodic continued fraction may always he expressed as an irrational 7iumber. Ill 1 Ex. Express - — - — - — as an irrational number. 1 + 3 + 1 + 3+.-. Let X denote the value of the fraction ; then, ^^ 1 1 ^ 3 + a; ^ S-\-x 1 + 3 + a; 3 + a;+l 4 + a;' Clearing of fractions, 4 ic + aj^ = 3 + ic, or x^-\-Sx = S. Solving this equation, x = -^+^^ + 12 ^ -3 + V21. It is evident that the positive sign must be taken before the radical. PROPERTIES OF CONVERGENTS 651. In §§ 652 to 657, inclusive, we shall suppose the con- tinued fraction to be «.+ 1111 ^2+ %+ a„+ a„ + iH And we shall let j?^ denote the numerator, and q,.th.e denomi- nator, of the rth convergent (§ 647), when expressed in its simplest form. 652. To determine the laio of formation of the successive convergents. The first convergent is %. The second is aj + i = %^^2+J:. // CONTINUED FRACTIONS 453 Thethirdis 0, + ^- l=a,+ -^i- = ^lMi±^L±±3. ^2+ «3 a2«3 + l a2«3 + l The third convergent may be written in the form ( aia2 4-l)^.3 + cti , a2«3 + 1 ' in which we observe that : I. Tlie numerator equals the numerator of the preceding con- vergent, multiplied by the last denominator taken, plus the numera- tor of the convergent next hut one preceding. II. The denomiyiator equals the denominator of the preceding convergent, multiplied by the last denominator taken, plus the denominator of the convergent next but one preceding. We will now prove by Mathematical Induction that the above laws hold for all convergents after the second, when expressed in their simplest forms. Assume that the laws hold for all convergents as far as the ?ith inclusive. The nth convergent is^ = aiH ••• — . qn ^2+ a3+ «« Then, since the last denominator is a„, we have JPn = a«2)n-i+i^«-2, andg„ = a„g„_i + g„_2. . (1) Whence, ^ = %Pn-i+Pn-2 ^ (2) The (n + l)th convergent is ,11 11 «2+ «3+ ««+ ««+l which differs from the nth only in having a„4- , or ^»^"+i"^ , in place of a^. ^"^^ ^«+i Substituting ^n^^+i-hl ^^^ ^^ -^ ^2), we have ^nCtn+l + l Pn-l -1-Pn-2 Pn+1 __ <*n+l ^ Qn+i M«±L±io -i-^ ^n-1 + Qn-2 454 ADVANCED COURSE IN ALGEBRA Then, 9n+l Cln+1 («n Qn-l + Qn-z) + ?„-! ««+l2^«+i?n- ^ by (1). (3) It is evident that the second member of (3) is the simplest form of the (n + l)th convergent, and therefore Pn+l = an+iPn+Pn-l, ^nd Qn+l = Ctn-^lQn + Qn-V These results are in accordance with laws I and II. Hence, if the laws hold for all convergents as far as the nth inclusive, they also hold for the (n -\- l)th. But we know that they hold for the third convergent, and hence they hold for the fourth ; and since they hold as far as the fourth, they also hold for the fifth ; and so on. Hence, the laws hold for all convergents after the second. Ex. Find the first five convergents of 1 + ^-1.-1- 1 1+ 2+3+4+ ... The first convergent is 1, and the second is 1 + 1, or 2. Then, by aid of the laws just proved, the third is the fourth is the fifth is 2. 2 + 1 5 1. ^ + 1 "3' 5. 3 + 2 17 3. 3 + 1 ~10' 17 .4 + 5 73 10 .4 + 3 43 653. The difference between two consecutive convergents _ id ^n±l is -1 Qn+l qnQn+l The difference between the first and second convergents is ^ as) ^ ~ 0^2 Thus the theoi-em holds for the first and second convergents. continup:d fractions 455 Assume that it holds for the nth. and (71 + l)th convergents. That is, ^~-^' = — -— , ov p^q^+^r^p,+^q^==l. (1) Then, 7^- = -^ ."T-^^ — 4777 (§^^2) gn+l(a«+2gn+l + 5«) gn+ign+2 ^ ^ gn+l9«+2 Hence, if the theorem holds for any pair of consecutive con- vergents, it also holds for the next pair. But we know that it holds for the first and second conver- gents, and hence it also holds for the second and third ; and since it holds for the second and third, it also holds for the third and fourth ; and so on. Therefore, the theorem holds universally. As an example of the theorem, the difference Ipetween the fourth and fifth convergents, in the example in § 652, is 17 73 ^ 731 - 730 ^ 1 10 43 10x43 10x43* 654. It follows from § 653 that p^ and q^ can have no com- mon divisor except unity; for if they had, it would be a divisor of i>n^n+i~i5n+i^»j 0^ uuity, wMch is impossible. Therefore, all convergents formed in accordance with the laws of § 652 are in their lowest terms. 655. Tlie even convergents are greater, and the odd conver- gents less, than the fraction itself. I. The first convergent, ai, is less than the fraction itself, since is omitted. a2H II. The second, a^ -\ — , is greater, because its denominator Oj ^ <^2 is less than ctg -\ — — , the denominator of the fraction. c^s ~r * * * 456 ADVANCED COURSE IN ALGEBRA III. The third, ai -\ — , is less, because, by II, the -1 ^a ~f" % 11 denominator 00+— is greater than a^ -] , the as ^ - a3+a, H denominator of the fraction; and so on. Hence, the first, third, •••, convergents are less, and the second, fourth, •••, convergents greater than the fraction itself. 656. Any convergent is nearer than the preceding convergent to the value of the /inaction itself. By §652, P^^<^n.^Pn..+Pn^ The fraction itself is obtained from its (n -f 2)th convergent by putting a^+2 H j- — in place of a„+2- Hence, denoting the value of the fraction by x, we have 1 ««4-2H-r «n+3+"- where m stands for a„+2 ' P'^^+P" ™p„,,+p„ Now, X • ^' m{Pn+iqn'^Pnqn+l} qn('^qn+i + qn) (§653). • (1) Also, X qn('^qn+i + qn) Pn+l ^ '^Pn+1 +Pn ^Pn+1 qn+1 mq^^^ + qn^ qn+i Pnqn+y^Pn+iqn ^n+l(^g«+l + qn) qn+l(inqn+l + ^n) (2) Since a^+2 is a positive integer, a„^2 H r — is > 1 ; that is,mis>l. «n+3 + - And since g„+i = a^+,q^ + $n-i (§ 652), q^^^ is > g«. CONTINUED FRACTIONS 457 Therefore, the fraction (2) is less than the fraction (1), for it has a smaller numerator and a greater denominator. Hence, the (n + l)th convergent is nearer than the wth to the value of the fraction itself. ^ 657. To determine limits to the error made in taking the nth convergent for the fraction itself With the notation of § &^Q>, the difference between the fraction itself and its nth convergent is m. 1 qni^Qn+l + qn) 'f^^ + S (1) Since m is >1 (§ 656), the denominator 9,/9„+i + — ) qnQn+i- Hence, the fraction (1) is > — ■. — - , and < That is, the error made in taking the ?ith convergent for the fraction itself lies between 1 and 1 qniQn+l + qn) qnqn+l As an example of the above theorem, the error made in taking the fourth convergent for the fraction itself, in the example in § 652, lies between - ill and , or — and 10(43 + 10) 10x43' 530 4^0 EXERCISE 108 Convert each of the following into a continued fraction, and find in each case the first five convergents. • J 118 3 145 6 118 ^ n^ ■ 91 ' * 612' ' 67l' ■ 1561' 2. 2o3. 4. 5.83. 6. ^. 8. ^. 179 611 5151 Convert each of the following into a continued fraction, find in each case the first four convergents, and determine limits to the error made in taking the third convergent for the fraction itself, 458 ADVANCED COURSE IN ALGEBRA 9. \/26. JO. V37. 11. a/8. 12. Vie. 13. 4- 14. V27. 15. v/63. 16. Vl9. 17 1 +\/46 18. V23. 19. 2\/7. 20. 3\/l4. Express each of the following in the form of a surd 1111 21 4+3+4+3+ 1111 2+5+2+5+ 25. a + 1 23. 4 + 1 1 2 + 2+ ... 24 1 + 1 1 1 1 1+6+1+6 + ••• 1 1 a + 2a + a + 2a + --. 26. The sidereal year is approximately 365.25636 days ; express the excess above 365 days as a continued fraction, and find its first four convergents. 27. A kilometer is approximately .62138 mile ; express this decimal as a continued fraction, find its fifth convergent, and determine limits to the error made in taking this convergent for the fraction itself. 28. A meter is approximately 1.09.863 yards ; express this decimal as ^ continued fraction, find its sixth convergent, and determine limits to the error made in taking this convergent for the fraction itself. 29. Express the greatest root of the equation 2 ic2 - 10 X = - 5 as a continued fraction, and find the first five convergents. Convert each of the following into a continued fraction, and find in each case the first four convergents : 30. V74. 31. -^- 32. \/a2 + a. 33. V57. V55 SUMMATION OF SERIES 459 XXXIV. SUMMATION OP SERIES 658. The Summation of an infinite literal series is the pro- cess of finding an expression frotn which the series may be developed. In § 630, we gave a method for finding the sum of an infinite geometric series. RECURRING SfRIES 659. Consider the infinite series 1 + 2 a? + 3 x'^ + 4 a;3 + 5 a;* + . . . . Here, (3 a^ - 2 x(2 x) + x\l) = 0, (4 a^) - 2 x(3 x') -\- x\2 x) = 0, etc. That is, any three consecutive terms, as for example 2x, Sx^, and 4 a?^, are so related that the third, minus 2 x times the second, plus a^ times the first, equals 0. 660. A Recurring Series is an infinite series of the form tto + OriOS + a^x^ -\ , where any r + 1 consecutive terms, as for example are so related that a„a;«+M«n-ia?"-') + gaj2(a„_2a:«-2)+ ... -f sa;'-(a„_,af»-'-) = ; p, q, '", s being constants. The above recurring series is said to be of the rth order, and the expression i +^^ + 3^. + ... + ,^. is called its scale of relation. The recurring series of § G59 is of the seconji order, and its scale of relation is 1 — 2 x + x^. An infinite geometric series is a recurring series of the first order. 460 ADVANCED COURSE IN ALGEBRA Thus, in the infinite geometric series 1 +a; + cc2 + x3+ •••, any two consecutive terms, as for example x^ and x^, are so related that (x^) — x(x^) = ; and the scale of relation is 1 — ic. 661. To find the scale of relation of a recurring series. If the series is of the first order, the scale of relation may be found by dividing any term by the preceding term, and sub- tracting the result from 1. If the series is of the second order, ao, a^, a^, a^, •••, its con- secutive coefficients, and l-\-px-\-qx^ its scale of relation, we shall have (a,+pa, + qao = 0, ttg +pa2 + qai = ; ^ ^ from which p and q may be determined. If the series is of the third order, aQ, a^, a^, a^, a^ a^, •••, its consecutive coefficients, and 1 -{- px -{- qx^ -{- ra^ its scale of rela- tion, we shall have ' 03 -f- pa2 + qai -{- rao = 0, • a^-{-pas + qa2 + rai = 0, . as +pa4 + (/a3 H- m2 = ; from which p, q, and r may be determined. It is evident from the above* that the scale of relation of a recurring series of the rth order may be determined when any 2r consecutive terms are given. To ascertain the order of a series, we may first make trial of a scale of relation of three terms ; if the result does not agree with the series, try a scale of four terms, five terms, and so on until the correct scale of relation is found. If the series is assumed to be of too high an order, the equa- tions corresponding to the assumed scale will not be indepen- dent. (Compare § 269.) 662= To find the sum of a recurring series when its scale of relation is known. Let 1 -\-px-{- qxP be the scale of relation of the series SUMMATION OF SERIES 461 Denoting the sum of the first n terms by S,^, we have Then, pxS^ = pa^x -i-pa^of -\ hpo^„-2^""^ + i>«„_i«", and qx^S„ = qa(fic^ H h qa^^^sif-^ + qa„_2^ + go^«-li»"+^ Adding these equations, and remembering that, by virtue of the scale of relation, the coefficients of x^, oif, •••, aj"~^ become 0, and we have Sn(l+px-{-qx^ = Oo + («i +i>«o) ^ + (i>««-i + gotn-2) a?" + ga„_ia;"+^ Whence, o _ gp + («i +pao) a; + (pan-i + g<^n-2) a^" + qan-i^'"'^'^ . n\ l+px + qaf ' ^^^ which is a formula for the sum of the first n terms of a recurring series of the second order. If X is so taken that the given series is convergent, the expression ^^^^^ j^ ^^^_^) ^n ^ ga„_ia;"+i approaches the limit when n is indefinitely increased (§ 542), and (1) becomes ^ ^ a, -^ (a, -^ pa,) x , .^. l-{-px + qa^ ' ^ ^ which is a formula for the value (§ 540) of a recurring series of the second order. If q = Oy the series is of the first order, and therefore Oh + mo = ; whence, ^ which is a formula for the value of a recurring series of the first order. (Compare § 530.) In like manner, we shall find the formula ^ _ ftp + («! -j-paj)) X + («o + pai 4- qoo) x^ .^. l-\-px-{- qx^ H- rx^ for the value of a recurring series of the third order. 462 ADVANCED COURSE IN ALGEBRA It will be observed, in formulae (1), (3), and (4), that the denominator is the scale of relation. A recurring series is formed by the expansion, in an infinite series, of a fraction, called the generating fraction. The operation of summation reproduces the fraction ; the process being just the reverse of that of § 564. Ex. Find the sum of the series 2 + a; + 5 0^2 + 7 ic^ + 1 7 a;4 + . . . . To determine the scale of relation, we first assume the series to be of the second order (§ 661). Substituting aQ = 2, ai = 1, ag = 5, Wg = 1, in (1), § 661, |5+ i> + 2g = 0, l7 + 5i9+ g = o. Solving these equations, p = — 1, g = — 2. To ascertain if 1 — a; — 2 a;^ is the correct scale of relation, consider the fifth term. Since 17 a;^ + (- x) (7 aj^) + (- 2 x") (5 x^) is 0, it follows that 1 — a; — 2 a?^ is the correct scale. Substituting the values of do, ai, p, and q in (2), o^ 2 + (l-2)a; ^ 2 - a; l-x-2x' l_a;-2a;2' The result may be verified by expansion. The series expresses the value of the fraction only for such values of x as make the series convergent. To find for what values of x the given series is convergent, we proceed as in § 572 ; we find by the method of § 567, 2-x _ 1 ,1 l-x-2a;2 \-2x \ + x = (1 + 2 cc + 2%2 +...) + (! _a; + cc2 ). The nth term of the given series is [2'*-i + (— 1)*'~^] x"-i. The ratio of the (n + l)th term to the nth term is or ^-— [2»+(- l)^]a;^ SUMMATION OF SERIES 463 This approaches the limit 2 x, when n is indefinitely increased. Then, the series is convergent if x is numerically <- (§ 555, I.). EXERCISE 109 In each of the following find the generating fraction, and the expres- sion for the nth term, and determine for what values of x the series is convergent : - 1. 4 - X + 7 ic2 - 5 a;3 + 19 ic* + —. 2. 1 - 13 x - 23 x2 - 85 5c3 - 239 x* + .... 3. 1 + 5 X + 21 x2 + 85 a:3 + 341 x* + .... 4. 5 ~ 13 X + 35 x2 - 97 x3 + 275 x* + .... 6. 3 + 10 X + 36 x2 + 136 x^ + 528 x* + .... 6. 1 - 2 X + x2 + 22 x3 - 191 x* + .-. 7. 3 + X + 33 x2 + 109 x3 + 657 x* + •••. 8. 1 + 31 X - 19 x2 + 391 x3 - 619 x* + .... In each of the following find the generating fraction, and continue the series to two more terms : 9. 1 + 2 X - 3 x2 + 6 x3 - 7 X* + 10 x5 - 11 x6 + .... 10. 1 - 2 X - x2 - 7 x3 - 18 x* - 59 x5 - 181 x^ + .... 11. 2 - 11 X + 15 x2 + 20 x3 - 133 x^ + 231 x^ + 130 x^ + .... THE DIFFERENTIAL METHOD 663. If the first term of a series be subtracted from the second, the second from the third, and so on, a series is formed which is called the Jlrst order of differences of the given series. The first order of differences of this new series is called the second order of differences of the given series ; and so on. Thus, in the series 1, 8, 27, 64, 125, 216, .••, the successive orders of differences are as follows : 1st order, 7, 19, 37, 61, 91, ••.. 2d order, 12, 18, 24, 30, •••. Sd order, 6, 6, 6, 4th order, 0, 0, 464 ADVANCED COURSE IN ALGEBRA The Differential Method is a method for finding any term, or the sum of any number of terms of a series, by means of its successive orders of differences. 664. To find any term of the series %, a^i %, 0^4, •••, a„, Ojn+ly '"' The successive orders of differences are as follows : 1st order, Og — tti, a-g — as? cii — ds, '", «n+i — ^n; **'• 2d order, ag — 2a2 + a], (14— 2o3 + a2, •••. 3d order, (14 — 3 ag + 3 ^2 — ^d • • • 5 etc. Denoting the first terms of the 1st, 2d, 3d, •••, orders of dif- ferences by di, d2, c?3, •••, respectively, we have di = a2 — (h', whence, otg = % + ^i- ^2 = ttg — 2 a2 + «! ; whence, ttg = — ai + 2 tta + c?2 = — oti + 2 % + 2 c?i + (?2 = <^ + 2 c?i + (^2- dg = a4 — 3 ag + 3 as — % ; whence, a4 = a^ — 3 tta + 3 ag -h c?3 = tti -h 3 dj + 3 ^2 + c?3 ; etc. It will be observed, in the values of as, a^, and a^, that the coefficients of the terms are the same as the coefficients of the terms in the expansion by the Binomial Theorem of a + a; to the ^rs^, second, and third powers, respectively. We will now prove by Mathematical Induction that this law holds for any term of the given series. Assume the law to hold for the nth term, a„ ; then the coef- ficients of the terms will be the same as the coefficients of the terms in the expansion by the Binomial Theorem of a + ic to the (n — l)th power ; that is, , , ., , , (n-l)(w-2) , [^ ^ (n-l)(n-2)(n-3) ^^^..._ ^^^ If the law holds for the wth term of any series, it must also hold for the nth. term of the first order of differences. SUMMATION OF SERIES 465 Or, a„^, - a. = d, -\-(n-l)d, + (^-l)(^-2) ^j^ _^ ...^ ^2) Adding (1) and (2), we have <»„+i = ai + [(»-l)+l](«. + ^[(»-2) + 2]d2 This result is in accordance with the above law. Hence, if the law holds for the nth term of the given series, it holds for the {n -f l)th term ; but we know that it holds for the fourth term, and hence it holds for the fifth term; and so on. Therefore, (1) holds for any term of the given series. If the differences finally become zero, the value of an can be obtained exactly. 665. To find the sum of the first n terms of the series ai, cfca, ag, a^, a„ .... (1) Let S denote the sum of the first n terms. Then S is the (n + l)th term of the series 0, ai, ai + aa, ai + aa + agj •••• (2) The first order of differences of (2) is series (1) ; whence, the rth order of differences of (2) is the same as the (r — l)th order of differences of (1). Then, if di, do, ••., represent the first terms of the 1st, 2d, •••, orders of differences of (1), a^, cZ^, c?2, .••, will be the first terms of the 1st, 2d, 3d, •.•, orders of differences of (2). Putting (Xi = 0, di = «!, (^2 = c?i? etc., in (3), § 664, ^ = na, + ^%lA)d, + "("-|H^^-^) d2+->. (3) 666. Ex. Find the twelfth term, and the sum of the first twelve terms, of the series 1, 8, 27, 64, 125, .... 466 ADVANCED COURSE IN ALGEBRA Here, n = 12, a^ = 1. Also, di = 7, ^2=12, (^3 = 6, and f^4 = (§663). Substituting in (1), § 664, the twelfth term = 1 4- 11 . 7 + llli^ . 12 + ^^ '\^'^ . 6 = 1728. Substituting in (3), § 665, the sum of the first twelve terms = 12 + iilll . 7 + l^^iill^ . 12 + 1241:^ . 6 = 6084. 667. Piles of Shot. Ex. If shot be piled in the shape of a pyramid with a tri- angular base, each side of which exhibits 9 shot, find the num- ber in the pile. The number of shot in the first five courses are 1, 3, 6, 10, and 15, respectively ; we have then to find the sum of the first nine terms of the 'series 1, 3, 6, 10, 15, •••. The successive orders of differences are as follows : 1st order, 2, 3, 4, 5, •••. 2d order, 1, 1, 1, •••. 3d order, 0, 0, .... Putting ?i = 9, «! = 1, di = 2, c\ = 1 in (3), § 665, EXERCISE no 1. Pind the first term of the sixth order of differences of the series 3, 5, 11, 27, 67, 159, 375, .... 2. Find the 15th term, and the sum of the first 15 terms, of the series 1, 9, 21, 37, 57, .... 3. Find the 14th term, and the sum of the first 14 terms, of the series 5, 14, 15, 8, -7, .... 4. Find the sum of the first n multiples of 3. 6. Find the nth term, and the sum of the first n terms, of the series 2, - 1, 1, 8, 20, .... SUMMATION OF SERIES 467 6. If shot be piled in the shape of a pyramid with a square base, each side of which exhibits 25 shot, find the number in the pile. 7. 