IN MEMORIAM FLOR1AN CAJOR1 SECOND COURSE IN ALGEBRA A SERIES OF MATHEMATICAL TEXTS EDITED BY EARLE RAYMOND HEDRICK THE CALCULUS By ELLERT WILLIAMS DAVIS and WILLIAM CHARLES BRENKE. ANALYTIC GEOMETRY AND ALGEBRA By ALEXANDER ZIWET and Louis ALLEN HOPKINS. ELEMENTS OF ANALYTIC GEOMETRY By ALEXANDER ZIWET and Louis ALLEN HOPKINS. PLANE AND SPHERICAL TRIGONOMETRY WITH COMPLETE TABLES By ALFRED MONROE KENYON and Louis INGOLD. PLANE AND SPHERICAL TRIGONOMETRY WITH BRIEF TABLES By ALFRED MONROE KENYON and Louis INGOLD. ELEMENTARY MATHEMATICAL ANALYSIS By JOHN WESLEY YOUNG and FRANK MILLETT MORGAN. PLANE TRIGONOMETRY By JOHN WESLEY YOUNG and FRANK MILLETT MORGAN. COLLEGE ALGEBRA By ERNEST BROWN SKINNER. ELEMENTS OF PLANE TRIGONOMETRY WITH COM- PLETE TABLES By ALFRED MONROE KENYON and Louis INGOLD. ELEMENTS OF PLANE TRIGONOMETRY WITH BRIEF TABLES By ALFRED MONROE KENYON and Louis INGOLD. THE MACMILLAN TABLES Prepared under the direction of EARLE RAYMOND HEDRICK. PLANE GEOMETRY By WALTER BURTON FORD and CHARLES AMMERMAN. PLANE AND SOLID GEOMETRY By WALTER BURTON FORD and CHARLES AMMERMAN. SOLID GEOMETRY By WALTER BURTON FORD and CHARLES AMMERMAN. CONSTRUCTIVE GEOMETRY Prepared under the direction of EARLE RAYMOND HEDRICK. JUNIOR HIGH SCHOOL MATHEMATICS By W. L. VOSBURGH and F. W. GENTLEMAN. This book is issued in a form identical with that of the books announced above, as is the First Course in Algebra by the same authors. SECOND COURSE IN ALGEBRA BY WALTER BURTON FORD n PROFESSOR OF MATHEMATICS, THE UNIVERSITY OF MICHIGAN AND CHARLES AMMERMAN THE WILLIAM McKINLEY HIGH SCHOOL, ST. LOUIS gorfe THE MACMILLAN COMPANY 1920 All rights reserved COPYRIGHT, 1920, BY THE MACMILLAN COMPANY. Set up and electrotyped. Published January, 1920 Nortoooti J. 8. Gushing Co. Berwick & Smith Co. Norwood, Mass., U.S.A. PREFACE IN the present volume, which is intended as a sequel to the author's First Course in Algebra, the following features may be noted : (a) The first seven chapters furnish a systematic review of the ordinary first course up to and including the subject of simultaneous linear equations. In these opening chapters the aim is to state briefly and concisely those fundamental principles which are basal in all algebraic work and with which the pupil already has some acquaintance. The method of treatment is largely inductive, being based upon solved problems and other illustrative material rather than upon any attempt at proofs or formal demonstrations. The various principles are explicitly stated, however, at all points in the form of rules, which are clearly set off in italics. Upon this plan the pupil is rapidly and effectively prepared for un- dertaking the newer and more advanced topics which follow. (6) Chapters VIII-IX (Square Root and Radicals) are essentially a reproduction of the corresponding chapters in the First Course, but all of the problems are new. This is true also of the early parts of Chapter X (Quadratic Equa- tions). These are topics which usually present more than average difficulty ; hence, even for pupils who have studied them carefully before, a complete treatment of them is de- sirable in the second course. The introduction and use of tables of square and cube roots at this point ( 43) is to be especially noted. It seems vi PREFACE clear that pupils should be made familiar with such tables at an earlier date than formerly. One strong reason for this is that a constantly increasing number of students pass directly from the high schools into technical pursuits where facility in the manipulation of tables of all kinds is especially desirable. (c) Part II, comprising Chapters XI-XX, presents the usual topics of the advanced course. The order of arrangement follows, so far as possible, that of the difficulty of the various subjects, and the whole has been prepared with a view of introducing a relatively large number of simple illustrative examples drawn from nature and the arts. Throughout the development, however, due emphasis has been given to those fundamental disciplinary values which should be preserved in any course in mathematics. Among the unusual features, it may be observed that the detailed consideration of exponents and radicals has been delayed until logarithms are about to be taken up. These topics in their extended sense have, in fact, but little to do with algebra until that time. Again, the chapter on logarithms is unusually full and com- plete. All the essential features of this relatively difficult but increasingly important subject are presented in detail. In the past, much has ordinarily been left for the teacher to explain. (d) Functions, Mathematical Induction (including the proof of the Binomial Theorem), and Determinants have been grouped together under the title Supplementary Topics. In fact, these subjects lie on the border line between the second course and the college course. Only the elements of each are taken up, but there is enough to show its im- portant bearing in algebra and to pave the way for its further development in the college course. For example, PREFACE vii the study of functions is so presented that it at once amplifies material to be found in the earlier chapters of the book and brings in new material which is connected with the graphical study of the theory of equations. Thus it serves as an in- troduction to the latter subject as presented in the usual college texts. As in the authors' other texts, a star (*) has been placed against certain sections that may be omitted if desired with- out destroying the continuity of the whole. WALTER B. FORD. CHARLES AMMERMAN. TABLE OF CONTENTS PART I. REVIEW TOPICS CHAPTER PAGE I. FUNDAMENTAL NOTIONS 1 II. SPECIAL PRODUCTS AND FACTORING ... 17 III. HIGHEST COMMON FACTOR AND LOWEST COMMON MULTIPLE 26 IV. FRACTIONS 29 V. SIMPLE EQUATIONS . 36 VI. GRAPHICAL STUDY OP EQUATIONS .... 41 VII. SIMULTANEOUS EQUATIONS SOLVED BY ELIMINATION 50 VIII. SQUARE ROOT 59 IX. RADICALS 67 X. QUADRATIC EQUATIONS 78 PART II. ADVANCED TOPICS XI. LITERAL EQUATIONS AND FORMULAS ... 95 XII. GENERAL PROPERTIES OF QUADRATIC EQUATIONS . 108 XIII. IMAGINARY NUMBERS 116 XIV. SIMULTANEOUS QUADRATIC EQUATIONS . . . 122 XV. PROGRESSION 141 XVI. RATIO AND PROPORTION 166 XVII. VARIATION . . . . . . . .178 XVIII. EXPONENTS 193 XIX. RADICALS 205 XX. LOGARITHMS 216 viii TABLE OF CONTENTS ix PART III. SUPPLEMENTARY TOPICS CHAPTER PAGE XXI. FUNCTIONS . . .... 245 XXII. MATHEMATICAL INDUCTION BINOMIAL THEOREM 255 XXIII. THE SOLUTION OF EQUATIONS BY DETERMINANTS 265 PART IV. TABLES TABLE OF POWERS AND ROOTS 275 TABLE OF IMPORTANT NUMBERS 296 INDEX . . .297 LAPLACE (Pierre Simon Laplace, 1749-1827) Famous in mathematics for his researches, which were of a most advanced kind, and especially famous in astronomy for his enunciation of the Nebular Hypothesis. Interested also in physics and at various times held high polit- ical offices under Napoleon. SECOND COURSE IN ALGEBRA PART I. REVIEW TOPICS f CHAPTER I FUNDAMENTAL NOTIONS 1. Negative Numbers. In the First Course in Algebra it was shown how negative as well as positive numbers may be used, the one being quite as common as the other in every- day life. Thus +15 (or simply 15) means 15 above 0, while -15 means 15 below 0. Similarly, +$25 (or simply $25) means a gain or asset of $25, while $25 means a loss or debt of the same amount. Negative numbers are compared with each other in much the same way as positive numbers. Thus, just as in arith- metic 4 is less than 5, 3 is less than 4, etc., until finally we say that is less than 1, so we continue this idea in algebra by saying that 1 is less than 0, 2 is less than 1, etc. The whole situation regarding the size of numbers is vividly brought out to the eye in the figure below : I I I I I I I I I I I I I I I I i I II I | I I I -12-11-10-9-87-6-5-4-3-2-1 1 2 3 4~ ~S 1 8 9 10 11 12 FIG. 1. Here the positive (+) numbers are placed in their order to the right of the point marked 0, while the negative ( ) numbers are placed in their order to the left of that same point. This figure shows all numbers (positive and negative) arranged in their increasing order from left to right. t Chapters I-X (pp. 1-94) furnish a review of the First Course, The remaining chapters deal with more advanced topics. B 1 2 SECOND COURSE IN ALGEBRA [I, 1 In Fig. 1, only the positive and negative integers and zero are actually marked. A complete figure would show also the positions of the fractions. Thus is located at the point halfway between the and the 1 ; 2^ is located at the point one third the way from 2 to 3 ; f is at the point f the dis- tance from to 1 ; 5 is at the point f the distance from 5 to 6 ; and so on for all fractions. By the numerical value (or absolute value) of a negative number is meant its corresponding positive value. Thus the numerical, or absolute, value of 3 is +3, or simply 3. NOTE. The numerical value of any positive number is the number itself. EXERCISES In each of the following exercises, state which of the two numbers is the larger. First locate the number at its proper place on the line shown in 1. 1. 7, 10. 6. |, f. 9. 3i, If. 2. 7, -10. 6. f,-f. 10. -3*, -If. 3. -7, 10. 7. -f, f. 11. -i 0. 4. -7, -10. 8. -f, -f. 12. -.3, -.05 2. Operations with Numbers. The following facts will be recalled from the First Course in Algebra^ (a) To add two numbers having like signs, add their absolute values (1) and prefix the common sign. Thus (-5)+(-6) = -ll. (b) To add two numbers having unlike signs, find the dif- ference between their absolute values and prefix the sign of the one whose absolute value is the greater. Thus (+3)+(-5) = -2. t References to the authors' First Course in Algebra are given in this book by page number. I, 2] FUNDAMENTAL NOTIONS 3 NOTE. If we have more than two numbers to add together, as for example (+4)+(-7)+(+5)+(-6)+(-l), the customary way is to add all the positive parts together, then all the negative parts, and finally to add the two results thus obtained. Thus, in the example just mentioned, the sum of the positive parts is 4 +5 = 9, while the sum of the negative parts is ( 7) + ( 6) +( 1) = 14. The final result sought is, therefore, (+9) +( -14) = -5. (c) To subtract one number from another, change the sign of the subtrahend and add the result to the minuend. Thus (-7) - (-5) = (-7)+(+5) = -2. (d) To multiply one number by another, find the product of their absolute values, and take it positive if the two numbers have the same sign, but negative if they have unlike signs. Thus (+3) (+2) = +6, and (-3) (-2) = +6; but (-3) (+2) = -6 and (+3) (-2) = -6. (e) To divide one number by another, find the quotient of their absolute values, and take it positive if the numbers have the same sign, but negative if they have unlike signs. Thus (+8H(+2) = +4, and (-8)-K-2) = +4; but (-8)^(+2) = -4, and (+8)- ( -2) = -4. EXERCISES Determine the value of each of the following indicated ex- pressions. 1. (+4) + (+7). 8. (+25) + (-32) + (- 2. (-4) + (+7). 9. (+7) -(+4). 3. (+4) + (-7). 10. (+7)-(-4). 4. (-4) + (-7). 11. (-7)-(+4). 5. (+9) + (-27). 12. (-7)-(-4). 6. (-32) + (+16). 13. (+34) -(-63). 7. (-3) + +2) + (-l). 14. -54--32. SECOND COURSE IN ALGEBRA [I, 2 16. [HINT. By (c) of 2, this may first be changed into the form (+6) +(+4) +(-2) +(+5).] 16. 17. (-23) + (+32)-(- 18. (+3) -(-4). 22. (+24) -5- (-6). 19. (-4) -(+3). 23. (-36) -5- (+6). 20. (+5) (+4). 24. (-55) -(-11). 21. (-5) -(-4). 25. (f)-(-i). 3. Use of Letters in Algebra. Algebra is distinguished from arithmetic not only because of its use of negative num- bers, but also because of its general use of letters to represent numbers. This is useful in many ways. In particular, it enables us to solve problems in arithmetic which would other- wise be very difficult. The following facts and definitions will be recalled from the First Course in this connection. The sum of any two numbers, as x and y, is represented by x+y. The difference between any two numbers, as x and y (meaning the number which added to y gives x), is represented bys-y. The product of any two numbers, such as x and y, is written in the form xy. It has the same meaning as xXy, or x - y. Either of the numbers thus multiplied together is called a factor of the product. The quotient of x divided by y is expressed either by x -5- y, or by -, or by x/y. The product x x is represented by x 2 and is read x square; similarly, x x x is represented by x 3 and is read x cube. More generally, x x x to n factors is represented by x n and is read x to the nth power. The letter n as thus used in x n is called the exponent of x. I, 3] FUNDAMENTAL NOTIONS 5 The symbol Vx denotes that number which when squared gives x. It is called the square root of x. Similarly, \/x is ailed the cube root of x and denotes that number which when cubed gives x. In general, ~\/x is called the nth root of x, and denotes that number which when raised to the nth power gives x. The letter n as thus used in Vx is called the index of the root. Whenever one or more letters are combined in such a way as to require any of the processes just described, the result is called an expression. Thus 2 x +3 y, ax-bxy, 6 ran -3V^+2Vn, and 2 x+y z -x* + xyz are expressions. An expression is read from left to right in the order in which the indicated processes occur. Indicated multiplica- tions and divisions are to be carried out, in general, before the indicated additions and subtractions. Thus 2 x+y 2 x 3 +xyz is read "Two x plus y square minus x cube plus xyz' 1 EXERCISES Read each of the following expressions : 1. 2x 2 . 6. 6. _|_ y w 2 3. a-2o&+6>. 7. VS+^. n. x 3 /2. s-y 4. m s -n 3 . 8. v^+v^. 12. \/a w +& r . 13. Express each of the following ideas in letters. (a) The sum of the squares of x and y. (b) The difference between m cube and n cube. (c) Three times the product of mn diminished by twice the quotient of x divided by the square root of y. 6 SECOND COURSE IN ALGEBRA [I, 3 14. The fact that the area of any rectangle is equal to the product of its two dimensions (length and breadth) is ex- pressed by the formula A=ab. Express similarly in words the meaning of each of the following familiar formulas : (a) A =| bh. (Formula for the area of a triangle.) (b) A =irr 2 . (Formula for the area of a circle.) (c) C = 2irr. (Formula for the circumference of a circle.) (d) h 2 = a?+b 2 . (Theorem of Pythagoras concerning any right triangle.) (e) V = % Tir 3 . (Formula for the volume of a sphere.) (/) A = 4 TIT*. (Formula for the area of a sphere.) 4. Evaluation of Expressions. Whenever the values of the letters in an expression are given, the expression itself takes on a definite value. To obtain this value, we must work out the values of the separate parts of the expression and then combine them as indicated. EXAMPLE. Find the value of the expression o+2-bc-c a+b+c when a = 1, b = 2, and c = 3. SOLUTION. Giving a, 6, and c their assigned values, the expres- sion becomes l 2 +2- (-2). 3-3 3 _l+(-12)+(-27)_-38_ ~~ : "" In evaluating expressions, it is useful to remember the following general facts, which result from (d) of 2 : (a) The sign of the product of an even number of negative factors is positive. Ex. (-2). (-3)- (-1). (-4) = +24. I, 4] FUNDAMENTAL NOTIONS -f- (b) The sign of the product of an odd number of negative factors is negative. Ex. (-2)- (-3)- (-l) = -6. (c) A negative number raised to an even power gives a positive result, but if raised to an odd power gives a negative result. Ex. ( -2) 4 = +16, but ( -2) 3 = -8. (d) An odd root of a negative number is negative. Ex. ^27 =-3; v^32=-2'; ^J= 1. EXERCISES Evaluate each of the following expressions for the indi- cated values of the letters. 1. a 2 +2a&+6 2 , when a=l, b=-l. 2. 4x z y+4:xy*+xyz, when x= 1, y = 2, z= 3 3. ?!Z^ w henm = 2,n = 6. 2 mn 4. 2v / 2x-3v / 9Y, when a; =18, 2/= -3. 5. , whenz=-l, 2/=-3, and a = l. , when m = 2, n= -2, x = 5, y = 6. 8. By means of the formulas in Ex. 14, p. 6, find (a) The area of the circle whose radius is 3 feet. [HINT. Take7r=3i.] (6) The circumference of the circle whose radius is 1^ feet. (c) The volume of the sphere whose radius is 5 inches. (d) The area of the sphere whose diameter is 1 yard. 8 SECOND COURSE IN ALGEBRA [I, 5 5. Definitions. A monomial (or term) is an expression not separated into parts by the signs + or , as 5 x z y. A binomial is an expression having two terms, as 3 x 2 4 yx. A trinomial is an expression having three terms, as Any expression containing only powers of one or more letters is called a polynomial. A polynomial is said to be arranged in descending powers of one of its letters if the term containing the highest power of that letter is placed first, the term containing the next lower power is placed second, and so on. Thus, if we arrange \ x 3 +-J- x 5 1 -\-x 3 x 4 in descending powers of x, it becomes -J x 6 3 x 4 -f -J- x 3 -fa; 1. Similarly, a polynomial is said to be arranged in ascending powers of one of its letters if the term containing the lowest power of that letter is placed first, the term containing the next higher power is placed second, and so on. Thus, if we arrange \ x 3 -\-\ z 6 1 -fa? 3 x 4 in ascending powers of z, it becomes -l+x+%x 3 -3 z 4 +| x 6 . Whenever a term is broken up into two factors, either factor is known as the coefficient of the other one. Usually the word is used to designate the factor written first. Thus, in 4 xy, 4 is the coefficient of xy ; in ax, a is the coefficient of x, etc. A common factor of two or more terms is a factor that oc- curs in each of them. Thus 5 x, ax, x 2 , and x 3 have the common factor x. Whenever two or more terms have a common literal factor, they are said to be like terms with respect to that factor. Thus 5 x, ax, x, and 2 x are like terms with respect to x ; and 2a(xy) and 3 b(x y) are like terms with respect to x y. I, 7] FUNDAMENTAL NOTIONS 9 6. Addition and Subtraction of Expressions. The follow- ing rules will be recalled from the First Course, pp. 49-57. (a) To add like terms, add the coefficients for a new coeffi- cient and multiply the result by the common factor. Thus 3 x +5 x -4 x = (3 +5 -4)z =4x. Similarly, m(xy)+n(xy) = (m+ri)(xy}. (b) To add polynomials, write like terms in the same column, find the sum of the terms in each column, and connect the results with the proper signs. Thus, in adding 3a+4b+2c, 5 a +3 b -2 c, and 7 a -9 6-5 c, the work appears as follows : 3a+46+2c 5 a +3 6-2 c 7 a-9 b-5c 15 a 26-5c. Ans. (c) To subtract a term from another like term, change the sign of the subtrahend and add the result to the minuend. Thus 8 x*y - ( -3 x*y) =8 x*y +3 x*y = 11 x*y. (d) To subtract one polynomial from another, change all signs in the subtrahend and add the result to the minuend. Thus, in subtracting 4 mn 2 nr +3 p from 5 ran 4 nr 4 p, what we have to do is to add 4 ran +2 nr 3 p to 5 ran 4 nr 4 p. Adding these (see the preceding rule for addition of polynomials) gives ran 2 nr 7 p. Ans. NOTE. If two or more expressions can be arranged according to the descending powers of some letter ( 5), it is usually best to do so before attempting to add, subtract, or perform other operations upon them. 7. Parenthesis ( ), Bracket [ ], Brace J j, and Vincu- lum . These are symbols for grouping terms that are to be taken as one single number or expression. Thus 4 x (x +3 y z) means that x +3 y z as a whole is to be subtracted from 4 x. 10 SECOND COURSE IN ALGEBRA [I, 7 The following rules will be recalled : (a) A parenthesis preceded by the sign + (either expressed or understood] may always be removed without any other change. (b) A parenthesis preceded by the sign -- may be removed provided the sign of each term in the parenthesis be first changed. Thus 2 a+3 6 + (z-3 y+z) = 2 a+3 b+x-3 y+z but 2 a+3 b-(x-3y+z)=2a+3b-x+3y-z. EXERCISES 1. State the common factors in each of the following expressions. (a) -4 x, -5 x, 6 x. (c) a(z+l) 2 , 6(z+l), c(;r+l) 3 . (b) rs, 3 rs, - 10 r 2 s. (d) 2 mn(a+b), 4 m 2 n(a-b), 8 ran 2 . 2. Add 7a+66-3cand4a-76+4c. 3. Add2x+3y 2xy, 7 xy 4x 9?/, andTx 5xy 4y. 4. Add z-S-T^+lSz 3 , 4+14 z 3 -!! x-x\ and [HINT. See Note, 6.] 5. Add 4(m+n)-3(g-r) and 4(m+n)+6(5-r). 6. Add 10(a+6)-ll(6+c), 3(a+6)-5(c+d), and 7. Add2mx+3nx qx, nx+lqxry, and py qz-\-3w. 8. Subtract 3 x 2 i/-fz from 5 x y-\- z. 9. Subtract 4 x 8 - 8 - 13 x 2 + 15 a; from 6 x 2 + 19 x 3 - 4 +12*. [HINT. See Note, 6.] 10. From 13 a+5 6-4 c subtract 8 a+9 6+10 c. 11. From 2 a+3 c+d subtract a 6+c. 12.' From 3 x 2 +7 x+ 10 subtract -x 2 -x-6. 13. Subtract l-a+a 2 -a 3 from 1-a 3 . 14. From the sum of x 2 4 xy+y* and 6 x 2 2 xy+3 y 2 sub- tract 3 a; 2 5xy+7 y 2 . Do it all in one operation if you can. I, 8] FUNDAMENTAL NOTIONS 11 15. From the sum of 2 s+8 4 w and 3 s 6 t+2 w take the sum of 8 s+9 Z+6 w and 4 s-7 t-4 w. Remove the parentheses and combine terms in each of the following expressions. 16. a-(2a+4)-(5a+10). 17. 6x+(5x-\2x+ll). [HINT. Remove the innermost group 'sign first.] 18. x-\x-(x-3x)\. 19. 2Qz-[(2z+7w)-(3z+5w)]. Find the values of the following when o = 4, 6 = 3, c= 2, and d= I. 20. 10c 2 -(3a+6+d). [HINT. Simplify the expression as far as possible before giving to a, 6, c, and d their special values.] 21. 3d-\a-(c-b)\. 22. o-Hc-(3d-6)j+3Ka-c)-7(&+-2 TTT. 6. -16zyz 2 -=-4z2/ 2 z. 9. 3a6(a+6) 2 7. 4a 4 6 2 c 3 ^20a 2 6 2 c. 10. (9 m 3 np+18 wn 3 p)-=-3 mn. 11. (6 z 2 ?/z+12 :n/ 2 z-24 zi/z 2 ) -5- (-3 xyz). In each of the following divisions, find the quotient, also the remainder if there is one. Check your answer for each. 12. (3z 2 -2z-l)-=-(z^l). 13. (15x 2 +x-2)-^-(3x-l). 14. (4 s 4 - 1 * ^ i . .4. r 9.. 9 ^O. . 24. ^JLuLOflfc 25t 9 ^ 4 - 30 ^ 2a + 2 5a 2 . 26. [HINT. 1 a 2 6 2 + 2 a6 = 1 o i.- (a-6) 2 .] ~ 27. (a-x 2 ) 2 -(6-i/ 2 ) 2 . 9. m 2 -4mn+4n 2 -16. * 28 10. 2xy-x -y 2 +l. 2 9. x 2 -9 t/ 2 +a:+3 y. 11. l+9c 2 +6c. 30. a 3 -2a 2 6+4a6 2 -86 3 . 1O Q /y3 ^_ Q /Y I /I /y4 /| /yi2 QO />4 /1 7 /v*2 Lv O */ O i/ \ A |/ A v O*i i/ """" A I */ CHAPTER III HIGHEST COMMON FACTOR AND LOWEST COMMON MULTIPLE 14. . Prime Factor. A number which has no factor except itself and unity is called in arithmetic a prime number. Such a number when used as a factor is called a prime factor. Thus the prime factors of 15 are 3 and 5. The word prime factor is similarly used in algebra. Thus we say that the prime factors of 3 abc are 3, a, 6, and c. The prime factors of 18 x 2 y are 2, 3, 3, x, x, and y. The prime factors of o 2 6(a 2 b 2 ) are a, a, b, a 6, and a +b. (See Formula V.) 15. Finding Common Factors. As soon as we have fac- tored each of several expressions into its prime factors, we can readily pick out their common factors (5). Thus, in finding the common factors of abc, a?b, a& 2 , and 3 ab, we write abc = a b c, a z b = a a b, ab 2 = a b b, 3 ab = 3 a b. The common factors are, therefore, a and 6, since these occur in each expression and they are the only factors thus appearing. 16. Highest Common Factor. The product of all the common prime factors of two or more expressions is called their highest common factor. It is called the highest be- cause it contains all the common factors, and the usual ab- breviation for it is H. C. F. EXAMPLE 1. Find the H. C. F. of lOz 2 !/ 3 , 2xy 2 , and 18sV- SOLUTION. Resolving each into its prime factors, 10zV=2- 5- x- x- y y y, 2xy*=2- x- y y, 18zV =2- 3'3-x-X'X-yy. The factors common to the three expressions are thus seen to be 2, x, y, y. The H. C. F. is, therefore, 2 x - y - y, or 2 xy*. Ans. 26 Ill, 17] COMMON FACTORS AND MULTIPLES . 27 EXAMPLE 2. Find the H. C. F. of 3 z 2 +3 x-18, 6 z 2 +36 z+54, and 9 z 2 -81. SOLUTION. 3z 2 +3z-18=3(z 2 -fz-6)=30c+3)(z-2). (See (c), 11.) 6z 2 +36z+54=6(z 2 +6-z+9)=2. 3(z+3)(z+3). (See (d), 11.) 9z 2 -81=9(z 2 -9)=3- 3(x+3)(x-3). (See (6), 11.) The common factors being 3 and (x +3), the H. C. F. is 3(x +3.) Ans. In general, to find the H. C. F. of two or more expressions : 1. Find the prime factors of each expression. 2. Pick out the different prime factors and give to each the lowest exponent to which it occurs in any of the expressions. 3. Form the product of all the factors found in step 2. NOTE. Since the H. C. F. of several expressions consists only of factors common to them all, it is always an exact divisor of each of the expressions. It is therefore called in arithmetic "the great- est common divisor" and is represented by G. C. D. EXERCISES Find the H. C. F. of each of the following groups. 1. 12, 18. 7. a 2 +7a+12, a 2 -9. 2. 16,24,36. 8. x 2 -y 2 ,(x-y) 2 ,x 2 -Zxy+2y*. 3. x 2 y, xy 2 . 9. ra 2 +4 m+4, m 2 -6 m-16. 4. a 2 b, ab 2 , a 2 b 2 . 10. 3 ?/ 2 -363, y 2 -7 T/-44. 6. 2 x*y, 6 x*y, 14 x*y 2 z 4 . 11. 2 a 2 +4 a, 4 a 3 +12 a 2 +8 a. 6. a 2 -6 2 , a 2 -2a&+6 2 . 12. 4 -i/ 4 , x*+x z y+xy 2 +y*. 13. 3 r 5 +9 r 4 -3 r 3 , 5 rV+15 rs 2 -5 s 2 , 7 ar 2 +21 ar-7 a. *14. a 3 -6 3 , a 2 -6 2 , a-b. *15. x 3 +l, x 2 -x+l. 17. Common Multiple. In arithmetic a number which is exactly divisible by two or more given numbers is called a common multiple of them. The word common multiple is similarly used in algebra. Thus 4 x 2 y 2 is a common multiple of x and y ; it is also a common multiple of 4 and x. Similarly, a 2 b 2 is a common multiple of a b and a +6. 28 SECOND COURSE IN ALGEBRA [III, 18 18. Lowest Common Multiple. The lowest common multiple of two or more numbers or expressions is that multiple of them which contains the fewest possible prime factors. Its abbreviation is L. C. M. The following examples illustrate what the L. C. M. means and how to obtain it. EXAMPLE 1. Find the L. C. M. of 10 a 2 b, 16 a 2 6 3 , and 20 a 3 6 4 . SOLUTION. Separate each expression into its prime factors. Thus 10a 2 6=2- 5- a- a- 6, 16a 2 6 3 =2- 2- 2-2- a- a- 6- 6- 6, 20a 3 6 4 =2- 2-5-a-a-a.6-6.b-6. The L. C. M. is 2 - 2 - 2 2 5 a a - a 6 6 - 6 - 6, or 80 a 3 6 4 , since this contains all the factors of each of the three expressions, and at the same time is made up of fewer factors than any other similar expression that can be found. EXAMPLE 2. Find the L. C. M. of z 4 -10z 2 +9 and SOLUTION. Therefore the L. C. M. is (z- In general, to find the L. C. M. of two or more expressions : 1. Find the prime factors of each expression. 2. Pick out the different prime factors, taking each the great- est number of times it occurs in any one of the expressions. 3. Form the product of all the factors found in step 2. NOTE. From the manner in which the L. C. M. is formed, it must be exactly divisible by each of the given numbers, or expressions. EXERCISES Find the L. C. M. of each of the groups of expressions in the exercises on p. 27. CHAPTER IV FRACTIONS 19. Definitions. Any expression of the form a/6 is called a fraction. It means the number, or expression, which when multiplied by b gives a. The part above the line, or a, is called the numerator, while the part below the line, or 6, is called the denominator. The numerator and denominator taken together are called the terms of the fraction. 20. Equivalent Fractions. It is often desirable to change the form of a fraction without changing its value. Such changes all depend upon the following principle. The numerator and denominator of a fraction may be multi- plied or divided by the same number, or expression, without changing the value of the fraction. Thus 3^3- 2^6. 4 4-2 8' 4 a_4 a a_4 a 2 . 5 5 a 5 a' a+b (a+6) 2 21. Changes of Signs in Fractions. There are three signs to be considered in a fraction ; the sign of the numerator, the sign of the denominator, and the sign of the fraction itself. _ o Thus in H -- , the three signs in the order just mentioned are 4 -, +, +, while in --^7 they are +, -, -. D 29 30 SECOND COURSE IN ALGEBRA [IV, 21 Since a fraction is merely an indicated division, the law of signs for division ( 2 (e), p. 3) must hold at all times, so that we arrive at the following rule. Any two of the three signs of a fraction may be changed with- out altering the value of the fraction. Thus _i_3 = +:d* = -3 = 3 +4 -4 +4 -4* Likewise a _ a _ a _ a b -b ~ 5 _&' Care must be taken, however, in changing the sign of the numerator or denominator of a fraction when polynomials are present. For example, if the numerator is a polynomial, we can change the sign of the whole numerator only by chang- ing the sign of every term in it. A similar statement applies when the denominator is a polynomial. Thus, Q+26+c _ a 2bc^ a +2 b+c 2a-3b-2c 2a-3b-2c -2o+36+2c* Observe carefully the reason for every change of sign here. 22. Reduction of Fractions to Lowest Terms. A fraction is reduced to its lowest terms when its numerator and de- nominator have no common factor except 1. To reduce a fraction to its lowest terms, factor numerator and denominator, then divide each by all their common factors. E EXAMPLE 2. a 2 -Ha +24 = (a- a 2 -a+6 (a- IV, 23] FRACTIONS 31 23. Lowest Common Denominator. The lowest common denominator of two or more fractions is the lowest common multiple ( 18) of their denominators. To reduce several fractions to their lowest common denomi- nator : 1. Find the L. C. M. of the denominators. 2. For each fraction, divide this L. C. M. by the given de- nominator, and multiply both numerator and denominator by the quotient. EXAMPLE. Reduce the following fractions to equivalent fractions having their lowest common denominator : and SOLUTION. Factoring the denominators, the fractions may be written The L. C. M. of these denominators is (a; +3) (a; -3) (z +6). In order to give the first fraction this L. C. M. as its denominator, multiply its numerator and denominator by x+G (this being the L. C. M. divided by the denominator of the first fraction). In order to give the second fraction this L. C. M. as its denomina- tor, multiply its numerator and denominator by x 3 (this being the L. C. M. divided by the denominator of the second fraction). The desired forms are, therefore, (x+2)(x+Q) d (x +3) (x -3) (x +6) (x +3) (x -3) (x +6) Observe that these fractions are respectively equivalent ( 20) to those with which we started, but these have denominators that are alike, which was not the case with the original forms. 32 SECOND COURSE IN ALGEBRA [IV, 23 EXERCISES Write each of the following fractions in three other ways without changing the value. 1 5 6 3 a-b_ . 3a-4 -6* ' l-x ' c-d ' (2z-l)i Reduce each of the following expressions to lowest terms. 240 ' * ' 320 (a6+l) 2 10 xy 4z 2 -2/ 2 O. * 11. OA /y>2/)/2 ni O /y. o\j ju y y i jc 36 xr 3 19 s 2 -6s+8 * ^r r~r* J-^. 72^^ s 2 -5s+6 -9a 2 5 2 c 2 1Q r 4 -6r 2 +5 o. -3-: r77~r~* !* 54 aW r 2 -6r+5 28xy 2 -12x 2 y Reduce all of the fractions in each of the following groups to the lowest common denominator. 16 2 f ^ . go 2 -^ __ L, ' 4'6'2 ' 6 2 'a 2 -6 2 ' a+b 17 5 ?. 21 ' a 6 ' a r y*> x -y ' ( a +6) 2 ' l'a 2 -6 2 1 1 23. . a;-2 rc-1 o:+3 Z4. x 2 -2x-8' x 2 -3x-10'a; 2 - IV, 24] FRACTIONS 33 24. Addition and Subtraction of Fractions. The follow- ing rule, which is the same in algebra as in arithmetic, will be recalled from the First Course, p. 151. To add, or subtract, fractions : 1. Reduce the fractions to equivalent fractions having their lowest common denominator (L. C. D.). 2. Add, or subtract, each numerator according to the sign before the fraction and write the result over the L. C. D. 3. Reduce the resulting fraction to its lowest terms. EXAMPLE. Simplify a-1 a+1 a 2 -! SOLUTION. Qn2 + _3_a 2 _ = 4 a+4_a 2 -3 a+2 + ^g 2 _ a-1 a + 1 a 2 -! a 2 -! a 2 -! a 2 -! _4a+4-(a 2 -3a+2)+3a 2 _2a 2 +7a-+2 a 2 -! a 2 -! EXERCISES Simplify each of the following expressions. ~3~ ~4~ 2' x y *' ab z ab a? 2 - 2 f + ?" 4 -^- 2J1 ^ + I- 6 -H+r 7 _ | _.. 12 2x-2 3(z+l) 3z-3 a 2 +2a6+6 2 a?-2ab+b 2 l3 - 9 _ . 14 _ . ab a+b n+4 1-n n 2 +3 n-4 10. ___ -+-~ 15. x-3+ r 2 -2r-8 r-4 r+2 x 2 +3 z+9 11 1 1 16 _x__ 1+a^ 2 ' x-l x 2 +x+l 34 SECOND COURSE IN ALGEBRA [IV, 25. Multiplication and Division of Fractions. Fractions are multiplied in algebra as in arithmetic by taking the product of the numerators for the numerator, and the product of the denominators for the denominator, canceling wherever possible. EXAMPLE 1. 4 xy* 9 ab 3 4 Canceling like factors from numerator and denominator, this reduces to ^- Ans. 66 2 EXAMPLE 2. a 2 +2 a -8 _ a 2 +5 a-6 a 2 +6 a a 2 +a-12 .a(a-3) In algebra we divide one fraction by another as we do in arithmetic, by inverting the divisor and proceeding as in multiplication. EXAMPLE. a 2 -b 2 . a-b a 2 -b 2 a-b A EXERCISES Perform each of the following multiplications. 3., 2 5 rs 2 1*8 m 3 n 3 7 o 1^2 O o 5. A a k 2 1 6 c ' a 2 +& 2 2/frl /d_ O' 1 ^ P\ / CM/ * */ M J 3 xy 5 a 2 2 z 2 +4 Z7/+2 i/ 2 5 . 6 a \ 10 b 8 8 r+2 g-1 4s-6 1 ' 8 a 3 ' ' 2s-3 *2r+2* r+2" IV, 25] FRACTIONS 35 9 . 10 . . - ___ -. x ?/ 6 a +?/ a 6 Perform the following divisions. 4^2 ' * 9*3 a(a+6) '2 0-6 12 10 & . 12 s r 2 -9s 2 * ' 15< 13. 11 y y 2 r 2 -4s 2 r 2 -rs-6 a 2 +2a, (a+2) 2 a 2 -2a* (a-2) 2 ' Perform the indicated operations and simplify each of the following expressions. 16 \x+y x-yj \x+y xyj is ' 3,5 Li. 19. 3_4 V m m-3\m-2 x y 21. fi- 1 1 1 22 . -X 1+05 CHAPTER V SIMPLE EQUATIONS 26. Preliminary Considerations. Suppose we wish to divide 64 into two parts such that if one part be divided by 5 and the other by 7 the sum of the quotients shall be 10. Such a problem as this can be done only with some difficulty by arithmetic, but it is a simple task by algebra. SOLUTION. Let x represent one part. Then 64 x will be the value of the other part. From the statement of the problem, we are to have Let us multiply both sides of this equality by 35 (as we may do without destroying it), thus clearing it of fractions. This gives 7 x +320 -5 x =350. Subtracting 320 from both sides of this last equality (as we may do without destroying it), and replacing 7 x 5 x by its value 2 x, we obtain 2 x= 30. Hence (dividing both sides by 2) we have z = 15. The two parts sought are therefore 15 and 64 15, or 49. Ans. CHECK. + = 3 + 7 = 1 - A statement of equality, like any of those above, wherein a single unknown letter occurs, and occurs to no higher power than the first, is called a simple equation. It is also known as a linear equation, or an equation of the first degree. The process of finding the value of the unknown letter is called solving the equation. The value of the unknown letter is called the solution, or root, of the equation. V, 27] SIMPLE EQUATIONS 37 27. Principles Useful in the Solving of Equations. In solving an equation we may at any point in the process add the same amount to both sides, or subtract the same amount from both, or we may multiply both by the same amount, or divide both by the same amount, as was illustrated in 26. Derived from these are the following useful principles. (a) A term may be transposed (carried over) from one side (or member) of an equation to the other provided its sign is changed. Thus, in the equation 3 a: 4 =2 we may transpose the term 4 to the second member, giving 3 x =2+4, or 3 x =6. This is equiva- lent to adding 4 to both members of the given equation. The solving of equations is greatly simplified by a free use of this principle of transposing terms. Thus, in solving 3 x 4=x+2, we may transpose the 4 from the first member to the second and at the same time transpose the term x from the second member to the first, giving z+3 x =2+4, or 2 x =6. Therefore x =3. Ans. (b) A term which appears in both members of an equation may be canceled. Thus, by canceling the 3 from both members of the equation 2 x +3 = 10 +3, we have simply 2 x = 10, and hence x =5. Note that to cancel a term in this way merely amounts to subtracting it from both members of the given equation. (c) The signs of all the terms in an equation may be changed. Thus -5z+3=z-9may be written 5 x -3= -x+9. Note that to change all signs in this manner amounts to multiplying both members of the given equation by 1. (d) An equation may be cleared of fractions by multiplying both members by the lowest common denominator of all the fractions. 38 SECOND COURSE IN ALGEBRA [V, 27 EXAMPLE. Solve the equation - = 6 SOLUTION. The L. C. M. of the denominators is 12. Multiplying both members by 12, 4(z-2) -3(3-3) =72-6(s-l). Removing parentheses (7), 4z-8-3z+9=72-6z+6. Transposing, 6x+4:c _ 3 1=72+6+8 _ 9> 7* =77. Therefore a: = 11. Ans. EXERCISES Solve the following, using the principles stated in 27. Check your answer for the first five. 2. ! +6 -T- 33-1 2 3-3 , 6z+5 /v. C K * _Q 2 3 4 6 2z 12 1 3+1 2 4 5 7. 8. Q 5 8 1 2 6 ' x x* 33 1 3+1 3(3+1) 1 1 12 4-6y 2-3i 1 ,r+l r-2 r 2_ r _ 2 r -2 r+1 ^ rv *k I -L *v " O io - jn-ija 12 ' a;-4 x-8 z 2 -12z+32 V, 27] SIMPLE EQUATIONS 39 13. If 10 be subtracted from a certain number, three fourths of the remainder is 9. What is the number? 14. Divide 38 into two parts whose quotient is 1 2 2 -. 15. Divide 96 into two parts such that f of the greater shall exceed f of the smaller by 6. 16. A man started on a journey with a certain sum of money. He spent -J of it for car fares and ^ of it for hotel bills. When he returned home he found he had $9. How much did he start with ? 17. I have $100 in one bank and $75 in another one. If I have $45 more to deposit, how shall I divide it among the two banks in order that they may have equal amounts? 18. A motor boat traveling at the rate of 12 miles an hour crossed a lake in 10 minutes less tune than when traveling at the rate of 10 miles an hour. What is the width of the lake? [HiNT. Time = Distance -5- Rate.] 19. A freight train goes 6 miles an hour less than a passen- ger train. If it goes 80 miles in the same time that a passen- ger train goes 112 miles, find the rate of each. 20. A tank can be filled by one pipe in 10 hours, or by an- other pipe in 15 hours. How long will it take to fill the tank if both pipes are open ? [HINT. Let x = the number of hours. Then l/x = the part both can fill in 1 hour. But, ^ = the part the first pipe can fill in 1 hour, and T 1 g-=the part the second pipe can fill in 1 hour. Hence we must have ! = _1_L_JL~| x 10 15* J 21. How long will it take two pipes to fill a tank if one alone can fill it in 5 hours and the other alone in 12 hours? 40 SECOND COURSE IN ALGEBRA [V, 27 22. Two pipes are connected with a tank. The large one can fill it in 3 hours ; the small one can empty it in 4 hours. With both pipes open, how long before the tank will fill ? 23. A does a piece of work in 4 days, B in 6 days, and C in 8 days. How long will it take them working together? 24. A can do a piece of work in 16 hours, and B can do it in 20 hours. If A works for 10 hours, how many hours must B work to finish? 25. A's age is ^ that of his father's. 12 years ago he was i as old as his father. How old is each now? 26. A boat goes at the rate of 12 miles an hour in still water. If it takes as long to go 27 miles upstream as 45 miles downstream, what is the rate of the current? 27. An aviator made a trip of 95 miles. After flying 40 miles, he increased his speed by 15 miles an hour and made the remaining distance in the same time it took him to fly the first 40 miles. What was his rate over the first 40 miles? 28. A 5-gallon mixture of alcohol and water contains 80% alcohol. How much water must be added to make it con- tain only 50% alcohol? [HINT. .50(z+5)=5X.80. Explain.] 29. How much water must be added to 65 pounds of a 10% salt solution to reduce it to an 8% solution ? 30. A train 660 feet long running at 15 miles an hour will pass completely through the Simplon tunnel in Switzerland in 49^ minutes. How long is the tunnel ? DESCARTES (Rent Descartes, 1596-1650) Profound student and ranked as one of the greatest leaders of all time in both mathematics and philosophy. He invented representation by graphs and was thus led to the discovery and development of the branch of mathe- matics called Analytic Geometry. He was also much interested in medicine and surgery. CHAPTER VI GRAPHICAL STUDY OF EQUATIONS 28. Definitions. Let two lines XX' and Y Y' be drawn on a sheet of squared (coordinate) paper, XX' being hori- zontal and Y Y' vertical. Two such lines form a pair of coordinate axes. The point where they intersect is called the origin. Consider any point, as P, and draw the perpendiculars PA and PB extending from P to the two axes Y Y' and XX' respectively. PA is then called the abscissa of P and PB is called the ordinate of P. The abscissa and ordinate taken together are called the coordinates of P. Y A A P i I (3 4) (-2 ,3) n 1 X' - 3 B X J 4 - 2- 1 3 < ) n S * (3 ) R -4 ( -3 -4) Y' FIG. 5. Thus the point P in the figure has its abscissa equal to 3 and its ordinate equal to 4. All abscissas on the right of Y Y' are considered positive, while all abscissas on the left of Y Y' are considered negative. Thus the abscissa of Q is -2 ; that of R is -3 ; that of S is +3. Similarly, all ordinates above XX' are considered positive, while all ordinates below XX' are considered negative. Thus the ordinate of Q is +3 ; that of R is -4 ; that of S is -2. 41 42 SECOND COURSE IN ALGEBRA [VI, 28 In reading the coordinates of a point, the abscissa is always read first and the ordinate second. Thus, in the figure, the point P is briefly referred to as the point (3, 4) ; similarly, Q is the point ( 2, 3) ; R is the point ( 3, 4) ; and S is the point (3, 2), etc. In practice, XX' is commonly called the x-axis, and Y Y' is called the y-axis. EXERCISES ' [The pupil will find it convenient to use the prepared coordinate paper such as usually may be secured at the stationery stores.] 1. Draw axes on a sheet of coordinate paper and then locate (plot) the following points : (2,4); (-3, -1); (2, -4); (2J, -3); (-2J, -2J); (-4, +4); (0, -5); (4,0); (0,0). 2. The part of the plane within the angle XO Y (see figure in 28) is called the first quadrant, the part within the angle YOX' is called the second quadrant, the part within X'O Y' is called the third quadrant, etc. Hence, state in which quad- rant a point lies when (a) its abscissa is positive and its ordinate negative, (&) its abscissa and ordinate are both negative, (c) its abscissa is negative and ordinate positive. 3. What can be said of the position of a point whose ordi- nate is positive ; whose abscissa is negative ? 4. A certain street runs due east and west. It is met by another street which runs due north and south, thus form- ing a " four corners." Taking the meeting place of the cen- ter-lines of the two streets as origin, and the east and north directions as positive, what are the coordinates of a flagpole which stands due northwest from the origin at a distance of 50 feet from the center-line of each road? Answer the same when the pole is 45 feet due west of the crossing point. VI, 29] GRAPHICAL STUDY OP EQUATIONS 43 5. Plot the following three points and then see if it is possible to draw a straight line that will pass through all of them: (1,5); (0,3); (-1,1). Do the same for the three points (2, 3) ; (!,!); (5, 0). 29. Graph of an Equation. We have seen in Chapter V that if we have any linear equation containing a single un- known letter, as for example the equation 2 x l = 3(x 1), we can always solve it ; that is we can find the value of x. Suppose now that we have a linear equation in which two unknown letters, x and y, appear, that is an equation in which no term contains both x and y nor any higher power of either of them than the first, as for example (1) x+y = 5. The meaning of such an equation and the interesting facts about it are best brought out by graphical methods in ways which we shall now explain. In the first place, it is to be observed that such an equa- tion is satisfied by a great many pairs of values for x and y. For example, the pair of values 0=1, y = 4) satisfies the equation, because when we put these values for x and y respectively in the equation, it becomes 1+4 = 5, which is true. Again, the same is seen to be true of the pair (x = 2, 2/ = 3) (explain) ; and, similarly, the same is true of any one of the pairs (x = , y = f), (x = Q, y=-l), (x = 8, y=-3), etc. In fact, we can obtain as many such x, y pairs as we wish, each pair having the property that the z-value and the ?/-value taken together satisfy the given equation. If we place x =3 in the equation above, we have 3 +y =5 and this, when solved for y, gives y =2. Thus (x =3, y =2) is a pair such as mentioned above. Similarly, we can assign to x any value we wish (positive or negative) and find from the equation the corresponding value of y, thus forming a new pair of values of x and y. 44 SECOND COURSE IN ALGEBRA [VI, 29 Whenever an equation contains two (but no more) un- known letters, such as x and y, any pair of values for x and y that satisfy it is called a solution of the given equation. It follows from what has been shown above that every such equation has an indefinitely large number of solutions. Returning to the equation x-\-y = 5, let us consider again the special solutions which we noticed on page 43 : (x = 2, 2/ = 3), (z = i, 2/ = f), (x=-l, i/ = 6), (x = 8, 2/=-3). Following the ideas brought out in 28, each of these may now be plotted a*s a point, using x as abscissa and y as ordi- nate. Upon locating these points carefully, it will be seen that they all lie on one and the same straight line, as indi- cated in the figure below. \ Y \ \< i- -) \ (2 3) + X A \ \ X \ (6 -U \ \ ( 8, 3) \ FIG. 6. This line is called the graph of the equation x+y = 5. It may be shown that every solution of the given equation gives rise when plotted to some point upon this line, and vice versa, every point upon this line has an a>value and a i/-value which, when taken together, form a solution of the given equation. 