1'ind the 13th term, and the sum of the first 13 terms, of the series 1, 3,9,25, 57,111, -. 8. Find the 10th term, and the sum of the first 10 terms, of the series 4, -2,10, 4, -56,-206, ....' 9. Find the sum of the squares of the first n multiples of 2. 10. Find the n\h term, and the sum of the first n terms, of the series 1, -3,-13,- 17, -3, 41, .... 11. Find the number of shot in a pile of 9 courses, with a rectangular base, if the number of shot in the longest side of the base is 24. 12. Find the number of shot in a truncated pile of 10 courses, with a square base, if the number of shot in each side of the lower base is 16. 13. Find the number of shot in a truncated pile of 8 courses, with a rectangular base, if the number of shot in the length and breadth of the base are 20 and 14, respectively. 14. Find the 12th term, and the sum of the first 12 terms, of the series 1, 13, 49, 139, 333, 701, 1333, .... 15. Find the 9th term, and 'the sum of the first 9 terms, of the series 20, 4, - 36, - 132, - 356, - 820, - 1676, .... 16. Find the sum of the fourth powers of the first n natural numbers. 17. Find the number of shot in a pile with a rectangular base, if the number of . shot in the length and breadth of the base are m and w, respectively. 18. Find the number of shot in a truncated pile of n courses, with a triangular base, if the number of shot in each side of the lower base is m. INTERPOLATION 668. Interpolation is the process of introducing between the terms of a series other terms conforming to the law of the series. Its usual application is in finding intermediate numbers between those given in Mathematical Tables. The operation is effected by giving fractional values to n in (1), § 664. The method of Interpolation rests on the assumption that a formula which has been proved for an integral value of n, holds also when n is fractional. 468 ADVANCED COURSE IN ALGEBRA Ex. Given V5 = 2.2361, V6 = 2.4495, V7 = 2.6458, V8- 2.8284, ...; find V6.3. In this case the successive orders of differences are : .2134, .1963, .1826, .... -.0171, -.0137, .... .0034, .... Whence, d^ = .2134, d^ = - .0171, d^ = .0034, .... Now, the required term is distant 1.3 intervals from V5. Substituting n = 2.3 in (1), § 664, we have, approximately, V6:3 = 2.2361 + 1.3 x .2134 + ^f ^ f (- .0171) 1x2^ . 1.3 X •3x-.7^_()Q3^ 1x2x3 = 2.2361 + .2774 - .0033 - .0002 = 2.5100. EXERCISE III 1. Given log 26 = 1.4150, log 27 = 1.4314, log28 = l. 4472, log 29=1.4624, ... ; find log 26. 7. 2. Given v^91 = 4.49794, v/92 =4.51436, v^93=4.53066, \/94=4.54684, ...; find v'92.5. 3. The reciprocal of 35 is .02857 ; of 36, .02778 ; of 37, .02703 ; of 38, .02032 ; etc. Find the reciprocal of 36.28. 4. Given log 124 = 2.09342, log 125 = 2.09691, log 126 = 2.10037, log 127 = 2.10380, ... ; find log 125.36. 6. Given 21^ = 9261, 22^ = 10648, 23^ = 12167, 24^ = 13824, and 253 = 15625 ; find the cube of 2H. 6. Given log 61 = 1.78533, log 62 = 1.79239, log 63 = 1.79934, log 64 = 1.80618, ... ; find log 63.527. SUMMATION OF SERIES BY SEPARATION INTO PARTIAL FRACTIONS 318. 1. Find the sum of the first n terms of the infinite series i i i 2.3 3.4 4.5 ' and determine whether the series is convergent or divergent. SUMMATION OF SERIES 469 The nth. term of the above series is — ; • (n + l)(n + 2) Separating into partial fractions by the method of § 567, we have ^ 11 (n + 1) (n-\-2) 71 + 1 n-\-2 Then the sum of the first n terms of the given series is (|-|)-(l-i)K14)--(.-Ti-.-i2) 11 n 2 n-i-2 2(n-^2) 11 1 Since approaches the limit -, when 7i is indefinitely 2 n -\-2 2 increased, the series is convergent (§ 543). 2. Find the sum of the first n terms of the infinite series 1-3 2-4 3-5 ■ ' and determine whether the series is convergent or divergent. The nth term of the series is — — , which, by the method n{n + 2) of § 567, equals A -1^^. 2 n 2{n + 2) Then the sum of the first n terms is \2 6; V4 ^) \^ lOy^ [_2n 2(n + 2)J All the terms cancel except the first two positive, and the last two negative. Thus the sum of the first n terms is 1,11 1 3 1 1 2 4 2(n + l) 2(n + 2)' 4 2(n + l) 2(7i + 2) 3 The latter expression approaches the limit - when n is indefinitely increased. Hence, the series is convergent. 470 ADVANCED COURSE IN ALGEBRA EXERCISE 112 In each of the following, find the sum of the first n terms, and deter- mine whether the series is convergent or divergent : 2. 53. 10 4. 15 1-42.53.6 12 . 22 ^ 22 . 32 32 . 42 In each of the following, find the sum of the first n terms, and deter- mine for what values of x the series is convergent : - 1 + , ..A ... + . J. „, + .■■■ a;(x + l) (x+l)(a; + 2) (x + 2)(a: + 3) (l+x)(l + 2x) (l + 2x)(l + 3x) (l + 8x)(l + 4x) 8. Find the sum of the first n terms of : 1 + , .,, 1 ., „. + ■■■■ «(x + l)(a: + 2) (x + l)(a; + 2)(a; + 3) THEORY OF NUMBERS • 471 XXXV. THEORY OF NUMBERS 670. In the present chapter, the word number will signify a positive integer; and every letter will be understood as repre- senting a positive integer. One number is said to divide another when it is contained in it without a remainder ; in this case, the second number is said to be a multiple of the first. A prime number is a number which cannot be divided, with- out a remainder, by any number except itself and unity. One number is said to be prime to another wnen there is no number, except unity, which will divide each of them without a remainder. 671. If a number divides the product of two others^ and is prime to one of them, it must divide the other. Let the number n divide the product ab, and be prime to a. Since n divides ab, the prime factors of ab must be the same as those of n, with certain additional prime factors. But since n is prime to a, n and a have no common factor except unity. Then, the prime factors of b must be either the same as those of n, or with certain additional prime factors ; and n divides b. 672. If a prime number divides the product of any number of factors, it must divide some f alitor of the product. Letp be a prime number which divides the product abc •••. Then, the prime factors of abc ••• must hep, with certain additional prime factors. Then, some one of the numbers a, b, c, '•-, must have p as a prime factor, and therefore p divides some factor of the product. 673. It follows from § 672 that If a prime number divides a positive integral power of a num- ber a, it must divide a. 472 ADVANCED COURSE IN ALGEBRA 674. If the numerator and denominator of a fraction are prime to each other, the fraction is in its lowest terms. Let a be prime to b. If possible, let - = — , where b' is < b. ^ ab' Multiplying both members by b', a' = — • Whence, since a' is an integer, b divides ab'. But a is prime to b, and hence b divides b' (§ 671). But this is impossible if b' is < b. Hence, b' cannot be < b, and - is in its lowest terms. b 675. It was proved, in § 674, that b divides b'. Whence, b' = mb, where m is an integer. rnr, , ab' amb Then, a' = — = -—~ = am. b b Hence, if a is prime to b, and ~ = — , a' and b' are equimulti- ples of a and b. 676. A number can be resolved into prime factors in only one way. Let w be a composite number ; and suppose n — abc •••, where a, b, c, •••, are prime numbers. Suppose also, if possible, that n = a'b'c' •••, where a', &',• c', •••, is a different set of prime numbers. We have, a6c ••• = a'6'c' ••-. (1) Then, a divides a'b'c' •", and therefore divides some one of the numbers a', b', c', ••• (§ 672). Then, since a,b,c, -", and a', b', c', •••, are prime numbers, a must equal some one of the numbers a', b', c', •••. "-- Suppose, then, a = a' -, dividing the members of (1) by a and a', respectively, ^^... ^^/c'-..; and, as before, b equals one of the numbers b', c', •••. Proceeding in this way, we can prove each of the numbers a, bf c, •••, equal to one of the numbers a', b', c', •••. THEORY OF NUMBERS 478 677. To find the highest power of 2 which is contained in [14. Of the factors 1, 2, 3, ••-, 14, the numbers 2, 2 x 2, 3 x 2, ••., 7x2 contain 2 as a factor at least once. The numbers 2^, 2 x 2^, 3 x 2^ contain 2^ as a factor at least once. The number 2^ contains 2^ as a factor once. Then, the highest power of 2 in [14 is evidently 7+3 + 1, or 11. We will now consider the general case. To find the highest power of a, where a is any prime number, which is contaiyied in \n. Of the factors 1, 2, 3, •••, n, the numbers a, 2 a, 3 a, •••, con- tain a as a factor at least once. Let the last term of this series be pa. Then, pa either equals w, or is < n ; whence, n ^n p = -, ov p< — a a Therefore, p is the greatest integer in — Again, the numbers a^, 2a^, Sa^, •••, contain a^ as a factor at least once ; let the last term of this series be qa\ Then, as before, q = —^, or q<—' a^ a- Whence, q is the greatest integer in — • ar Continuing in this way, the highest power of a in \n is i> + g + r+..-, where p is the greatest integer in -, q in — , r in — . etc, a (T (T We will now solve the example of § 677 by this method. In this case, a = 2, w = 14 ; then, t* = 7, ^ = I, -^ t= I 7 7 The greatest integer in 7 is 7 ; in -, is 3 ; in -, is 1. 2 4 Then, the highest power of 2 in [14 is 7 + 3 + 1, or 11. 474 ADVANCED COURSE IN ALGEBRA 678. The product of any n consecutive integers is divisible by [n. Let the integers be m + 1, m -\- 2, •••, m -\- n. Multiplying both numerator and denominator by the prod- uct of the natural numbers from 1 to m, inclusive, the fraction i (m 4- 1) (m + 2) .-. (m 4- n) ^ 1^^ + ^ \n \m\n If a is any prime number, the exponent of a in \m -^ 7i is 7n -\- 11 p + g + r-f •••, where p is the greatest integer in , q in — =^, r in \ , etc. (§ 677). a'^ a^ The exponent of a in [m is pi + ^i + r^ H , where p^ is the ,,., . m m ■ m , greatest integer m — , g^ m — , r^ m — , etc. a a^ or The exponent of a in |^ is pg + Q'2 + ^2 + •*•? where p2 is the greatest integer m -, gg m — ? ^2 m — j ©tc. tt a^ a^ Then, the exponent of a in |m[9i is {pi +p^ + fe + ^'2) + •••• Now, the greatest integer in is either the sum of the greatest integers in — and -, or else it exceeds this sum by unity. That is, p equals Pi -{-P2, or Pi-\-p2-\- 1. Similarly, q equals gi + gg, or gi + g2 + 1 ; etc. Then, p ^ q ^ r -] is not < (pi + q^) + {P2 + ^2) + ••• ; that is, the exponent of a in \m + n is not < its exponent in \m\n. ^, Im + n Then, must be an integer. \m\n Whence, (m + 1) (m 4- 2) • • • (m + n) is divisible by \n. 679. If n is a prime number, the coefficieiit of any term, except the first and last, in the expansion by the Binomial Theorem of {a + xy, is divisible by n. THEORY OF NUMBERS 475 By § 287, the coefficient of the (r + l)th term of (a + a;)" is \r By § 678, n (n — 1) '•' (n — r -{- 1) is divisible by |r. If r has any value from 1 to n — 1, none of the factors of [r , can divide n] for, by hypothesis, n is a prime number. Then, \r must divide (w — 1) ••• (n — ?- + 1). Therefore, expression (1) is an integer, and divisible by n. 680. Fermat's Theorem. If 71 is a prime number, and m is prime to n, m"~^ — 1 is a multiple of 71. Let a, b, c, '•', k be any m numbers. Expanding by the Binomial Theorem, we have (a + 6 + c + ... + A;)" = [a + (& + c + - + k)Y = a" + na" ' (?> + c 4- • • • + fc) + ''^(^-^) a«-2 (6 4- c + . . . + A;)2 + -+(6 + c + ...+A;)". *- . (1) By § 679, each coefficient n, ^^^~ ^ ^ -.., is divisible by n. Then, every term in the expansion (1), which is not a power of a, b, c, ..., k, is a multiple of n; and (1) can be written (a + & + c 4- ••• + A;)" = (a» + &" + c" + ♦•• + A;") +_pn, where p is an integer. Putting a = b = c= •" =k=l, we have m'^ = m -i-pn, or m" — m =pn, or m (m"~^ — 1) =pn. Since m is prime to n, it cannot be a multiple of n. Then, m"7^ — 1 must be a multiple of n. 681. The result m^' — m^pn, of § 680, holds if m is not prime to n. That is, m" — m is a multiple^f n when 7i is prime, whether m is prime to n or not. 476 ADVANCED COURSE IN ALGEBRA EXAMPLES 682. 1. If two consecutive numbers are not multiples of 3, their sum is a multiple of 3. The numbers must be in the forms 3 m + 1 and 3 m + 2. Then their sum is 6 m + 3, which is a multiple of 3. It follows from the above that any number of the form n(n + 1) [w + (w + 1)], or n(n + 1) (2 n + 1), is divisible by 6. For either n or n + 1 must be even ; and if neither n nor w + 1 is a multiple of 3, their sum is a multiple of 3. 2. Every perfect square is in the form 5 n, ov 5n ± 1. Tor any number is in the form 5 m, or 5 m ± a, where a may be 1 or 2. (5 my is in the form 5 n. Again, {5 m ± ay equals a multiple of 5, plus al But a^ is either 1 or 4 ; and 4 = 5 — 1. Hence, (5 m ± ay differs by 1 from a multiple of 5, and is therefore in the form 5n-\-l or 5n — l. 3. If n is even, and not a multiple of 3, n^ -\-2 is divisible by 6. For n must be in the form 3 m ± 1, where m is odd. Then, n^ -\-2 = 9 m^ ±6 m-\-l + 2, which is divisible by 3. Again, n^-\-2 is even; hence, it is divisible by 6. 4. Every even power of an odd number is in the form -8 n + 1. Now, any odd number is in one of the forms 8 m ± 1, or 8 m ± 3. If this be raised to any even power, the result will be a multiple of 8 plus 1, or a multiple of 8 plus an even power of 3. Now any even power of 3 is in the form of a multiple of 8, plus 1. Hence, any even power of an odd number is in the form 8n + l. THEORY OF NUMBERS 477 5. Prove that n' — n is divisible by 42. By § 681, since 7 is prime, n'' — n is divisible by 7. Again, nJ — n = n {n^ — l) = n (n^ — 1) (71'^ -f n^ + 1) = (n - 1) n (n + 1) (n'^ + ri^ + 1). By § 678, (n -1) n (n + 1) is divisible by [3, or 6. Hence, -nJ — n is divisible by 7 and 6, or by 42. 6. Prove that the fourth power of any number is in the form 5 71 or 5 ?i + 1- If m is prime to 5, m'' — 1 is a niultiple of 5 (§ 680). That is, m* is a multiple of 5, increased by 1 ; and is there- fore in the form 5n -\-l. If m is not prime to 5, it contains 5 as a factor; and m^ is in the form 5 n. EXERCISE 113 1. If n is odd, and greater than 1, (n — 1) n (n + 1) is divisible by 24. 2. Prove that n^ — n is divisible by 30, if n is greater than 1. 3. Prove that w (w + 1) (n + 5) is a multiple of 6. 4. Find the highest power of 2 in |^. 5. Find the highest power of 3 in [80. 6. Prove that every perfect square is in the form 3 n, or 3 n + 1- 7. Prove that every perfect sixth power is in the form 7 n, or 7 w + 1. 8. Prove that, if w is odd, and not a multiple of 3, n^ + 5 is divisible byC. 9. Prove that, if n is odd, (71^ + 3) (w^ + 7) is divisible by 32. 10. Prove that any odd power of 7 is in the form 8 n — 1. 11. Prove that w^ — 5 n^ + 4 » is divisible by 120, if n is greater than 2. 12. Prove that, if w is odd, and greater than 1, n^ — n is divisible by 240. 13. Prove that every perfect cube is in the form 7 w, or 7 n i 1. 14. Prove that every perfect cube is in the form 9 n, or 9 w ± 1. 478 ADVANCED COUKSE IN ALGEBRA XXXVI. DETERMINANTS IS 683. The solution of the equations a^x + bi,y = Ci, a^x + b^ = C2, hfi — 61C2 CoO-i — Cia2 x = y a-J)2 — <^2^1 <^1^2 — ^2^1 The common denominator may be written in the form (1) This is understood as signifying the product of the upper left-hand and lower right-hand numbers, minus the product of the lower left-hand and upper right-hand. The expression (1) is called a Determinant of the Second Orde>. The numerators of the above fractions can also be expressed as deter- minants ; thus, ■ ^ 1, X. ^ll 02CI - 01C2 = ^ 02, &2 , and C2ai ^1' ^^ -cia2= ^ ^ • a2, C2 684. The solution of the equations r a,x + b^tj + c,z = di, a^ + ^22/ + G^ = d2, I agOJ + &3?/ + C32; = dg, .„ ^ d,h2C, - d,b,C2 + ^2?>3Ci - ^2&iC3 4- dsb,C2 - d,b2Ci C^A^S — CtAC'I + «2&3Cl — «2&lC3 + «3&lC2 " ^3 Vl with results of similar form for y and 2;. The denominator of (1) may be written in the form «!, 5i, Ci a2, b2, C2 • % &3, c? (1) (2) DETERMINANTS 479 This is understood as signifying the sura of the products of the numbers connected by lines parallel to a line joining the upper left-hand corner to the lower right-hand, in the follow- ing diagram, minus the sum of the products of the numbers connected by lines parallel .to a line joining the lower left-hand corner to the upper right-hand. Y 1 The expression (2) is called a Determinant of the Third Order. The numerator of the value of x can also be expressed as a determinant, as follows : as may be verified by expanding it by the above rule. 685. The numbers in the first, second, etc., horizontal lines, of a determinant, are said to be in the Jirst, second, etc., 7'ows, respectively ; and the numbers in the first, second, etc., vertical columns, in th^ Jirst, second, etc., columns. The numbers constituting the determinant are called its eletneyits, and the products in the expanded form its terms. Thus, in the determinant (2), of § 684, the elements are «i> <^2j %) ^tc, and the terms a^h^Cz, — a-J^^Cz, etc. EXERCISE 114 Evaluate the following determinants : "x-f-y, x-y\ 2 \a-h, -2a x — y, x-{- y\' * I 2 6, a — h 1. 1, 5, 2 3. 4, 7, 3 . 9, 8, 6 480 ADVANCED COURSE IN ALGEBRA 6, 4 , 7 2, - 3, 11 9, 0, 8 . 5. - 2, 4, 5 5, 3, 2 3, -1, -4 a, b, c a^ + y, 7. &, a, (Z . 8. X, c. d, a 2/, 1, X, a 1, y, & 1, ^, c ;2 X . + x ^> Verify the following by expanding the determinants : as, 11. aa, as, 686. If, in any permutation of the numbers 1, 2, 3, •••, w, a greater number precedes a less, there is said to be an inversion. Thus, in the case of five numbers, the permutation 4, 3, 1, 5, 2 has six inversions ; 4 before 1, 3 before 1, 4 before 2, 3 before 2, 5 before 2, and 4 before 3. Consider, now, the n^ elements 6l, Ci «s, 63, as wai, 61, Ci ai, mbu ci 62, C2 — C2, &2, a2 10. ?)ia2, 62, C2 = a2, W62, C2 63, Cs ci, 61, ai mas, 6s, Cs as, W63, Cs 61, c 62, C 63, c I 2 3 = ^ 62, C2 : ai , 63, Cs -hi ao, C2 as, Cs as. 62 63 ' ^2,l> ''2,2? ^-2,3? ag (1) XI? "'n,2j "■m,3? •"? ^»i The notation in regard to suffixes, in (1), is that the first suffix denotes the row, and the second the column, in which the element is situated. Thus, aj,^ ,. is the element in the /cth row and ?'th column. Let all possible products of the elements taken n at a time be formed, subject to the restriction that each product shall con- tain one and only one element from each row, and one and only one from each column, and write them so that the first suffixes shall be in the order 1, 2, 3, •••, n. This is equivalent to writing all the permutations of the numbers 1, 2, • ", w in the second suffixes. Make each product + or — according as the number of inver- sions in the second suffixes is even or odd. DETERMINANTS 481 The expression (1) is called a Determinant of the nth Order. The number of terms in the expanded form of a determinant of the nth order is |22^(§ 625). The elements lying in the diagonal joining the upper left- hand to the lower right-hand corner, are said to be in the principal diagonal; the term whose factors are the elements in the principal diagonal is always positive. The determinant (1) of § 686 is sometimes expressed by writing only the elements in the principal diagonal ; thus, I «i, 1) «2, 2, •••? «„,„!• When in the form (1) of § 686, the determinant is said to be in the Square Form. 687. It may be shown that the definition of § Q>^Q agrees with that of § 684. For consider the determinant ,, ,.^ <^l, 1) <^l, 2J <%, 3 ^2, Ij ^2, 2j ^2, 3 <^3, 1^ <^3, 2J <^3, 3 The products of the elements taken 3 at a time, subject to the restriction that each product shall contain one and only one element from each row, and one and only one from each column, the first suffixes being written in the order 1, 2, 3, are ^1, 1 ^2, 2 ^3, 3) <^1, 1 ^2, 3 ^3, 2j %, 2 ^2, 1 ^3, 3? ^1, 2 <^2, 3 ^3, 1? <^1, 3 <^2, 1 <^3, 2? and Oi 3 a2, 2 <^3, 1- In the first of these there are no inversions in the second suffixes ; in the second there is one, 3 before 2 ; in the third there is one; in the fourth, two; in the fifth, two; in the sixth, three. Then by the rule of § 686, the first, fourth, and fifth products are positive, and the second, third, and sixth are negative ; and the expanded form is ^,1^2,2^3,3 — ^,1 ^2,3^3.2 — ^1.2<^2,1%,3 ~f" %,2<^2,3^3,1 ~r C^l. 3 (^2, 1 %,2 — ^1. 3 ^2, 2 ^Z, 1) which agrees with § 684. 482 ADVANCED COURSE IN ALGEBRA 688. If, in any permutation of the letters 1, 2, 3, •••, n, two numbers he interchanged, the number of inversions is iyicreased or diminished by an odd number. If two consecutive numbers be interchanged, the number of inversions is increased or diminished by 1. Thus, if ••• aZcmft ••• is a permutation of 1, 2, 3, •••, n, the number of inversions in ••• ahmb ••• differs by 1 from the number vc\. •" amkb •"•, for the number of inversions so far as the numbers preceding k, and following m, is concerned is not affected by the interchange. Now let ••• akbc ••• efmg ••• be a permutation of the numbers 1, 2, 3, •••, 71, having |7 numbers between h and m. By interchanging m with /, then m with e, and so on in suc- cession with each of the p-\-l numbers to the left of m,, m may be brought to the left of k. Then, by interchanging k with b, then k with c, and so on in succession with each of the p numbers to the right of k, k may be brought to the right of/. Thus, k and m may be interchanged by p + 1 +jp, ov 2p-\-l interchanges of consecutive numbers ; that is, by an odd number of interchanges of consecutive numbers. It follows from this that, if k and m be interchanged, the number of inversions is increased or diminished by an odd number. 689. Any permutation of the numbers 1, 2, 3, •••, n can be obtained from any other by repeated interchanges of two numbers. If we commence in this way with the permutation 1, 2, 3, •••, n itself, after one such interchange there will be an odd number of inversions in the resulting permutation (§ 688). If in the latter we interchange two numbers, we shall have a permutation with an even number of inversions. Proceeding in this way, since the whole number of permuta- tions is an even number, we shall find that the number of per- mutations with an odd number of inversions equals the number with an even number of inversions. DETERMINANTS 483 It follows from the above that, in the expanded form of the determinant (1), of § 686, the number of positive terms equals the number of negative terms. 690. The expanded form of the determinant (1), § 686, may also be obtained by writing the seco7id suffixes in the order 1, 2, 3, ••♦, n, and making each product + or — according as the number of inversions in the Jirst suffixes is even or odd. For let the absolute value of any term, obtained by the rule of §686, be ai,pa2,.-an,.; (1) where p, g-, ••♦, r is a permutation of 1, 2, •••, n. This is obtained from the first term by changing second suffixes, 1 to p, 2 to g, •••,71 to r. . Since p, q,"',r is a permutation of 1, 2, •••,?i, (2) may be written ^ ^ ... ^ . and (1) may be obtained from this by changing first suffixes, p to 1, g to 2, •••, r to n. In these two methods, there is the same number of inter- changes of two suffixes, and the term (1) will therefore have the same sign. PROPERTIES OF DETERMINANTS 691. A determinant is not altered in value if its rows are changed to columns, arid its columns to rows. The truth of the theorem for a determinant of the third order may easily be seen by expanding the determinants «1, &i, Cl ai, a2, as a2, &2, C2 and bu &2, 63 as, &3, C3 Cl, C2, Cb in each of which, by § 684, the expanded form is aiboCs — aihsCi + a^bsCi — a^hic^ + as&iCg — a^biCi. Proof of the theorem. 484 ADVANCED COURSE IN ALGEBRA Consider the determinants ^2, IJ ^h, 2? Otj f 2, « and ^1,1? ^2, IJ ^2, 2J 'n,2 ^«, 1? ^n, 2j '•'} (^n, n ^1, n? ^2, nj • • • j <^n, n Since the second suffixes of the first determinant are the same as the first suffixes of the second, if the first determinant be expanded by the rule of § 686, and the second by the rule of § 690, the results will be the same. Therefore the determinants are equal. 692. A determinant is changed in sign if any two consecutive rows, or any tivo consecutive columns, are interchanged. The truth of the theorem when the first and second rows of a determi- nant of the third order are interchanged, may be seen by expanding the determinants and &2, &3, By § 684, the expanded forms are, respectively, «i&2C3 — aihsC2 + azbsCi — a^hiCs + a3&iC2 and a2&iC3 — aa&sCi + ai&3C2 — ai&2C3 + asbiCi one of which is the negative of the other. Proof of the theorem. Consider the determinants «3&2Cl, «3&iC2 ; ''1,1? ^r, IJ (^r ^?,2? ^M,1J and <^>-, 1> <^r, 2) C('r,n *'?,!> ^3,2? '^n.lj ^n,2} Cln,n the gth and rth rows of the first being, respectively, the rth and gth rows of the second. Let the absolute value of one of the terms of the first deter- minant be ^ ... ^ n ---a„„; (1) '1,8 <^5,«Ctr, where s, •••, i, w, •••; v is a permutation of 1, 2, •••, w, DETERMINANTS 485 Since the element in the qih row and '^th column oi the second determinant is a,.^<, and the element in the rth row and uth column a^^ ^, the absolute value of the corresponding term of the second determinant may be obtained from (1) by- replacing a^^t and a;,„ by a^,j and a,,^, respectively; that is, The latter expression is also the absolute value of one of the terms of the first determinant, since it has one and only one element from each row, and one and only one from each column ; and writing it so that the first suffixes shall occur in the order 1, 2, • • •, n, we have Cf'l,*"'Cl ^2,rj ^9 . (1) Let each, element in the rth column be the sum of m terms, ai,r=&i + Ci-h ••• 4-/1, a2,r=^2+C2H- ••• +/2, as follows : Let tti.p ••• ctj^r ••• ^«,« be the absolute value of one of the ele- ments of (1) ; then, Oi. • a„ It is evident from this that the determinant (1) can be ex- pressed as the sum of the determinants al,l> a2,u "', «l,n •", (^2,n ^n,l> -, K, "} Ctn,« + ••• + «1,1) <^, 1) ' '> Jni «2,n 488 ADVANCED COURSE IN ALGEBRA 697. If all the elements in one column, or in one roio, are multiplied by the same number, the determinant is multiplied by this number. The truth of the theorem, when all the elements in the first column of a determinant of the third order are multiplied by the same number, may be seen by expanding the determinant. Consider, for example, the determinant «2, &2, C2 . (1) Multiplying each element in the first column by m, we have (2) mai, &i, Cl ma25 62, C2 mas, &3, cs The expanded form of (2) is jnai(62C8 — &3C2) + ma2{hsCi — &1C3) + wasC&iCa — boCi), which is m times the expanded form of (1). It is evident from this that (2) equals (1) multiplied by m. Proof of the theorem. Consider the determinant "'1,1? ^2,1) } ^l,r> 'y Ct'2, r) f2,« ^»i,l) *•*> ^n (1) Multiplying each element in the rth column by m, we have «1,1J •••? '^(^l,r, ■-, «l,n <^2,1) •••> ^^2,r? •••? ^2,w a„ ma„ (2) Let Oj p--- a^ ^••« a„, s he the absolute value of one of the terms of (1). ■ Replacing a^,^ by ma^^rj ^^^ absolute value of the corre- sponding term of (2) is mai.p---a^,^ •••«„,«• It is evident from this that the determinant (2) equals m times the determinant (1). DETERMINANTS 489 698. If all the elements in any column, or roio, he multiplied by the same number, and either added to, or subtracted from, the corresponding elements in another column, or roio, the value of the determinant is not changed. Let the elements in the 7'th column of the following determinant be multiplied by m, and added to the correspond- ing elements in the gth column. (1) «1) '", «2J •••> <^rj •", «n K •••? K "'^ ^r, "-, K , ^IJ •*•? "'2? *•*? "'r? *••? '^n ain the determinant «i) •••? a^-\-ma„ •••, a^, •••, a K "-, ^ + ^i^r, •••, K ", &, „ ..., k^ + mK, ..., which, by §§ 696 and 697, is equal to (2) 6i, 7c„ iCj. + m ki, K But the coefficient of m is zero (§ 695). Whence, the determinant (2) is equal to (1). 699. Minors. If the elements in any m rows and any m columns of a determinant of the nth. order be erased, the remaining ele- ments form a determinant of the (n — m)th order. This determinant is called an mth Minor of the given deter- minant. Thus, ai, di, ei Ct's, ^3, 63 «6, ^5, 65 is a second minor of ^1} ^1} ^1} ^i> ^1 dgj t>2) ^2> ^*2j ^2 %> ^3> ^3) ^3> ^3 <*4j ^4> <^4j •^4? ^4 ^5) ^5; ^5? ^5> ^5 490 ADVANCED COURSE IN ALGEBRA It is obtained by erasing the second and fourth rows, and the second and third columns, of the latter determinant. 700. To find the coefficient of a^ ^ in the determinant ^1, IJ ^1, 2) ^2, 1? ^2, 2J ^2, n \ 1) ^n, 2j (1) By § 686, the absolute values of the terms which involve ai^i are obtained by forming all possible products of the ele- ments taken n at a time, subject to the restrictions that the first element shall be ai^ i, and that each product shall contain one and only one element from each row except the first, and one and only one from each column except the first. It is evident from this that the coefficient of aj^i in (1) may be obtained by forming all possible products of the following elements taken ti — 1 at a time, ^2, 2? ^2, ^3, 2j ^3, ^2, n subject to the restriction that each product shall contain one and only one element from each row, and one and only one from each column, writing the first suffixes in the order 2, 3, •••, n, and making each product + or — according as the number of inversions in the second suffixes is even or odd. Then by § 686, the coefficient of a^^ j is 2,2) ^2,3) ••*? <^2, i 1, 2) ^n, 3> * * *) "n that is, the minor obtained by erasing the first row and the first column of the given determinant. 701. By aid of § 700, a determinant of any order may be expressed as a determinant of any higher order. DETERMINANTS 491 Thus, ttl, K Ci «2, h, C2 = «3, K C3 1, 0, 0, 0, tti, bi, Ci 0, a2, h, C2 0, as, bsy C3 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, ai, 61, Ci 0, 0, a^, 62, C2 0, 0, as, bs, Cg etc. 702. Coefficient of any element of a Determinant. To find the coefficient of 63, in the determinant Oj, h, Ci ttg, h, C2 ttg, K C3 (1) By two interchanges of consecutive rows, the last row may be made the first ; thus, by § 6y2, the determinant equals (-ly ^3, h, C3 a„ &1, Ci a^. h, C2 (2) By interchanging the first two columns, the determinant (2) equals (-1)= By § 700, the coefficient of b^, in (3), is &3, «3, C3 K Oi, Cl K «2, C2 (3) (-ly ttj, Ci That is, the coefficient of the element in the third row and second column equals (—1)^+^ multiplied by that minor of (1) which is obtained by erasing the third row and second column. We will now consider the general case ; to find the coefficient of a^fc, f in the determinant (4) ttl.lJ '"} «l,r^ '"} (h,n «*.l, '"} «*,rj '*•> Ctk.n «», 1} ' "i Ctn,r, '") «n,«/ 492 ADVANCED COURSE IN ALGEBRA By k — 1 interchanges of consecutive rows, and r — 1 inter- changes of consecutive columns, the element a^^,. may be brought to the upper left-hand corner. Thus, by § 692, the determinant equals (- !)'-'(- ir' ^k.n a. a„ Then, by § 700, the coefficient of a^^ ^ is ^1,1? 1, » ''n, 17 But Hence, the coefficient of the element in the Z:th row and rth column equals (— l)*"^*", multiplied* by that minor of (4) which is obtained by erasing the A:th row and 7'th column. 703. By aid of § 702, a determinant of any order may be expressed in terms of determinants of any lower order. Thus, since every term of a determinant contains one and only one element from the first row, we have, ^4, &1, &2J = ai + Ci b„ Ci, (k a,, Cl, d. h, Cs, d. -h «3, Cs, d. K Ci, d. «« c« d. 5, 10 a, a, 6, a &, 6, 6, a &, a, a, a &, a, &, & DETERMINANTS 495 13. .: 14. i> 0, 1, 1, 1 1, 0, a2, 62 1, a\ 0, c2 1, b\ C2, • 15. 7, 10, 13, 3 14, 19, 27, 6 24, 33, 41, 10 31, 47, 64, 15 16. 5, -3, -2, 4, 1, -6, - 1, 4, 3, 0, 6, -4, 2 -5 2 1, 1, 1, 1 a, 6, c, d a2, 62^ c2, C?2 a3, 63, c3, # 17. Prove that a + 6 + c is a factor of the determinant (a + 6)2, 62, (6 + c)2, 18. Prove that a^b -\-c b^ (c + ay -d is a factor of the determinant a, 6, c, -d h a, — d, c c, — d, a, 6 d, c, 6, a 705. Let A,, B„ /r^, denote the coefficients of the ele- ments ar, b^, -, k^, respectively, in the deterjninant ^ij ^1) '"} ki «£, O2, "•, K"2 a„, b^, •••, lc„ (1) Then, since every term of the determinant contains one and only one element from the first column, the value of the deter- minant is ^^^^ ^ ^^^^ H- . . . + A^a,. In like manner, the value of the determinant also equals 706. With the notation of § 705, if m^ m^,"-, m„ are the elements in any column of the determinant (1), of § 705, except the first, ^^^^ _^ ^^^^ + . . • + Awn is the value of a determinant, which differs from (1) only in having mi, mg, •••, m„ instead of a^, a^, ••♦, a„ as the elements in the first column. 496 ADVANCED COURSE IN ALGEBRA Then, Ainii + ^2^2 H h ^n^n = ; for it is the value of a determinant which has two columns identical. In like manner, if mj, mg, •••, m„ are the elements in any column of the determinant (1), except the second, BiTrii + ^2^2 H h -S„m„ = ; and so on. SOLUTION OF EQUATIONS 707. Let it be required to solve the following system of n linear, simultaneous equations, involving n unknown numbers :^ aiXi + ... + g^aj, H h \x^ =Pi- (1) Cin^l + ••• + gn«r H h K^n = Pn- ■ (3) Let Qi, Qs? •••? Qn denote the coefficients of the elements qi, Qi) '"} Qn} respectively, in the determinant ^i> ••*> 9ij '"} ^1 jr^__ <^2> •••? Q2) '"} 1^2 Multiplying equations (1), (2), ••., (3) by Q^ Qa, —, Q„, respectively, and adding, we have Xi(QiC('i + Q2«2 + ••• + Q„a„)+ ... + ^XQiQi + ^292 4- ••• + Qn9»)+ - 4-aJ„(QA 4- ©2^2 + ••• + QnK)= QlPl + Q2P2 + - + QnPn- By § 706, the coefficient of each unknown number, except x^, is zero. By § 705, the coefficient of x^is D; also, the second member is the value of a determinant which differs from D only in having p^, p2, •••, p„ instead of q^, q^, -", g„ as the elements in the rth column. Denoting the latter by Z>^, we have A x^D = D^, and x^ = D DETERMINANTS 497 The determinant D is called the Discriminant of the given system of equations. 708. Ex. Find the value of y from the equations 2a; + 6?/-92=-23. .4£c — 2?/ — 52;= 9; The denominator of the value of y is the determinant 3, -6, 7 2, 6, -9 4, -2, -5 The numerator is obtained by putting for the second column the second members of the given equations. Therefore, 3, 28, 7 2, -23, -9 4, 9, -5 3, - 5, 7 2, 6, -9 4, - 2, -5 630 210 = -3. 709. Consider the following system of n homogeneous linear equations, involving n unknown numbe^rs: aja^i + h^X2 + CiO^g H h \x^_^ + \x^ = 0. a.jXi + b^2-{-c^3-\ h M«-i + ^2aJ« = 0. (1) ««a;i + b^X2 -\-CnXs-\ h KX^_-^ + A^n^^n = 0. One solution of the system is a;i = 0, iC2 = 0, •••, a/'„ = 0- We will ' now find what relations must hold between the coefficients in order that there may be other solutions. We may write the given system b^x^ 4- Ci»3 H h Mn = - (h^i- b^2 + c^3-^ h Mn = - Clffh' (2) [ bnX2 + Cn^g H h Kx^ = — a^^x^. We will now use the last 71 — 1 of these equations to express the values of X2, •••, x^ in terms of Xy. 498 ADVANCED COURSE IN ALGEBRA By § 707, 0-2^1, C2, — a^x^, c. •) K,n ^25 ^2? •••> "^2 ^n» ^nJ '"j "'n = -iCl a» c» -, *2 «M c« -, K 6„ Cj, ..., h. 6« c« ■■; K with results of similar form for iCg, •••, x^. Substituting these values in the first equation of system (1), and clearing of fractions, we have O2, C2, •••, A;2 ^n5 ^n? *••? "^n -&1 A^l ^2) ^2, •••, fcj ^n) ^n) '*•? "'m ^2^ ***> "'2? ^2 — C, 62J 0^2, ••*, /C2 ^n) ^n) •••> \ 'n> 7 '"nJ "'n By § 693, the coefficient of x^ equals 0. (3) Oj K •••, Zcg • . . -*i K -, K «2, C2, •••, k2 • • ' ' + Ci ««, c,, -, A:. ^2j ^2j ^2> •**; "'2 ^nj ^») ^n> ••*> "'n That is (§ 702), ^IJ ^1, •••, /Cj <^2J ^2> •*•? "^2 ^n> ^n? '") ""n (4) If the determinant (4) equals 0, equation (3) is satisfied when a^ has any real value. Then, if the discriminant of the given system vanishes, the system has an indefinitely great number of solutions. If any real number, m, be taken as a value of xi, the corresponding value of X2 must be a2, C2, • •, k2 an, Cn, • ; Tcn H C2, •• ., k2 bn, Cn, • ., kn DETERMINANTS 499 Similar considerations hold with respect to xs, •••, a;„. 710. Consider the following system of n linear equations involving n — 1 unknown numbers : ■b r aiXi ■i-biX2-\ h QiXn-i + A^i = 0. ^■k I a2a^l + &2a'2 H h g2^n-l + ^2 = 0. [anXi+bnX2-\ h gn^^n-l + ^» = 0. Multiplying the terms of each equation by u, we have a^uxi + biUX2 H h q\ux,^ -^ + TciU = 0. aswo^i + 62^^*2 H h ^2^''^n-i + A;2?t 0. (1) a^WiCj -h &„Wa72 H h gnU^n-l + ^n^ = 0. " This may be considered as a system of n homogeneous linear equations, involving the n unknown numbers, ux^, ux2, •••, ux^^i, u. By § 709, the condition which must hold in order that the system (1) may have possible solutions is %j bi, '••} 5'ij "'I Q2} 1^2 Kj ^n) fc^ = 0. This, then, is the condition which must hold in order that the equations of the given system may be consistent. 711. Multiplication of Determinants. Consider the determinant ttiCj + a2C?i, biCi 4- 62^1 aiC2 -\- a^d^, 61C2 + 62^2 By § 696, this can be expressed in the form 4- Orfil, 61C1 ^1^2? ^1^2 That is, by § 697, Ci, Ci C2, C2 a.c. b.A (XjC2, 02^2 a^d^, biCi ^2^2) ^1^2 4- aodi, 62^1 a2d2, M2 aj6i H-«A Cg? ^2 + «2^1 (X2, C2 + a2&2 0(2) ^2 600 ADVANCED COURSE IN ALGEBRA The first and last determinants are zero, since they have two columns identical. Then, by § 692, the given determinant equals a-p2 C2, d2 -aafti Ci, d^ Cg, ^2 = («! ^2-«2&] ) C2, ^2' = a2, &2 X C2, ^2 That is, the product of two determinants of the second order can be expressed as a determinant of the second order. Again, consider the determinant ajdi + asCi + tts/i, Ml + ^261 + &3/1J Mi + c^ei + Cs/i aid2 4- ^262 + ^sA hdi + 62^2 + &3/2J CiC^2 + ^262 + Cg/g «l<^3 + ^263 + tts/s; ?^1^3 + ^2^3 + ^3/3? ^1^3 + 0263 + 03/3 By § 696, this equals the sum of twenty-seven determinants, of which the following are types : V11 C3/1 ^2^2? C3/2 ) ^2^3, C3/3 Ci\d2, That is, by § 697, d\) ^ij f\ 0102^3 ^2) ^2> .^ d^y ^3? .^ Ct](X2y ctids, CfrfilC2 hdi, bid2, hds, di, 0261 02^2 CoCo dfi. 61 d2, ^2 ds, 63 1 1> Oidg, aidg, ai^iCj bid2, bids, C-^Oj2 Cids di, di, di d2, d2, do ds, ds, ds The eighteen determinants of the second type, and the three of the third type, are all zero by § 695. Hence, the given determinant equals ciihcs + azbsCi di, 61, /i di, fi, e. ei, d„ /i ^2, 62, /2 + «1?