30. Graph Determined from Two Points. In practice, the graph of a linear equation is drawn by locating two points upon it, and connecting them by a straight line. VI, 30] GRAPHICAL STUDY OF EQUATIONS 45 EXAMPLE. Draw the graph of the equation 5 x 4 y = 20. SOLUTION. Placing x =0 in the equation gives y = 5. Hence (0, 5) is a point on the graph. Placing y = in the equation gives x = 4. Hence (4, 0) is a point on the graph. Plotting these two points (0, 5) and (4, 0) and drawing (with ruler) the straight line through them, gives the required graph. Y V A f / (4 0) / x / / ^ / 4 / / (0 -5 ) b / FIG. 7. NOTE. As in the example just considered, it is often simplest to select as the first point the one whose abscissa is 0, and as the second point the one whose ordinate is'O. However, it is equally correct to take any two points whose coordinates satisfy the given equation. If the two points selected are too close to each other, it is difficult to draw the line accurately ; if this happens, plot a third point on the line at a considerable distance from the first two. EXERCISES 1. State (orally) what is true of each point on the line in the last figure. [HINT. Its abscissa and ordinate, taken as a pair of numbers, (x, y), form a . . . .] Draw the graph of each of the following linear equations. 2. 2x-y = 4:. 4. 2x+3y=12. 6. 4x = 3y-7. 3. 2x+y = 2. 5. x-3y = 3. 1. 2x = 3y. 46 SECOND COURSE IN ALGEBRA [VI, 30 8. If a person travels at the rate of 15 miles per hour, the distance s which he will have traveled at the end of t hours is given by the formula s = 15 t. Draw the graph of this equation, using the ^-values as abscissas and the s-values as ordinates. From your figure read off (approximately) how far he will have traveled at the end of (a) 2 hours ; (6) 3 hours; (c) 3^- hours; (d) 4-J- hours. [HINT. Take the Z-axis horizontal and the s-axis vertical, and let the unit length on each be about half an inch. In order to get the diagram into relatively compact form, allow each unit on the s-axis to represent 15 miles, taking each unit on the 2-axis to repre- sent 1 hour.] 9. A boy has $10 in the bank when he begins saving at the rate of $3 a month, adding this amount month by month to his account. Find graphically how many months must elapse before his account will amount to $22. [HINT. Let A represent the amount of the account at the end of t months. Then, A =10 +3 L (Why?) Now draw the graph of this equation, using ^-values as abscissas and A -values as ordinates, and taking for convenience* one unit on the A -axis to represent $2, while one unit on the t-axis represents 1 month. The problem then calls for that abscissa which goes with the ordinate A =22.] 10. A boy has $30 in the bank when be begins spending it at the rate of $4 a month. Find graphically how long it will be before he has but $2 left. [HINT. Use the same letters and units as in Ex. 9.] 11. A wheel is rotating at the rate of 10 revolutions a second when the power is shut off. The wheel slows down uniformly and comes to rest at the end of 30 seconds. Make a diagram from which you can read off how many revolutions the wheel was making at any given instant after the power was shut off and use your diagram to determine how many VI, 31] GRAPHICAL STUDY OF EQUATIONS 47 revolutions per second were being made at the end of (a) 6 seconds ; (6) 9 seconds ; (c) 18 seconds ; (d) 26 seconds. [HINT. Let r represent the number of revolutions per second at the end of t seconds. Then the conditions of the problem tell us that r = 10 when t=0, and that r=0 when =30. Thus we have two points on the graph, and we can draw the graph completely without even getting its equation.] 12. The temperature at which water freezes is 32 on the Fahrenheit scale, but it is on the Centigrade scale. The temperature at which water boils is 212 on the Fahrenheit scale, but it is 100 on the Centigrade scale. Make a dia- gram from which you can read off the Centigrade tempera- ture that corresponds to any given Fahrenheit temperature. [HINT. Let F represent the Fahrenheit reading and C the Centi- grade reading. Then C=0 when F= 32 and C=100 when F=212. This gives two points. The graph is the straight line joining these two points.] 31. Simultaneous Equations. Suppose that, instead of having a single linear equation containing the two unknown letters x and y (as in 29), we have two such equations; for example x+y = Q and 3 x-2 y= -2. Of all the pairs of values (x, y) that will satisfy the first equation and all the pairs (x, y) that will satisfy the second equation, is there a particu- lar pair (x, y) that will satisfy them both at the same time? We shall consider this question graphically. Draw the graphs of the two equa- tions on the same sheet of coordinate paper, using the same axes throughout. The lines thus obtained are seen to in- tersect each other in the point (2, 4). This means that the \ FIG. 8. 48 SECOND COURSE IN ALGEBRA [VI, 31 pair (x = 2, 2/ = 4) satisfies both equations at once, since it lies on both the graphs. This pair (x = 2, y = 4) is there- fore the pair desired, and it is the only such pair because two straight lines can intersect in but one point. That this answer is correct is readily seen by substituting this pair of values in the given equations. Thus, with x=2 and y = 4, the equations become 2+4 = 6 and 6 8= 2, which are true. The two equations above illustrate what is known as a set of simultaneous equations, and the particular pair of values (x = 2, i/ = 4) which we found would satisfy both the equations at one time, illustrates what is called the solution of the set. In general, two or more equations are said to be simultaneous if they are considered at the same time. In the present chapter we shall deal only with sets containing two unknowns, as in the preceding example. 32. Inconsistent Equations. Although two linear simul- taneous equations in x and y will in general have a solution (as in 31), there are cases where no solution can be found, and indeed none exists. For example, if. we draw the graphs of the equations x+y = 3, and x+y = 6, we see that the lines do not intersect ; in other words, they are parallel. Thus, there is no pair of values (x, y) that will satisfy both equations at once ; that is, there is no solution. Such a pair of simul- taneous equations is called inconsistent. EXERCISES Determine graphically which of the following sets of simul- taneous equations has a solution arid which does not. In \ Y \ \ \ a \ Xk X t \ \ o 3 \ \ \ X \ \ \ "^ u> > * * VI, 32] GRAPHICAL STUDY OF EQUATIONS , 49 case a solution exists, determine it and check your result by substituting it in the given equations. f 9. A man starts at a given time and walks along a certain road at the rate of 5 miles an hour. An hour later another man starts from the same place and travels in the same direction at the rate of 10 miles an hour. Find (graphically) how far from the starting point they will meet. [HINT. If s represents distance (in miles) traveled in t hours, the first man's motion is described by the equation s =5 t, while the second man's motion is described by the equation s = 10(Z 1). Now draw a pair of axes, and draw in the graphs of these two equa- tions, using lvalues as abscissas. The problem then calls for that value of s which belongs to the intersection of the two graphs.] 10. Use your diagram for Ex. 9 to answer the following question : After how much time will the two men meet ? 11. A man starts and walks along a certain road at the rate of 5 miles an hour. At the same instant another man starts out at a point on the same road 15 miles distant and travels toward the first man on a bicycle at the rate of 10 miles an hour. How far from the first man's starting point will they meet? How long will it take them? 12. B and C start to save money. B has $10 when they begin and saves at the rate of $3 a month, while C at the start owes $6 and saves at the rate of $7 a month. Find graphically how soon C will be able to cancel his debt and have savings equal to B's, and how much each will then have. CHAPTER VII SIMULTANEOUS EQUATIONS SOLVED BY ELIMINATION 33. Elimination by Substitution. The process of combin- ing two equations in two unknowns in such a way as to cause one of the unknowns to disappear is called elimina- tion. We shall consider first the method called elimination by substitution. The process is illustrated by the following example. EXAMPLE. Solve the system (1) 2x+3y = 2, (2) 5z-4i/ = 28. SOLUTION. From (1), (3) 2x=2-3y. Therefore (4) o o Substituting ^ for x in (2) gives (5) Clearing (5) of fractions, (6) 5(2 -3 y) -8 y = 56. 50 VII, 34] ELIMINATION 51 Simplifying, (7) W-15y-8y=5G. Collecting, (8) -23 y = 46. Therefore (9) y=-2. Substituting -2 for y in (4) now gives The required solution of the system (1), (2) is, therefore, (z=4, i/=-2). Ans. CHECK. Substituting x =4 and y = -2 in (1) gives 2(4)+3(-2), which is equal to 8 6, or 2, as (1) requires. Substituting x =4 and y = 2 in (2) gives 5(4) -4( -2), which is equal to 20+8, or 28, as (2) requires. To solve two simultaneous equations by substitution : 1. Solve either equation for one of the unknown letters in terms of the other one. 2. Place the result thus obtained in the other equation and solve it. 3. Having thus found one of the unknown letters, substitute its value in either of the given equations and solve for the other unknown letter. 34. Elimination by Addition or Subtraction. The only other method of elimination which we shall consider here is called elimination by addition, or subtraction. The process is illustrated by the following example. 52 SECOND COURSE IN ALGEBRA [VII, 34 EXAMPLE. Solve the system (1) 3s+4y=12, (2) 2z-5y = 54. SOLUTION. Multiplying (1) by 2, (3) 6z+8'2/=24. Multiplying (2) by 3, (4) 6z-15 y = 162. Subtracting (4) from (3) , (5) 232/=-138. Therefore 2/=-6. Substituting y = -6 in (1), 3 x -24 = 12, or, 3 x =36. Therefore x = l2. The required solution of the system (1), (2) is, therefore, (z = 12, y= 6). Ans. CHECK. Substituting 1? for x and 6 for y in (1), gives 3(12) +4( -6) =36 -24= 12, as (1) requires. Substituting 12 for x and 6 for y in (2), gives 2(12) -5( -6) =24+30=54, as (2) requires. NOTE. Instead of multiplying (1) by 2 and (2) by 3 and then subtracting them, thus eliminating x, we might just as well have multiplied (1) by 5 and (2) by 4 and added them, thus eliminating y. Either plan leads to the same solution for the given system. To solve two simultaneous equations by addition or subtrac- tion: 1. Multiply one, or both, of the given equations by such num- bers as will make the coefficients of one of the letters (say, y) numerically equal. 2. Subtract (or add) the two equations thus obtained, thus eliminating one of the unknown letters. 3. Solve the resulting equation for the letter it contains, and obtain the value of the other letter by substituting the value of the letter already found into either of the given equations. VII, 34] ELIMINATION EXERCISES 53 1. Solve by substitution : 2x+Zy=12, x+5y=13. 3. 4. < [2x-3y=-W. 5. < _ , / \ ^* ^ 4 * Solve by addition or subtraction : 6. [ A+B=-9, {10*-3 V .6, \7A-3B = 7. \ 7x+42/ = 8. Solve by either method : x 4 y 3 x y x-\-y 11. 9 Q J 2 15 ' 2~ 3 ' 3z+i/ 3o:-2/_ 2 3 11 3 [HINT. First clear of fractions.] 12. 2H-6, s_ 8 i. r ' M-n, x y 10 12 4 12 L 4:. HINT TO Ex. 16. Do not clear of fractions, but solve for - and --~| 2x-5?/=-l, -+ = 5, 13. x-2/ 3x+2y 7 23 15_30 = _ 1 V 14. 6x-8i/ 8x-20i/ ft 1 1 1 5 11 ' . 18 ' x+V=|- i r . ^ >> x-1 2/+1 2 , 3 _ 10 Xl y+l 54 SECOND COURSE IN ALGEBRA [VII, 35 35. Simultaneous Equations in Three Unknown Letters. We often meet with a system of three linear equations between three unknown letters. Such a system, like those already considered ( 33, 34), may be solved by elimination. EXAMPLE. Solve the system (1) x+y+z = 6, (2) 2z-2/+3z = 9, (3) x+2y-z = 2. SOLUTION. Eliminate one of the unknowns, say y, from (1) and (2). Thus (4) 3 x +42 = 15. [(l)+(2)] Eliminate the same unknown, y, from (2) and (3). Thus (5) 4 x -2 y +6 z = 18. (2)X2 (6) x+2y- 2 = 2. (3) 5x +52=20, (5) + (6) or (7) x +2 = 4. Equations (4) and (7) contain only x and z and hence may be solved for these letters, as in 33, 34. Thus (8) 3 x +4 z = 15. (4) (9) 3 x +3 z = 12. (7) X3 0=3. (8) -(9) Substituting z =3 in (7), we obtain z+3 =4. Therefore x = 1. Substituting 2=3 and z = 1 in (1), we find l+y+3 =6. There- fore y=2. The desired solution is, therefore, (x = l, y =2, 2=3). Ans. To solve three simultaneous equations : 1. Eliminate one of the unknown letters from any pair of the equations, then eliminate the same unknown from any other pair of the equations. 2. Solve the two equations thus obtained, as in 34. 3. This gives two of the letters, and the third may then be found by substituting the letters already found in either of the given equations. - > a VII, 35] H ELIMINATION \\VA-- tA\-u ^vr- to 55 EXERCISES Solve for x, y, and z each of the following sets of equations. +3^ = 14, \x+y-z = Q, 1. 2z+?/+22 = 10, 4. z-?/ = 4, -3^ = 2. z+z = 7. 2. 3. 5. M+i-2, x y z 2 -i+i = 7, x ?/ z [HINT. See hint to Exer- cise 16, p. 53.] APPLIED PROBLEMS 1. The sum of two numbers is 75 and their difference is 5. Find the numbers. [HINT. Let x be one of the numbers and y the other, and form two equations.] 2. One third of the sum of two numbers is 10, while one sixth of their difference is 1. Find the numbers. 3. The perimeter of a certain rectangle is 10 inches less than 3 times the base. If the base is 4|- times the height, what are the base and height? 4. Each base angle of a certain isosceles triangle is 66 more than the vertical angle. Find each angle. 5. A father's age is 1^- that of his son. Twenty years ago his age was twice his son's. How old is each? 6. Four years ago A's age was 2^ B's age. Four years hence A's age will be 1 T 8 T B's age. What is the age of each? 56 SECOND COURSE IN ALGEBRA [VII, 35 7. A part of $2500 is invested at 6% and the remainder at 5%. The yearly income from both is $141. Find the amount in each investment. 8. One sum of money is invested at 5% and another at 6%. The total yearly income from both investments is $53.75. If the rates should be reversed, the annual income would be increased by $2.50. Find the sums of money in- vested. 9. A and B together can do a piece of work in 12 days. After A has worked alone for 5 days, B finishes the work in 26 days. In what time could each do the work alone? [HINT. Let x = the time in which A can do it alone, y = the time in which B can do it alone. Then the part A can do in one day is -, and the part B can do in one day is -. So the equations become i + 1 = J_ an d 5 +? = l. Now solve as in Ex. 16, p. 53.] x y 12 x y 10. A and B can do a certain piece of work in 16 days. They work together for 4 days, when B is left alone and completes the work in 36 days. In what time could each do it separately? 11. A laborer agreed to stay on a farm for 100 days. For each day he worked he was to receive $2 and board, but for each idle day he was to forfeit 75 cents for his board. When the time expired, he received $180.75. How many days did he work? 12. An errand boy went to the bank to deposit some bills, some of them being $1 bills and the rest $2 bills. If there were 38 bills in all and their combined value was $50, how many of each were there? [HINT. Let rr=the number of $1 bills, and y =the number of $2 bills. Then their combined value was x +2 y dollars.] VII, 35] ELIMINATION 57 13. The receipts from the sale of 300 tickets for a musical recital were $125. Adults were charged 50 cents each, and children 25 cents each. How many tickets of each kind were sold? 14. A grocer wishes to make 50 pounds of coffee worth 32 t;ents per pound by mixing two other grades, one of which is worth 26 cents per pound and the other 35 cents per pound. How much of each must he use ? 15. One cask contains 18 gallons of vinegar and 12 gallons of water; another, 4 gallons of vinegar and 12 of water. How many gallons must be taken from each so that when mixed there may be 21 gallons, half vinegar and half water? 16. Two cities are 140 miles apart. To travel the distance between them by automobile takes 3 hours less time than by bicycle, but if the bicycle has a start of 42 miles, each takes the same time. What is the rate of the automobile, and what the rate of the bicycle ? 17. A boy rows 18 miles down a river and back in 12 hours. He can row 3 miles downstream while he rows but 1 mile upstream. What is his rate in still water, and what is the rate of the stream ? 18. A motor boat can run r miles an hour in still water. If it went downstream for s hours and took t hours to return, what was the total distance traveled, and what was the rate of the current? 19. The sum of three numbers is 20. The sum of the first and second is 10 greater than the third, while the difference between the second and third is 6 less than the first. Find the numbers. [HINT. Use the three letters x, y, z to represent the unknown numbers, and form three equations. Solve as in 35.] 58 SECOND COURSE IN ALGEBRA [VII, 35 20. A, B, and C have certain sums of money. B would have the same as A if A gave him $100 ; C would have four times as much as B if B gave him $100 ; and C would have twice as much as A if A gave him $100. How much has each? 21. I have $90 on deposit in bank A, $51 in bank B, and $75 in bank C. If I have $144 more to deposit, how shall I distribute it among the three banks so as to make the three deposits equal? 22. The perimeter of a certain rectangle is 16 feet. If the length be increased by 3 feet and the breadth by 2 feet, the area becomes increased by 25 square feet. What are the length and breadth? 23. A barrel of vinegar is to be bottled for selling and it is desired that some of the bottles be of pint size, others of quart size and ojthers of gallon size. In order that there be 52 bottles in all, and twice as many of the pint as of the quart size, how many of each will be necessary? [HINT. 1 barrel =32 gallons.] 24. For any pulley block, the relation between the weight to be raised and the pull necessary to raise it is pull = x+yX weight, where x and y are numbers that are different for different pulleys. In two experiments with a certain pulley block, a weight of 100 pounds was raised by a pull of 22 pounds, and a weight of 200 pounds was raised by a pull of 42 pounds. Find the values of x and y for this pulley. CHAPTER VIII SQUARE ROOT 36. Definitions. The square root of a given number is the number whose square equals that number. Thus 2 is the square root of 4 because 2 2 =4. Likewise, 3 is the square root of 9 because 3 2 =9, etc. The square root of a number is denoted by the radical sign V placed over it. Thus V4=2, V9=3, Vl6=4, etc. The process of finding the square root of a number is called extracting its square root. 37. Extracting Square Roots in Arithmetic. Many times we can pick out the square root of a number by inspection. Thus, A/Ill is seen to be 12 because 12 2 = 144. Similarly, Vl96 = 14. But in finding the square root of a large number, such as 74,529, we cannot ordinarily determine the answer by mere inspection. The process for such a case is illustrated below, and is explained on the next page. PROCESS. 7'45'29 | 273 Ans. 4 Trial divisor =2x20 =40 Complete divisor = 40 +7 =47 345 329 Trial divisor = 2 X270 = 540 Complete divisor = 540+ 3 = 543 59 1629 1629 60 SECOND COURSE IN ALGEBRA [VIII, 37 EXPLANATION. First separate the number into periods of two figures each, beginning at the right. That is, in the present case, write the number in the form 7 '45 '29. Find the greatest square in the left-hand period and write its root for the first figure of the required root. This gives the 2 ap- pearing in the answer. Square this figure (giving 4), subtract the result from the left- hand period and annex to the remainder the next period for new dividend. This gives the 345 appearing in the process. Double the root already found, with a annexed (giving 40) for a trial divisor and divide the last dividend (345) by it. The quo- tient (or, in some cases, the quotient diminished) forms the second figure, 7, of the required root. Add to the trial divisor the figure last found (7), giving the complete divisor (47). Multiply this complete divisor by the figure of the root last found (7), giving the 329 appearing in the process. Subtract this from the dividend, and to the remainder annex the next period for the next dividend. This gives the 1629 of the process. Proceed as before, and continue until a new dividend equal to is obtained. In the example above, this happens at once, giving 273 as the required root. This process is the one commonly used in arithmetic, and is stated here as a review. We shall see in 38 that a sim- ilar process may be used in extracting the square roots of expressions in algebra. In the example just solved, the root comes out exact be- cause 74,529 (whose root is being extracted) is a perfect square that is, it is like one of the numbers 1, 4, 9, 16, 36, etc. If we had started with a number which was not a per- fect square, the process would be the same except that we should not finally reach a new dividend which equals 0. In such cases, in fact, the process continues indefinitely, but if we stop it at any point, we have before us the desired root correct (decimally) up to that point. For example, in find- ing the square root of 550 correct to two decimal places, the process is as follows. VIII, 37] SQUARE ROOT 61 PROCESS. 5'50.00'00 | 23.45 Ans. (correct to two 4 decimal places) 2X20 = 40 40+3 = 43 150 129 2X230 = 460 460+4 = 464 2100 1856 2X2340 = 4680 4680+5 = 4685 24400 23425 975 NOTE. In the process above we have first written 550 in the form 550.0000. If we had written it with six zeros, that is 550.000000, and then carried the process forward until all these were used be- low, we should have obtained the root correct to three decimal places instead of two. In general, the root obtained would be correct to a number of decimal places equal to half the number of zeros added. Square roots of decimal numbers, such as 334.796, are ob- tained like those for whole numbers, except that in the begin- ning the separation of the number into periods of two figures each must be carried out both ways from the decimal point. Thus 334.796 would be written 3 '34.79 '60. Similarly, 3.67893 would be written 3 '.67 '89 '30. The extraction of the root is then carried out as in the process shown above. EXERCISES Find (by inspection or by the process shown in 37) the square root of each of the following numbers. 1. 49. 5. 576. 9. 8281. 13. f = |Xf.] 2. 81. 6. 1444. 10. 15,876. [HINT. 3. 64. 7. 4225. 11. 42,025. 14- M 4. 169. 8. 1681. 12. 95,481. 15. & 62 SECOND COURSE IN ALGEBRA [VIII, 37 Find the square root of each of the following numbers correct to two decimal places. 16. 567. 19. 17.76. 22. 3. 17. 633. 20. 13. 23. f. [HINT. Write as [HINT. Write f as 13'.00'00.] .75.] 18. 1305. 21. 2. 24. . 38. Extracting Square Roots in Algebra. (a) Monomials. The square root of a monomial can usually be seen by inspection. 36m% 2 =6 m z n (because (6 m 2 n) 2 =36 m 4 n 2 ). Similarly, (b) Trinomials. If a trinomial is a perfect square, its square root can be obtained by inspection. Thus suppose we wish to find the square root of 9 z 2 + 12 xy +4 y*. This trinomial is a perfect square because its terms 9 z 2 and 4 y* are squares and positive, while its remaining term, 12 xy, is equal to 2- A/9^ 2 - V41/ 2 . (See 11 (d), p. 20.) Hence the trinomial can be expressed in the form (3 x +2 yY, whence the desired square root of 9 x 2 + 12 xy +4 y z is 3 x +2 y. Similarly, V4s 2 -4s+l =2s-l because 4s 2 -4s+l is a perfect trinomial square, and as such is factorable into (2s-l)(2s-l) or (2s-l) 2 . (c) Polynomials. To find the square root of a polynomial of more than three terms we may follow a process much like that employed for finding square roots in arithmetic. This is illustrated in the following example. VIII, 38] SQUARE ROOT 63 EXAMPLE. Find the square root of 4x 4 -f-12x 3 -3x 2 -18x+9. PROCESS. 4x 4 +12x 3 -3x 2 -18x+9| 2x 2 +3x-3 4 x 4 Ans. Trial divisor, 4 x 2 Complete divisor, 4x 2 +3 x 12x 3 -3x 2 12 x*+9 x 2 Trial divisor, 4 x 2 -j-6 x Complete divisor, 4x 2 +6 x 3 -12x 2 -18x+9 -12x 2 -18x+9 EXPLANATION. Arrange the terms of the polynomial in the descending (or ascending) powers of some letter. In the example, the arrangement is in descending powers of x. Extract the square root of the first term, write the result as the first term of the root (giving the 2 x 2 in the answer), and subtract its square from the given polynomial (giving the 12 x 3 3 x 2 in the second line of the process). Divide the first term of the remainder by twice the root already found, used as a trial divisor. The quotient (3 x} is the next term of the desired root. Write this term in the root, and annex it to the trial divisor to form the complete divisor (the 4 x 2 +3 x of the process). Multiply the complete divisor by this term of the root, and subtract the product from the first remainder (giving the 12 x 2 18 x +9 of the process). Find the next term of the root by dividing the first term of the remainder by the first term of the new trial divisor. This gives the 3 of the answer. Form the second complete divisor and continue in the manner above indicated until a remainder of is obtained. In the example just considered, only one letter, as x, appears. A similar process may be employed, however, in all cases by first arranging the expression in descending (or ascending) powers of some one of the letters. For example, 4 x 4 +9 y* 12 x^y 3 + 16 x 2 + 16 - 24 y z , when arranged in descending powers of x, becomes 64 SECOND COURSE IN ALGEBRA [VIII, 38 EXERCISES Find (by inspection or by the process shown in 38) the square root of each of the following expressions. Check each answer by squaring it to see if the result thus obtained is the given expression. 1. 4 xY- 4. 81 a 8 6 6 c 10 . 7. 196 p lo q 12 . 10. a?x 2m . 2. 9a 6 6 4 . 5. 225 ra 8 n 4 . 8. m 2 nVr 12 . 11. 9 ra 2 r* 4p . 3. 25z 2 2/ 4 -3 6 . 6. 625 rW. 9. 529 r 10 s 14 . 12. m 2 *^. 13. z 2 +2z+l. 18. 9ra 2 -6raz-fz 2 . 14. z 2 -4z+4. 19. x 2 +xy+ y 2 . 15. 4m 2 +12mn+9n 2 . 20. 9 z 2 +66 z+121. 16. 4x 2 +4xy+y 2 . 21. 17. c 2 -4ac+4a 2 . 22. 23. 4 24. z 6 25. a; 4 [HINT. See remark at the close of 38.] 26. z 8 +2a 6 z 2 -a 4 z 4 -2a 2 z 6 +a 8 . [HINT. First arrange in descending powers of x.] 27. 9 28. 9 29. z 8 +4 x 7 -3 x 4 -20 x 5 -2 x 6 +4+4 x 2 - 16 x+32 x*. 39. The Double Sign of the Square Root. We know that 3 is the square root of 9 because 3 2 = 9. But we also have ( 3) 2 = 9. Therefore, 3 can also be regarded as a square root of 9. In other words, 9 has two square roots, +3 and 3, which are opposite in sign but otherwise the same. Similarly, 16 has the two square roots +4 and 4, and in general, a 2 has the two roots a and a. The double sign is sometimes used. Thus we say that the square root of 9 is 3. This is merely a brief way of saying that the two roots are +3 and 3. VIII, 40] SQUARE ROOT 65 In order to avoid all confusion, it is to be understood here- after that the radical sign V~ when placed over a number means the positive square root of that number. If it is de- sired to indicate the negative square root, it is done by the symbol V . Thus Vl6 means +4, while - Vl6 means -4. Similarly, Va means +Va. 40. Equations Containing Radical Signs. Equations con- taining radical signs may often be solved by squaring each member. This is equivalent to multiplying each member by the same amount, and hence is justified by 27. EXAMPLE 1. Solve the equation Vx 2 = 6. SOLUTION. Squaring both members gives z-2=36, whence x=38. Ans. CHECK. V38-2 = V36~=6. EXAMPLE 2. Solve the equation Vz 1 VE 4 = 1. SOLUTION. Transpose the Vx -4 to the right ; this gives Square both members, using Formula VI, 10 for finding (1 + Vc-4) 2 . This gives or Canceling x from both sides and transposing the 1 and 4 to the left, gives _ 2 =2 Vc-4, or Vz-4 = l, whence (squaring again) z-4 = ! 2 = l. Therefore x=5. Ans. CHECK. V5^1 - V5^4 = V4- VI =2-1 =1. 66 SECOND COURSE IN ALGEBRA [VIII, 40 NOTE. It is especially important to check all the answers ob- tained for equations containing radical signs, since the process of squaring both members sometimes leads to a new equation whose roots do not all belong to the first one. Thus, if we square both members of the equation x =5, we get x 2 =25 and this last equation has 5 as a root as well as 5. EXERCISES Solve each of the following equations and check each answer. 3. 4. Vx+7-Vx = [HINT. First transpose one of the radicals to the right side, as in Example 2, 40.] 9. 10. If 16 be added to 4 times a certain number, the square root of the result is 6. What is the number? 11. If 9 be added to the square of a certain number, the square root of the result is 5. What is the number? 12. The difference between the square root of a certain number and the square root of 11 less than that number is 1. Find the number. 13. Solve each of the following equations. (a) (6) CHAPTER IX RADICALS 41. Radicals. Suppose we have a square which we know contains exactly 2 square feet. How long is each of its four sides? In order to answer this, we naturally let x represent the desired length. Then we must have x -x = or 2 sq.ft. Therefore x = A/2 ft. Ans. This number A/2 cannot be determined exactly because it is the square root of a p IG 10 number, 2, which is not a perfect square. However, A/2 measures a perfectly definite length, as indi- cated in the figure. Its value, correct to two decimal places only, is 1.41. Such a number as A/2 is called a quadratic radical. This name is used in general to denote the indicated square root of a number. Thus V3, V7, Vzi, VI06, V213 are all radicals. The word radical is also used in connection with other roots than square roots. Thus VlO means the cube root of 10, that is the number whose cube is 10. Similarly, v^6 means the fourth root of 6, etc. All such numbers represent per- fectly definite magnitudes, as did \/2 in the figure above, even though we cannot express them exactly in decimal form. 67 68 SECOND COURSE IN ALGEBRA [IX, 41 In general, the nth root of any number a is written A/a, and this is known as a radical of the nth order. The number n is here called the index of the root, and the number a is called the radicand. NOTE. When no index is expressed, the index 2 is to be under- stood. Thus V3 means \/3. The same definitions apply also to algebraic expressions. Thus V3xy* and Vat+b 2 are radicals. 42. Rational and Irrational Numbers. Surds. The posi- tive and negative integers and fractions, and zero, are called rational numbers. If an indicated root of a number cannot be extracted exactly, that is, cannot be expressed exactly as one of these rational numbers, it is called a sure?. Any posi- tive or negative number that is not rational is called irrational Thus_ V3, ^10, V6 are surds ; but V, v^S, Vj^ are rational, since Vg=3, ^8=2, EXERCISES Determine which of the following radicals are surds ; and state the index and the radicand of each. 1. V7. 2. V8. 3. Vl6". 4. V^T 5. Vff . 6. V75. 7. V^S. [HINT. -8 = (-2) 3 .] 8. Vis. 10. Vs. 12. V20. 14. Vs^y. 9. V27. 11. Vl6. 13. V32. 15. Vg m 6 (a+6) 9 . 43. Value of Radicals. Use of Table. To determine the value of a radical correct to two or more decimal places usually calls for a rather long process. (See 37, p. 61.) In order to save time and labor, the values of those radicals which are needed most in ordinary life (the square and cube IX, 43] RADICALS 69 roots) have been printed in a table and placed for convenient reference at the end of this book. For the sake of complete- ness, the second and third powers of numbers are also printed in the table. Just how to use this table is described on page 275, which the pupil should now read carefully. Below are a few illustrative examples. EXAMPLE 1. Find \/7, using the table. SOLUTION. The top number in the third column on page 290 (table) gives V? =2.64575. This value is correct up to the last decimal figure given, that is to the fifth place. Thus the answer may be written "^ = 2.64575+, the sign + indicating that this value for V7 is correct up to the last decimal place stated. EXAMPLE 2. Find vT, using the table. SOLUTION. The top number in the sixth column on page 290 (table) gives ^7 = 1.91293+. Ans. EXAMPLE 3. Find A/70. SOLUTION. The top number in the fourth column on page 290 (table) gives V70 =8.36660+. Ans. EXAMPLE 4. Find v/70. SOLUTION. V70 = 4.12129+, from the seventh column, page 290. EXAMPLE 5. Find ^700. SOLUTION. v'TOO =8.87904+, from the eighth column, page 290. EXERCISES By means of the table, determine the approximate values of the following radicals. 1. V6. 2. V60. 3. ^6. 4. v"60. 5. ^500. 6. VeT. [HINT. 61=6.10X10. So use the information given for 6.10 in the table.] 70 SECOND COURSE IN ALGEBRA 7. S/6L 8. V92". 9. -v/920. 12. \/78^2 13. A/782. SOLUTION. 782 = 7.82 X 100. Therefore V782 is the same as V7.82 except that the decimal point in the root must be moved one place farther to the right. (See p. 275.) Now, V7.82 = 2.79643 (table), so V782 =27.9643. Ans. 14. A/561. [HINT. See Solution of Ex. 13.] 15. A/779. 16. A/895". 17. ^6120. [HINT. 6120 =6.12 XlOOO. (See p. 276.)] 18. 19. [HINT. .67 =^ X6.7 (Now see p. 276.)] 20. V.0676 APPLIED PROBLEMS Use the tables in working the following problems. 1. If the sides of a right triangle are 3 inches and 2 inches long, respectively, what is the length of the hypotenuse? [HINT. If x be the hypotenuse, then x 2 =3 2 +2 2 = 13.] 2. A baseball diamond is a square 90 feet on a side. How far is it from home plate to second base ? 3. If the diagonal of a square is 13 feet long, how long is each side? 4. The dimensions of a certain rectangular field are 103 feet by 337 feet. In going from one corner to the opposite corner, how much shorter is it to go by the diagonal than to go around? IX, 441 RADICALS 71 5. How long must the radius of a circle be in order that the area be 12 square inches? (See Ex. 14 (6), p. 6. 6. What is the length of the edge of a cube if the volume is 357 cubic inches? 7. If the volume of a sphere is 440 cubic inches, how long is its radius ? (See Ex. 14 (e), p. 6.) 8. In the accompanying figure how long should the radius of the inner semi- circle be in order that the area inclosed ^ g - Q D may be 132 square feet? AB = CD= 2 feet FIG. 11. 9. The area A of a triangle in terms of its three sides, a, b, and c, is A = Vs (s a) (s b) (s c) , where s = The sides of a triangle are respectively 2i 6 inches, 7 inches, and 9 inches long. What is the area ? 10. Two circular cones have altitudes, h, which are the same, but their bases have different radii. What is the ratio of the longer radius to the shorter if the volume of the one cone is three times that of the other? [HINT. The formula for the volume of a cone is V = irr 2 /&, where h represents the altitude and r is the radius of the base.] 44. Simplification of Radicals. We know that the square root of the product of two numbers is the same as the product of their square roots. For example, V4x25 is the same as A/4 X V25, because both are equal to 10 (the first being VIM, or 10, and the second being 2 X5, or 10). In the same way, v / 8x3 = v / 8XV / 3, or simply 2^3. In fact, we have the following general formula. 72 SECOND COURSE IN ALGEBRA [IX, 44 Formula I. Vafc = Va V&. Again, Vf is the same as - because both are equal to f . V9 (Explain.) Similarly, v^f may be written - , or So V 8 2 in general we have the following formula. Formula II. V 1 /? Formulas I and II enable us to simplify many radical ex- pressions, as is illustrated in the following examples. EXAMPLE 1. Simplify V63. SOLUTION. Using Formula I, we have V63 = V9X7 = V9 X V? =3 V?. Arcs. EXAMPLE 2. Simplify \/32. SOLUTION. ^32=4/8X4=^/8X^4=2^4. Ans. EXAMPLE 3. Simplify /8 Vg V4x2 Vix^/2 2\/2 SOLUTION. \~ = : = ^ = - = ^27 V27 \/9x3 .V9XV5 3V3 EXAMPLE 4. Simplify V20 a 6 ". SOLUTION. V20 a 6 = V4a 6 x5 = V4^ x V5 =2 a 3 Vs. Ans. EXAMPLE 5. Sunplify v/ SOLUTION. ) X (9 x 2 ) = v'S * Z* ^6 2 2 2 2 NOTE. It will be observed that in each of the examples above the process of simplification consists in removing from under the IX, 44] RADICALS 73 radical sign the largest factor of the radicand which is a perfect square, or perfect cube, as the case may be. Thus, in Example 1, the radicand, 63, was first broken up into factors in such a way that 9 (which is a perfect square) appears clearly to the eye. Similarly, in Example 2 (where we are dealing with a cube root) we first write the radicand, 32, in a form which brings out conspicuously its factor 8, which is a perfect cube. The first step in all such examples is, therefore, to get the radicand properly broken up into factors. This requires good judgment, but becomes very easy after slight practice and experience. EXERCISES 1. By Formula I, p. 72, we have A/20 = A/4x5 = A/4 X A/5 = 2A/5. Look up the values of A/20 and A/5 in the table and thus prove that A/20 is the same as 2 A/5. 2. Show that Formula I, p. 72, gives A/54 = 3A/(T and verify the correctness of this result by use of the table, as in Ex. 1. Simplify each of the following radicals. (See Note in 44.) 3. A/18. 6. A/125. 9. v/54. 4. A/24. 7. A/108. 10. A^Sl. 6. A/Tl2i 8. A^32. 11. V||. [HINT. First use Formula II, 44.] 12 - ^5S- 19. A^27 x*y*z\ 20. \/ 16 M4 . o3/ 6 a?bVab. Ans. 4(a 2 -6 2 ) 74 SECOND COURSE IN ALGEBRA [IX, 44 Write each of the following in a form having no coefficient outside the radical sign. 23. 2>/3. SOLUTION. By Formula I, 44, 2^3 = Vix \/3 = Vl2. Ans. 24. 2V2". 25. 5V5^ 26. 2v^ 27. W2. A/9 A/9 [HINT. Write W2 =-rr =-^, and apply Formula II, 44.] 3 Vg 33. 2gVx-y. 38. fV3. . Ans. 34. ra^Xn. 39. f^S. 35. 2aV 2 -6 2 . 40. fVf. 36. ^L- 41. (a-6)V2c". 42. o- 32. SmnVmn. 37. r\'-- 45. Addition and Subtraction of Similar Radicals. When- ever two radicals having the same index have also the same radicand (or can be given the same radicand by simplifica- tion) they are called similar radicals. Thus 2 V2 and 3 V2 are similar radicals ; so also are V2 and V32 since the last of these may be simplified into 4V2, by 44. Like- wise, V3 a 2 x and V3 b 2 x are similar, being equal, respectively, to a VWx and 6 V3 x, thus coming to have the same radicand. Whenever similar radicals are added or subtracted, the result may be expressed in a single term. Thus 4V3+5V3 = (4+5)\/3=9\/3. Ans. _ Again, 3V32-2V8=3X4V2-2X2V2~=12V2-4V2 = (12-4)V2=8V2. Ans. Likewise, 2 V4~cM> + V9 a z b - Vl6 a*b =2 2 avT+3 a*N/6_-4 aV6 = (4 a+3 a-4 a)V6=3 aVF. Ans. IX, 46] RADICALS 75 NOTE. The pupil is especially warned that in general we cannot write Va+6 = Va + Vb. Thus when a =4, 6=9 this would give Vl3=2+3=5, which is clearly false. Similarly, we cannot write V~a~b = Va-Vb. Thus V25^9=Vl6=4, but V25-V9=5-3 = 2. EXERCISES Combine the following radicals. 1. V8+V18+V32. Check your answer by use of the table; that is show that \/8+ VT8+V32, as computed by the table, has the same value as your answer, similarly computed. 2. V108+V27-V75. 3. v'm-h^lG-v^. 4. V72+V32-V50". Check your answer as in Ex.1. Check as in Exs. 1 and 2. 7. 32 a 2 - 8. 9. 46. Multiplication of Radicals. We may multiply one radical, as Va, by another of the same index, as \/b, by For- mula I of 44. If n 2, this gives as an important special case Va Vb = 76 SECOND COURSE IN ALGEBRA [IX, 46 EXAMPLE 1. Find the product of V2^and Vl8. SOLUTION. V% > Vl8 = V2 18 = V36 =6. Ans. EXAMPLE 2. Multiply V3+A/5 by 2 A/3- \/5. SOLUTION. V3+ V5 2V3- V5 2- 3+2VI5 - VI5-5 6+ 5-5 = l + Vl5. Ans. EXAMPLE 3 . Multiply Va + Va b by Va Va b. SOLUTION. Va + Va b Va- Va-b a + Va 2 ab Va 2 ab (o-b) a (a 6) =a a+b = b. Ans. EXERCISES Find the following products, simplifying results as far as possible. 1. V3 - \/27. 7. V7 . \/I 2. V8 - Vl2. 8. VI - VTOl. 3. V6-V4. 9. (V3-V^)(\/3+>/2). 4. \/7 V9. 10. (2v / 3-V2)(2\ / 3+v / 2). 5. VI6 - V3 - V2. 11. (V6- V3) 2 . 6. \/f . V|. 12. 13. (3\/3+2V5)(V3-3\/5). 14. (V2+V3 + V5)(V2-V3). 15. Va-.Vtf. 18. 16. Vo6 V^ 3 . 19. (Va-\/6) 2 . 17. Vtf-V- 20. IX, 47 RADICALS 77 21. 22. Find the value ofz 2 -4z-lifa: = 2+ \/5. 23. Find the value of z 2 +3z-2if 24. Does A/3+2 satisfy the equation x? 4 z+1 = ; that is, is the equation true when x = A/3+2 ? Answer the same question when x = A/3 + A/2. 47. Division of Radicals. We may divide one radical, as Va, by another of the same index, as V6, by the Formula II of 44. If n = 2, this gives as an important special case *-4 EXERCISES Express each of the following quotients as a fraction under one radical sign, and reduce your answer to simplest form. A/15 Ans. 10. V81+9 a 2 +a 4 ^- Va 2 -3 a+9. CHAPTER X QUADRATIC EQUATIONS 48. Quadratic Equation. An equation which contains the unknown letter to the second (but no higher) power is called a quadratic equation, or briefly, a quadratic. Thus the equations 2 x 2 4 x = I and ^ x 2 +x 3 are quadratics, but 2 x -3 =0 and 4 x 3 -5 x 2 +x =2 are not. 49. Pure Quadratic. When the quadratic contains the second power only of the unknown letter, it is called a pure quadratic. Thus 2 x 2 27 =0 and ax" 2 = be are pure quadratics, but x 2 4 x =2 and x 2 +bx+c=Q are not. 50. Affected Quadratic. When the quadratic contains both the first and second powers of the unknown letter, it is called an affected quadratic. Thus z 2 +3z=7 and x z -\-2ax=a z are affected quadratics, but 2 x 2 -7 =0 and 5 x 2 -16 o 2 6 2 =c 2 are not. 51. Solution of Pure Quadratics. The following example will suffice to show how the solution of any pure quadratic may be obtained. EXAMPLE . Solve 2 x 2 - 30 = 0. SOLUTION. Transposing and dividing through by 2 gives^ 2 = 15. Taking the square root of both members gives x = = Vl5. Ans. To get the approximate value of Vl5, we may consult the table, where we find Vl5 =3.87298+. The answer may, therefore, be written in the form x = 3.87298+. CHECK. 2 (Vl5) 2 - 30 =2X15 -30 =30 -30=0, as required. 2( - Vl5) 2 -30 =2 X 15 -30 =30 -30 =0, as required. 78 X, 51] QUADRATIC EQUATIONS 79 NOTE. Strictly speaking, when we extract the square root of both members of the equation x 2 = 15 we get 3 = VlS. But to say that x = == Vl5 means the same as +x = == Vl5, so it suffices to write simply x = == Vl5 to cover all cases. An examination of the example above shows that we have the following rule. To solve a pure quadratic, solve for x 2 , then take the square root of the result. There will be two solutions, the one being the negative of the other. EXERCISES Solve each of the following equations, checking your answer for the first five. If you meet with a radical, find its approximate value by use of the table. 1. z 2 -81 = 0. 11 3 _ 1 j 1 2. 3* 2 -192 = 0. ' *+ 5 2 *~ 5 ' 12 JL ! _Z_L 1 ' ' 3. 4z 2 +8 = 10* 2 -16. 4. 3x 2 -15 = 0. x+3 ,2x-l =Q ~ 5. 3, 2 -!6 = 0. _ 6. 3z 2 -17 = 0. -' - ? ^_^ =19 x-2 x+3 [HINT. See 40.] 17. 10. (o;+l) 2 -2(x+l)=4. 18. V25-6 x+ V25+6 a; = 8. 80 SECOND COURSE IN ALGEBRA [X, 51 APPLIED PROBLEMS 1. What numbers are equal to their own reciprocals? 2. One side of a right triangle measures 3 inches and the hypotenuse measures 7 inches. Find (approximately) the length of the other side. [HINT. Work by algebra, making use of the principle that in any right triangle the square of the hypotenuse equals the sum of the squares of the two sides.] 3. What is the length of the longest umbrella than can be placed in the bottom of a trunk the inside of which is 33 inches long by 21 inches wide? 4. A certain square has a side which is three times as long as the side of another square. If the difference of their areas is 72 square feet, how long is the side of each ? 5. Find the mean proportional between 25 and 9; also that between 17 and 21. In what particular is the latter one essentially different from the first one ? 6. It is proved in geometry that whenever a perpen- dicular is drawn from a point on a semicircle to the base, as PQ in Fig. 12, its length is a mean Q proportional between the segments AP and PC of the base ; that is, AP = PQ PQ PC IfAP = 8 inches and PC = 10 inches, how long is PQ ? 7. Determine the formula for (a) The side of the square whose area is a. (b) The radius of the circle whose area is a. X, 52] QUADRATIC EQUATIONS 81 (c) The radius of the sphere whose area is a. [HINT. See Ex. 14, p. 6.] (d) The diameter of the base of a circular cone whose volume is v and whose altitude is h. (HINT. The volume of a circular cone is equal to the area of its base multiplied by -^ its altitude, i.e., V = ^Trr 2 h.] 8. The distance s, measured in feet, through which an object falls in t seconds when dropped vertically downward is given by the formula s = %gt*, where = 32 (approximately). Hence, determine (approximately) how long it will take a stone to drop to the bottom of a mine 300 feet deep. 9. One of the sides of a certain triangle is m units long. What is the formula for the corresponding side of a similar triangle whose area is n times as great? [HINT. It is shown in geometry that if two geometric figures are similar ; that is, have the same shape but are of different sizes, then the square of any line in the first figure is to the square of the cor- responding line in the second figure as the area of the first figure is to the area of the second.] 10. A map of the United States is uniformly enlarged in such a way as to cover twice as much area on the paper as before. By what factor should the scale of the map be now multiplied? 