>3C2 d„ U e.2 + a2&iC3 62, ^2, /s ^3, eg, /s ds, fs, 63 637 ^3J J^ ei, fi, d. A, ^ij ei fi, ^1, di ^2, ./2) do + a,b^C2 .4 ^2J 62 + (^sbiOi f2, ^2, d2 es, f„ ds /sj d^, 63 fi, 63, dg DETERMINANTS 601 By § 692, the abpve equals or, &1, Ci &2, C2 X ?>3, C3 + ag&iCg — ag^^Ci) (^, 61, /i ^2J 62, /g • C^3J 63? /3 ^i, ei, /i d^, ^2, /2 ds, 63, /3 That is, the product of two determinants of the third order can be expressed as a determinant of the third order. EXERCISE 116 Solve the following by determinants : x + 2y -\-Ss= 4. 1. \ 3z+5y+ z = 18. 3. 4a; + y-\-2z = 12. 4. 8x-3y- 7;^ = 85. x + 6y-4z = - -12. 2x — 6y+ z = 33. 6. Express 4, -3 -7, 8 X 5, -6 10, 7 1, 2,-6 4, - 6. Express 0,-1, 1 X -1, - 2, -3, -5 -7, 5, 0, 1 >^4'" 7. Express -5, 2,-1 -1, -8, 41 *> X + y+ z = l. ax + by + cz = d. a^x + b^y 4- C^z = <^. «+ y+ 2+ M= 1. 2x + Sy-4-z + 5w=- 31. 3x-4y + 52; + 6w=-22. .4x + 5y-60- u=:-13. as a determinant. 1, -3 4, -5 0, 2 as a determinant. 8. Express the square of a, 5, c 6, c, a c, a, 6 3 I as a determinant. 5 I pare § 701.) a determinant. (Com- 502 ADVANCED COURSE IN ALGEBRA XXXVII. THEORY OF EQUATIONS 712. Every equation of the nth degree, with one unknown number, can be written in the form where the coefficients may be positive or negative, integral or fractional, rational or irrational, real, pure imaginary, or com- plex, or zero. If no coefficient equals zero, the equation is said to be CoTn- plete; otherwise, it is said to be Incomplete. We shall call (1) the General Form of the equation of the nth degree. 713. It will be proved in the Appendix that every equation of the above form has at least one root, real, pure imaginary, or complex. 714. (Converse of § 183.) If a is a root of the equation X^+P^X^-^ + ... +Pn-iX +P. = 0, (1) the first member is divisible by x — a. If a is a root of (1), a" +Pia**"^ H hi>„ = 0. But by § 139, a"+pia**~^ H hPn is the remainder obtained by dividing the first member of (1) by a; — a. Then, the first member of (1) is divisible by a; — a. 715. Number of Roots. An equation of the nth degree has n roots, and not more than n. Let the equation be X^ 4- PlX""-"- +P2X--' + • • • -\-Pn-lX +Pn = 0. (1) By § 713, this equation has at least one root. Let a be this root ; then by § 714, the first member is divisible by ic — a, and the equation may be put in the form {x - a) (x--' + q^--"" + . . . + qn-ix + g„) = 0. THEORY OF EQUATIONS 603 By § 182, the latter equation may be solved by placing a; — a = 0, and x^-^ + q^^-'' + • • • + q^-xX + g, = 0. (2) Equation (2) must also have at least one root. Let h be this root ; then (2) may be written {x - h) (x--' + r,x--' + - + r^-ix + r,) = 0, and the equation may be solved by placing x-b = 0, and x**~^ + rgic"-^ + ••• + r^^iX + r^ = 0. Continuing the process until 7i — 1 binomial factors have been divided out, we shall arrive finally at an equation of the first degree, ^j - A; = ; whence, x = k. Therefore, the given equation has the 7i roots a, b, •••, k. The roots are not necessarily unequal; compare note, § 183. 716. Depression of Equations. It follows from § 715 that, if m roots of an equation of the nth degree are known, the equation may be depressed to another equation of the (n — m)th degree, which shall contain the other n — m roots. Thus, if all but two of the roots of an equation are known, these two may be obtained from the depressed equation by the rules for quadratics. -' Ex. Two roots of the equation 9 x^ - 37 ic' - 8 a; + 20 = are 2 and — f ; what are the others ? By § 714, the first member of the given equation is divisible by {x - 2) (3a; + 5), or 3aj2 - a; - 10. Dividing 9 «* - 37 a^ - 8 a; + 20 by 3 ar^ - a; - 10, the quotient is3a^4-a:-2. Then the depressed equation is 3a;2 + a;-2 = 0. 2 Solving by the rules for quadratics, x = - or — 1. o 504 ADVANCED COURSE IN ALGEBRA EXERCISE 1!7 1. One root of ic^ - a:2 - 32 x + 60 = is 5 ; find the others. 2. One root of 3 x^ - 6 x^ - 34 a: + 24 = is - 3 ; find the others. 3. One root of 24 a;^ - 60 a;2 + 9 x + 20 = is - - ; find the others. 4. One root of 27 x^ + 54 a;^ - 141 x + 28 = is ^ ; find the others. 5. Two roots of 9 x* + 6 x* - 69 x^ - 20 x + 100 = are -.2 and ^; find the others. 6. Two roots of 20 x* - 177 x^ + 256 x^ + 492 x + 144 = are 4 and 3 ; find the others. 4' 7. Two roots of x* + ax^ - 22 a^x^ _ 16 aH + 96 a* = are - 3 a and 4 a ; find the others. 8. One root of a;3 _^ (•^ + 6) a;2 _ (^2 _ 4 ^ _ 11) ^ _ ^3 _ 2 n2 + 5,71 + 6 =0 is - w - 3 ; find the others. 717. Formation of Equations. It follows from § 715 that if the roots of are a, 6, •••, h, the equation may be written in the form {x — a){x — 'b) "' {x — h) = 0. Hence, to form an equation which shall have any required roots, Subtract each root from x, and place the product of the resulting^ expressions equal to zero. Ex. Form an equation which shall have the roots 1, -, and — -• 2' 3 By the rule, (x -l)(x- ^ fx-\-^=: 0. Multiplying the terms of the second factor by 2, and of the third by 3, we have (x-T)(2x-l)(Sxi-5)=0. Expanding, 6 x^ + x'-12 x + 5 = 0. THEORY OF EQUATIONS 505 EXERCISE 118 Form equations having the roots : 1 ^ -7. ^' 5' 6 8. 4, 4, -1 -1 2. 1, -4, 6. 3. -2±V'6. 9. m, '^i/^ 4 4. 2, - 3, 5, 0. 10. 3±V2, -3±V2, 0. 5. 2 a - 5, 3 a + 4. 6. -1, -2, 7, -8. 11. a, -b, ^, -I- a b '• ^' «' 1' 1- .„ 4±2V3 -2±V3 3 ' 3 ■ 18 2V-: 2±\/-5 _2\/-2±\/-5 2 2 14. n + 1, -n-2, -71 + 3, ?i - 4. 718. Composition of Coefficients. By § 717, the equation whose roots are a and b is (x — a)(x — b) = 0, or x^—(a-\- h)x -\-ab = 0. The equation whose roots are a, b, and c is (x — a){x — b)(x — c)=Oy or ic^— (a + 6+c)x^+(a6 + ac4-6c)a; — a&c = 0. The equation whose roots are a, b, c, and d is (x — a)(x — 6)(ic — c)(£c — d) = 0, or iB* — (a + 6 + c + d)ar^ + (ab -{- ac -^ ad -\- be -\-bd + cd)x^ — (abc + aftd + acd + 6cd)a; + abed = 0. In the above expanded forms, we observe the following laws : T7ie coefficient of the second term is equal to minus the sum of all the roots. TJie coefficient of the third term is equal to the sum of the products of the roots, taken two at a time. The coefficient of the fourth term is equal to minus the sum of the products of the roots, taken three at a time; etc. I The last term is equal to plus or minus the product of all the i roots, according as the number of roots is even or odd. ^^ 506 ADVANCED COURSE IN ALGEBRA -^ We will now prove by Mathematical Induction that the above laws hold for any number of roots. Assume them to hold for n roots ; a, h, c, d, •••, k, I, m. That is, (x — d)(x — b)(x — c) ••• {x — m) = x'''-\-p^x''-^-{-p^''-^-\-p^x''-^-\- •.. 4-Pn; where pi=— (a + 64-c+ ••• 4-Zc + ?4-m); p^z= ah -{- ac -\- he -\- ••• -\-lm\ Ps— — (ahc -\- ahd + acd + • • • + Mm) ; Pn= ± <^hcd ••• kbn, according as n is even or odd. If we introduce an additional root, r, we equate to zero the product of x^ -\-piX''~'^ +^2^"*"^ +P3X^~^ + ••• -\-Pn by x — r. This gives the equation Here, the coefficient of the second term is -(a + & + c+ ••• +A;+Z + m + r). Of the third term ah -{- ac ■\- he -\- ••• -{-Im + ria + h -\-c-\- ••• +m). Of the fourth term — {ahc-\- ••• -\-ldm) — 7'{ah-\-aG-\- ••• + ^m). The last term is =F ahcd • • • mr, according as n is even or odd ; or ± ahcd -•- r, according as n + 1 is even or odd. These results are in accordance with the above laws. Hence, if the laws hold for n roots, they hold for ?i-|-l roots. .But we know that they hold for four roots, and hence they hold for five" roots; and since they hold for five roots, they hold for six roots ; and so on. Hence, the laws hold for any number of roots. ' 719. It follows from § 718 that, if an equation of the nth. degree is in the general form, 4 If the second term is wanting, the sum of the roots is 0. THEORY OF EQUATIONS 607 ;f the last term is wanting, at least one root is 0. ^ We also see that if the last term is integral, it is divisible by every integral root. nT 720. If all but one of the roots of an equation of the nth degree in the general form are known, the remaining root may be found by changing the sign of the coefiicient of the second term of the given equation, and subtracting the sum of the known roots from the result ; or, by dividing the last term of the given equation if n is even, or its negative if n is odd, by the product of the known roots. If all but two are known, the coefficient of the second term of the depressed equation (§ 716) may be found by adding the sum of the known roots to the coefficient of the second term of the given equation; and the last term of the depressed equation may be found by dividing the last term of the given equation by plus or minus the product of the known roots according as n is even or odd. Ex. Two roots of the* equation 9 aj^ — 37 a;^ — 8 a; + 20 = are 2 and — -; what are the others ? t) We first put the equation in the general form by dividing each term by 9. It then becomes ^* — -^ ^^ — q ^ + -^ = 0. j7 y «7 Since there is no ic^ term, the coefficient of the second term isO. Then the coefficient of the second term of the depressed 5 1 equation is + 2 — - , or - • o o The coefficient of the last term of the depressed equation . 20 / 10\ 2 The depressed equation isa^-f-a; — - = 0. Solving, a? = Q or — 1. o / J 508 ADVANCED COURSE IN ALGEBRA EXERCISE 119 1, Two roots of 16 x^ - 37 x - 21 = are - 1 and - ; find the other 4 root. • 2. Three roots of 4 a;* - 17 x^ - 64 x'^ + 257 ic - 60 = are 3, - 4, and 6 ; find the other root. 3. Four roots of 2 ac^ _|. 3 ^4 _ 30 x^ - 65 a;2 + 18 ic + 72 = are 1, - 2, Q 4, and — : find the other. ' 2' 4. Five roots of 36 x^ - 475 x^ - 431 cc^ + 91 x^ + 71 cc - 12 = are — 1, — 3, — , -, and - ; find the other. ''236 5. One root of x^ + 8 x2 - 23 x - 210 = is - 6 ; find the others. 6. Two roots of 6 x* + 7 x^ - 37 x'^ - 8 x + 12 = are 2 and -; find the others. 7. Three roots of x^ - 6 x* - 27 x^ + 148 x^ + 204 x - 720 = are 2, — 3, and — 4 ; find the others, 8. Two roots of x* - (2 a2 + 5) x^ + 6 ax + a* _ 5 ^2 + 4 = are a - 1 and — a + 2 ; find the others. 721. Symmetrical Functions of the Roots. The expressions for pi, P2, "-jPni iii § ''^18, are symmetrical (§ 146) functions of the roots of the equation, and can be expressed as follows (compare § 150) : Pi= ~ ^a. P2 = Sa6. ^3 == — "^abc, etc. Other symmetrical functions of the roots can be expressed in terms of the coefficients. Ex. If a, h, and c are the roots of the equation find the values of 2a^, and ^a-b. We have ^a'^o? ^W -\- (?={a^-b -^cf -2(ah -{-hc + ca) = CXay -2{^ab) = {-p,y -2p2=p,'-2p2, Again, %a^b = a^b + a^c + b^c + b'a + c^a + c^b = {a 4- b -\- c) (ab -\- be -{- cd) — 3 abc = {-Pi)P2 + Sp, = 3ps -PiP2- ^ THEORY OF EQUATIONS 609 EXERCISE 120 If a, 6, and c are the roots of 7? -^-pxx^ +P2X +^3 = 0, find 1. si. 2. S— . 3. 2a3. 4. Za^\ 5. S— . a ab o2 If a, 6, c, and d are the roots of x* + piic^ 4- pix"^ + p^x +p^ = 0, find 6. si. 7. S— . 8. Sa^fec. 9. 2^2;,. 10. Sa8. a ao , > [p^ 722. Fractional Roots. An equation in the general form with integral coefficients cannot have as a root a rational fraction (§ 198) in its loivest terms. Let the equation be x^ +^91^-1 +i)2aj"-2 ^ ... ^p^^x +p^ = 0, where pi, p2, •••, Pn are integral. If possible, let -, a rational fraction in its lowest terms, be b a root of the equation ; then, . Multiplying each term by 6""^ and transposing, ^=.-(p^a--^ -^p,a^n + ... +Pn-iab--' -\-pJ>''-')- By hypothesis, a and b have no common divisor; hence a" and b have no common divisor (§ 673). We then have a rational fraction in its lowest terms equal to an integral expression, which is impossible. Therefore, the equation cannot have as a root a rational fraction in its lowest terms. 723. Complex Roots. If a complex number is a root of an equation in the general form, with real coefficients, its conjugate (§ 425) is also a root. Let the equation be a;»-f pia;"-^+ ... -\-pn-iX+p^ = 0, (1) where pi, «.., ^\ are real numbers. 510 ADVANCED COURSE IN ALGEBRA Let a -f hi, where a and h have the same meanings as in § 415, be a root of the equation ; then, (a 4- uy +pla + Uy-^ + ••• +P«-i(« + &0+Pn = 0. Expanding by the Binomial Theorem, we have by § 411, +Pi[a"-^ + (w - l)a^-^U - (^-l)(^-^) t^n-352 -| + -4-Pn-i(ot + &0+i>-=0- (2) Collecting the real and imaginary terms, we have a result of the form P+Qi = 0. (3) Here, P stands for the sum of all the terms containing a alone, together with all the terms containing even powers of t j Qi for all terms containing odd powers of i. In order that equation (3) may hold, we must have P=0, and Q = 0(§418). Now substituting a — hi for x in the first member of (1), it becomes (a _ uy +p,{a - Uy-^ + - +P„-i(a - hi) -\-p^. (4) Expanding the powers of a — hi, we shall have a result which differs from the first member of (2) only in having the even terms in each expansion, or those involving i as a factor, changed in sign. Then, collecting the real and imaginary terms, the expression (4) equals P-Qi] where P and Q have the same meanings as before. But since P=0 and Q = 0, P-qi = 0. Hence, a — hi is a root of (1). The above demonstration holds without change when a equals zero ; thus the theorem holds for any pure imaginary number (§ 418). 724. It follows from §§ 713 and 723 that every equation of odd degree has at least one real root; for an equation cannot have an odd number of complex roots. THEORY OF EQUATIONS 511, 725. The product of the factors of the first member of (1), § 723, corresponding to the conjugate complex roots a-\-b V— 1 and a — 6 V— 1, is [x-(a 4- 6 V^n:)][a; - (a- 6 V^] = (x- af--{b^'^iy = ix - af + h''', and is therefore positive for every real value of x. TRANSFORMATION OF EQUATIONS 726. To transform an equation into another which shall have . the same roots with contrary signs. Let the equation be X^ +i>l«^"' +P2«""' + — +Pn-lX +Pn = 0. (1) Substituting — ?/ for x, we have {-yY+Pi(-yr-'+P2(-yT-'-h-"+Pn-i(-y)-^Pn = o. Dividing each term by (— 1)", we have Or, y--p^^f-l^p^y-^ ±Pn-i2/Ti?n = 0; (2) the upper or lower signs being taken according as n is odd or even. - - -- - -- > \ It follows from (1) and (2) that the desired transformation may be effected by changing the signs of the terms of odd degree, if the degree of the equation is even, or the signs of the terms of ^ even degree and of the independent term^ if the degree of the ' equation is odd. v'*^ ' "'' ^ ' - ^^ ^ ' ' "^ The above rule applies whether the equation is complete or incomplete. Ex. Transform the equation x^ — 10x-\-4: = into another which shall have the same roots with contrary signs. By the rule, the transformed equation is a^ — 10 a; — 4 = 0. 727. To transform an equation into another whose roots shall be respectively m times those of the first. Let the equation be a;" + 2)iX"-^ + p2a;"-2 -\ \- p,^_^x 4- Pn = 0. o 512 ADVANCED COURSE IN ALGEBRA Putting mx = y, that is, ^ for x, we have m Multiplying each term by m**, 2/" +i)im2/^~' +i?2my-' + •• • +i>«_im"-^2/ +Pnm'' = 0. ^ jij Hence, to effect the desired transformation, multiply the *^ ijiecond term by m, the third term by m^, and so on. Ex. Transform the equation £c^ + 7ic^ — 6 = into another whose roots shall be respectively 4 times those of the first. Supplying the missing term with the coefhcient zero, and applying the rule, we have x3 + 4.7ar + 42.0ic-43.6 = 0, ora53_|.28a^-384 = 0. ^M 728. To trayisform an equation with fractional coefficients into another whose coefficients shall be integral, that of the first term being unity. The transformation may be effected by transforming the equation into another whose roots shall be respectively m times those of the first (§ 727), and then giving m such a value as will make every coefficient integral. By giving to m the least value which will make every coeffi- cient integral, the result will be obtained in its simplest form. Ex. Transform the equation oi? —-\ = into an- 3 36 108 other whose coefficients shall be integral, that of the first term being unity. By § 727, the equation 3 36 108 has its roots respectively m times those of the given eqiiation. It is evident, by inspection, that the least value of m which will make every coefficient integral, is 6. Putting m = Qj we have ix?-2x^-x + 2 = 0, whose roots are 6 times those of the given equation. THEOllY OF EQUATIONS 518 i^^' 729. To transform an equation into another whose roots f shall he respectively those of the first increased by m. '^^. ^ 7>\ -fff^ Let the equation be a;»+i>ia^ rV-i:- + Pn-i^ +Pn = 0. (1) Putting x-\-m — y, that is, y — m for x, we have (?/-m)"+|>i(2/-m)^-i+ ... +i)n-i(y-m)+p„ = 0. (2) Expanding the powers oi y — m hj the Binomial Theorem, and collecting the terms involving like powers of y, we shall have a result of the form rj. q,y--' + - + Qn-iV + gn = 0, (3) whose roots are respectively those of the given equation in- creased by m. Ex. Transform the equation a^ — 7cc + 6 = into another whose roots shall be respectively those of the first increased by 2. Substituting i/ — 2 for x, (2/-2)3_7(2/-2)+6 = 0. Expanding, and collecting the terms involving like powers of 2/, we have f _Qy2 _^^y ^i2 = 0. 730. If m and the coefficients of the given equation are integral, the coefficients of the transformed equation may be conveniently found by the following method. Putting x-\-m for y in (3), we obtain (x^rny^q^(x-{-m)^-^+ ... + g._i (x -f ??i) + g„ = ; (4) which must, of course, take the same form as (1) on expanding the powers of x + m, and collecting the terms involving like powers of x. Dividing the first member of (4) by a; + m, we have - (x + 7ny-^ + q,(x + my-^-{- -•• -\-qn-^(x-{-m) -^q^-i (6) as a quotient, with a remainder q^. Dividing (5) hj x-{- m, we have the remainder q^-i ; etc. II 514 ADVANCED COURSE IN ALGEBRA Hence, to obtain the coefficients of the transformed equation : Divide the first member of the given equation by x + m; the remainder will be the last term of the required equation. Divide the quotient just found by x-\-m; the remainder will be the coefficient of the next to the last term of the transformed equa- tion; and so on. Ex. Transform the equation a^ — 7a; + 6 = into another whose roots shall be respectively those of the first increased by 2. Dividing a;^ — 7 a; -f- 6 by aj + 2, we have the quotient a^ — 2 a; — 3, and the remainder 12. Dividing x^ — 2 x — ^ by x-{-2, we have the quotient x — 4, and the remainder 5. Dividing a; — 4 by a; + 2, we have the remainder — 6. Then, the transformed equation is i^_6a;2_^5aj + 12 = 0. Compare Ex., § 729. 731. To transform an equation into another whose roots shall be those of the first diminished by m, we change y — mto y -f m in the method of § 729, and x-\-m to x — m in the rule of § 730. 732. To transform the equation where 2h 'is not zero, into another whose second term shall be wanting. Expanding the powers of y — mm the first member of (2), § 729, and collecting the terms involving like powers of y, we y"" -\-{2\—mn)y'^ + ••• =0. If m be so taken that pi — mn — 0, whence m = — , the coeffi- cient of ?