11. Find three consecutive integers such that the square of the second plus the product of the other two equals 31. [HINT. Let x be the second integer. Then the first and last integers will be x 1 and x + 1 respectively.] 52. Solution of Affected Quadratics by Factoring. It is a familiar principle of arithmetic that the product of two numbers is zero if either of the numbers is zero ; that is, if either factor is zero. For example 2X0=0, 0X4=0, (-3) X0=0, etc. G 82 SECOND COURSE IN ALGEBRA [X, 52 This principle is frequently used to solve affected quadratic equations. EXAMPLE. Solve by factoring the equation # 2 +8 x = 48. SOLUTION. Transposing all terms to the left, we have Factoring, z 2 +8z-48 = (z-4)(z + 12). [ 11 (c)] Thus the given equation becomes This equation will be satisfied, according to the principle men- tioned above, whenever the factor x 4 equals zero or the factor z + 12 equals zero, that is in case x 4=0 or a; + 12=0. Solving these two simple equations gives z=4 and x= 12, which must therefore be the desired solutions. CHECK. When x = 4 the left side of the original equation becomes 4 2 +8x4, which reduces to 16+32 =48, as the equation demands. When x = 12 we have in like manner (-12) 2 +8x(-12) = 144-96=48. We thus have the following rule. To solve quadratics by factoring : 1. Transpose all terms to the left so as to have on the right. 2. Factor the left member of the resulting equation. 3. Place each factor equal to and solve the resulting simple equations. The two results are the solutions required. EXERCISES Solve each of the following equations by factoring, checking your answer in the first five. 1. z 2 -7z+10 = 0. 3. z 2 +8z=-15. 2. z 2 -5z=-6. 4. z 2 +7z-30 = 0. X, 53] QUADRATIC EQUATIONS 83 5. z 2 6. l- 7. [HINT. Write as 8. x 2 -l = 3 5 then apply the principle in 52.] ~z+4 = 13. 3(z+l)(z-3)+4(z-3)=0. 10 . j- = __ i __ 5. 14- (z+l) 2 +3(z+l)+2 = 0. x 2 x 2 [HINT. Solve first for x+L] 53. Solution of Any Quadratic by Completing the Square. We often meet with a quadratic, such as x z -\-7 x 5 = 0, which we cannot solve as in 52 by factoring. The difficulty here is that we cannot factor readily x 2 +7 x 5. However, this quadratic and all others (whether solvable by factoring or not) can be solved by a certain process known as completing the square. How this is done will be best understood from a careful study of the following examples. EXAMPLE 1. Solve z 2 + 6 x = 16. SOLUTION. The first member of this equation, or x 2 +6 x, would become' a trinomial square [ ll(d)] if 9 were added to it. Our first step, therefore, is to add 9 to both members of the given equa- tion, thus "completing the square" in the first member and giving us the equation X 2 _j_6 x _jig = 25, or (x +3) 2 =25. Taking the square root of both members of the last equation is now an easy process and gives x +3 = 5. Therefore we must either have #+3 =5, or x+3 = 5. Solving the last two equations gives as the desired solutions x=2andx= 8. Ans. CHECK. Substituting 2 for x in the first member of the given equation gives 2 2 +6X2, which reduces to 4 + 12 = 16, as desired. Similarly, with x equal to 8, the first member of the given equa- tion becomes (-8) 2 +6X(-8), or 6448, which reduces to 16 as required. 84 SECOND COURSE IN ALGEBRA [X, 53 EXAMPLE 2. Solve x 2 - 8 x+ 14 = 0. SOLUTION. Transposing, x 2 8 x = 14. Completing the square by adding 16 to both sides gives z 2 -8z + 16=2, or (z-4) 2 =2. Taking the square root of both members, x-4=V2. Solving the last two equations, z=4 + V2andz=4- V2. Ans. CHECK. With x=4 + V2, the first member of the given equation becomes (4 + V2) 2 -8(4 + V2) +14. By Formula VI of 10 this may be written (16+8V2+2)-8(4 + V2)+14, or 16+8V2~+2-32-8V2 + 14. Here the 8>/2 and the -8^2 cancel, while the rest of the expres- sion (namely 16+232 + 14) reduces to 0, as required. Likewise, when x has its other value, namely x =4 V2, the first member may be shown to become 0. NOTE. Since the solutions obtained above for Example 2 con- tain the surd V2, they cannot be expressed exactly (see 37), but we_can express their values approximately. Thus, the table gives V2 = 1.41421+ so that the two solutions become 4 + 1.41421+ and 4-1.41421+, which reduce to 5.41421+ and 2.58579+. Ans. EXAMPLE 3. Solve 3 z 2 +8 x = 15. SOLUTION. Dividing through by 3 so as to have +1 as the co- efficient of x 2 , the equation becomes z 2 +fz=5. Completing the square by adding (-|) 2 (or -^-) to both sides gives or, Taking the square root of both members, x + * = Therefore, the two solutions are z=-f+iV61 and z These two answers may be written together in the condensed form x =-^( 4 V61) and by looking up the value of V61 in the tables, these values of x may be determined approximately, as in- dicated in the Note to Example 2. X, 54] QUADRATIC EQUATIONS 85 54. Summary and Rule. It is now to be observed care- fully that in each of the three examples just considered ( 53) the first step in the solution consists in reducing the given equation to the type form where p and q are given numbers. Thus, in Example 3, we first put the equation 3 x 2 +8 x = 15 into the form z 2 +f z=5. Here p=f, and q =5. The next step is to complete the square. This is done in each case by adding to both members the square of half the coefficient of x, that is we add (p/2) 2 to both members. Thus, in Example 3, we had p=f, so we added (|-) 2 to both members. After this, the equation is such that we can extract the square root of the left member, and when we do so and equate results, we obtain two simple equations, each yielding a solution of the given quadratic. This may now be summarized in the following rule. To solve any quadratic : 1. Reduce the equation to the form 2. Complete the square by adding (p/2) 2 to both members. 3. Extract the square root of both members of the new equa- tion and equate results. This yields the two solutions desired. EXERCISES Solve each of the following equations, checking your answer in the first five. 1. z 2 -5z = 14. 6. 8z = z 2 -180. 2. z 2 -20z = 21. 7. z 2 +22z=-120. 3. z 2 -12z+20 = 0/ 8. 2/ 2 = 10-3i/. 4. x 2 -2x = ll. 9. z 2 - 6. z 2 -3z-5 = 0. 10. 6z 2 86 SECOND COURSE IN ALGEBRA [X, 54 11. 2z 2 + 12. 1-3 * = 2* 2 . 17. ^ 13. 2 (z-f-4)=42. x 14. (3x-2) 2 = 6z+ll, 18. -^ 15. x+ = 16. . z+5 z-2 55. Solution of Quadratics by the Hindu Method. A simple way preferred by many for completing the square in any quadratic is the one called the Hindu method. It con- sists of two steps : 1. Multiply both members by four times the coefficient of x 2 . 2. Add to both members of the new equation the square of the original coefficient of x. EXAMPLE. Solve 2 x 2 - 3 x = 2. SOLUTION. Multiplying through by 4 times the coefficient of x\ that is by 8, gives 16 x 2 -24 x = 16. Adding the square of the original coefficient of x to both sides, that is adding ( 3) 2 , or 9, to both sides, gives 16z 2 -24z+9=25. The first member is now a perfect square, being equal to (4 x 3) 2 . Therefore, extracting square roots, we obtain 4 x -3 =5 and 4 x -3 =-5. Solving the last two equations gives x=2 and x = J. Ans. EXERCISES Solve each of the following quadratics by any method. 1. 9z 2 +6z = 35. 4. 16z 2 -7o;-123 = 0. 2. 4z 2 -12z = 27. 6. 1 3. 4z 2 -z-3 = 0. 6. 2 X, 56] QUADRATIC EQUATIONS 87 7. 2z 2 +6z = f 17. 3z 2 +z-200 [HINT. First multiply both ^ r numbers by 2.] 18. x+- = -- 8. 3z 2 -2z = 5. 9. 3z 2 +7z-110 = 0. 19. -- ^ = r o <-i f\ 3C O X 10. 5z 2 -7z=-2. 11. l-3z = 2z 2 . 20. 12. 4* 2 -3*-2 = 0. 13. 2a: 2 +3x = 27. 21 14. 3x 2 -7x+2 = [HINT. Proceed as in 40.] 15. 4 x 2 - 17 z =-4. 22 - Vx+1 Va;-2 = A/2 a;- 5. 16. 8z = z 2 -180. 23. Vx-l+VlQ-x=3. 56. Solution by Formula. Every quadratic is an equation of the type form ax 2+ bx + c = 0> where a, b, and c are given numbers. We may solve this equation as it stands by the process of 54. Thus, ax?-\-bx= c. Dividing through by a, *M*--4 a a Adding 6/(2 a) 2 to both sides ( 54) gives x* \ b x I f b b * c a \2aJ \2aJ a 4 a 2 a 4 o 2 Extracting the square root of both members, x+ === /fr 2 -4ac^ Vb 2 -4ac 2 a "V 4 a 2 2 a Transposing the term 6/(2 a), we thus have the following formulas for the two roots : x .-+vy=a and x= - b -v^=47c. 2a 2 o 88 SECOND COURSE IN ALGEBRA [X, 56 NOTE. Observe that these formulas express the values of x in terms of the known letters a, 6, and c, as should be the case. Thus, in any example, we have only to put the given values of a, 6, and c into the formulas in order to have at once the desired values of the two roots. EXAMPLE. Solve by formula 2 x 2 3 x = 2. SOLUTION. This equation may be written 2z 2 3# 2=0. Hence the values of a, b, and c in this case are as follows: a =2, b = 3, c = 2. Placing these values in the formulas gives as the roots _ -3) 2 -4(2) ( -2) 2-2 and a ,-(-3)-V(-3)-4(2)(-2). 2 2 Simplifying, * = and ,- The two roots are, therefore, z 2 and x= ^. Ans. (Com- pare solution in 55.) EXERCISES 1. Solve by the formula Exs. 13-17, p. 87. 2. Solve by the formula the equation 3 x 2 6 x+2 = 0. [HINT. The roots are found to be x = 1 +^ V3 and x = 1 -| Vs. This quadratic thus has roots which necessarily contain radicals. From the table of square roots at the end of the book, we find V3 = 1.73205. Hence, the roots, correct to five decimal places, are 1+ 1.73205 and 1 _1.73205 > which reduce respectively to 1.57735 3 3 and 0.42265. Ans.] 3. Solve by the formula the equation x 2 5 z+3 = 0, and use the tables if necessary to determine the roots decimally. 4. Solve by the formula the equation 4 x 2 3 x 2 = 0, ex- pressing the roots decimally correct to five places. X, 56] QUADRATIC EQUATIONS 89 APPLIED PROBLEMS [The method of solving quadratics by formula ( 56) is usually the most direct.] 1. The square of a certain number is 4 less than five times the number. Find the number. [HINT. Remember that there should be two solutions.] 2. Divide 20 into two parts whose product is 96. 3. One side of a right triangle is 2 inches longer than the other. If the hypotenuse is 10 inches long, how long are the sides? [HINT. Letting x represent the shorter side, the equation here becomes x 2 + (x+2) 2 = 100, and in solving this we find that one of the solutions is negative. But a negative solution can have no mean- ing in such an example as this, so we keep only the positive solution. This frequently happens in applied problems involving quadratics, so the pupil must always be on his guard to keep only such solu- tions as can actually fit a given problem.] 4. A gardener spades a bed 30 feet long by 20 feet wide. He then decides to double its size by adding a border of uni- form width throughout. How wide must ,_. the border be made? 5. In Example 4 suppose that instead of doubling the area, the gardener wishes merely to add 200 square feet to it. Show that the strip added around the outside must then be made a little over 1.86 feet wide. 6. A circular swimming pool is surrounded by a walk 4 feet wide. If the area of the walk is one fourth that of the pool, find (approximately) the radius of the pool. (Take 90 SECOND COURSE IN ALGEBRA [X, 56 FIG. 14. is marked P A G 7. If a train had its speed diminished by 10 miles an hour, it would take it 1 hour longer to travel 200 miles. What is p x B the speed ? -c ^ . L 8. A circular curbing touches the line of \ the street curbing on each of two streets / that meet at right angles. In Fig. 14, the middle point of the circular curbing is marked M. The point at which the straight curbings would meet if extended If PM = 5 ft., find the radius (x in Fig. 14) of the circular curbing, f 9. The figure represents a pattern fre- quently used in window designs, consisting of a square A BCD with a semicircle EFG mounted upon it, the diameter GE of the semicircle being slightly less than one of the sides of the square. If the shoulders AG and DE are each 1 foot long, how long must each side of the square be made in order that the total lighting surface shall be 88 square feet? 10. A soap bubble of radius r is blown out until the area of its outer surface becomes double its original value. Show that the radius has thus been increased by an amount h given by the formula h = r(V2-l). [HINT. See Ex. 14 (/), p. 6.] 57. Graphical Solution of Quadratics. Consider the quadratic x 2 3 x 4 = 0. Let us represent the left member by the letter y ; that is, let us place FIG. 15. Now, if we give to x any value, this equation determines a t This problem suggests a practical plan for finding the radius of circular curbings when the center of the circle cannot be reached. X, 57] QUADRATIC EQUATIONS 91 corresponding value for y. For example, if x = Q, then ?/ = 2 -3xO-4=-4. Again, if x=l, then i/=l 2 -3xl-4 = 6. The table below shows a number of x-values with their corresponding ^/-values, determined in this way. When x = 1 2 3 4 5 . 6 -1 -2 -3 then y = 4 -6 -6 -4 6 14 6 14 The graph is now obtained by draw- ing an z-axis and a 2/-axis, as in 28, then plotting each of the points x, y which the table contains, and finally drawing the smooth curve passing through all such points. The form of the graph thus obtained is indicated in the adjoining figure. Observe that this graph is not a straight line and is there- fore different in character from the graph of a linear equation. (See 29.) And it is especially important to notice that it cuts the x-axis in two points whose ^-values are 1 and 4, respectively. These two values of x determined in this purely graphical way are the two solutions of the given quadratic, x 2 3 x 4 =0, for they are those values of x that make y 0, that is that make z 2 -3z-4 = 0. The graphical study which we have just made of the special quadratic x 2 3 x 4 = leads at once to the following more general statements. Every quadratic has a graph which -is obtained by first placing y equal to the left member of the equation (it being understood that the right member is 0) , then letting x take a series of values FIG. 16. 92 SECOND COURSE IN ALGEBRA [X, 57 and determining their corresponding y-values, plotting the points, x, y, thus obtained and finally drawing the smooth curve through them. The x-values of the two points where the graph cuts the x-axis will be the roots of the given quadratic. EXERCISES Draw the graphs of each of the following quadratics, and note where each cuts the z-axis. In this way determine graphically the solutions, and check the correctness of your answer by actually solving by one of the methods explained in 54-56. 1. z 2 -z-2 = 0. 2. x 2 -7x+l2 = Q. 3. x z + 4. z 2 -5z=-6. [HiNT. Remember to write first as x 2 5 x +6 =0.] 5. x*+3x=W. 6. 2x 2 +3x = 9. 58. Quadratics Having Imaginary Solutions. Consider the quadratic x 2 = 1. This is a pure quadratic ( 49) and hence can be solved immediately by merely taking the square root of each member. This gives as the required solu- tions x=-\-V 1 and x=\^\. But V 1 means the number whose square is 1, and there is no such number among all those (positive or negative) which we have thus far met. In fact, we know that the square of any number, whether the number be positive or negative, is positive [ 2(d)]. There- fore, in any such case as this, we say that the solutions are imaginary, and we speak of the numbers themselves which, like V 1, enter into algebra in this way, as imaginary numbers. They are imaginary, however, only in the sense that they have not been encountered before. X, 59] QUADRATIC EQUATIONS 93 As an example of an affected quadratic having imaginary roots, let us consider the equation x 2 6x+15 = 0. When we proceed to solve this by the method of completing the square, as in 54, the work is as follows. Transposing, we have x 2 -6x= -15. Adding 9 to both sides to complete the square, z 2 -6z+9= -6, or (z-3) 2 = -6. Extracting the square root of both sides, x-3= V^6. Therefore the solutions are =3- V^Q. Ans. Both of these solutions_are seen to be imaginary because they contain the expression V Q. 59. Definitions. A number like 3+v^6 or 3-V-6 is frequently called a complex number in distinction to such a number as V 6, which is called a pure imaginary. Thus, a complex number is a combination of a positive or negative number with a pure imaginary. All numbers considered in the chapters preceding this (including irrationals) are called real numbers in distinction from the imaginary numbers just described. Thus, the solu- tions of all quadratics considered in 54-56 are real instead of imaginary. EXERCISES Find (by solving) whether the solutions of the following quadratics are real or imaginary. 1. z 2 +9 = 0. 3. 2z 2 +2z+3 = 0. 6. 2. z 2 -6z+10 = 0. 4. 3z 2 +2z = 4. 6. 94 SECOND COURSE IN ALGEBRA [X, 60 60. Determining Graphically Whether Solutions are Real or Imaginary. It was shown in 58 that the solutions of the quadratic x 2 6 x= 15 are imaginary. Let us now see what corresponds to this fact in the graph. The table below shows several values of x and their corre- sponding ^/-values, as determined from the given equation When x = -1 1 w_ 2 3 4 5 6 then y = 22 15 7 6 7 10 15 FIG. 17. Plotting the various points (x, y) thus obtained and drawing the curve through them gives the graph indicated in the accompanying figure. This graph is es- sentially different from those met with in 57 in one particular, namely it does not cut the x-axis. A similar result holds for the graph of every quadratic whose solutions are imaginary. Therefore, in order to tell whether the solutions of any given quad- ratic are real or imaginary, we need only draw its graph and note whether or not it cuts the z-axis. If it does, the solu- tions are real ; if it does not, the solutions are imaginary. EXERCISES Find by drawing the graph whether the roots of each of the following quadratics are real or imaginary. 1. z 2 +2z+3 = 0. 3. z 2 -2z+3 = 0. 5. 6z 2 +5z+l=0. 2. z 2 +2z-3 = 0. 4. 3z 2 +4z+l = 0. 6. 2z 2 -3z+4 = 0. LAGRANGE (Joseph Louis Lagrange, 1736-1813) Famous for his discoveries in all branches of mathematics and regarded as the greatest mathematician of the 18th century. In algebra he gave much attention to the study of equations and determinants, extending and unify- ing the work of previous mathematicians in these fields. PART II. ADVANCED TOPICS CHAPTER XI LITERAL EQUATIONS AND FORMULAS 61. Literal Equations. Equations in which some, or all, of the known numbers are represented by letters are called literal equations. The known letters are generally repre- sented by the first letters of the alphabet, as a, b, c, etc. Literal equations are solved by the same processes as numeri- cal equations,. EXAMPLE. Solve the following literal equation for x : ax = bx+7 c. SOLUTION. Transposing, ax bx=7 c. Combining like terms, (a-6)z=7c. Dividing by (a 6), *=-?X Ans. ab CHECK. Substituting the answer for x in the given equation, Multiplying by (a b), 7 ac =7 be +7 c(a -b) = 7 be +7 ac -7 be. Transposing, 7 ac 7 ac=7 be -7 be. Simplifying, 0=0, which is a correct result. 95 96 SECOND COURSE IN ALGEBRA [XI, 61 It is to be carefully observed that a literal equation is said to be solved for the unknown letter, as x, only when that letter has been expressed in terms of the other (known) letters. Thus, in the example above, we obtained x in terms of a, 6, and c. This when once done, is what we mean by the solution. NOTE. If a literal equation is satisfied no matter what values be given to the letters appearing in it, it is called an identity. Thus x 2 a 2 = (x a) (x +a) is an identity. This fact is often expressed by means of the symbol =. Thus, x 2 a?=(x o)(x+o). Likewise, (z-a) 2 = z 2 -2 ax+a 2 , etc. EXERCISES Solve for x in the following, checking your answer in the first five. 1. x-a = b. 9 *+b = *+a. 2. ax-l = b. a 3. ax+bx = c. - a , b _ 4. 3z+6 = z-3&. cx 5. 4(3&-z) = 3(26+z). 6. (x-a)(x-b)=x(x+c). 7. -J-- a. 12. 1+x x-3 a, 6_o 13. Divide a into two parts # # whose quotient is m. 14. If A can do a piece of work in a days, and B can do it in b days, how long will it take them working together? (See Exs. 23, 24, p. 40.) [HINT. These are simultaneous equations, to be solved for the two unknowns x and y in terms of a and b.] XI, 61] LITERAL EQUATIONS AND FORMULAS 97 16. I ax-by = 2, { 3 ax-f2 by = ab, ax by = ab. 18. x y a 1-1=1. x y b [HINT. Solve first for l/x and l/y. See Ex. 16, p. 53.] a_b _ _., x y~ lj a 19. x y 20. ax z -c=l. [HINT. See 51. Ans. x = * >/ ] 21. az 2 -fa 3 = 5a 3 -3az 2 . 22. !+ = i. x a x 23. 24. (x+a)(x+b)+4(x+a)=Q. [HINT. Solve by factoring.] 25 /2 ax = 2 a 2 [HINT. See 54, p. 85. Ans. x =2 a, or x = -a.] 29. x = 4 ax 2 a . 30. Vx a+Vb x = Vb a. H 98 SECOND COURSE IN ALGEBRA [XI, 62 62. Formulas. If a person travels for 10 hours at the rate of 15 miles an hour, the distance he travels is 15 XlO = 150 miles. Stated in general (algebraic) language, we can say in the same way that if a person travels for t hours at the rate of r miles an hour, the distance s he travels is s = rt. This is a literal equation expressing the value of s in terms of r and t. If we wish, we can solve it for t, giving t = s/r, and what we now have is t expressed in terms of s and r. Or, we can solve the original equation for r, giving r = s/t, and this expresses r in terms of s and t. These examples illustrate the important fact -that in nearly all branches of knowledge, especially in engineering, geometry, physics, and the like, there are general laws which are expressed by means of mathematical formulas. Such formulas are merely literal equations in which two or more letters appear, and it is often desirable to solve them for some one letter in order to express its value in terms of the others. EXERCISES 1. The area A of a rectangle whose dimensions (length and breadth) are a and b is given by the formula A=ab. Solve this for a ; also for b. In each case state in terms of what letters your answer is written. 2. The formula for the area A of a triangle whose height (altitude) is h and whose base is a is A = % ah. Solve for a ; also for h. 3. Solve for b in the formula (Formula for the area A of a trapezoid whose FIG. is. bases are B and b and whose altitude is h.) XI, 62] LITERAL EQUATIONS AND FORMULAS 99 4. Solve for r in the formula A=irr 2 . (Formula for the area of a circle whose radius is r.) 5. The interest / which a principal of p dollars will yield in t years at r % is determined by the formula 100 Solve this for r and use your result to answer the following question : What rate of interest is necessary in order that $50 may yield $6 interest in 2 years' time ? [HiNT. Solving for r gives at once r = 1007 pt ' Now see what the right member of this equation becomes when 7=6, p=50, and t=2.] 6. Using tne interest formula of Ex. 5, solve it for t and use your result to answer the following question : How long will it take $600 to yield $63 interest if invested at 6%? 7. The velocity of sound v, in feet per second, is given by the formula v= 1090+1. 