/""^ will be zero. Hence, the desired transformation may be effected by sub- stituting in the given equation y — ~ in place of x. THEORY OF EQUATIONS 616 733. Synthetic Division. The operation of division, in examples like that of § 730, may be conveniently performed by a process known as Synthetic Division. Let it be required to divide a^ — 12 x- + 29 x — 21 by a; — 3. Using detached coefficients (§ 104), we have 1 _ 12 + 29 - 21 1-3 1-3 1-9 + 2, - 9 - 9 + 27 + 2 + 2-6 -15, Remainder. We may omit the first term of each partial product, for it is merely a repetition of the term immediately above. Also, the second term of each partial product may be added to the corresponding term of the dividend, provided we change the sign of the second term of the divisor before multiplying. The work now stands : 1-12 + 29-21 + 3 - 9 -27 + 2 + 6 1+3 1-9 + 2 -15 The first term of the divisor being unity in all applications of § 730, it may be omitted ; and the first terms of the succes- sive dividends constitute the quotient. Raising the oblique columns, the operation will stand as follows : Dividend, 1 -12 +29 -21 | +3 Partial products, _ + 3 —27 + 6 Quotient, 1 — 9 + 2,-15 Remainder. The complete result is obtained as follows : 616 ADVANCED COURSE IN ALGEBRA Multiplying the first term of the dividend by 3, and adding the result to the second term of the dividend, gives the second term of the quotient. Multiplying the latter by 3, and adding the result to the third term of the dividend, gives the last term of the quotient. Multiplying the latter by 3, and adding the result to the last term of the dividend, gives the remainder. Hence, the quotient is x^ — 9x-\-2, and the remainder — 15. If the term involving any power of x is wanting, it must be supplied with the coefficient before applying the rule. The work of transforming, by Synthetic Division, the equation ^_7a;4.6 = into another whose roots shall be respectively those of the first increased by 2, will stand as follows (compare § 730) : 1 +0-7 + 6 | -2 -2 +4 + 6 - 2 - 3 +12, 1st Rem. -2 +8 - 4 + 5, 2d Rem. -2 - 6, 3d Rem. Thus, the transformed equation is a^_6a^ + 5i» + 12 = 0. EXERCISE 121 Transform each of the following into an equation which shall have the same roots with contrary signs : 1. x*- 8x3 -7x2 + 3x + 4 = 0. 2. x*^ + 6x* - 2x - 5 = 0. 3. Transform x^ + 10 x^ + 5 x — 7 = into an equation whose roots shall be, respectively, 5 times those of the first. 4. Transform x* — 4 x^ + 2 x^ + 3 = into an equation whose roots shall be, respectively, — 6 times those of the first. 6. Transform Sx^ + Sx^ — 2 = into an equation whose roots shall 3 be, respectively, - those of the first. THEORY OF EQUATIONS 617 6. Transform 7ic* + 6x3 — 75a; + 125 = into an equation whose roots shall be, respectively, those of the first multiplied by 5 Transform each of the following into an equation with integral coeflB- cients, that of the first term being unity : 4 32 18 54 ^ 72 8. xB + ^-ll = 0. 10. :^+^ + l^-J- = 0. 25 40 5 125 225 11. Transform x^ — 5x^ -{■ 6x + 11 =0 into an equation whose roots shall be, respectively, those of the first diminished by 5. 12. Transform x^ — 4:x'^ — 3x — 29 = into an equation whose roots shall be, respectively, those of the first increased by 6. 13. Transform x^ + 13 x^ _ §2 = into an equation whose roots shall be, respectively, those of the first increased by 2. 14. Transform x^ + 2 x^ + x^ — 7 x + 31 = into an equation whose roots shall be, respectively, those of the first diminished by 1. 15. Transform 2c* — 3 ic^ + 8 cc^ _ iq = into an equation whose roots shall be, respectively, those of the first increased by 3. 16. Transform x^ + Qx"^ + 6x + 19 = into an equation whose roots shall be, respectively, those of the first diminished by 4. 0r}(^ DESCARTES' RULE OF SIGNS 734. If an equation of the nth degree is in the general form (§ 712), a Permanence of sign occurs when two succes- sive terms have the same sign, and a Variation of sign occurs when two successive terms have opposite signs. Thus, in the equation a;^ — 3aj* — a^-|-5ic-|-l=0, there are two permanences and two variations. 735. Descartes' Rule of Signs. No equation J tchether complete or incomplete, c an Jmve a greater number of positive roots tha7i it has variations of sign; and no complete equation can have a greater number of nega- tive roots than it has permanences of sign. Let an equation in the general form have the following signs: + + _ + , the missing terms being supplied with zero coefficients. + + - - + + - 4- + + 1 m 2 3 4 + 5 - 6 7 — m 8 9 4- 10 518 ADVANCED COURSE IN ALGEBRA If we introduce a new positive root a, we multiply this by aj — a (§ 717) ; writing only the sig7is which occur in the process, we have 12 3 456789 4-+0-+00-- (1) + ~ (2) where m signifies a term which may be +, 0, or — . . Now, in (1), let a dot be placed over the first minus sign, then over the next plus sign, then over the next minus sign, and so on. The number of dots shows the number of variations ; thus, in (1) there are three variations. In the above result, we observe the following laws : I. Directly under each dotted term of (1) is a term of (2) having the same sign. Thus, the terms numbered 4, 5, and 8, in (1) and (2), have the same sign. II. The last term of (2) is of opposite sign to the term directly under the last dotted term of (1). The above laws are easily seen to hold universally. By the first law, however the term marked m is taken, there are at least as many variations in the first eight terms of (2) as in (1) ; and by the second law, there is at least one variation in the remaining terms of (2). Hence, the introduction of a new positive root increases the number of variations in the equation by at least one. If, then, we form the product of all the factors correspond- ing to the negative and imaginary roots of an equation, multi- plying the result by the factor corresponding to each positive root introduces at least one variation. Hence, the equation cannot have a greater number of posi- tive roots than it has variations of sign. <«-6frw^4- ^ t^^^ ^^f^ f>a^^ .J'>^' ^ l^*^*- -v«>./7 f^ . A^5 f liEORY OF EQUATIONS ' ' ' 519 ^ To prove the second part of Descartes' Rule, let — ?/ be substituted for x in any complete equation. Then since the signs of the alternate terms commencing with the second are changed (§ 726), the original permanences of sign become variations. But the transformed equation cannot have a greater number of positive roots than it has variations. Hence, the original equation cannot have a greater number of negative roots than it has permanences. In all applications of Descartes' Kule, the equation must contain a term independent of x ; that is, no root must equal zero (§ 182) j for a zero root cannot be regarded as either positive or negative. "y 736. It follows from the last part of § 735, and from § 726, that in any equation, complete or incomplete, the number of negative roots cannot exceed the number of variations in the equation which is formed from the given equation by changing the signs of the terms of odd degree. ^r^l2n. In any complete equation, the sum of the number of permanences and variations equals the number of terms less one, or the degree of the equation. That is, the sum of the number of permanences and variar tions equals the number of roots (§ 715). \ Hence, if the roots of a complete equation are all real, 1 the number of positive roots equals the number of variations, '"^ and the number of negative roots equals the number of per- ' manences. An equation whose terms are all positive can have no posi- ) w tive root ; and a complete equation whose terms are alternately / positive and negative can have no negative root. 738. Ex. Determine the nature of the roots of a^ + 2a; + 5 = 0. There is no variation, and consequently no positive root. Changing the sign of the independent term, we have a^ + 2a;-5 = 0. 520 ADVANCED COURSE IN ALGEBRA Here there is one variation ; and therefore the given equa- tion cannot have more than one negative root (§ 736). Then since the equation has three roots (§ 715), one of them must be negative and the other two imaginary. If two or more successive terms of an equation are wanting, it follows by Descartes' Rule that the equation must have imaginary roots. EXERCISE 122 If the roots of the following are all real, determine their signs : 1. x^ + x'^-Ux-U4: = 0. 3. x*-a;3-19a;2 + 49x-30 = 0. 2. 4x^-2Zx2 + Ux + 6 = 0. 4. 5ic4-43x3^112ic2-68x-48 = 0. 5. x* - 4 x^ - 23 x2 + 54 .r + 72 = 0. 6. x5-lla;* + 33a:3+llx2-154x-120 = 0. 7. 2 a;5 + 29x* + 119x3 + 159x2 + 7 X -60 = 0. Determine the nature of the roots of the following: 8. x3- 2x2- 3 = 0. 11. x5 + 4x3-l=0. 9. 2x* + 5x2 + 4 = 0. 12. 3x6-5 = 0. 10. x5 + 32=0. 13. x7 + 3x4 + 5x2+ 2 = 0. 14. Prove that the equation x^ + x^ — x2 + 3 = has at least two imagi- nary roots. LIMITS TO THE ROOTS ^■T*^ 739. To find a superior limit to the positive roots of an equation. The following examples illustrate the process of finding a superior "limit to the positive roots of an equation. 1. Find a superior limit to the positive roots of Grouping the positive and negative terms, we can write the first member in the form a?{x-Z) + 2{x~^. (1) It is evident that if x equals or exceeds 3, the expression (1) is positive. THEORY OF EQUATIONS 521 Hence, no root of the given equation equals or exceeds 3, and 3 is a superior limit to the positive roots. 2. Find a superior limit to the positive roots of x^ - 15 x' - 10 a.' + 24 = 0. 2 x^ X* We separate the first term into the parts and — , and write the first member in the form f^^l5A + (^-10x\+24.,ov'^(2x'-4.5) + ^(x^-^^^ It is evident from this that no root can be so great as 5} hence, 5 is a superior limit to the positive roots. ■ If we had written the first member in the form (t _ 15 xA + (- - 10 xW 24, or ^(x^ - 30) + -(x^- 20) + 24, we should have found 6 as a superior limit to the positive roots. Thus, separating x* into -^ and ^, instead of ^ and — , gave a smaller limit. .Ji 740. To find an mferior limit to the negative roots of an equation. First transform the equation into another which shall have the same roots with contrary signs (§ 726). The superior limit to the positive roots of the transformed equation, obtained as in § 739, with its sign changed, will be an inferior limit to the negative roots of the given equation. Ex. Find an inferior limit to the negative roots of a^ + 2ar^H-5a^-7 = 0. Changing the signs of the x^ term and the independent term (§ 726), we have a^ + 2a^-5a^ + 7 = 0. (1) We 'can write the first member in the form It is evident from this that no root of (1) can be so great as 2;. hence, — 2 is an inferior limit to the negative roots of tho given equation. 522 ADVANCED COURSE IN ALGEBRA By grouping the x^ and x"^ terms, in (1), we obtain a smaller limit than if we group tlie x^ and x:^ terms. EXERCISE 123 In each of the following, find a superior limit to the positive roots, and an inferior limit to the negative : 1. x3 + 3x2 + x-4 = 0. 4. 3x4 -5x2 -8a:- 7 = 0. 2. x4 + 5x3-15x-9 = 0. 5. x^ - 4x* + 6x3 + 32x2 - 15x+3 = 0. 3. x4 + 3x2-5x-8 = 0. 6. 2x5 + 5x* + 6x3 - 13x2 -25x + 4 = 0. 7. In the equation x^- 2x2-^ 3x+ 1 =0, prove 3 a superior limit to the positive roots, and - 2 an inferior limit to the negative. 8. In the equation 2 x^ + 5 x2 - 7 x - 3= 0, prove - 4 an inferior limit to the negative roots, and find a superior limit to the positive. 9. In the equation x> + 3 x^ - 9 x2 + 12 x - 10 = 0, prove 3 a superior limit to the positive roots, and - 6 an inferior limit to the negative. LOCATION OF ROOTS ^f^ ? /tj^ 741. If two real yiumhers, a and b, not roots of the equation f(x) = x^ 4-Piaj"-' + • • • + Pn-lX +i>n = 0, ivhen substituted for x in f(x), give results of opposite sign, an odd. number of roots off(x) = lie between a and b. Let a be algebraically greater than b. Let d, '"y ghQ the real roots of /(«) = lying between a and b, and h, "•, k the remaining real roots. Then, by § 717, f(x)=^{x-d)-'{x-g)-(x-h).-.(x-k)-F{x)', (1) where F(x) is the product of the factors corresponding to the complex roots off(x) =0. Substituting a and b for x in (1), we have (§ 251), f(a) = (a-d) ... (a-g) - (a-h) ... {a-h)'F(a\ . . and f(b) = {b-d) ... (fi-g) - (b-h) ... {b-7c)'F(b). Since each of the numbers c?, ...,(/ is less than a and greater than b, each of the factors a — d,-",a — gis positive, and each of the factors b —d, •••, b — g negative. THEORY OF EQUATIONS 523 Again, since none of the numbers h, •••, k lie between a and h, the expression (a — li) • • • (a — k) has the same sign as (h — h) ■■■{b-k). Also, F(a) and F(p) are positive; for the product of the factors corresponding to a pair of conjugate complex roots is positive for every real value of x (§ 725). But by hypothesis, /(a) and f(h) are of opposite sign. Hence, the number of factors b — d, •••, b — g must be odd ; that is, an odd number of roots lie between a and b. If the numbers substituted differ by unity, it is evident that the in- tegral part of at least one root is known. Ex. Locate the roots of a^ + a^ — 6ic — 7 = 0. By Descartes' Rule (§ 735), the equation cannot have more than one positive, nor more than two negative roots. The values of the first member for the values 0, 1, 2, 3,-1, — 2, and — 3 of a; are as follows : x = 0', -7. x = 2', -7. x=-l', -1. x=-S', -7. x = l', -11. x==3', 11. x=-2; 1. Since the sign of the first member is — when x = 2, and ' H- when a; = 3, one root lies between 2 and 3. >^ The others lie between — 1 and — 2, and — 2 and — 3, re- spectively. In locating roots by the above method, first make trial of the numbers 0, 1,2, etc., continuing the process until the number of positive roots deter- mined is the same as has been previously indicated by Descartes' Rule. Thus, in the above example, the equation cannot have more than one positive root ; and when one has been found to lie between 2 and 3, there is no need of trying 4, or any greater positive number. The work may sometimes be abridged by finding a superior limit to the positive roots, and an inferior limit to the negative roots of the given equation (§§ 739, 740), for no number need be tried which does not fall between these limits. EXERCISE 124 Locate the roots of the following : 1. x3 + 4 x2 - 6 = 0. 2. ic8 - 7 x2 + 6 X + 6 = 0. 524 ADVANCED COURSE IN ALGEBRA 3. x* + 3 x3 - 4 X - 1 = 0. 4. x* + x3 - 19 x2 - 17 X + 1 = 0. 5. Prove that the equation x^ -\-2x^ -}- 5x + 6 = has either one or three roots between — 1 and — 2. 6. Prove that the equation x^ — 5 x^ — 7 x — 2 = has a root between 2 and 3, and at least one between and — 1. 7. Prove that the equation x* — 3x3 + x2 — 3x — 4 = has a root be- tween and — 1, and at least one between 3 and 4. ^yiJ^ 742. Location of Roots by Synthetic Division. With the notation of § 741, if f(a) and/(&) are of opposite sign, an odd number of roots oif(x) = lie between a and b. Now by § 251, f(a) = a- +p,a^-' + •-. +Pn-ia +Pn, (1) and fgj) = -b-^p^h--^Jr"'+Pn-,h^-pn. (2) Also, (1) and (2) are the remainders obtained by dividing a;" + p^x""-^ H h p^_^x -\-p^ hj x — a and x—b, respectively (§ 139). Hence, if, when f(x) is divided by a? — a and x—b, the remainders are of opposite sign, an odd number of roots of f(x) = lie between a and b. The remainders may be obtained by Synthetic Division. Ex. Locate the roots of x^ + x"^ — 5 x — A = 0. By Descartes' Eule, the equation cannot have more than one positive, nor more than two negative roots. Dividing oi^-\-x- — 5x — 4: by x, the remainder is — 4. (3) Dividing the first member successively by ic — 1, x — 2, x — 3, x-\-l, x-i'2, and x-\-3, we have 1 4_i _5 _4 Lj. (4) 1 2-3 2 -3 -7 1 +1 _5 _4 |_2 (5) 2 6 2 3 1-2 1 -!_ 1 _ 5 _ 4 |_3 (6) 3 12 21 4 7 17 +1 -1 -5 -5 -4|-1 5 1 (7) + 1 -2 -1 -5 2 -3 -4|-2 6 2 (8) + 1 -3 -2 -5 6 1 -4|-3 -3 -7 (9) THEORY OF EQUATIONS 525 In (5) and (6), the remainders are — 2 and + 17, respectively ; hence one root lies between 2 and 3. In (3) and (7), the remainders are — 4 and + 1, respectively; hence a root lies between and — 1. In like manner, a root lies between — 2 and — 3. The above process is nothing more than a convenient way of applying the test of § 741. It has moreover the advantage over the method of direct substitution that, when the integral part of a root has been found, the v^rork performed is identical with the first part of Horner's method (§ 794) for deter- mining additional root-figures ; thus, in the above example, the work in (5) is identical with the first three lines of the determination by Horner's method of the root of the given equation lying between 2 and 3. The note to § 741 applies with equal force to the method of § 742. EXERCISE 125 Locate the roots of the following : 1. a;3 + 3 x2 - 7 ic + 2 = 0. 3. a;* - 4 a;^ + 6 x - 2 = 0. 2. a;3 + 4 x2 + ic - 3 = 0. 4. x* - 7 x^ + x + 4 = 0. 6. Prove that the equation x^ + 6x + 'i = has one root between and — 1. 6. Prove that the equation x* + 2 x^— 5 x^ — 4 x - 6 = has a root be- tween 2 and 3, and at least one between — 3 and — 4. 743. The methods of §§ 741 and 742, though simple in principle, and easy to apply, are not sufficient to deal with every problem in location of roots. Let it be required, for example, to locate the roots of By § 724, the equation has at least one real root. By Descartes' Eule, it has no positive root. By § 740, — 3 is an inferior limit to the negative roots. Putting X equal to 0, — 1, — 2, — 3, respectively, the corre- sponding values of the first member are 1, 1, 1, and —5, respectively. Then, the equation has either one root or three roots between —2 and —3; but the methods already given are not sufficient to determine which. 526 ADVANCED COURSE IN ALGEBRA Sturm's Theorem (§ 758) affords a method for determining completely the number and situation of the real roots of an equation. It IS more difficult to apply than the methods of §§ 741 and 742, and should be used only in cases which the latter cannot resolve. -r 744. Graphical Representation. The graph of an expression of higher degree than the second, with one unknown number, may be found as in § 465. Ex. Find the graph of x^-2x'-2x-{-S. Fiity = x'-2x^-2x-\-3. lix=0,y=3. Ux=S, y=6. If x=l, y=0. If x=-l, y=2. If x=2, y= -1. If x= -2, y=-d. etc. The graph is the curve ABC, which extends in either direc- tion to an indefinitely great distance from XX'. 745. Graphical Location of Roots. The principle of § 280 holds for the graph of the first mem- ber of an equation of higher degree than the second, with one unknown number. Thus, the graph of § 744 intersects XX' at oj = 1, between x = 2 and x = 3, and between x = —l and x = — 2. And the equation a;^ — 2a^ — 2aj + 3 = has one root equal to 1, one between 2 and 3, and one between — 1 and — 2. This may be verified by solving the equation ; the factors of the first member are x — 1 and x'^ — x — S. This method of locating roots is simply a graphical represen- tation of the process of § 741, and is subject to the limitations stated in § 743. If the graph does not intersect XX', the equation has no real root. THEORY OF EQUATIONS 527 The note to § 741 applies with equal force in the graphical method of locating roots. EXERCISE 126 Locate the roots of the following equations graphically : , 1. a;3-3a;-l=0. 3. x^-1 x'^-\-12x-6=0. 5. x3-8a:2+19x-12=0. 2. x*+2ic2+3=0. 4. x^+7 x'^+Ux+S=0. 6. x^-Sx'^-2x+6=0. 7. xH2x3-6a;2_7a;+6=0. ^r^^ ^ T* DIFFERENTIATION 746. Derivatives. In any function of x (§ 250), let x-\-h be substituted for x ; subtract from the result the given function, and divide the remainder by h. The limiting value of the result as h approaches the' limit zero, is called the derivative of the function with respect to x. Let it be required, for example, to find the derivative of jc3 _ 2 ^2 + 5 with respect to x. Substituting x + h for x, and subtracting from the result the given function, we have (x -]-hy-2(x + hy + 6- (x3 - 2 x2 + 5) = Sx'^h + Sxh'^-hh^-4:Xh-2 h?. Dividing this result by h, we have 3 x2 + 3 a;;^ + 7i2 - 4 X - 2 /i. (1) The limiting value of (1) as h approaches 0, is 3 x^ — 4 x. Hence, the derivative oi x^ — 2 x'^ ■\- b with respect to a; is 3 x" — 4 «. The process exemplified above is called Differentiation. 747. In general, let u represent any function of x; and suppose that, when x is changed to a; + ^, u becomes u + h\ Then, the derivative of u with respect to x is expressed as follows (compare § 265), lira V {u + h^ — u \ 528 ADVANCED COURSE IN ALGEBRA It follows from the above that —u= ,^^!^^ [-1; (1) dx ^^ = 0^ d . where — u stands for the derivative of u with respect to x. dx 748. The process of differentiation is facilitated by means of the following formulse, in which a represents any constant, n any positive integer, and w, v, w, •••, any functions of x : I. ^x = l. dx 11. ±(u-\-a)=^u. dx dx III. —(au) = a — u. dx dx dx dx dx dx V. (UVW '••) = (VW •••) U-{-(uW •••) V+"'. dx ^ dx ^ dx VI. —(u^) = nu^-^ — u. dx dx yil. — (ax'') = nax^-\ dx 749. In proving the formulae of § 748, we shall suppose that, when x-\-h is substituted for x, u is changed to % + h', v to V + h", w to w-{- h"\ etc. Proof of I. (x + h)-x l_^ ^' J That is, the derivative with respect toxofx itself is unity. Proof of II. d /„ I «N_ lim [ (u-{-h' + a)-(u-^a) l_ lim pn_ d -^u + a)-^^^^ I ]-h:^olh\-d^''' by § 747, (1). That is, the derivative with respect to x of a function of x plus a constant equals the derivative of the function of x. By S 747, ±. = ^% THEORY OF EQUATIONS 529 For example, ^(So^-5)=-f{3 x^. Proof of III. A dx /^,,x _ lim \~ a(u + h') — au ~\ _ lim [ahn That is, the derivative with respect to x of a constant times a function of x equals the constant times the derivative of the function of x. For example, — (3 ar^ = 3 — (a^). dx dx Proof of IV. -—(ll-\-V-\- IV -{-•'•) dx ^ lim nu-\-h'-{-v + h"'^W-hh"'-\-"')~(u-\-v-{-W-{-'-) ~\ h = 0\_ h J ^ lim r h' + h"-\-h"'+''' l h = 0\_ h J =;To[l']+."ro[g+."r„[^]+-(^254) d , d , d , dx dx dx That is, the derivative with respect to x of the sum of any 7iumber of functions of x equals the sum of their derivatives. Proof of V. Consider first the case of two factors. A dx (^^)= lim r (u-\-h')(v + h")-uv l ^ lim r uh"^(v + h")h n h = 0\_ h J =-;ro[l'>;ro[^+'^"]x."ro[g. by §§ 254, 257. 530 ADVANCED COURSE IN ALGEBRA As h approaches the limit 0, /i" also approaches 0, and there- fore the limiting value oi v -\- h" is v. Whence, — (uv) = u-^v-j-v-—u- (1) dx dx dx Consider next the case of three factors. — (uvw) = — l{uv) ' w'] = w— (uv) -\-uv-—w, by (1) dx (xx (xx itx / d , d \ , d =:W{U V -}-V U ]-\- UV W. \ dx dx J dx d , d , d = vw — u-{-uw — V -}-uv — w. dx dx dx In like manner the theorem may be proved for any number of factors. That is, tJie derivative with respect to x of the product of any number of functions of x equals the sum of the results obtained by multiplying the derivative of each factor by all the other factors. For example, — [{x + 1) x'^ = {x + 1) ^~ (x^ ^x'A(x-\- 1). dx /-IN dx dx Proof of Yl. If we suppose v, w, •••, in V, to be all equal to u, and that the number of factors is w, we have by V, A (u") = it«-i — u + w"-^ — u-\ to n terms = tiw""^ — u. dx dx dx dx For example, — [(x' + 1)'] = S(x' + If ~ {a? + 1). dx ax Proof of Nil. By III, — (ax'') = a — (ic") = anx^-^ — x, by VI, dx dx dx = awa;""^, by I. That is, the derivative with respect to x of a constant times any positive integral poiver of x equals the constant, times the exponent of the power, times x raised to a power whose exponent is less by 1. THEORY OF EQUATIONS 531 For example, — (3 x') = 12 o^. (XX Ex. Find the derivative with respect to x of By II and IV, ^(2a^-5x' + 7x-6) = — (2af)-—(5x') + —(7x) dx dx dx ^ ^ dx = 6a;2_10aj + 7, by VII. EXERCISE 127 Find the derivative with respect to a; of : 1. bx^-\-Tx. 5. 4a;6-7a;* + 8a:3-ic. 2. 3 x8 - a;2 + 3. 6. x^ - 4 ic* + 6 ic^ + 7 a;2 _ 6. 3. aj'i + 5 ic3 - 12 X - 4. 7. 2 x^ + 3 x* - x^ + 8 x2 + 5 x. 4. x5 + 9x4-4x2. 8. 5x6-2x5-4x3+11x2 + 8. 750. Successive Differentiation. If u is any function of x, the derivative of the derivative of u is called the Second Derivative of u with respect to x, and is d? represented by —^u- (XX The derivative of the second derivative of u is called the TJiird Derivative of u with respect to x, and is represented by d^ — u: etc. dx^ ' Ex. Find the successive derivatives with respect to x of 3a^-9x'-12x + 2, We have, — (3 o^ - 9 aj^ _ 12 a; +- 2) = 9 a;^ - 18 « - 12. dx ;^(3aj3- 9a^ - 12 a; + 2) = 18» - 18. aou -^(3a^- 9a^- 12 a; +- 2) = 18. dx^ — r3a^-9ar^-12a;+-2)=0: etc. dx* ^ 532 ADVANCED COURSE IN ALGEBRA It will be understood hereafter that when we speak of the derivative of a function of x, the first derivative is meant. EXERCISE 128 Find the successive derivatives with respect to x of : 1. 4a;2 + 7x-3. 4. 7 a;* + a;^ + 9 x2. 2. 2x3-ll(c2 + 4. 5. 3a;5 + 2a;4 + 6a:2-5. ^3. x^-bx^~2x. 6. x6 - 4 x5 - 10 ic3 + 13 x. 751. Graphical Representation of Derivatives. Let PQ be the graph of any function of x, f(x). Let P be any point on the graph having the abscissa x, and Q another point having the abscissa X -^ h. Draw PM and QN perpendicular to XX' J and PP perpendicular to Q]^. Then, PM represents f{x), and QN represents f{x-\- Ji). Whence, QB represents f(x + h) — Then, the ratio /(^ + ^0 -/(«') jg represented by S^- h PR If we denote the derivative off(x) with respect to x hy f'(x), fix) = ^"-^ \f±±Kj^zlM]^ (§ 747). ^ ' Then, f{x) is represented by the limiting value of the ratio ■^— as h approaches the limit ; that is, as Q approaches P. Psi Hence, if PT is tangent to the graph at P, meeting QN /T70 produced at T, f(x) is represented by the ratio --^• PR The latter ratio is the tangent of the angle TPB. 752. By application of the principles of § 751, we can deter- mine the points where any graph is parallel to XX'. TR For, if the graph is parallel to XX' at P, the ratio — — -t Jill is- evidently equal to zero ; so that f\x) is zero at that point, Y o^^^^-~~ ^ T R ^ / _ -rr M h N ^ Y' THEORY OF EQUATIONS 533 If then we solve the equation f\x) = 0, we shall determine in what points the graph of f(x) is parallel to XX'. Consider, for example, the graph of § 744. Here,/'(a;) = 3aj2-4a;-2. Solving the equation 3a^— 4ic— "2 = 0, we have ^_2±V4 + 6_2±V10 JC — — * 3 3 Then, one point is ^-^^^^ to the right of TY', and the other ^^^ ~ ^ to the left. 3 Let the pupil determine the points where the graphs of the first mem- bers of the equations in, Exs. 1 to 6, inclusive, in Exercise 126, are parallel to XX'. MULTIPLE ROOTS 753. If an equation has two or more* roots equal to a, a is said to be a Multiple Root of the equation. In the above case, a is called a double root, triple root, quad- i ruple root, etc., according as the equation has two roots, three roots, four roots, etc., equal to a. 754. Let the equation jpoaj** -f-piOJ^-i 4- ••• +Pn-lX +Pn = (1) have m roots equal to a. By § 717, the first member can be put in the form (x^arf(x)', (2) where f(x) is the product of the factors corresponding to the remaining roots of (1), and is therefore a rational and integral expression of the (n — m)th degree with respect to x. By § 748, V, the derivative of (2) with respect to x is (a,_a)»£/(»,)+/(a,)|-[(»;-a)»]; or, (x - aYf'{x) + m{x - a)'^'f{x), by § 748, VI. (3) 534 ADVANCED COURSE IN ALGEBRA It is evident that the expression (3) is divisible by (ic— a)*""^ ; and therefore the equation formed by equating it to zero will have m — 1 roots equal to a. Hence, if any equation of the form (1) has m roots equal to a, the equation formed by equating to zero the derivative of its first member will have m— 1 roots equal to a. It follows from the above that, to determine the existence of multiple roots in an equation of the form we proceed as follows : Find the H. C. F. of the first member and its derivative. If there is no H. C. F., there can be no multiple roots. If there is a H. C. F., by equating it to zero and solving the re- sulting equation, the required roots may be obtained. The number of times that each root occurs in the given j equation exceeds by one the number of times that it occurs in ^ the equation obtained by equating the H. C. F. to zero. Ex. Find all the roots of a;4_6a^ + 12 aj2- 10 aj + 3 = 0. (1) The derivative of the first member is 4a^-18a^ + 24a;-10. The H, C. F, of this and the first member of (1) is x'^-2 x+\ Solving the equation a^ — 2x-f-l = 0, the roots are 1 and 1. Hence, the multiple roots of (1) are 1, 1, and 1. Subtracting the sum of 1, 1, and 1 from 6, the remaining root is 3. EXERCISE 129 Find all the roots of each of the following equations : 1. x8-5a;2 + 3a;+9 = 0. 3. a;8- 12a;+ 16 = 0. 2. a;8+3aj2-9a;-27 = 0. 4. 18a;8 + 16a;2 _ 4x - 4 = 0. 5. a;4 + 2 a;8 - 11 a;2 -12 a; + 36 = 0. 6. a;*- 11 a;8 + 36x2. -16 a; -64=0. THEORY OF EQUATIONS 535 • 7. a:5 - 18 a;3 + 4 x2 + 57 X + 36 = 0. 8. a:4 + 11 x3 + 33 x2 + 6« - 50 = 0. 9. x^ -Sx* -x^ + 7 x^ -4 = 0. 10. x^ + 6x^ + Ux^ -\- 2x'^ - 12x - 8 = 0. 755. We will now construct the graph of the first member of (1), § 754. The first member can be put in the form (^ _ i)3(^. _ 3)^ (1) This shows that the graph cuts XX' at X = 1 and x = 3. Since {x — iy(x — 3) is positive when « is < 1, or > 3, and negative when x ^~~o is between 1 and 3, the graph is above XX ' when x is < 1, or > 3, and below when x is between 1 and 3. By § 748, V, the derivative of (1) is (X. ^ dx^ 3) + {x Ux' ly = {x-lf + {x-3)'^x-lf = {x- = {x-l)\A.x-10). Equating this to zero, we have a; = 1 or l)2(a;-l + 3a;-9) Then the graph is parallel to XX' at x= 1 and x=-\ Li 752). If any equation, with one unknown number, has a multiple root, the graph of its first member is tangent to XX}. If the root is a triple root, the curve crosses the axis of X at the point of tangency, and reverses its direction at that point. If the root is a double root, the curve is entirely above, or entirely below XX' at the point of tangency. (Compare § 467.) Let the pupil construct the graphs of the first members of the equations in Exs. 1, 2, 3, 4, 9, and 10, Exercise 129 ; finding all the points where the graphs are parallel to XX'. 756. An equation of the form a;" — a = can have no multi- ple roots ; for the derivative of a;'* — a is ?ia;"-^ and a;" — a and W na;""^ have no common factor except unity. 536 ADVANCED COURSE IN ALGEBRA Therefore, the n roots of x"' = a are all different. It follows from the above that every expression has two dif- ferent square roots, three different cube roots, and in general ?i different nth roots. STURM'S THEOREM 757. We will now demonstrate two theorems which are used in the proof or application of Sturm's Theorem (§ 758). I. Ifxbe taken sufficiently great, any term of the expression may be made to numerically exceed the sum of all the following terms. For the ratio of the (r -f- l)th term to the sum of all the fol- lowing terms is Pi"^ or P'- (2^ By taking x sufficiently great, the denominator of (2) can be made numerically as small as we please ; hence, the ratio of p^x""'' to the sum of the following terms can be made numeri- cally as great as we please. II. If X be taken sufficiently small, any term of the expression PoX"" + pix""-^ H f- p,,_jx + _p„ may be made to numerically exceed the sum of all the preceding terms. Eor the ratio of the (r + l)th term to the sum of all the preceding terms is ^i^ , or ^ (3) By taking x sufficiently small, the denominator of (3) can be made numerically as small as we please ; hence, the ratio of p^'*'"' to the sum of the preceding terms can be made numeri-j cally as great as we please. THEORY OF EQUATIONS 537 758. Sturm's Theorem. Let f{x) = X- + p^x^-^ + . . . 4- p^_ja; + p„ = (1) be an equation from which the multiple roots have been removed (§ 754). Let fi{x) denote the derivative of fix) with respect to x (§ 746). > Dividing /(a;) by/i(a;), we shall obtain a quotient Qi, with a remainder of a degree lower than that of fix). Denote this remainder, with the sign of each of its terms changed, by fX^), and divide fix) by fix), and so on ; the operation being precisely the same as that of finding the H. C. F. of f{x) and fix), except that the signs of the terms of each remainder are to be changed, while no other changes of sign are permissible. Since, by hypothesis, f{x) = has no multiple roots, f(pS) and fix) have no common divisor except unity (§ 754) ; and we finally obtain a remainder /^(ic) independent of x. The expressions /(x), /i(a7), /2(aj), "',fn{x) are called Sturm/s Functions. The successive operations are represented as follows : f(^) = Qifi{^)-f2(x), (2) f(x) = Q,fXx)-f{x), (3) fix) = Q,f{x) -fix), (4) fn-2(x) = Qn-lfUx) -fni^). We may now enunciate JSturm^s TJieorem. If two real numhers, a and b, are substituted in place of x in Sturm^s Functions, and the signs noted, the difference between the number of variations of sign (§ 734) in the first case and that in the second equals the number of real roots of fix) = lying between a and b. The demonstration of the theorem depends upon the fol- lowing principles : I. Two consecutive functions cannot both become for the same value of x. * 538 ADVANCED COURSE IN ALGEBRA For if, for any value of x, f-^{x) = and /^{x) == 0, then, by (3), f^{x) = ; and since f^ix) = and f.^{x) = 0, by (4), f^{x) = ; continuing in this way, we have finally fn(x) = 0. But by hypothesis,/,, (a;) is independent of x, and consequently cannot become for any value of x. Hence, no two consecutive functions can become for the same value of x. II. If any function, except f(x) and fn{x), becomes for any value of X, the adjacent functions have opposite signs for this value of X. For if, for any value of x, f{x) = 0, then, by (3), we must have/i(a;) = —f{x) for this value of x. Therefore, fi{x) and f^{x) have opposite signs for this value of X ; for, by I, neither of. them can equal zero. III. Let c be a root of the equation f^{x) = 0, where f(x) is any function except /(cc) and /„(«.'). By II, fr-i(x) and fr+i(x) have opposite signs when x = c. Now let h be a positive number, so taken that no root of fr_i(x) = or fr+i{x) = lies between c — h and c + h. Then as x changes from c — h'toc-\- h, no change of sign takes place in/._i(a.") orf^-^(x) ; while f(x) reduces to zero, and changes or retains its sign according as the root c occurs an odd or even number of times mf(x) = 0. Therefore, for values of x between c — h and c, and also for values of x between c and c + h, the three functions /,._i(a?), /.(a;), and/.+i(a;) present one permanence and one variation. Hence, as x increases from c — A to c + /i, no change occurs in the number of variations in the functions/,._i(a;),/,.(fl?), a,ndf.+i(x) ; that is, no change occurs in the number of variations as x in- creases through a root of f(x) — 0. IV. Let c be a root of the equation f(x) — ; and let h be a positive number, so taken that no root of fi{x) = lies between c—h and c -\- h. Then, as x increases from c — h to c -{- h, no change of sign takes place in /(a.-) ; while /(x) reduces to zero, and changes sign. THEORY OF EQUATIONS 539 Putting x = c — h in (1), we obtain /(c - h) = (c - hf + p,(c - hy-' + . . . + P„-i(c -h)-\- p„. Expanding by the Binomial Theorem, and collecting the terms involving like powers of h, we have /(c -h)=^ c" +i>ic"-^ + ••• +Pn-lC +i?„ - ;i[nc«-i + (^ - l)i>ic"-' + - +i>„-i] 4- terms involving h^, h^, ••-, /i". But since c is a root off(x) = 0, we have by (1), c« +pic"-i + ... +i>„_ic +i)„ = 0. Also, it is evident that the coefficient of — h is the value of fi(x) when c is substituted in place of x ; therefore, f(c — h) = — hfi(c) + terras involving h^, h^, -•-, /i". (5) In like manner it may be shown that /(c 4- /i) = -f- /i/i(c) -h terms involving h^, W, .-., /i". (6) Now if h be taken sufficiently small, the signs of the second members of (5) and (6) will be the same as the signs of their first terms, — hfi{c) and + lifiic), respectively (§ 757, II). Hence, if li be taken sufficiently small, the sign of /(c — h) will be contrary to the sign of /i(c), and the sign of /(c + h) the same as the sign of /i(c). Therefore, for values of x between c — h and c, the functions f{x) and fi{x) present a variation, and for values of x between c and c -\-h they present a permanence. Hence, a variation is lost as x increases through a root of /(a;) = 0. We may now demonstrate Sturm's Theorem; for as x in- creases from h to a, supposing a algebraically greater than b, a variation is lost each time that x passes through a root of f(x) = 0, and only then ; for when x passes through a root of fj^x) = 0, where /(a;) is any function except f(x) and /„(x), no change occurs in the number of variations. Hence, the number of variations lost as x increases from 6 to a equals the number of real roots of f(x) = included between a and b. 540 ADVANCED COURSE IN ALGEBRA ♦ 759. It is customary, in applications of Sturm's Theorem, to speak of the substitution of an indefinitely great positive number for x, in an expression, as substituting + oo for x ; and the substitution of a negative number of mdefinitely great absolute value as substituting — co for x. The substitution of +00 and —00 for x in Sturm's Func- tions determines the number of real roots oi fix) = 0. The substitution of +00 and for x determines the number of positive real roots, and the substitution of — oo and the number of negative real roots. Since Sturm's Theorem determines the number of real roots of an equation, the number of imaginary roots also becomes known (§ 715). 760. If a sufficiently great number be substituted for x in the expression F(X) = Po^*^ + PiX'^-l + • • • + Pn-lX + Pn, the sign of the result will be the same as the sign of its first term, poX"" (§ 757, I) ; hence. If -\- CO be substituted for x in F(x), the sign of the result will be the same as the sign of its first term. If —CO be substituted for x in F{x), the sign of the residt will be the saine as, or contrary to, the sign of its first term, according as the degree of F(x) is even or odd. 761. We will now consider an example. Let it be required to determine the number and situation of the real roots of y^^^) = a^l 2a^ - x'+l = 0. Here, fi{x) = 3 x^ — 4: x — 1. In the process of finding /^(x), fs(x), etc., any positive numeri- cal factor may be omitted or introduced at pleasure ; for the sign of , the result is not affected thereby ; in this way fractions may be avoided. In this case, we multiply f(x) by 3 to make its first term divisible by 3 x^. THEORY OF EQUATIONS 541 3iB2 # 4a;- -l)3x'-6a^- 3a;-}-3(aj 3 a:3 _ 4 ^ _ ^ -2a;2- 2x + 3 3 -6fl^- 6a;4-9(-2 -6a:2^ 8a; + 2 7)_14a;4-7 (z^2x + l .