140-32), where t is the tem- perature of the air in Fahrenheit degrees. Find (a) The velocity of sound when the temperature is 75. (6) The temperature when sound travels 1120 ft. per sec. 8. Derive formulas for each of the following statements. (a) The number N of turns made by a wagon wheel d feet in diameter in traveling s miles. (6) The number N of dimes in m dollars, n quarter dollars, and q cents. 9. An automobile travels for T hours at the rate of v miles per hour. By how much must this rate be increased in order to make the same journey in t minutes less time? 10. A has $a and B has $6. Between them they give $c to a certain charity, after which the amounts of money they have are equal. How much does each contribute? 100 SECOND COURSE IN ALGEBRA [XI, 63 63. Law of the Lever. If two weights are balanced at the ends of any (uniform) bar, as shown in the figure, we have an example of a lever. The point of support, F, is called the fulcrum. If we let W and w be the values (in pounds or ounces or any FIG 19 other convenient unit) of the two weights, while D and d stand for the distances respectively of W and w from F, then, when- ever the balance is perfect, we have the formula D F d A fw W w D Sometimes a single weight W is balanced by a force, p, usually called a power. This may happen in several ways, as indicated by the following figures. In all such cases, if we I D- (2) let W represent the weight, p the power, D the distance from W to the fulcrum, there exists the following formula when- ever the balance is perfect : W = d P D XI, 64] LITERAL EQUATIONS AND FORMULAS 101 This is called the general law of the lever. By clearing the equation of fractions, it may be written in the form WD=pd. Translated into words, this last relation means that the weight times the weight arm equals the power times the power arm. It is in this form that the law is usually remembered by engineers. EXERCISES ON THE LEVER 1. If the fulcrum of a 5-foot crowbar is placed 1 foot from the end, what weight can be lifted by a man weighing 180 pounds ? [HINT. Here we have Fig. 19 with W = ?, w (or p) = 180 pounds, D=lfoot, d =4 feet.] 2. The figure represents a simple form of pump. If the pump handle AF is 16 inches long, while the piston-arm FC is 3 inches long, what will be the upward pull at C when there is a 9-pound downward push at A? 3. A certain lever, after being balanced, has r pounds added to the weight W. Determine (in terms of W, p, D, d, and r) how much the power, p, must be increased to keep the balance perfect. 64. Gear Wheel Law. Whenever a gear wheel having T teeth revolves at the rate of N revolutions per minute and turns another similar wheel having t teeth at the rate of n revolutions per minute, there exists at all times dur- ing the motion the formula T = n. Fro. 22. t N FIG. 21. 102 SECOND COURSE IN ALGEBRA [XI, 64 This is the gear wheel law. Clearing this equation of frac- tions, it becomes TN=tn. Translated into words, this last relation means that the number of teeth in one wheel multiplied by its rate of turning is equal to the number of teeth in the other wheel multiplied by its rate of turning. EXERCISES ON GEAR WHEELS 1. If in Fig. 22 the small wheel has 30 teeth and is making 96 revolutions per minute, how many teeth must the large wheel have in order to revolve 16 times per minute? 2. In order that the large wheel revolve three fourths as fast as the small one, how must the wheels be made ? 3. If the large wheel in 64 be made larger by the addi- tion of r teeth to its run, determine the amount by which the speed of the smaller wheel will be thereby increased. [HINT. Let x represent the unknown amount and find a formula for x in terms of T, t, N, n, and r.] 65. Other Useful Formulas. In addition to the formulas already mentioned, the following from plane and solid geometry and from elementary physics are often used. 1. The area A of an equilateral triangle of side a (Fig. 23) is _V3 2 FIG. 23. : ~^~ fl ' where A/3 = 1.732 approximately. 2. The area A of a regular hexagon of side a (Fig. 24) is 3\/3 fl 2 . 2 FIG. 24. XI, 65] LITERAL EQUATIONS AND FORMULAS 103 3. The area A of any triangle in terms of its three sides a, 6, and c (Fig. 25) is A = Vs(s-a)(s-b)(s-c), where s = a FIG. 25. 4. The area A of the sector of a circle when the intercepted arc is a and the radius is r (Fig. 26), is FIG. 26. 5. The length of the diagonal d of a rectangular block whose dimensions b\ are a, 6, and c (Fig. 27) is 6. The volume V of a circular cylinder of altitude h and radius of base r (Fig. 28) is F=irr 2 /i. 7. The volume V of a circular cone of altitude h and radius of base r (see Fig. 29) is FIG. 28. V=Trr*h. 8. The volume V of a pyramid of altitude h and base B (Fig. 30) is 9. The volume V of a spheri- cal segment (or slice of a sphere between two parallel cutting planes), where h is the altitude, and a and b the radii of the two bases (Fig. 31) is Fio. 29. FIG. 30. FIG. 31. 104 SECOND COURSE IN ALGEBRA [XI, 65 10. The surface S of the zone (or portion of the surface of a sphere lying between two parallel cutting planes), where h is the distance between the cutting planes and r the radius of the sphere (Fig. 31), is S = 2*117/1. 11. The length of belt I required to go around two wheels whose diameters are D and d and whose centers are at the distance a apart is FIG. 32. (a) In case the belt does not cross itself, (6) In case the belt crosses itself once, / = 2 12. The force F, measured in pounds, with which a body weighing W pounds pulls outward (centrifugal force) when traveling with a velocity of v feet per second in a circle of radius r (Fig. 33) is determined by the formula FIG. 33. r 32 13. The pressure P exerted by a letter press (Fig. 34) is determined by the formula D 2irrF where F is the value of the force applied at FIG. 34. XI, 65] LITERAL EQUATIONS AND FORMULAS 105 the wheel, r is the radius of the wheel, and h is the distance from one thread of the screw to the next one. 14. The weight W which can be raised by means of a toothed wheel and screw such as indicated in Fig. 35 is deter- mined by the formula where P represents the pressure applied at the handle and where R, r, d, and I are the dimensions indicated in the figure. Other important formulas from elementary geometry and from physics may be found in Chapter XVII. EXERCISES 1. Find by Formula 3 of 65 the area of the triangle whose three sides are respectively 5 inches, 5 inches, and 8 inches long. 2. Show that in the case of an equilateral triangle of side a the expression for A in Formula 3, 65, reduces, as it should, to that for A in Formula 1, 65. 3. Show that in case the two wheels in Fig. 32 have the same diameter D, the formula for the length of belt re- duces in case (a) to the simple form l=TrD+2 a, and in case (6) to Z = 2VD 2 +a 2 +7rD. 4. How much leather (surface measure) will it take for a belt 6 inches wide to connect two pulleys whose diameters are 5 feet and 1 foot, respectively, the distance between centers being 10 feet ? [HINT. Viol = 10.2, approximately.] 106 SECOND COURSE IN ALGEBRA [XI, 65 5. A pail of water weighing 5 pounds is swung round at arm's length at a speed of 10 feet per second. If the length of the arm is 2 feet, find (a) the pull at the shoulder when the pail is at the uppermost point of its course, (b) when at the lowest point of its course. Also, find the least velocity which the pail can have without the water dropping out at the top point of the course. 6. What pressure is exerted by a letter press in case the force applied at the wheel is 10 pounds, the diameter of the wheel is 1^ feet, and the threads of the screw are % inch apart ? 7. In the device shown in Fig. 35, show that if the distance d between two adjacent threads be halved and the number of teeth on the wheel be correspondingly doubled to fit the new gear, other parts remaining the same, the weight W that can be raised with a given pressure P will be doubled. 8. The volume V of the frustum of a cone or pyramid made by a plane parallel to the base is given by the formula where B and b denote the areas of the lower and upper bases, and h denotes the altitude. FIG. 36. Determine the volume of the frustum of a circular cone the radii of whose bases are 4 inches and 3 inches, the altitude being 5 inches. XI, 65] LITERAL EQUATIONS AND FORMULAS 107 Show that in case the upper and lower bases are equilateral triangles whose sides are a and b respectively, the formula becomes [HINT. See Formula 1, 65.] 9. Show by means of Formula 3, p. 103, that if the three sides of any triangle be extended to twice their original length, the area will become quadrupled. 10. If, in the first of the two cases represented in Fig. 32, the diameter of each wheel be increased by the same amount, say a, show that the length of belt will thereby become in- creased by the amount ira. 11. If, in the second of the cases represented in Fig. 32, the diameter of the large wheel be increased by any given amount and at the same time the diameter of the small wheel be diminished by the same amount, show that pre- cisely the same length of belt as before will fit over them tightly. 12. A body is moving along a circle with a velocity of 3' feet per second. Show that in order for the centrifugal force (12, p. 104) which it is exerting to be doubled, the velocity must be increased by about 1J feet per second. By how much would the velocity have to be diminished in order that the centrifugal force become halved ? 13. By means of Formulas 9 and 10 ( 65) obtain the formulas for the area and volume of a whole sphere. Com- pare with Ex. 14, (e), (/), p. 6. CHAPTER XII GENERAL PROPERTIES OF QUADRATIC EQUATIONS 66. The Classification of Numbers. A real number is one whose expression does not require the square root of a negative quantity, while an imaginary number is one whose expression does require such a square root. (See 59, p. 93.) Thus 1, 3, -7, , f, -f , \/2, 1+ \/3 are real numbers, while -S, V- i, 2 + A/-3, are imaginary numbers. In case a real number can be expressed in the particular form ^ where p and q are integers (positive or negative) it is q called a rational number. The number zero is also included among the rational numbers. (See 42, p. 56.) Thus l, f , f , 5, 73, 10, ^J, are rational numbers. In case a real number cannot be expressed in the particu- lar form just mentioned, it is called an irrational number. Thus V2, V3, Vf , ^2, -vXJ", V^ 1 + A/6, are irrational num- bers. Imaginary numbers are either pure imaginaries, such as V^3, or complex numbers ( 59, p. 93), such as 1 + V 3. 108 XII, 67] GENERAL PROPERTIES OF QUADRATICS 109 These divisions and subdivisions may be summarized into a table as below : n , f Rational Real{ T .. , r f ., Irrational Numbers of Algebra T . . j Pure Imaginanes Imaginary { 1 , y r , \ Complex Numbers NOTE. The rational numbers are themselves subdivided into three sub-classes : the single number zero, the integers (positive or negative), and the rational fractions. The latter are those rational numbers which, like f , cannot be expressed as integers. All the numbers used in arithmetic are positive real numbers. The negative real numbers together with the imaginary numbers owe their existence to algebra. 67. Determining the Character of the Roots of a Quad- ratic. It is often desirable to determine the character of the roots of a given quadratic, that is, whether the roots are real or imaginary ; and if real, whether they are rational or irrational, etc. Thus the roots of 2 z 2 -7 x+l = are (by 56, p. 87), 7=t=V ^. 4 Since 41 is positive, these roots are real numbers ( 66). Since 41 is not a perfect square, the roots are irrational ( 66). Since VH is added to 7 in the one root and subtracted from 7 in the other root, the roots are unequal. Thus the character of the roots in this case is described by saying that they are real, irrational, and unequal. There is, however, a much shorter method than the one illustrated above for determining the character of the roots of a given quadratic. Thus we know ( 56) that the two roots of any quadratic, namely, any equation of the form ax 2 -\-bx-\-c = Q, _ and _ 2a 2a 110 SECOND COURSE IN ALGEBRA [XII, 67 An examination of the form of these expressions gives at once the following rule. RULE. For any given quadratic, ax*+bx+c = Q, in which the coefficients a, b, c are real numbers, the two roots will be (1) Real and unequal if b 2 4 ac is positive. (2) Real and equal if b 2 4 ac = 0. (Both roots then reduce to 6/2 a.) (3) Imaginary ifb 2 4acis negative. Moreover, if the coefficients a, b } c are rational numbers, the two roots will be (4) Rational, if 6 2 4 ac is a perfect square; irrational if b 2 4ac is not a perfect square. Because of the manner in which the character of the roots thus comes to depend upon the value of b 2 4 ac, this expres- sion is called the discriminant of the given quadratic. EXAMPLE 1. Determine the character of the roots of 2z 2 -3z-9 = 0. SOLUTION. Here a =2, b =3, c= 9. Hence the value of the discriminant, or 6 2 -4ac, is ( -3) 2 -4(2)( -9) =9+72=81 =9 2 . Hence, by (1) and (4) of the rule, the roots are real, unequal, and rational. EXAMPLE 2. Determine the character of the roots of SOLUTION. Here a =3, b=2, c = l. Hence 6 2 4 ac=4 12= -8. Hence, by (3) of the rule, the two roots must be imaginary. EXAMPLE 3. Determine the character of the roots of 4z 2 -20z+25 = 0. SOLUTION, a =4, 6= -20,c=25. Hence 6 2 - 4 ac= 400 -400=0. Therefore, by (2) of the rule, the roots are real and equal. The common value which the two roots have may be found if desired by actually solving the equation. It turns out to be f . XII, 68] GENERAL PROPERTIES OF QUADRATICS 111 EXERCISES Determine (without solving) the character of the roots of each of the following equations. 1. 2z 2 -3z+l=0. 7. x*+x=-l. 8. 9. 10. 11. 12. *68. Character of Roots Considered Geometrically. We have seen in 57 that whenever a quadratic has its two roots real and unequal, its graph will cut the #-axis in two distinct points. On the other hand, if the roots are imaginary, the graph of the equation will not cut the a>axis at all ( 60). Suppose now that we have a quadratic whose two roots are equal to each other, for example (1) 4z 2 -12z+9 = 0. Here the discriminant is (-12) 2 - 4(4) (9) = 144 -144 = 0, so that the roots must be equal ( 67). If we now proceed to draw the graph of (1) in the usual way by placing 2/ = 4z 2 12 z+9, then forming a table of x, y values, etc., it appears that the . graph corresponding to (1) just touches the x-axis instead of cutting through it. This, in fact, is what we should expect, since the equality of the roots virtually means that there is but one root, and this can be possible only when the graph merely touches (is tangent to) the x-axis. FIG. 37. 112 SECOND COURSE IN ALGEBRA [XII, 68 In order to illustrate in one single diagram all the facts mentioned thus far about the graph, it is instructive to take such an equation as (2) z 2 -2z+c = 0, and let c take different values, thus obtaining various quad- ratics. For example, if we choose c 2, the quadratic equation becomes and if we draw the graph of this, we find that it cuts through the z-axis at two points, thus in- dicating that its solutions are real and unequal. Likewise, we find a similar result when we let c= 1, and when c = 0, though the various graphs are themselves different. But if we choose c = l and pro- ceed as before, the graph of the corresponding quadratic no longer cuts through the o>axis, but merely touches it, thus indicating real and equal roots. Finally, for such values of c as 2, 3, or 4, the quadratics (2) come to have graphs which do not cut the x-axis, thus indicating that they have imaginary roots. The effect of changing c is thus merely to slide one and the same curve vertically up and down the coordinate paper. The figure shows the positions of the curve corresponding to c = 2, c = 1 , and c = 4. The pupil is advised to draw in for himself the positions of the curve for c = 1, c = 0, c = 2, and c = 3. FIG. 38. XII, 69] GENERAL PROPERTIES OF QUADRATICS 113 69. The Sum and Product of the Roots. We have seen that every quadratic is an equation of the form (1) and we have also seen ( 56) that the roots, or solutions, of this equation are and -6-V& 2 -4 oc 2a It is now to be observed that if we add these solutions together the radical cancels and we obtain the simple result -2b_ b 2 a a Again, if we multiply the two solutions together, we obtain a -final result which is very simple in form. Thus _ (-fr) 2 - (Vb 2 -4 ac) 2 _6 2 - (6 2 -4 ac) _4 ac_ c 4 a 2 4 a 2 ~4~a?~a' These results may be summarized in the following rule. RULE. In the general quadratic equation ax 2 -\-bx-\-c = Q, (1) The sum of the two solutions is b/a. (2) The product of the two solutions is c/a. EXAMPLE. Find the sum and the product of the solutions of the equation 3x 2 2 z+6 = 0. SOLUTION. Here a = 3, 6= 2, c = 6. r> Hence the sum of the solutions is , or f, and the product of the solutions is % = 2. l 114 SECOND COURSE IN ALGEBRA [XII, 69 EXERCISES State (by inspection) the sum and the product of the solu- tions of each of the following equations. Check your answer in Exs. 1, 2, 3, 4 by actually solving the equations and de- termining the sum and the product of the solutions. 1. 2z 2 +5z-7 = 0. 5. 6z 2 +7z = 42. 2. 3x*-7x+2 = Q. 6. 3. 5z 2 -2z = 16. 7. 4. 3z = 200-z 2 . 8. x 2 +px = q. 9. Show that in the quadratic x 2 -\-rnx +n = the sum of the roots is m and the product of the roots is n. This general result is important and may be stated in words as follows : // in a quadratic the coefficient of x 2 is 1, the sum of the solu- tions will be the coefficient of x with its sign changed, while the product of the solutions will be the remaining term. Explain and illustrate by means of the equation z 2 10 z+12 = 0. 10. Apply the result stated in Ex. 9 to determine the sum and the product of the solutions of the following equations: (a) z 2 -5z+7 = 0. (e) x 2 -(a+b)x+ab = Q. (b) z 2 -4z = 10. (/) 2z 2 +3z-4 = 0. (c) x 2 %x = 2. [HINT. First divide through (d) z 2 -V2z+V3 = 0. by 2.] 70. Formation of Quadratics Having Given Solutions. EXAMPLE 1. Form the quadratic -whose solutions are 1 and -i-. SOLUTION. If x = l, then x -1=0; if x = -J-, thenz+ = 0. Hence the equation (x-l)(z+) =0, or z 2 -^ x-=Q, will be satisfied when either z = lorz= ^-( 52). The desired quadratic is therefore z 2 1- x J- =0, or 2 x 2 -x 1 =0. Ans. XII, 70] GENERAL PROPERTIES OF QUADRATICS 115 Similarly, if the given solutions are any numbers, as a and 6, the equation having these as solutions is obtained by subtracting a from x and 6 from x, then multiplying the two factors thus obtained together and placing the result equal to ; that is, the desired equation is (x a)(x b) = EXERCISES Form the quadratics whose roots are 1. 2, 3. 8. 2. -2, -3. 9. 3m, -5m. 3. 4, f 10. 2a-6, 2 a+b. 4. 12, -5. 11. 3+V2, 3-V2. 5. i, -. 12. 2-V5, 2+ V5. 6. V2, V3. 13. 2V3. [HINT. Write answer in the 14. \ (3 V) . form * 2 -(V2 + V3)z + V6 =0.] lg i(_iV2). 7. V8, -\/2. 16. m(22V5). 17. Show that if in any quadratic, as az 2 +6z+c = 0, one root is double the other, then the relation 2 b 2 = 9 ac must exist among the coefficients, a, 6, and c. [HINT. Let r be one root. Then, by what the problem assumes, the other root is 2 r. Now form the quadratic having r and 2 r as roots, and examine its coefficients.] 18. Show that if in any quadratic, as ax 2 -\-bx-\- c = 0, one root is three times the other, then we must have 4 ac = 6 2 9 a 2 . 19. Find the relation which must hold between a, b, and c in order that the roots of the quadratic ax 2 +bx+c = Q may be to each other in the ratio 2:3. [HINT. Use the Rule of 69.] CHAPTER XIII IMAGINARY NUMBERS 71. Preliminary Statement. Just as V2 means the num- ber whose square is 2 ; that is, (V2) 2 = 2, so V 2 means the number whose square is 2; that is, (V 2) 2 = 2. The latter case differs essentially, however, from the former, be- cause we cannot conceive of any number, positive or nega- tive, whose square gives a negative result, like 2. In fact, this would seem to contradict the law of signs [ 2 (d)], according to which the square of either a positive or negative quantity is always positive. The explanation is that V 2 belongs to an altogether new class of numbers, called im- aginary numbers, so that we cannot expect to think of them in the way just mentioned ; namely, as though they were real numbers. Imaginary numbers first came to our notice in 58, where we found that the very simple quadratic x 2 = 1 has the two imaginary roots x = V 1 and x V 1. Imaginary numbers have certain definite properties. They may be added, subtracted, multiplied, divided, etc., in ways which will be explained in the present chapter. 72. The Imaginary Unit. Every pure imaginary number ( 66) may be expressed as the product of a certain real num- ber multiplied by V 1. For this reason, V 1 is called 116 XIII, 72] IMAGINARY NUMBERS 117 the imaginary unit, and for convenience is represented by the letter i. Thus V^ V27( - 1) = V27 V^I = A pure imaginary number when thus written as a real number multiplied by i is said to be expressed in terms of i. EXERCISES Write each of the following expressions in terms of i. 5. V^ 1. V-IQ. 2. V-25. 3. V 18. SOLUTION OF Ex. 9. 10. v^J". 11. V^T. 12. 13. - 14. SOLUTION OF Ex. 14. ll^ J?7 V^ 4 ' 4 i. Ans. 16. = 118 SECOND COURSE IN ALGEBRA [XIII, 73 73. Addition and Subtraction of Pure Imaginary Num- bers. EXAMPLE. Add V^9 and V-25 and express the result in terms of i. SOLUTION. EXERCISES Simplify each of the following expressions, obtaining the result in terms of i. -81 + -64+-100. -16a 2 + ~-100a 2 - V-81a 2 . 9. V - 25 zy + v - 225 x*y 2 + V - 625 x 2 y 2 - 10. V-32m 2 -hV-20m 2 -V-27m 2 . 11. V-24 /i 2 / 12. V 74. Simplification of Complex Numbers. EXAMPLE. Simplify 2+V-9. SOLUTION. 24-^^9=2+3^. ^Ins. In this exercise we have a real number, 2, to which is added the pure imaginary number, V 9, thus giving a complex number ( 59). EXERCISES Simplify each of the following complex numbers. 1. 2. 6-2V 3- XIII, 75] IMAGINARY NUMBERS 119 7. Show that the quadratic equation x 2 4z+13 = has as its solutions x = 2 3 i. [HINT. Solve as in 56.] 8. Find the solutions of the equation 4(2 x 5) =x 2 . 75. Multiplication of Imaginary Numbers. EXAMPLE 1. Multiply V^3 by V^4. SOLUTION. V^3 V^4 = V$i Vi = Vl2- i*= Vl2(V-l)2 = -Vl2=-2V3. Ans. Note that the process of multiplication consists in first expressing each number in terms of the unit i, then making use of the fact that (since i = V 1) we may write 1 for i 2 . EXAMPLE 2. Multiply 2+^3 by 4-~^3. SOLUTION. (2 + V^3) (4 - V~^3) = (2 + Vs i) (4 - VJTt). 8+4 V3i 8+2 V3i-3(-l)= 11+2 V3i. Ans. EXERCISES Find each of the following indicated products. 9. 10. 11. 12. 13. 14. 15. a 16. 120 SECOND COURSE IN ALGEBRA [XIII, 76 76. Division of Imaginary Numbers. EXAMPLE 1. Find the value of V^27-r- V^3. so -- --- -* EXAMPLE 2. Find the value of 6 -^V^3. SOLUTION. _^=-^_ =_^_ = ^_ = ^M= _ 2 Vs i. ^ V V 2 -V EXAMPLE 3. Find the value of 3 -5- (1 + V^~2) . SOLUTION. 2 --- L - 3(1 -V2i) . ^ l-2i 2 1+2 Note that in Example 2 we rationalized the denominator of - % by multiplying both numerator and denominator -Va by the value A/3. Likewise, in Example 3 we rationalized Q the denominator of - - by multiplying both numerator and denominator by 1 V2t. In general, any fraction hav- ing a denominator of the form a-\- v b may be rationalized by multiplying both numerator and denominator by a V 6, which is called the conjugate imaginary of a+ V 6. EXERCISES Find the following indicated quotients, rationalizing the denominator in each. 4. 2-i-V^6. 5. V% -r- V^2. IMAGINARY NUMBERS 121 11. *77. Geometric Representation of Complex Numbers. All real numbers (positive or negative) can be represented as points on a line, as explained in 1. Similarly, all complex numbers may be repre- sented as points in a plane, and it is convenient for many purposes to regard them in this way. Thus, the complex number 5 +4 i may be looked upon as lying at the point (5, 4), that is at the point whose abscissa is 5 and whose ordinate is 4. (See 28.) Likewise, the complex number -2+3 i lies at (-2, 3) ; the number 3 2 i lies at (3, 2) ; and, in general, the number x+yi, where x and y are any (real) numbers, lies at the point (x, y). Whenever a plane is used in this way to represent com- plex numbers, it is called a complex plane. The re-axis is called the axis . Y & 1 e f X4 y 8 3 1 5- -4i -5 i \ . Q - / / / ,/ < / / \ I/ Ax S f r als ^ Sw X ^ 3-2i FIG. 39. of reals and the ?/-axis is called the axis of pure imaginaries (because the pure imaginary part of each complex number is to be measured parallel to it). The straight line joining any point x+iy to the origin is called the radius vector of that point, or number. Observe that from this point of view all real numbers become represented by the points on the axis of reals, while all pure imaginary numbers become represented by the points on the axis of pure imaginaries, the other points of the plane being then taken up by what are properly the complex numbers. CHAPTER XIV SIMULTANEOUS QUADRATIC EQUATIONS I. ONE EQUATION LINEAR AND THE OTHER QUADRATIC 78. Graphical Solution. In Chapter VI we have seen how we may determine graphically the solution of two linear equations e&ch of which contains the two unknown letters x, y. The method consists in first drawing the graph of each equation, then observing the x and the y of the point where the two graphs intersect (cut each other). The particular pair of values (x, y) thus obtained constitutes the solution. We often meet with simultaneous equations which are not both linear, as for example the two equations (1) x-y = l, (2) z 2 +2/ 2 = 25. Here the first equation is linear ( 26) but the second is not. In order to solve them, that is in order to find that pair (or pairs) of values of x and y which satisfy both equations, we may proceed graphically as follows. The graph of (1) is found (as in 29) to be the straight line shown in Fig. 40. To draw the graph of (2), we first solve this equation for y. Thus y 2 = 25 - x 2 . Therefore (3) 2/= 122 XIV, 78] SIMULTANEOUS QUADRATICS 123 We now work out a table of pairs values of x and y that will satisfy (3), that is we give x various values .in (3) and solve for the corresponding y value. (Compare 60.) The result is shown below. Observe that to the value x = there cor- Whenz = + 1 +2 +3 +4 +5 then y = V25 5 W24 4.8 V21 4.5 VlQ 4 V9 3 Vo respond the wo values T/ = 5 ; similarly to x = 1 correspond the two values y= 4.8 (approximately), etc. Moreover, if we assign to x the negative value, x=l, we find in the same way that corresponding to it y has the two values y = 4.8. Likewise, for x = 2 we find y = 4.5, etc., the values of y for any negative value of x being the same each time as for the corresponding positive value of x. FIG. 40. Plotting all the points (x, y) thus obtained and drawing the smooth curve through them, the graph is as shown in Fig. 40. This curve is a circle, as appears more and more clearly as we plot more and more points (x, y) belonging to the equation (3). 124 SECOND COURSE IN ALGEBRA [XIV, 78 NOTE. The form of (3) shows that there can be no points in the graph having x values greater than 5, for as soon as x exceeds 5 the expression 25 x 2 becomes negative and hence V25 x 2 becomes imaginary, and there is no point that we can plot corresponding to such a result. Similarly, it appears from (3) that x cannot take values less than 5. Thus the graph can contain no points lying outside the circle already drawn. Returning now to the problem of solving (1) and (2), we know (31) that wherever the one graph cuts the other we shall have a point whose x and y form a solution of (1) and (2), that is we shall have a pair of values (x, y) that will satisfy both equations at once. From the figure it appears that there are in the present case two such points, namely (z = 4, 2/ = 3) and (x=-3, y=-4). Equations (1) and (2) therefore have the two solutions (z = 4, 2/ = 3) and (x= 3, y 4). Ans. CHECK. For the solution (x = 4, y=3) we have x y=4 3 = 1, and x 2 +y z = 16 +9 = 25, as required. For the solution (x = 3, y = 4) we have x -y = -3 - ( -4) = 1, and z 2 +?/ 2 = 9 +16 = 25, as required. The following are other examples of the graphical study of non-linear simultaneous equations. EXAMPLE 1. Solve the system (4) (5) SOLUTION. The straight line representing the graph of (4) is drawn readily. To obtain the graph of (5), we have 9 y 2 = 100 4 x 2 . Hence y 2 =(100 -4 z 2 ) =|(25 -x 2 ), and therefore (6) XIV, 78] SIMULTANEOUS QUADRATICS Corresponding to (6), we find the following table : 125 When z= + 1 +2 +3 +4 +5 greater than +5 then y= ifV/25 (5) 3.3 3^ / 24 1(4.8) 3.2 |V / 21 (4.5) 3.0 |X / 16 (4) 2.6 |V^ (3) 2 1^0 imaginary imaginary imaginary For any negative value of x, the y- values are the same as for the corresponding positive value of x. (See the solution of (1 ) and (2) .) The graph thus obtained for (6), or (5), is an oval shaped curve. It belongs to a general class of curves called ellipses. \ FIG. 41. The two graphs are seen to intersect at the points (x = 4, y =2) and (x =-5,y = 0). Therefore the desired solutions of (4) and (5) are (x=4, y=2) and (x= 5, 2/=0). Ans. EXAMPLE 2. Solve the system (7) 2x-y=-2, (8) Z2/ = 4. SOLUTION. The graph of (7) is the straight line shown in Fig. 42. To obtain the graph of (8), we have (9) y=, 126 SECOND COURSE IN ALGEBRA [XIV, 78 from which we obtain the following table : When x = 8 7 6 5 T 4 3 t 2 1 * i ST i T 16 i 5" then y = i * t 1 2 4 8 12 20 This table concerns only positive values of x, but it appears from (9) that for any negative value of x the appropriate y value is the negative of that for the corresponding positive value of x. The graph thus obtained for (9), or (8), consists of two open curves, each indefinitely long, situated as in Fig. 42. These taken together (that is, regarded as one curve) form what is known as a hyperbola (pronounced hy-per'bo-la). The part (branch) lying to the right of the i/-axis corresponds to the above table, while the other branch corresponds to the negative x values. /\ FIG. 42. The two graphs are seen to intersect in the points (a? = l, y=4) and (x= -2, y= -2). Therefore the desired solutions of (7) and (8) are (x = l, j/=4) and (x = 2, y = 2). Ans. NOTE. Ellipses and hyperbolas are extensively considered in the branch of mathematics called analytic geometry a study which may be pursued after a course in plane geometry and a course in algebra equivalent to that in this book. It is usually taught during the first year of college mathematics. XIV, 78] SIMULTANEOUS QUADRATICS 127 EXAMPLE 3. Consider graphically the system (10) (11) SOLUTION. The graph of (10) is found in the usual manner, and is represented by the straight line in Fig. 43. The graph of (11) has already been worked out (see discussion of (2)), being a circle of radius 5 with center at the origin. The peculiarity to be es- pecially observed here is that these two graphs do not intersect. This means (as it naturally must) that there are no real solutions to the system (10) and (11) ; in other words, the only possible solu- tions are imaginary. FIG. 43. Likewise, whenever any two graphs fail to intersect, we may be assured at once that the only solutions their equations can have are imaginary. The system (10) and (11) and other such systems will be considered further in the next article. EXERCISES Draw the graphs for the following systems and use your result to determine the solutions whenever they are real. 4 z 2 +9 i/ 2 = 100. [HINT. See Example 1, 78. 128 SECOND COURSE IN ALGEBRA [XIV, 78 3. J*-20=-l, 6 \y = x*. [HINT. See Example 2, 78.] O. 5~ q , n Jj o u u. 79. Solution by Elimination. Let us consider again the system (1) and (2) of 78. (1) x-y=l, Instead of solving this system graphically, we may solve it by elimination, that is by the process employed with two linear equations in 33. Thus we have from (1) (3) y = x-l. Substituting this value of y in (2), thus eliminating y from (2), we obtain or, dividing through by 2, (4) z 2 -z-12 = 0. Solving (4) by formula ( 56), gives as the two roots 1+7 = 4, and (-l)-V(-l)-4(l)(-12)_l-Vl+48 = l-7^_q 2 22 When x has the first of these values, namely 4, we see from (3) that y must have the value ?/ = 4 1, or 3. XIV, 79] SIMULTANEOUS QUADRATICS 129 Similarly, when x takes on its other value, namely 3, we see that y has the value y= 3 1, or 4. The solutions of the system (1) and (2) are, therefore, (^ = 4, ^ = 3) and (x= -3, y= -4). Ans. Observe that these results agree with those obtained graphically for (1) and (2) in 78. Further applications of this method are made in the examples that follow. EXAMPLE 1. Solve the system (5) 2x-\-y = 4, (6) z 2 +2/ 2 = 12. SOLUTION. From (5), (7) y=-2x. Substituting this expression for y in (6), we find or (8) 5z 2 -16z+4 = 0. The two roots of (8), as determined by formula ( 56), are = - ( - 16) =fc V ( - 16) 2 -4(5) (4) = 16 V256 -80 = 16 =*= vT76 2(5) 10 10 _16^4VII^8=>=2Vn 10 5 The first of these values, namely x= (8+2vTF)/5, when sub- stituted in (7), gives as its corresponding value of y, j,.4_ia+4vH_4-|V] The second value, namely x = (8 2VTT)/5, when substituted in (7), gives as its corresponding value of y, 16-4VT1 4+4VTT , 5 Hence the desired solutions are x 8+2VTT 8-2VTI 5 and 4-4VH 5 ' 4+4^11 y 5 ' y " 5 130 SECOND COURSE IN ALGEBRA [XIV, 79 To obtain the approximate values of the numbers thus obtained we have VTI =3.31662 (tables), and hence the above solutions re- duce to the forms x= 2.9266, (x =0.2734, y=- 1.8533, \y = 3.4533. These are the solutions of the system (5), (6), correct to four places of decimals, which is sufficient for ordinary work. NOTE. It may be remarked that, while the graphical method of solution described in 78 is very instructive in showing how many solutions a given system will have, and what their geometric significance is, it does not usually afford a ready means of deter- mining the exact values of the solutions. This is illustrated in the example just solved, where, if the graphs of (5) and (6) be drawn, they will intersect at points whose x and y contain the surd Vll (as the above solution shows), and it would be difficult to measure off any such values accurately on the scale of the diagram. In fact, it would be practically impossible to determine graphically the solutions of (5) and (6) correct to three or even two places of decimals, yet this degree of approximation was easily obtained above by the method of elimination. For such reasons, it is preferable, whenever one is concerned only with finding the values of solutions, to proceed from the beginning by the method of elimination. EXAMPLE 2. Solve the system (9) (10) SOLUTION. From (9), (11) ^ = 10-z. Substituting this expression in (10), x 2 + (100 -20 z +z 2 ) = 25, or (12) 2 x 2 - 20x+75=0. Solving (12) by formula, we find its solutions to be and Since these z-values contain the square root of the negative quantity -200, they are imaginary (58). The ^-values are also XIV, 79] SIMULTANEOUS QUADRATICS 131 imaginary, as appears by substituting the z-values just found into (9), which gives the results The desired solutions of the systems (9), (10) are therefore x=- 10 -=- and 10 10 y = 10 This result should now be contrasted with what we saw in Example 3 of 78 regarding this same system (9) and (10). There we found graphically that the solutions must be imaginary be- cause the graphs failed to intersect, but we could not find the actual imaginary numbers which form the solutions. This we have now been able to do, however, by the method of elimination. The method of elimination enables one to determine imaginary as well as real solutions in all similar cases. EXERCISES Solve each of the following systems by the method of elimi- nation, and, in case surds are present, find each solution correct to two places of decimals by use of the tables. f x+y=-l, 1. 2. f x-2y=-l, 6. 7. f x-2y = 2, O i <> * A 9 Of 8 i 9. x-2y = 3. *-2y*=-S, x-2y=-3. f 3o; 2 -a-5 = 15' xy=-Q. 2x+y = 2, xy=-Q. f 10. I x-\-y x y 6' 132 SECOND COURSE IN ALGEBRA [XIV, 80 II. NEITHER EQUATION LINEAR *80. Two Quadratic Equations. In each of the systems considered in 78, 79 one of the two given equations was linear. However, the same methods of solving may often be employed in case neither equation is linear. In such cases four solutions may be present instead of two. EXAMPLE 1. Solve the system (1) (2) z 2 -? 2 =15. SOLUTION. Here only x z and y 2 appear and we begin by finding their values. Thus, multiplying (2) through by 16 and adding the result to (1), we eliminate y 2 and find that 25 z 2 =400, or (3) z 2 = 16. Substituting this value of x in (2), we find (4) From (3) and (4) we now obtain (5) x= and y ==*=!. Forming all the pairs of values x, y that can come from (5), we obtain as our desired solutions (x = 4, y = 1) ; and (x= 4, y= 1). Ans. CHECK. Each of these pairs of values of x and y is immediately seen to satisfy both (1) and (2). Let the pupil thus check each pair. When considered graphically, equation (1) gives rise to an ellipse (compare 78, Ex. 1), while (2) gives a hyperbola situated as shown in the diagram. These two curves intersect in four points which correspond to the four solutions just obtained. FIG. 44. XIV, 80] SIMULTANEOUS QUADRATICS 133 EXAMPLE 2. Solve the system -l2. (7) (8) xy SOLUTION. Here we cannot proceed as in Example 1 because we cannot find readily the values of x 2 and y*. But if we multiply (8) by 2 and add the result to (7), we obtain (9) x 2 +2 xy +y* = l. Taking the square root of both members of (9) gives (10) Similarly, multiplying (8) by 2 and subtracting the result from (7), x 2 - 2 xy +?/ 2 = 49, and hence (11) X -y=7. Taking account of the two choices of sign in (10) and (11), we see that they give rise to the four simple (linear) systems (a) x+y = l, x-y=7; (6) x+y=-l, x-y=7; (c) x+y = l, x-y=-7; (d) x+y=-l,x-y=-7', Thus we have replaced the original system (7) and (8) by the four simple systems (a), (6), (c), and (d), each of which may be immediately solved by elimination, as in 33, 34. Since the solu- tions of (a), (6), (c), (d) are respectively (x=4, y= -3), (x=3, y = 4), (x = -3, y =4), and (x = 4, y =3), we conclude that these are the desired solutions of (7) and (8). Ans. The graphical significance of these solutions is shown in Fig. 45, where the circle z 2 +?/ 2 = 25 is cut by the hyperbola xy= 12 in four points that correspond to the four solutions just found. CHECK. That these four solutions each satisfy (7) and (8) appears at once by trial. FIG. 45. 134 SECOND COURSE IN ALGEBRA [XIV, 80 While no general rule can be stated for solving two equations neither of which is linear, the following observation may be made. Unless the equations can be solved readily for x 2 and y 2 (as in Example 1), the system should first be carefully examined with a view to making such combinations of the given equations as will yield one or more new systems each of which can be solved (as in Example 2) by methods already familiar. All solutions obtained in this way should be checked, since false combinations of the x- and ^-values are frequently made by beginners when the work becomes at all complicated. EXERCISES Solve each of the following systems, and draw a diagram for each of the first three to show the geometric meaning of your solutions. = 25, f = 7. \ f [HINT FOR Ex. 4. First add, then subtract the two equations, thus showing that the given system is equivalent to two others, namely Now solve each of these systems as in 80.] z 2 +92/ 2 = 85, , ( x- y+2xy=-2Q. CARDAN (Girolamo Cardan, 1501-1576) An equation of the third degree (cubic equation) has three roots and these can be found only by methods which are more powerful than those employed in the study of quadratics. Cardan was the first to obtain and publish a method for solving such equations. His methods are also sufficient to solve any pair of simultaneous quadratics, but are too advanced to be given in this book. XIV, 81J SIMULTANEOUS QUADRATICS 135 *81. Systems Having Special Forms. The systems of equations considered in ^ 79, 80 illustrate the usual and more simple types such as one commonly meets in practice. It is possible, however, to solve more complicated systems provided they are of certain prescribed forms. We shall here consider only two such type forms. I. When one (or both) of the given equations is of the form where the coefficients o, 6, c are such that the expression ax 2 +bxy+cy 2 can be factored into two linear factors. EXAMPLE. Solve the system (1) x*+2x-y = 7, (2) x 2 -xy-2y*=0. SOLUTION. Here we see that (2) is of the form mentioned above, since x 2 xy2 y z can be factored (as in 12 (e)} into (x 2 y)(x+y}. (2) may thus be written in the form (3) (x-2y)(x+y)=Q. It follows ( 52) that either x-2y=Q, or x+y=Q. Hence the system (1), (2) may be replaced by the two following systems : and fz 2 +2o;-2/ = 7, I x+y=0. Each of these two systems may now be solved as in 80, and we thus find that the solutions of the first system are (x = 2, y = l) and ( x =-%,y= J) while the solutions of the second system are and _ y=i(3+V37). The desired solutions of (1) and (2) consist, therefore, of these four solutions just obtained. Ans. 136 SECOND COURSE IN ALGEBRA [XIV, 81 II. When both the given equations are of the form ax z +bxy+cy*=d, where a, 6, c, and d have any given values (0 included). EXAMPLE. Solve the system (4) x*-xy+y* = 3, (5) x*+2xy=5. SOLUTION. Let v stand for the ratio x/y ; that is, let us set y Then (6) x=vy. Substituting in (4), (7) 0V Substituting in (5), (8) Solving (7) for y*, (9) Solving (8) for y 2 , (10) ?/ 2 Equating the values of y z given by (9) and (10), v 2 +2v v 2 - Clearing of fractions, (11) 2v z -llv+5 = Solving (11) by formula (56), ^H=faVl21-40 4 44 Therefore v = 5, or v =%. Substituting 5 for v in (9), or (10), Hence = i or L Vf C V7 Substituting ^ for v in (9) or (10), y z =4. Hence y = +2, or -2. XIV, 82] SIMULTANEOUS QUADRATICS 137 The only values that y can have are, therefore, 1/V?, -1/V?, 2, and -2. Since x_=vy (see (6)), the value of x to go with y = l/V7 is x=5(l/v / 7)=5/V7. _Similarly, when y= -1/V? we have x =5( I/VT) = 5/V?. Likewise, when ?/=2 (in which case v = -^, as shown on p. 136) then z =^ 2 = 1, and when y 2, then z=i(-2) = -l. Therefore the only solutions which the system (4), (5) can have are(x=5/V7, y = l/V7) ; (x = -5/S/7, = -1/V?) ; (x = 1, y = 2) ; OT= 1, i/= 2); and it is easily seen by checking that each of these is a solution. Ans. 82. Conclusion. Every system of equations considered in this chapter has been such that we could solve it by finally solving one or more simple quadratic equations. We have examined only special types, however, and the student should not conclude that all pairs of simultaneous quadratics can be solved so simply. MISCELLANEOUS EXERCISES Solve the following simultaneous quadratics. The star (*) indicates that the exercise depends upon 81. 3. * y 2 -- = [HINT TO Ex. 4. First eliminate xy between the two equa- tions so as to obtain a linear equation between x and y.] z 2 +2z-2/ = 5, I 2 z 2 -3 x+2 y = S. \ , lx 2 -i/ 2 = 9. i [HINT TO Ex. 6. Divide the first equation by the second.] 138 SECOND COURSE IN ALGEBRA - [XIV, 82 10 xy-y=l2. : 15. APPLIED PROBLEMS In working the following problems, let x and y represent the two unknown quantities, then form two simultaneous equations and solve them. If surds occur, find their approximate values by the tables. 1. The sum of two numbers is 13 and the difference of their squares is 91. Find the numbers. 2. A piece of wire 48 inches long is bent into the form of a right triangle whose hypotenuse is 20 inches long. What are the lengths of the sides? (See Ex. 14 (d), p. 6.) 3. If it takes 26 rods of fence to inclose a rectangular garden containing ^ of an acre, what are the length and breadth ? 4. Figure 46 shows two circles just touching (tangent to) each other, the smaller one being outside the larger one. If their combined area is 15f square feet and the distance CC' between the two centers is 3 feet, find the radius of each circle. FIG. 46. 5. Work Ex. 4 in case the circles touch on the inside of the larger one, taking the shaded area to be 110 square feet and CC' to be 5 feet. FIG. 47. XIV, 82] SIMULTANEOUS QUADRATICS 139 6. Do positive integers exist differing by 5 and such that the difference of their squares is 45 ? If so, find them. 7. Answer the question in Ex. 6 in case the differ- ence of the squares is taken to be 10, other conditions re- maining the same. 8. The area of a certain triangle is 160 square feet, and its altitude is twice as long as its base. Find, correct to three decimal places, the base and the altitude. (See Ex. 14 (a), p. 6.) 9. The area of a rectangular lot is 2400 square feet, and the diagonal across it measures 100 feet. Find, correct to three decimal places, the length and breadth. 10. The mean proportional between two numbers is 2 and the sum of their squares is 10. What are the numbers ? (See Ex. 6, page 80.) 11. The dimensions of a rectangle are 5 feet by 2 feet. Find the amounts (correct to two decimal places) by which each dimension must be changed, and how, in order that both the area and the perimeter shall be doubled. 12. Two men working together can complete a piece of work in 6 days. If it would take one man 5 days longer than the other to do the work alone, in how many days can each do it alone? (Compare Ex. 9, p. 56.) 