•./,(x) = 2a;-l. 3a^_ 4a;-l 2 f . i^^^HL^-y^^' 2a;-l)6a;-'- 8a;-2(3aj 6x'- Sx 1: ' ^-' - 6x-2 f,.. M. 2 -10a;-4(-5 - 10 a; + 5 -9 .•./8(a;) = 9... Substituting — oo for ic in fix), fi{x), fsfx), and fs(x)y the signs are — , +, — , and +, respectively (§ 760); substituting for X, the signs are +, — , — , +, respectively; and sub- stituting + 00 for a;, the signs are all 4-. Hence, the roots of the equation are all real, and two of them are positive and the other negative (§ 759). We now substitute various numbers to determine the situa- tion of the roots : /W /iW f2(x) Mx) a! = -oo, — + — + 3 variations. x = -l, — + — + 3 variations. x = 0, + — — + 2 variations. x = \, — — • + + 1 variation. x = 2, — + + + 1 variation. a; = 3, + + + + no variation.' a; = oo, + 4- + + no variation. ? We then know that the equation has one root between and — 1, one between and 1, and one between 2 and 3. l^ 542 ADVANCED COURSE IN ALGEBRA Y ^— ' 762. It will be found useful to con- struct the graphs of f(x) and fi(x), in the example of § 761. The graph of f(x) is the ■ curve ABC', cutting XX' at A, between x = and x = — 1, at B, between x = and x = lf and at C, between x = 2 and x = S. The graph of fi(x) is the dotted curve DE', cutting XX' at D, between x = and x = — l, and at E between x = l and x = 2. To find the abscissas of D and E, we solve the equation 3x'-4.x-l = 0. 2 ± 2.64+ Then, x = 2±V4 + 3 1.54+ or -.21+. 3 3 If we put x = — .21, in f(x), the result is positive. This shows that D is between O and A. The graph illustrates in an excellent way the truth of § 758, IV; that as x increases from a value just below to a value just above a root of f(x) = 0, no change takes place in the sign of fi(x), while f(x) reduces to zero and changes sign. 763. As X increases from -co to +go, f(x) and/i(cc) change signs alternately, for they are always unlike in sign just before f(x) changes sign (§ 758, IV) ; hence, if the roots of f(x) = and fi{x) = are all real, a root of fi{x) = lies between every two adjacent roots of f(x) = 0. That a root of fi(x) = lies between every pair of adjacent roots of f(x) = 0, is admirably shown in the figure of § 762. 764. We will give one more illustrative example. Determine the number and situation of the real roots of f(x) = 4oX^ — 6x — 5 = 0. Here, fi(x) = 12 a^ — 6 ; or, 2 ic^ — 1, omitting the factor 6. 2x^-l)4:a^-6x-5(2x 4.a^-2x — 4 a? — 5 .-. /six) = 4 a; +-, 5. THEORY OF EQUATIONS 543 2^- 1 2 4a; + 5)4a;2_ 2{x 4a^ + 5x — 5x- 2 4 . — 20 a;- 8(- -5 — 20 a;- 25 17 .•./,(;») = -17. The last step in the division may be omitted ; for we only need to know the sign of fsix); and it is evident by inspection, when the remainder — 5x — 2 is obtained, that the sign of fs^x) will be — . /W /.(^) M^) M^) a; = — 00, — + — — 2 variations. x^O, — — + — 2 variations. x = l, — + + — 2 variations. x = 2, + + + — 1 variation. X=(X), + + + — 1 variation. Therefore, the equation has a real root between 1 and 2, and two imaginary roots. In substituting the numbers, it is best to work from in either direc- tion, stopping when the number of variations is the same as has been previously found for +oo or — oo, as the case may be. EXERCISE 130 Determine the nature of the roots of the following : 1. a;8 + 2 a;2 - a; - 1 = 0. 5. a;* - 8 a:2 - 8 x + 1 = 0. 2. a;8 + 3 a; - 5 = 0. 6. x* + 2 x^ - 5 a;2 - 10 x - 3 = 0. 3. a:8-5x2-f 2x + 6=0. 7. x* + 3x8 - 3x + 1 = 0. 4. a;8 + x2-15x-28 = 0. 8. x* + 4x8 + 2x2 - 6x - 7 = 0. 765. Continuity. A function of x, f(x), is said to be Continuous at x = a when an indefinitely small change in a produces an indefinitely small change in /(a). 544 ADVANCED COURSE IN ALGEBRA Consider the rational integral function of x fix) =PqX''+P^X''-^ + ... ^-Pn-iX -\-p^. Putting a-\-hin. place of x, we have /(a + h) =p,(a + hy +i)i(a + hy-' + ... +Pn. Expanding by the Binomial Theorem, f(a + h) =poa^-\-p,a^-'- + - +Pn-ia +Pn + terms involving h, h^, "-, h"" =f(a) + terms involving h, h^, .••, ^^ Then, f(a + h) —/(a) = terms involving h, W, •••, /i". If h be taken indefinitely small, f(a-\-h)—f{a) will be indefi- nitely small ; for the coefficients of h, Jr, • • ., li"- are finite. Hence, an indefinitely small change in a produces an indefi- nitely small change in /(a). That is, a rational integral function of x is continuous at every value ofx. It follows from the above that the graph of a rational integral function of ic is a continuous line, without breaks. We have assumed this in the figures of Chap. XIV, §§ 465, 467, 482, and 483, and all the figures of Chap. XXXVII. 766. We will illustrate a discontinuous function of a; by a figure. Consider the fraction ^ x — 1 Put2/=-^- x — 1 As X increases from —oo to 1, 2/ is negative, and increases indefinitely in ^ absolute value. As X increases from 1 to oo, ?/ is posi- tive; and commencing with an indefi- y'-fi nitely great value, decreases indefinitely. The graph consists of two branches, AB and CD ; and it is evident that, at ic = 1, an indefinitely small change in x pro- duces an indefinitely great change in y. h —X SOLUTION OF HIGHER EQUATIONS 545 XXXVIII. SOLUTION OF HIGHER EQUATIONS COMMENSURABLE ROOTS We shall use the term commensurable root, in Chap. XXXVIII, to signify a rational root expressed in Arabic numerals. 767. By § 722, an equation of the nth degree in the general form (§ 712), with integral numerical coefficients, cannot have as a root a rational fraction in its lowest terms. Therefore, to find all the commensurable roots of such an equation, we have only to find all its integral roots. Again, by § 719, the last term of an equation of the above form is divisible by every integral root. Hence, to find all the commensurable roots, we have only to ascertain by trial which integral divisors of the last term are roots of the equation. The trial may be made in three ways : I. By substitution of the supposed root. II. By dividing the first member of the equation by the unknown number minus the supposed root (§ 183) ; in this case, the operation may be conveniently performed by Syn- thetic Division (§ 733). III. By Newton's Method of Divisors (§ 769). In the case of small numbers, such as ± 1, the first method may be the most convenient. The second has the advantage that, when a root has been found, the process gives at once the depressed equation (§ 716) for obtaining the remaining roots. If the number of divisors is large, the third method will be found to involve the least work. Considerable work may sometimes be saved by finding a superior limit to the positive roots, and an inferior limit to the negative roots (§§ 739, 740); for no number need be tried which does not fall between these limits. 546 ADVANCED COURSE IN ALGEBRA Descartes' Rule of Signs (§ 735) may also be advantageously employed to shorten- the process. Any multiple root should be removed (§ 754) before apply- ing either method. Ex. Find all the roots of x^ - 15 a;'^ + 10 a; + 24 = 0. By Descartes' Rule, the equation cannot have more than two positive, nor more than two negative roots. We can write the first member x^ (x^ — 15) + 10 ic + 24 ; then no root can be as great as 4 (§ 739). Changing the sign of the x term, the first member becomes s;4_i5a;2_-L0a; + 24. We write this in the form (see Ex. 2, § 739), |\2a^2_45)_^^(^_30) + 24. o o Then no negative root can be as small as — 5 (§ 740). The integral divisors of 24 which are less than 4, and greater than — 5, are ±1, ±2, ±3, and — 4. By substitution, we find that 1 is not, and that — 1 is, a root of the equation. Dividing the first member by a; — 2 and « — 3 (§ 733), we have 1 4>0 -15 4-10 +24 [2 ^ 1 4-0 -15 +10+24 [3 2 4-22-24 3 9 -18-24 2-11-12, Rem. 3-6-8, Rem. The work shows that 2 and 3 are roots of the given equation ; and since the equation cannot have more than two positive roots, these are the only positive roots. The remaining root may^be found by dividing 24 by the product of — 1, 2, and 3 (§ 720), or by the same process as above. Dividing the first member by a; + 2, a; + 3, etc., we have 1 +0 -15 +10 +24 [^ 1 +0 -15 +10 +24 L-3 -2 4 22 -64 -3 9 18 -84 -2-11 32-40 -3-6 28-60 SOLUTION OF HIGHER EQUATIONS 547 1+0-15+10+24 I -4 _ 4 16 -4-24 -4160 The work shows that the remaining root is — 4. 768. By § 728, an equation of the nth degree in the general form, with fractional coefficients, may be transformed into another whose coefficients are integral, that of the first term being unity. The commensurable roots of the transformed equation may then be found as in § 767. Ex. Find all the roots of 4 ar' - 12 iK^ _^ 27 a; - 19 = 0. Dividing through by the coefficient of x^, we have a^_3a^2^27»_19^0 ' ^. 4 4 Proceeding as in § 728, it is evident by inspection that the multiplier 2 will remove the fractional coefficients ; thus the transformed equation is a^ _ 2 . 3aj2^ 22 . ?I^- 2« . 1?= 0, , 4 4 or, ar^-6aj2 + 27a;-38 = 0; (1) whose roots are those of the given equation multiplied by 2. By Descartes' Eule, equation (1) has no negative root. The positive integral divisors of 38 are 1, 2, 19, and 38. Dividing the first member by x — 1, x — 2, etc., we have l_6+27-38Ll 1-6+ 27- 38 [_2 1 - 5 22 2 - 8 38 -5 22 -16 _4 19 The work shows that 2 is a root of (1). The remaining roots may now be found by depressing the equation; it is evident from the right-hand operation above that the depressed equation is x- — 4 a; + 19 = 0. Solving this, a; = 2 ± V— 15. Thus the roots of (I) are 2 and 2 ± V- 15. 648 ADVANCED COURSE IN ALGEBRA Dividing by 2, the roots of the given equation are land 1 ± -V— 15. z ]/ 769. Newton's Method of Divisors. If a is an integral root of the equation x^ + Pia;"-i H h p^_2x2 + p^^-^x + p^ = 0, where pi, •••, p,i are integers, then Transposing, and dividing by a, ^ = -Pn-,-Pn-2a p,a^-'-a--'', (1) a from which it is seen that — must be an integer. a We may write (1) in the form Pn a -\- Pn-y= — Pn-i'^ Pia''-'— a' p Kepresenting ^^ +p„-i by g,,_i, and dividing by a, from which it is seen that ^^ must be an integer. a Proceeding in this way, it is evident that, if a is a root of the equation, each of -the numbers or -^^, •••, (Id (1 or — > must be an integer, and — + 1 must equal 0. a a a ^ We then have the following rule : Divide the last term of the equation by one of its integral divisors, and to the quotient add the coefficient of x. Divide the result by the same divisor, and, if the quotient is an integer, add to it the coefficient of x^. Proceed in this manner with each coefficient in succession; then, if the divisor is a root of the equation, each quotient will be integral, and the last quotient added to unity will equal zero. SOLUTION OF HIGHER EQUATIONS 549 If a fractional quotient is obtained at any stage, the cor- responding divisor is not a root of the equation. Ex. Find all the roots oi x^ - a? — 7x^ + x -\- Q> = 0. By Descartes' Rule, the equation cannot have more than two positive, nor more than two negative roots. The integral divisors of 6 are ±1, ±2, ±3, and ± 6. By actual substitution, we find that 1 and — 1 are roots. We will next ascertain if 2 is a root ; a convenient arrange- ment of the work is shown below : 1 -1 -7 +1 +6[2_ 2 3 -5 4 The operation is carried out as follows : Dividing 6 by 2, gives 3 ; adding 1, gives 4. Dividing 4 by 2, gives 2 ; adding — 7, gives — 5. Dividing — 5 by 2, the quotient is fractional ; therefore, 2 is not a root. l-l-7+l+6[3_ l_l_7-M+6 [-2 -1 -2 __1 __2 -1 3 1-3 0-3-6 3 2-6-2 In these cases, each quotient is integral, and the last quotient added to unity gives j therefore, 3 and — 2 are roots. There is no need of trying + 6 in this example, for we know that the equation cannot have more than two positive roots. EXERCISE 131 In each of the following, find all the commensurable roots, and the remaining roots when possible by methods already given : 1. x8 _ 9a;2 + 23ic - 15 = 0. 4. «» + 4 a;2 _ 9a; - 36 = 0. 2. x^- 8x2+ 5a; + 14 = 0- 5. Sx" + 4a;a - ISx + 6 = 0. 3. x8 + 12x2 + 44x + 48 = 0. 6. 4x8 + 16 x2- 7x - 39 = 7. X* + 10 x8 + 36 x2 + 60 X + 24 = 0. 8. x* - 5 x8 + 20 X - 16 = 0. 9. X* - 15 x» + 65 x2 - 105 X 4- 54 = 0. 550 ADVANCED COURSE IN ALGEBRA 10. X* + 8 x3 + 11 x2 - 32 X - 60 = 0. 11. x* - 2 a;3 - 17 a;2 + 18 X + 72 = 0. 12. 4 x* - 12 x3 - 9 x2 + 47 X - 30 = 0. 13. 6 X* - 7 x3 - 37 x2 + 8 X + 12 3:: 0. 14. x5 + 8 X* - 7 x3 - 103 x2 + 69 X + 18 = 0. 15. 3 X* + 2 x3 - 18 x2 + 8 = 0. 16. x* + x3 - 6 x2 + 16 X - 32 = 0. RECIPROCAL OR RECURRING EQUATIONS 770. A Reciprocal Equation is one such that if any number is a root of the equation, its reciprocal is also a root. It follows from the above that, if - be substituted for a: in a X reciprocal equation, the transformed equation will have the same roots as the given equation. 771. Let a;" H-Pia;"-1 -{-p^X""-^ -\ \-Pn-2X^ +Pn-lX+Pn = (1) be a reciprocal equation. Putting - for x, the equation becomes X Clearing of fractions, and reversing the order of the terms, Pn^"" + Pn-l^^""' + Pn-'P^'"''^ H h i?2«^ + Pl^? + 1 = 0. Dividing through by p„, Pn Pn Pn Pn Pn By § 770, this equation has the same roots as (1) ; and hence the following relations must hold between the coefficients of (1) and (2), P. = ^,P. = ^^ ••.P.-. = f,P„-. = |-',P.=f (3) Pn Pn Pn Pn Pn From the last equation, p^ = l\ whence, p,, = ± 1. I I SOLUTION OF HIGHER EQUATIONS 551 Then the equations (3) become Pl=±Pn-l, P2=±Pn--2, '"] all the upper signs, or all the lower signs, being taken together. We then have four varieties of reciprocal equations : 1. Degree odd, and coefficients of terms equally distant from the extremes of the first member equal in absolute value and of like sign ; as, ic^ — 2a^ — 2a;-fl = 0. 2. Degree odd, and coefficients of terms equally distant from the extremes of the first member equal in absolute value and of opposite sign ; as, 3ar^ + 2a5^--i»^-|-a^ — 2a; — 3 = 0. 3. Degree even, and coefficients of terms equally distant from the extremes of the first member equal in absolute value and of like sign ; as, cc'^ — 5a^4-6a^ — 5cc + l = 0. 4. Degree even, and coefficients of terms equally distant from the extremes of the first member equal in absolute value and of opposite sign, and middle term wanting; as, 2a;« + 3a^-7a;* + 7a^-3ic-2 = 0. On account of the properties stated above, reciprocal equa- tions are also called Recurring Equations. 772. Every reciprocal equation of the first variety may be written in the form p^"" + Pvxf-^ + p^''-'^ H h P^ + Pvc + i?o = 0, or, plx- + 1) +Pi{x--' +x) + p^ix""-' + a^) + - = 0; (1) or, pIx- 4- 1) + i>ia;(a;"-2 + 1) + p^{x--^ + 1) _^ . . . = ; the number of terms being even. By § 142, since n is odd, each of the expressions oj^-f 1, rf.n-2 _|_ -^^ g^g^ jg divisible by a; + 1. Therefore, — 1 is a root of the equation. Dividing the first member of (1) by a; + l, the depressed equation is p^ix''-'^ - a;"-2 + a;"-3 \-x^-x +1) + Pi(a;''-^ - x^-^ + »"-'* h «^ — a^ + a?) + ^2(35""^ - «""* + a^""* -f. a;4 _ a^ _^ ^) ^ ... ^ 0. 55^ ADVANCED COURSE IN ALGEBRA Or, i?oa^'*-^ + (i>i-Po)aJ"-^+ (7?2-JPi+Po)aJ'*"^+ ... which is a reciprocal equation of the third variety. 773. Every reciprocal equation of the second variety may be written in the form p^"" + piic''"^ -\- p.;f(f'~^ 4- . . . _ p^y^ — p^x — po = 0, or, p.iof - 1) + p^{x--^ -x)+ pIx^--" - x2) + . . . = 0, (1) or, poix^ - 1) + Pixix'^-'' - 1) + P'fo\x''-' - 1) + . . . = 0. Since each of the expressions x"" — 1, ic"~^ — 1, etc., iS divisi- ble by a; — 1, + 1 is a root of the equation. Dividing the first member of (1) by ic — 1, the depressed equation is Poix""-^ + x""-" + x""-^ H h a^ 4- ic + 1) + i>2(a5^"^ 4- a?"""* + aj""^ + ••• + a;* + a^ + aj2) + ... = 0, or, j5oa;^~H (i>i+i>o)a5'"^+ (P2+i>i+i>o)a3""^+ ••• + (i^2 + i>i + Po) a^ + (i>i + Po) a^ + i?o = ; which is a reciprocal equation of the third variety. 774. Every reciprocal equation of the fourth variety may be written in the form p,(x^ - 1) + p,(x^-^ - x) -^p,(x-'^ _ a;2) 4- ... = 0, (1) or, po(^'»-l)+i>ia^(^''"'-l)+P2aj'(i»'^-'-l) + -=0; the number of terms being even (§ 771). Since each of the expressions x^ — 1, o?*""^ — 1, etc., is divisible by x^ — 1, both 1 and — 1 are roots of the equation. Dividing the first member of (1) by x^ — 1, the depressed equation is Pq{x^-'^ + x^'"^ + x""-^ H h a^^ + a^ + 1) -f-Pi(af*-^ + x""-^ + X''-'' H h a^ + a^ + a;) -\-p2{xr~'^ + x""-^ +.a;"-« H h a?^ + a;^ + aj^) + ••• = 0, or, poaJ"-'' + Pia;"-' + (i?2 +i>o) a;**-" + .- 4- (i>2 +Po)a^ +PiaJ +Po = 0; which is a reciprocal equation of the third variety SOLUTION OF HIGHER EQUATIONS 553 775. Every recijyi'ocal equation of the third variety may he reduced to an equation of half its degree. Let the equation be Dividing through by a?*", the equation may be written 4-Pm-/«+^Vp„ = 0. (1) Put X -{- -=z y. X Then, a^4-i = ('aj + -y-2 = 2/2_2; = 2/(2/'-32/)- (2/^-2) = 2/''-42/^ + 2; etc. In general, an expression of the rth degree with respect to y. Substituting these values in (1), the equation takes the form 776. It follows from §§ 772 to 775 that any reciprocal equa- tion of the degree 2 m + 1, and any reciprocal equation of the fourth variety of the degree 2 m + 2, can be reduced to an equation of the mth degree. 777. Ex. Solve2aj^-5aj^-13a:34-13iB2_^5a._2^0. The equation being of the second variety, one root is 1 (§ 773). 554 ADVANCED COURSE IN ALGEBRA Dividing by a; — 1, the depressed equation Is a reciprocal equation of the third variety. Dividing by x\ 2(x^ + i^ _ 3 fa? + -^ - 16 = 0. Putting a; + - = 2/, and a?2 + 1 =/ - 2 (§ 775), we have X X 2(f-2)-3y-16 = 0. Solving this equation, j^ = 4 or Taking the first value, ic + - = 4, or x^ — 4:X = —1. X Whence, a; = 2 ± V3. 1 5 Taking the second value, x-\-- = , or 2 a^ -\-5x = — 2. X 2 Whence, a; = — 2 or ' 2 The roots of the given equation are 1, — 2, , and 2 ± V3. A That 2 + V3 and 2 — \/3 are reciprocals may be shown by mulMplying them ; thus, (2 + V3) (2 - V3) = 4-3 = 1. ^ , , _, . EXERCISE 132 Solve the following : 1. 4 a;3 + 21 a;2 -f 21 X + 4 = 0. 3. cc^ - 5 x2 - 5 x + 1 = 0. 2. ic3 + 4 a;2 - 4 X - 1 = 0. 4. 6 x* + 13 ic^ - 13 x - 6 = 0. 5. 24 x* - 10 5c3 - 77 ic2 - 10 X + 24 = 0. 6. x5 + 2 x4 - 6x8 + 5 x2 - 2 X - 1 = 0. 7. 5x5 - 56 x* + 131 x3 + 131 x2 - 56 X + 5 = 0. 8. 3 x5 + 4 X* - 23 x3 - 23 x2 + 4 X + 3 = 0. 9. 6 x5 - 7 X* - 27 x3 + 27 x2 + 7 X - 6 = 0. 10. 10 x6 - 19 x5 - 19 x* + 19 x2 + 19 X - 10 = 0. 778. Binomial Equations. A Binomial Equation is an equation of the form x'^ = a. Binomial equations are also reciprocal equations, and, in certain cases, may be solved by the method § 777. SOLUTION OF HIGHER EQUATIONS 555 Putting X = ay, the equation a;** = ± a" becomes ?/" = ± 1 ; which is a form to which every binomial equation may be reduced. In § 460, methods were given for the solution of the bi- nomial equations ^^^ = ±1, x^=±l, and x^=±l. The forms x^= ±1 may be solved by the method of § 777. Binomial equations of any degree may be solved by a method involving Trigonometry. EXERCISE 133 Solve the following : 1. x^ = 1. 2. x^ = - 1. 3. x^ = a^. 779. The Cube Roots of Unity. By Ex. 1, § 460, the roots of the equation a?^ = 1 are 1, =^±^, and ^izLV^. The third root is the square of the second ; for / _ 1 + V^Ts Y _ 1 - 2 V^^ - 3 _ -1-V^ The second root is also the square of the third. * Hence, if the second root be denoted by a>, the three cube roots of unity are 1, w, and w^ Since w^ = 1, they are 1, w, and — or — 780. If the second root be denoted by a, the three roots are a, am, and aw^ ; for these equal o>, ^ or 1, respectively. In like manner, if the third root be denoted by a, the three roots are a, aw, and a and aw^ CUBIC EQUATIONS 781. A Cubic Equation is an equation of the third degree (§ 113), containing but one unknown number. 556 ADVAK^CED COtlESE IN ALGEBRA 782. By § 732, the cubic equation where pi is not zero, may be transformed into another whose second term shall be wanting by substituting ^ — ;^' for x. Hence, every cubic equation can be reduced to the form a^ 4- aa? + 6 = 0. 783. Cardan's Method for the Solution of Cubics. Let it be required to solve the equation ^ -\- ax-\-h~0. Putting X = y+^j the equation becomes or, y^j^z^j^{^y^^){y + z)^-b = (}. We may give such a value to z that Syz + a shall fequal zero. Whence, z = — ^ (1) oy Then, 2/' + 5' + & = Q. (2) Substituting the value of z from (1) in (2), we have f--^^ + b = 0, or if + bf = ^' ■ ^ 21 f J ^ y 27 This is an equation in the quadratic form (§ 468). Solving by the rules for quadratics, we have Theiiby(2), ^ = -/-6 = -|T^|' + g- (4) Taking the upper signs before the radicals, in (3) and (4), and substituting in the equation x = y -{-z, we have The lower signs before the radicals give the same value oix. The other two roots may be found by depressing the given equation (§ 716). SOLUTION OF HIGHER EQUATIONS 657 784. Ex. Sol ve the equation a:^ + 3 a^ — 6 cc + 20 = 0. We first transform the equation into another whose second term shall be wanting. Putting x = y-\ = y-X{^ 782), we have o 2/3-^2/' + 32/-l.+ 32/^-6?/ + 3-62/ + 6 + 20 = 0, or, 2/3_9^_,_28 = 0. To solve the latter equation, we substitute a = — 9 and 6 = 28 in (5), § 783. Thus, 2/ = V_ 14 + V196 - 27 + V_14-V196-27 = ^^^ + V^^^ = _l_3 = -4. Therefore, flji=2/ — 1 = — 5. Dividing the first member of the given equation by a; + 5, the depressed equation is a;2_2a;.+ 4 = 0. Solving, x = l± V^^. Thus, the roots of the given equation are — 5 and 1 ± V~3. EXERCISE 134 Solve the following : 1. x3-245c-72 = 0. 6. ic3 + 6x2 + 27x-86 = 0. 2. ic8 _ 12 X + 16 = 0. 7. a;3 + 9 x2 + 12 ic - 144 = 0. 8. a;8 + 72 x + 152 = 0. 8. ic^ + x2 - 3 x + 36 = 0. 4. x8 - 12 x2 + 21 X - 10 = 0. 9. x3 - 2 x2 - 15 X + 36 = 0. 5. x3 - 3 x2 + 48 X + 52 = 0. 10. x^ - 4 x2 + 8 x - 8 = 0. 11. Find one root of x^ H- x — 2 = 0. A cubic equation having, a commensurable root is solved more easily by the method of § 767 than by Cardan's rule. 785. If h is any one of the cube roots of — o+vi '^on^ Z '4 Zi and 7c any one of the cube roots of "~9~\/x"^97' ^^® three cube roots of the first expression are h, hw, and ho)^, and the three cube roots of the second are k, koi, and fcw^ (§ 780). 558 ADVANCED COURSE IN ALGEBRA This would apparently indicate that a; has nine different values. But by (1), § 783, yz — — -; that is, the product of the terms o whose sum is a value of x must be — - • o Hence, the only possible values of x are h-{-7c, hu>-{-k(i)^, and h -r • (Compare § 786, 3.) Putting x = 2 m cos A, the equation becomes 8 m^ cos^ A — 2 am cos ^ — 6 = 0; or, 4 cos^ ^ — -^ cos ^ — - — - = 0. m- 2 m^ It is proved in Trigonometry (see Ex. 16, p. 35, Wells^ Com- plete Trigonometry) that 4 cos^ J. = cos 3 ^ + 3 cos A. Whence, cos 3 ^ + 3 cos ^ — — , cos A — = 0. m^ 2m^ Or, cos3^ + (^3-^Vos^l--^ = 0. (1) V my 2 m^ SOLUTION OF HIGHER EQUATIONS 561 We may take m so that 3 — -^, = ; then m = -J- • (2) Then (1) becomes cos 3A = Substituting in this the value of m from (2), cos3^ = |^|. (3) Since, by hypothesis, -r < t^? we have — x -^ < 1. 4 27 4 a^ Taking the square root of both members of the inequality, Then, the value of 3^ in (3) is possible, since its cosine ig <^ ;|^^ Let z be the least positive angle whose cosine equals j:\—^' Then, one value of 3 ^ is ;$;. ^ Two other values are 2 7r-{-z and 2 tt — 2; ; for the cosines of these angles are equal to the cosine of z. Then, SA = z, or 2'7r±z; and a; = 2mcos^ = 2^^cos| or 2^cosC^±^\ where z is given by the equation cos z = o\/~i* An equation of the third degree cannot have more than three different roots; so that these are the only values of x. 789. Ex. Solve the equation aj3_4a;-2 = 0. i Here, a = 4, 6 = 2; V27 — • By logarithms, log cos 2; = - (log 27 — log 64). 562 ADVANCED COURSE IN ALGEBRA log 27= 1.4314 log 64 = 1.8062 2 )19.6252-20 log cos 2= 9.8126-10 Then, z = 4:9° 29.3'. Thus, ^ = 16° 29.8', o Then, the values of x are : 16 cos 16° 29.8', 3 4 ^15 cos (120° + 16° 29.8') = aM cos 136° 29.8' -V -sin 46° 29.8', o and yj— cos (120° - 16° 29.8') = yM cos 103° 30.2' = -^^sin 13° 30.2'. log -^ = 1 (log 16 - log 3) = ^ (1.2041- .4m)=.3635. (1) log cos 16° 29.8' = 9.9817 - 10. ' (2) log sin 46° 29.8 ' = 9.8606 - 10. (3) log sin 13° 30.2' = 9.3683 - 10. (4) Adding /2), (3), and (4) to (1), the logarithms of the abso- lute values of x are 0.3452, 0.2241, and 9.7318 - 10. The numbers corresponding to these are 2.214, 1.675, and .5393. Then, a; = 2.214, -1.675, or -.5393. SOLUTION OF HIGHER EQUATIONS 563 If the given equation had been r"^ — 4 x + 2 = 0, we should have had /27 a = 4, & = — 2 ; and cos z would have been —a/ — >64 /27 |27 If cos ^ = - ^— , COS (tt - S) = - COS = -y| — . We should then have found ir -z-W 29.3', and z = 130° 30.7'. The three values of x would then have been : J^ cos 43° 30.2', and yj^ cos (120° ± 43° 30.2'). We should have found x = 1.675, - 2.214, or .5393. In any case where a and b are positive, z is acute, and the equation has one positive and two negative roots ; if a is positive and b negative, z is obtuse, and the equation has two positive and one negative root. EXERCISE 136 Solve the following : 1. x3 - 4 a; - 1 = 0. 3. a;^ + 6 a;2 _ ^ - 1 = 0. 2. x3-6x + 3 = 0. 4. x3-3x2-2x + l=0. BIQUADRATIC EQUATIONS 790. A Biquadratic Equation is an equation of the fourth degree, containing but one unknown number. Euler's Method for the Solution of Biquadratics. By § 732, every biquadratic equation can be reduced to the f oi^ni x' + aa^-^bx + c = 0. (1) Let x = u-{-y-{-z', then, a^ = u- -\- y^ -^ z^ -{- 2 uy -{-2 yz ■^2zu, or, x" - (u'' + y' + z') = 2(uy + ^JZ + zu). Squaring both members, we have x^-2x' (^2 4- / + 2") H- (u^ + 2/' + zy = ^(uy-{-yz-\- zuf = 4 (uY + i/z^ + ;2 V) + 8 uyz (u-hy + z). Substituting x for u-^y -{- z, and transposing, x*—2x^ (u^ 4- 2/2 -f 2;^) — 8 ayzx + (u' -hy'-h ^^y - 4 (uhf + 7/V + z'u"^ = 0. 664 ADVANCED COURSE IN ALGEBRA This equation will be identical with (1) provided b = — Suyz, (2) and c=:(u' + tf-^ z'f - 4 (uY + 2/'^' + ^'^')- Then, ,^2 _^ / 4- ^2 ^ _ ^, and AV=— • «'-c Also, uY + 2/V + 2:2^2 ^ v_ — ^ ^ ^ — = __ = —j^' If, now, we form the cubic equation the values of t will be u% if, and ;22 (§ 7I8). Hence, if the roots of the cubic equation j3 + «j2 + «!zii^j_^^0 (3) be I J m, and n, we shall have u = ± V^, y=± Vm, and z = ± \^n. Kow a; = 1^ + 2/ 4- :2 ; and since each of the numbers u, y, and z has two values, apparently x has eight values. But by (2), the product of the three terms whose sum is a value of X must be 8 Hence, the only values of x are, when b is positive, — V^ — Vm — Vn, — V^ + Vwi 4- Vn, V^ — Vm4-V^, and V/+Vm— V?i; and when b is negative, vT + Vm + Vw, V^ — Vm — V?ij — Vl H- Vm — Vn, and — V^ — Vm + Vn. Equation (3) is called the auxiliary cubic of (1). 791. Ex. Solve the equation x* — 46 or^ — 24 a; -h 21 = 0. Here, a = -46, & = -24, c = 21. SOLUTION OF HIGHER EQUATIONS 565 Whence, ^'7/^ = 127, and — = 9. 16 ' 64 Then the auxiliary cubic is t^ —23t^-{- 127 « — 9 = 0. By the method of § 767, one value of t is 9. Dividing the first member by ^ — 9, the depressed equation isf-Ut-^l = 0. ,__ Solving, ^=:7± V49-1 = 7±4V3. Proceeding as in § 392, v^e have V(7±4V3)=V(4±2V12 + 3) = 2±V3. Then since b is negative, the four values of x are 3 + 2 + V3 + 2-V3, 3-2-V3-2+V3, -3 + 2 + V3-2 + V3, and -3-2- V3 + 2- V3. ^ That is, 7, -1, -3 + 2V3, and -3-2V3. c , t. r 11 . EXERCISE 137 Solve the following : 1. x^-mx^-\-S0x + SS4:=0. 2. x4 - 44 a;2 + 16 X + 192 = 0. 3. a:* - 40 a;2 + 64 X + 128 = 0. 4. X* - 54 a:2 - 216 x - 243 = 0. 5. x4 - 22 x2 - 12 X + 48 = 0. 6. X* + 4 x3 - 4 x2 - 37 X - 42 = 0. 792. Discussion of the Solution. The auxiliary cubic of af* -j- ax^ -{- bx -\- c = is t' + 1^ + ^^^t - 1^ = (§ 790). (1) ^ lb o4 Since the last term is essentially negative, the equation must have either three positive, one positive and two nega- tive, or one positive and two pure imaginary or complex roots (§718). Transforming the equation into another whose coefficients shall be integral, that of the first term being unity (§ 728), we have t^ + 2at'' + {ct'-^c)t-b'' = 0, (2) whose roots are respectively 4 times those of (1). 566 ADVANCED COURSE IN ALGEBRA Denoting 2 a, a^ — 4iC, and — b^ by a', b', and c', respectively, it is necessary, before we can determine the nature of the roots of (2), to compute the values of 3 6' — a'-' and 2a'^ — 9a'b'-{- 27 c' (§787). •Now, 3 6'-a'2 = 3(a2_4c)-4a2=_(a2+l2c), and 2a'^-9a'6' + 27c' = 16a-^-18a(a2-4c)-2762 = -(2a^-72ac + 2Tb^). Then it follows from § 787 that : 1. If a--\-12 c is negative, or if a^-\-12 c is positive and 4 (a^-|- 12 cy less than (2 a^ — 72 ac + 27 6-)^, the auxiliary cubic has one positive and two pure imaginary or complex roots. If V7=i>j Vm = g + rV — 1, and V^ = g — rV — 1, the roots of the biquadratic are — p±2q andjp±2rV— 1, or p±2q and — p±2rV — 1, according as b is positive or negative. That is, the biquadratic has two real and two pure imagi- nary or complex roots. 2. If o?-\-\2c is positive, and 4 (a^ + 12 cf is equal to (2 o? — 72 ac + 27 6^)^, the cubic has two roots equal. If Vn = Vm, the roots of the biquadratic are — V7 ± 2 Vm, Vl, and V7, or VT ± 2 Vm, — VT, and — V^ according as b is positive or negative. That is, the biquadratic has two roots equal. J 3. If a^ H- 12 c is positive and 4(a2 + 12 cf greater than (2 o? — 72 ac + 27 6^^, the cubic has either three positive, or ona positive and two negative roots. In the first case, the roots of the biquadratic are all real ; in the second case, they are all pure imaginary or complex. 4. Ifa2+12c = 0and2a3-72ac + 2762 = 0,thenc = -:^. Substituting from the third equation in the second, Sa'-^27b'==0, ora=-?6*; whence, a^ - 4 c = ^' = 3 6*. ^b'== SOLUTION OF HIGHER EQUATIONS 567 In this case, the auxiliary cubic becomes '-!*-fe*-£=».»('-f)-»> 2 and each of its roots is equal to — • The roots of the biquadratic are — , -^j -^j and -~, or 3 6^ 6* 6* , 6^ ,. , . , —-J7, — 17> and — — , according as b is positive or z Z 2d . Z negative; that is, the biquadratic has three roots equal. 5. If a^ — 4 c = and 5 = 0, the biquadratic becomes and its roots are ±^—^ and ±\/— -• That is, the biquadratic has two pairs of equal roots. INCOMMENSURABLE ROOTS 793. We will now show how to find the approximate nu- merical values of those roots of an equation which are not commensurable (§ 767). 794. Homer's Method of Approximation. Let it be required to find the approximate value of the root between 3 and 4 of the equation a:3-3ic2-2icH-5 = 0. We first transform the equation into another whose roots shall be respectively those of the first diminished by 3 (§ 733).' 1 _3 _2 +5 13^^. ii-\ __? _9 Zl? '-''-^^^ -2 -1 3 9 3 7 3 6 568 ADVANCED COURSE IN ALGEBRA The transformed equation is y^ -\- 6 y- -{- 7 y — 1 = 0. (1) We know that equation (1) has a root between and 1. If, then, we neglect the terms involving y^ and y^, we may- obtain an approximate value of y by solving the equation 7?/ — 1=0; thus, approximately, y = .l and a; =,3.1. Transforming (1) into an equation whose roots shall be respectively those of (1) diminished by .1, we have 1 +6 +7 -1 \jl 6 +7 -1 .1 .61 6.1 7.61 .761 - ,239 a .62 6.2 8^3 .1 6^ The transformed equation is 23 _^ e_3 ^2 _^ g 23 2 _ .239 = 0. (2) Neglecting the z^ and z^ terms, we have, approximately, z = -^ = .02. 8.23 Thus, the value of x to two places of decimals is 3.12. Transforming (2) into an equation whose roots shall be respectively those of (2) diminished by .02, we have 1 +6.3 +8.23 -.239 |^ .02 .1264 .167128 6.32 .8.3564 -.071872 .02 .1268 6.34 8.4832 .02 6.36 The transforuied equation is u^ + 6.36 u' + 8.4832 u - .071872 = 0. Dividing .071872 by 8.4832, we have .008 suggested as the fourth figure of the root. SOLUTION OF HIGHER EQUATIONS 569 Thus, the value of x to three places of decimals is 3.128. The process may be continued until the value of the root has been found to any desired degree of precision. The work is usually arranged in the following form, the coefficients of the successive transformed equations being denoted by (1), (2), (3), etc. : 1 -3 - -2 + 5 13.128 3 - -2 (1) -6 -1 3 3 (1)" 9 7 (2) .761 - .239 3 .61 .167128 (1) 6 7.61 (3) - .071872 .1 .62 6.1 (2) 8.23 .1 .1264 6.2 8.3564 .1 .1268 (2) 6.3 (3) 8.4832 .02 6.32 .02 6.34 .02 (3) 6.36 We derive from the above the following rule for finding the approximate value of a positive incommensurable root : Find by §§,741 or 742, or by Sturm^s TJieorem (§T58), the integral part of the root. (Compare § 743.) Transform the given equation into another ivhose roots shall be respectively those of the first diminished by this integral part. Divide the absolute value of the last term of the transformed equation by the absolute value of the coefficient of the first power of the unknown number, and write the approximate value of the residt as the next figure of the root. 570 ADVANCED COURSE IN ALGEBRA Transform the last equation into another whose roots shall be respectively those of the first diminished by the figure of the root last obtained, and divide as before for the next figure of the root; and so on. In practice, the work may be contracted by dropping such decimal figures from the riglit of each column as are not needed for the required degree of accuracy. In determining the integral part of the root, it will be found convenient to construct the graph of the first member of the given equation. 795. To find the approximate value of a negative incommen- surable root, transform the equation into another which shall have the same roots with contrary signs (§ 726), and find the corresponding positive incommensurable root of the transformed equation. The result with its sign changed will be the required negative root, 796. In finding any particular root-figure by the method of § 794, we are liable, especially in the first part of the process, to get too great a result ; the same thing occasionally happens when extracting square or cube roots of numbers. Such an error may be discovered by observing the signs of the last two terms of the next transformed equation ; for since each root-figure obtained as in § 794 must be positive, the last two terms of the transformed equation must be of opposite sign. If this is not the case, the last root-figure must be diminished until a result is obtained which satisfies this condition. Let it be required, for example, to find the root between and — 1 of the equation oc^-\-4rX^ — 9x — 5 = 0. Changing the signs of the x^ term and the independent term (§ 726), we have to find the root between and 1 of the equa- tion a^-4a^- 9 x-f5 =0 (§ 795). Dividing 5 by 9, we have .5 suggested as the first root-figure ; but it will be found that in this case the last two terms of the first transformed equation are — 12.25 and — .375. SOLUTION OF HIGHER EQUATIONS 571 This shows that .5 is too great ; we then try .4, and find that the last two terms of the first transformed equation are of opposite sign. The work Jf finding the first three root-figures is shown below : (1) -4 - 9 + 5 |.469 .4 - 1.44 -4.176 -3.6 - 10.44 (1) .824 A - 1.28 - .713064 -3.2 (1) - 11.72 (2) .110936 .4 -2.8 - .1644 - 11.8844 .06 - .1608 -2.74 (2) - 12.0452 .06 -2.68 .06 (2) -2.62 The required root is — .469, to three places of decimals. 797. In case too synall a number is taken for the root-figure, the number suggested for the next root-figure will be greater than .09. Let it be required, for example, to find the root between and 1 of the equation a^-2iB2 + 3a;-l = 0. Dividing 1 by 3, we have .3 suggested as the first root- figure. 1 -2 + 3 -1 ^ .3 - .51 .747 -1.7 2.49 - .253 .3 - .42 -1.4 2.07 .3 -1.1 The number suggested by the next division is greater than .1 J showing that too small a root-figure has been taken. 572 ADVANCED COURSE IN ALGEBRA 798. If the coefficient of the first power of the unknown number in any transformed equation is zero, the next figure of the root may be obtained by dividing the absolute value of the last term by the absolute value of the coefficient of the square of the unknown number^ and taking the square root of the result. For if the transformed equation is / + a/ -{-b = 0, it is evi- dent that, approximately, ay^-\-b = 0, or i/ = •^/ _ _. ^^ a We proceed in a similar manner if any number of consecu- tive terms immediately preceding the last term are zero. Horner's method may be used to find any root of a number approxi- mately ; for to find the nth root of a is the same thing as to solve the equation x" — a = 0. 799. If an equation has two or more roots which .have the same integral part, the first decimal root-figure of each must be obtained by the method of §§ 741 or 742, or by Sturm's Theorem. If two or more roots have the same integral part, and also the same first decimal root-figure, the second decimal root- figure of each must be obtained by the method of §§ 741 or 742, or by Sturm's Theorem ; and so on. Horner's method may be used to determine successive figures in the integral, as well as in the decimal, portion of the root. If all but one of the roots of an equation are known, the remaining root may be found by changing the sign of the coefiBcient of the second term of the given equation, and subtracting the sum of the known roots from the result (§ 720). EXERCISE 138 Find the root between : 1. 1 and 2, of a:^ _ 9 ^2 + 23 x - 16 = 0. 2. 4 and 5, of x3 - 4 x2 - 4 x + 12 = 0. 3. and -^ 1, of x^ + 8x2 - 9x - 12 = 0. 4. - 2 and - 3, of x8 - 3x2 - 9 X + 4 = 0. 5. 3and4, of x3-6x2 + 15x-19 =0. 6. and 1, of X* + x3 + 2x2 - X - 1 = 0. SOLUTION OF HIGHER EQUATIONS 573 7. 2 and 3, of x* - 3 x3 + 4 ic - 5 = 0. 8. - 1 and - 2, of X* - 2 ic3 - 3x2 + ic - 2 = 0. Find all the real roots of the following : 9. x3 + 2 x2 - X - 1 = 0. 13. x8 - x2 + 2 X - 1 = 0. 10. x8-2x2 - 7x- 1 =0. 14. x3-x2-15x + 28 = 0. 11. x3-5x2 + 2x + 6 = 0. 15. x*-6x2 + llx + 21 = 0. 12. x* + 2 x3 - 5 = 0. 16. X* - 6 xH 6 x2+ 8 x + 1 = 0. Find the approximate values of the following : 17. v^. 18. v^21. 19. v^. 20. -^ i W mi "^^1% '-^^^ 4^^ Rii'-^ ^ :-'^ .^„.,,.. 1- LD MAY 1 5 B6Z agib % APR 22 1963 JANS 1970 fi """-^M-i-in'm RETURNED TO DEC 2 1972 LOAN AHC MAY t -' m\ ^zm^ JAN 1^ i.qy? 8 7 3 73-irAW8 8 LD 21A-50m-9,'58 (6889sl0)476B General Library University of California Berkeley Sia./n^3 /^9 w THE UNIVERSITY OF CAUFORNIA UBRARY AV} .^s j^r&^^!^ G^^sv^iIS^ cforsv^X^