13. The fore wheel of a carriage makes 28 revolutions more than the rear wheel in going 560 yards, but if the circumference of each wheel be increased by 2 feet, the difference would be only 20 revolutions. What is the circumference of each wheel? 14. A sum of money on interest for one year at a certain rate brought $7.50 interest. If the rate had been 1% less and the principal $25 more, the interest would have been the same. Find the principal and the rate. 140 SECOND COURSE IN ALGEBRA [XIV, 82 15. A man traveled 30 miles. If his rate had been 5 miles more per hour, he could have made the journey in 1 hour less time. Find his time and rate. 16. Figure 48 shows a circle within which a diameter A B has been drawn. At a certain point P on AB the perpen- B dicular PG measures 4 inches, while at the point Q, which is 1 unit from P, the perpendicular QF measures 3 inches. How long is the diameter AB1 [HINT. Let x = AP,y = PB. Find x and y, using the fact stated in Ex. 6, page 80, then take their sum.] 17. Show that the formulas for the length I and the width w of the rectangle whose perimeter is a and whose area is b are 18. Find the formulas for the radii of two circles in order that the difference of the areas of the circles shall be d and the sum of their circumferences shall be s. and fi ..?=ii Ans . 4 7TS 4 TTS 19. Find two fractions whose sum is f , and whose differ- ence is equal to their product. 20. The diagonal and the longer side of a rectangle are together five times the shorter side, and the longer side exceeds the shorter side by 35 yards. What is the area of the rectangle? CHAPTER XV PROGRESSIONS I. ARITHMETIC PROGRESSION 83. Definition. An arithmetic progression is a sequence of numbers, called terms, each of which is derived from the preceding by adding to it a fixed amount, called the common difference. An arithmetic progression is denoted by the abbreviation A. P. Thus 1, 3, 5, 7, is an A. P. Each term is derived from the preceding by adding 2, which is therefore the common difference. The dots indicate that the sequence may be extended as far as one pleases. Thus the first term after 7 would be 9, the next one would be 11, etc. Again, 5, 1, 3, 7, 11, is an A. P. Here the common difference is 4. EXERCISES Determine which of the following are arithmetic progres- sions ; determine the common difference and the next two terms of each of the arithmetic progressions. 1. 3, 6, 9, 12, -. 4. 30, 25, 20, 15, 10, . 2. 3, 5, 8, 12, .... 6. -1, -1^, -2, -2, . 3. 6, 4, 2, 0, -2, -4, . 6. a, 2 a, 4 a, 5 a, . 7. a, a+3, a+6, a+9, . 8. a, a-\-d, a+2 d, a+3 d, a+4 d, . 9. x-4y, x-2y, x-y, . 10. 3x+3y, Qx+2y, 9 x+y, . [HINT. The common difference may always be determined by subtracting any term from the term immediately preceding.] 141 142 SECOND COURSE IN ALGEBRA [XV, 83 11. Write the first five, terms of the A. P. in which (a) The first term is 5 and the common difference is 2. (6) The first term is 3 and the common difference 1. (c) The first term is 3 a and the common difference is b. 84. The nth Term of an Arithmetic Progression. From the definition ( 83) it follows that every arithmetic progres- sion is of the form a, a+d, a+2 d, a+3 d, a+4 d, . Here a is the first term and d the common difference. Observe that the coefficient of d in any one term is 1 less than the number of that term. Thus 2 is the coefficient of d in the third term; 3 is the coefficient of d in the fourth term, etc. Therefore the coefficient of d in the nth term must be (n 1). Hence, if we let I stand for the nth term, we have the formula EXAMPLE. Find the llth term of the A. P. 1, 3, 5, 7, . SOLUTION. We have a = l, d =2, n = ll, 1= ? The formula gives Z = a + (n-l)d = l+10X2 = 1+20 = 21. Ans. EXERCISES 1. Find the llth term of 3, 6, 9, 12, . 2. Find the 13th term of 6, 10, 14, 18, . 3. Find the 20th term of 4, 2, 0, -2, -4, . % 4. Find the 15th term of -1, -1, -2, -2J, . 5. Find the 10th term of x-y, 2 x 2 y, 3 x-3 y, . 6. When a small heavy body (like a bullet) drops to the ground it passes over 16.1 feet the first second, 3 times as far the second second, 5 times as far the third second, etc. How far does it go in the 12th second? 7. If you save 5 cents during the first week in January, 10 cents the second week, 15 cents the third week and so on, how much will you save during the last week of the year? XV, 85] PROGRESSIONS 143 8. What term of the progression 2, 6, 10, 14, is equal to 98? [HINT, a = 2, d=4, n=?, /=98.] 9. What term of 3, 7, 11, 15, is equal to 59? 10. The first term of an A. P. is 8 and the 14th term is 47. What is the common difference? 85. The Sum of the First n Terms of an Arithmetic Pro- gression. Let a be the first term of an A. P., d the common difference, I the nth term. Then the sum of the first n terms, which we will call S, is (1) s = a+(a+d) + (a+2d) + (a+Zd)+''.+ (l-d)+l. This value for S may be written in a very much simpler form, as we shall now show. Write the terms of (1) in their reverse order. This gives (2) S = l+(l-d) + (l-2d) + (l-3d)+---+(a+d)+a. Now add (1) and (2), noting the cancellation of d with d t 2 d with -2 d, etc. The result is or 2S = n(a+l). Therefore S = -(a+/). 2 This is the simple form for S mentioned above. If we re- place I by its value a+(nl)d (84), this result takes the form 2 EXAMPLE. Find the sum of the first 12 terms of the A. P. 2,6, 10, 14, . SOLUTION, a = 2, d=4, n = 12. Therefore, by the second form for S in 85, we have 4} = 6(4+44} = 6X48 =288. Ans. 144 SECOND COURSE IN ALGEBRA [XV, 85 EXERCISES Find the sum of each of the following arithmetic progres- sions. 1. The first ten terms of 3, 6, 9, 12, . 2. The first fifteen terms of -2, 0, 2, 4, . 3. The first thirteen terms of 1, 3^, 6, -. 4. The first ten terms of 1, -1, -3, -5, . 5. The first n terms of 1, 8, 15, . 6. How many strokes does a common clock, striking hours, make in 12 hours? 7. A body falls 16.1 feet the first second, 3 times as far the second second, 5 times as far the third second, etc. How far does it fall during the first 12 seconds? 8. Find the sum of all odd integers, beginning with 1 and ending with 99. 9. If you save 5 cents during the first week in January, 10 cents during the second week, 15 cents the third week, and so on, how much will you save in a year? 10. The first term of an A. P. is 4 and the 10th term is 31. What is the sum of the 10 terms? [HINT, a = 4, ft = 10, Z=31. Now use the first of the formulas in 85.] 11. The first term of an A. P. is and the 12th term is 11-J-. What is the sum of the 12 terms? 12. Figure 49 shows a series of 16 dotted lines which are equally distant from each other. If the highest one is 6 inches long and the lowest one is 3 feet long, what is the sum of all their lengths? Fl - 49 - [HINT. The lines form an' A. P. since their lengths increase uni- formly.] XV, 85] PROGRESSIONS 145 13. The rungs of a ladder diminish uniformly from 2 feet 4 inches long at the base to 1 foot 3 inches long. at the top. If there are 24 rungs, what is the total length of wood in them? 14. Find the sum of the circumferences of 10 concentric circles if the radius of the innermost one is ^ inch and the radius of the outermost one is 4 inches, it being under- stood that the circles are equally spaced from each other. FIG. 50. 15. Figure 51 shows a coil of rope in the ordinary circular form, containing 12 com- plete turns, or layers. If the ength of the innermost turn is 4 inches and the length of the outermost turn is 37 inches, how long is the rope? FIG. 51. [HINT. Regard each turn as a circle, thus neglecting the slight effect due to the overlapping at the beginning of each turn after the first.] 16. If in Fig. 51 the length of the innermost turn is a inches and that of the outermost turn is b inches, and the number of turns is n, what represents the total length of the rope? 17. A small rope is wound tightly round a cone, the number of complete turns being 24. Upon unwinding the rope from the top, the lengths of the first and second turns are found to be 2J inches and 3^ inches respec- tively. How long (approximately) is the rope? FIG. 62. 146 SECOND COURSE IN ALGEBRA [XV, 86 86. Arithmetic Means. The terms of an arithmetic pro- gression lying between any two given terms are called the arithmetic means between those two terms. Thus, the three arithmetic means between 1 and 9 are 3, 5, 7, since 1, 3, 5, 7, 9 form an arithmetic progression. Whenever a single term is inserted in this way between two numbers, it is briefly called the arithmetic mean of those two numbers. Thus, the arithmetic mean of 2 and 10 is 6, because 2, 6, 10 form an arithmetic' progression. A formula for the arithmetic mean of any two numbers, as a and b, is easily obtained. Thus, if x is the mean, then a, x, b forms an A. P. Therefore, we must have x a = bx. Solving for x, this gives x _ab 2 Thus we have the following theorem : The arithmetic mean of two numbers is equal to half their sum. NOTE. The arithmetic mean of two numbers is also called their average. EXAMPLE. Insert five arithmetic means between 3 and 33. SOLUTION. We are to have an A. P. of 7 terms in which a =3, I =33, and n =7. We begin by finding d. Thus, Z = a + (n-l)d ( 84) so that 33=3+6d. Solving, d=5. The progression is therefore 3, 8, 13, 18, 23, 28, 33 and hence the desired means are 8, 13, 18, 23, 28. Ans. EXERCISES 1. Insert three arithmetic means between 7 and 23. 2. Insert four arithmetic means between 5 and 10. 3. Insert seven arithmetic means between ^ and 25f . 4. What is the arithmetic mean of 8 and 30 ? 6. What is the arithmetic mean of J and -J-? XV, 86] PROGRESSIONS 147 / FIG. 53. B 6. Show that the first formula for S obtained in 85 may be stated as follows : " The sum of n terms of an arith- metic progression is equal to n multiplied by the arithmetic mean of the first and nth terms." 7. A BCD is any trapezoid (that is, any four-sided figure having its bases A B and DC parallel to each other). The line EF, called the median, joins the middle point of the side AD to the middle point of the side BC, and it is shown in geometry that the length of ^ this line EF will always be the arith- A metic mean of the lengths of the bases AB and CD. Hence answer the following questions. (a) If the bases are 10 inches long and 2 inches long, respectively, what is the length of the median? (b) If the lower base is 14 inches long and the median 8 inches long, -how long is the upper base? (c) If the upper base is 3 feet long and the median 4 feet long, how long is the lower base? (d) If the bases are a inches long and b inches long, re- spectively, what represents the length of the median? 8. The figure shows the frustum of a cone and the frus- tum of a pyramid, and in each case the " mid-section " has FIG. 54. been drawn in (that is, the section made by a plane which passes midway between the bases AB and BC). It is shown in solid geometry that in all such cases the perimeter of the 148 SECOND COURSE IN ALGEBRA [XV, 86 mid-section will always be the arithmetic mean of the perimeters of the two bases. Hence answer the following questions. (a) If the perimeters of the two bases are 30 inches and 10 inches respectively, what is the perimeter of the mid-section ? (6) In the frustum of a cone, the radius of the upper base is 2 inches and that of the lower base 8 inches. What is the perimeter of the mid-section? 87. The Five Elements of an Arithmetic Progression. In any arithmetic progression there are the five elements, a, d, I, n, Sj defined in 84, 85. If any three of these are given, we can always find the other two by means of the formulas in 84, 85. EXAMPLE 1. Given a =-, ft = 30, S = 2l. Find d and I. SOLUTION. From 85, we have S =(a +Z). Hence 21i = 15( Solving, I = lii. Now, I = a + (n - l)d. Hence 1J = -i +29 d. Solving, d=^. EXAMPLE 2. Given a = 3, d = 4, = 300. Find n and I SOLUTION. S=%{2 a + (n-l)d}. Hence 300=S6 + O-1)- 4}. 2 2i Therefore 600 = n{4n+2}; 4 n 2 +2 n-600=0; 2n 2 +n-300=0. Solving the last (quadratic) equation by formula (56), gives or _ 4 44 Since n is the number of terms and therefore a positive integer, it follows that n = 12. (See Hint to Ex. 3, p. 89.) To find Z, we now use the formula I = a + (n l)d. Thus 1=3+11 -4=3+44 = 47. XV, 87] PROGRESSIONS 149 EXERCISES 1. Given a = 3, n = 25, = 675, find d and /. 2. Given a= -9, n = 23, Z = 57, find d and S. 3. Given = 275, 1 = 4:5, n = ll, find a and d. 4. Find w and d when a = - 5, I = 15, S = 105. 5. Find a and n when 1 = 1, d=%, =-20. 6. How many terms are there in the arithmetic pro- gression 2, 6, 10, 70? 7. Given a, I, and n, derive a formula for d. 8. Given a, d, and Z, derive a formula for n. 9. Given a, n, and S, derive a formula for Z. 10. Given d, I, and S, derive a formula for a. 11. Find an A. P. of 14 terms having 13 for its 6th term and 25 for its 10th term. 12. Find an A. P. of 16 terms such that the sum of the 6th, 7th and 8th terms is 16J-, and the sum of its last two terms is -28. 13. Find three integers in arithmetic progression such that their sum is 24 and their product 384. 14. The figure represents one of the four sides of a steel tower such as is commonly seen at wireless telegraph stations. It is desired to make one of these towers so that each girder, such as A B, will be 2 feet longer than the one just above it, as CD. How many girders will the tower have (counting all four sides) in case the total amount of girder steel used is to be only 864 feet and the lowest girders are each to FIG. 55. haye a lenth of 20 feet? 150 SECOND COURSE IN ALGEBRA [XV, 88 II. GEOMETRIC PROGRESSION 88. Definitions. A geometric progression is a sequence of numbers, called terms, each of which is derived from the preceding by multiplying it by a fixed amount, called the common ratio. A geometric progression is denoted by the abbreviation G. P. Thus 2, 4, 8, 16, 32, is a G. P. Each term is derived from the preceding by multiplying it by 2, which is therefore the common ratio. Again, 10, 5, +-JJ-, -, is a G. P. whose common ratio is -g- The next two terms are +f , -y 5 . EXERCISES Determine which of the following are geometric progres- sions, and find the common ratio and the next two terms of each geometric progression. 1. 3,6, 12,24,48, . 2. 4, 12, 48, 75,---. 3. *,i,i,A, ; 4. -1,2, -4,8, -16,".. 5. a, a 2 , a 3 , a 4 , -. 6. 2x, 4z 3 , 8z 5 , 16 z 7 , . 7. a, ar, ar 2 , or 3 , or 4 , . 8. a, aV 2 , a 3 r 4 , a r 6 , . 9. (0+6), (a+6) 3 , (a+6) 5 , (o+&) 7 , -. in m 2 m 4 ra 6 m 8 n 3 ' n 4 ' n ^ n J ' 11. Write the first five terms of the G. P. in which (a) The first term is 4 and the common ratio is 4. (6) The first term is 3 and the common ratio is 2. (c) The first term is a and the common ratio is r. XV, 89] PROGRESSIONS 151 89. The nth Term of a Geometric Progression. From the definition in 88 it follows that every geometric progression is of the form a, ar, ar 2 , ar 3 , ar 4 , -. Here a is the first term, and r the common ratio. Observe that the exponent of r in any one term is 1 less than the number of that term. Thus 2 is the exponent of r in the third term ; 3 is the exponent of r in the fourth term, etc. Therefore, the exponent of r in the nth term must be (n l). Hence, if we let Z stand for the nth term, we have the formula EXAMPLE. Find the 7th term of the G. P. 6, 4, f, SOLUTION. We have a =6, r=, n=7, l=? The formula gives I = or-i = 6 X (|) ' =2 X3 x| =| = Ans. EXERCISES 1. Find the ninth term of 2, 4, 8, 16, . 2. Find the eighth term of ^, ^, 1, . 3. Find the ninth term of 1, 2, 4, 8, . 4. Find the tenth term of 4, 2, 1, -J, . 5. Find the eighth term of -*-, , f , 6. Find the eleventh term of ax, a?x 2 , a 3 z 3 , a 4 # 4 , . 7. Find the tenth term of 2, A/2, 1, -. 8. What term of the G. P. 3, 6, 12, 24 is equal to 384? 9. What term of the progression 6, 4, f is equal to |4 ? 10. For every person there has lived two parents, four grandparents, eight great grandparents, etc. How many ancestors does a person have belonging to the 7th genera- tion before himself, assuming that there is no duplication? Answer also for the 10th generation. 152 SECOND COURSE IN ALGEBRA [XV, 89 11. If you save 50 cents during the first three months of the year and double the amount of your savings every three months afterward, how much will you save during the last three months of the second year ? 12. From a grain of corn there grew a stalk that produced an ear of 100 grains. These grains were planted and each produced an ear of 100 grains. This was repeated until there were 5 harvestings. If 75 ears of corn make a bushel, how many bushels were there the fifth year? 90. The Sum of the First n Terms of a Geometric Progres- sion. Let a be the first term of a geometric progression, r the common ratio, I the nth term. Then the sum of the first n terms, which we will call S } is (1) /S = a+ar+ar 2 +arH ----- \-ar n - 2 +ar n ~ l . This value for S may be written in a very much simpler form, as we shall now show. Multiply both members of (1) by r. This gives (2) rS = ar+ar 2 +arH ----- \-ar n ~ l -\-ar n . Now subtract equation (2) from equation (1), noting the cancelation of terms. This gives S-rS = a-ar n . Solving this equation for S gives l-r This is the simple form for S mentioned above. It is to be observed also that since I = ar n ~ l , we may write rl = ar n . Putting this value of ar n into the form just found for S, we obtain as a second expression for S the following formula. fl-r? XV, 90] PROGRESSIONS 153 EXAMPLE. Find the sum of the first six terms of the G. P. 3, 6, 9, 12, .-. SOLUTION, a =3, r = 2, n=6. To find S. 1> -g-j ^j -fr, '" SOLUTION, a =3, r = ^. Since r is numerically less than 1, we have by the formula of 92, < a 3 39., 158 SECOND COURSE IN ALGEBRA [XV, 92 EXERCISES Find the sum to infinity of each of the following progres- sions, and state in each case what your answer means, draw- ing a diagram similar to Fig. 57 to illustrate. 2 Q 3 3 3 O, -I". T'B't ~a~Ti . [HINT. r= i and hence is numerically less than 1. The formula of 92 therefore applies.] 4. 5, .5, .05, .005, . 6. 1 x+x 2 x 3 + whenx=f. 7. V2, 1,4=' I .., I 2 _2\/2 4 1 3' 3 vr 9' '"* 10. A pendulum starts at A and swings to B, then it swings back as far as C, then forward as far as D, etc. If the first swing (that is, the circular arc from A to B) is 6 inches long and each suc- ceeding swing is five sixths as long as the one just preceding it, how far will the pendulum bob travel before coming to FIG. 58. rest? 11. At what time after 3 o'clock do the hands of a watch pass each other? [HINT. We may look at this as follows : The large (minute) hand first moves down to where the small (hour) hand is at the be- XV, 93] PROGRESSIONS 159 ginning, that is through 15 of the minute spaces along the dial. Meanwhile the small hand advances T ^ as far or |f of a minute space. This brings the small hand to the position indicated by the dotted line in the figure. The large hand next passes over this ^f of a minute space. Meanwhile the small hand again advances ^ as far, which is -f of a minute space. The large hand next covers this ^f^ of a minute space, but the small hand meanwhile advances ^ as far, or yyf-g- of a minute space, etc. Thus, the successive moves of the large hand, counting from the first one, form the G. P. 15, if, ^, T ^, .... The sum of this to infinity will be the total distance passed over by the large hand before the hands pass.] 93. Variable. Limit. We have seen ( 92), in connection with the geometric progression 1, -g-, -J-, -J-, ', that the sum of its first n terms is a quantity which, as n increases indefinitely, comes and remains as near as we please to the exact value 2. The usual way of stating this is to say that as n increases, the sum of the first n terms approaches 2 as a limit. The sum of the first n terms is here called a variable since it varies, or changes, in the discussion. A similar remark applies to all the infinite geometric progressions which we have consid- ered. In every case the sum to infinity is the limit which the sum of the first n terms, considered as a variable quantity, is approaching. NOTE. It may be asked whether the sum of the first n terms of the G. P. 1, \, \, -|, could ever actually reach its limit 2. The answer is that it may or it may not, depending upon circumstances. Thus, if we think of the terms, beginning with the second, as being added on at the rate of one a minute we could never reach the end of the adding process, since the number of the terms is inexhaustible and hence the minutes required would have no end. In other words, the sum of the first n terms could never reach its limit on this plan. 160 SECOND COURSE IN ALGEBRA [XV, 93 But suppose that instead of this we were to add on the terms with increasing speed as we went forward. For example, suppose we added on the ^ in ^ a minute, then the -J in \ of a minute, then the i in -1- of a minute, etc. On this plan we would actually reach the limit 2 in 2 minutes of time. Here the constantly increasing speed of the adding process exactly counterbalances the fact that we have an indefinitely large number of terms to add, with the result that we reach the end of the process in the definite time of 2 minutes. This idea is practically illustrated in Ex. 11, p. 159, where the hands of the watch would never pass each other at all except for the fact that the successive moves of the large hand, which constitute the terms of the progression 15, ^-|, -^, xrls"' ' * * are added on in less and less time as the process goes on, each being added on in ^ the time occupied by the one just before it. The question of whether a variable can reach its limit is inti- mately connected with the famous problem considered by the Schoolmen in the Middle Ages and known as the problem of Achilles and the tortoise. In this problem, Achilles, who was a celebrated runner and athlete, starts out from some point, as A, to overtake a tortoise which is at some point, as T, the tortoise being famous for the slow rate at which it crawls along. Both start at the same in- stant and go in the same direction, as indicated in the figure. A T FIG. 60. Achilles soon arrives at the point T, from which the tortoise started, but in the meantime the tortoise has gone some distance ahead. Achilles now covers this last distance, but this leaves the tortoise still ahead, having again gained some additional distance. This continues indefinitely. How, therefore, can Achilles ever overtake the tortoise? The Schoolmen never quite answered this question satisfactorily to themselves. The secret of the difficulty lies in the fact that, as in the other problems mentioned above, the successive moves which Achilles makes are done in shorter and shorter inter- vals of time, with the result that, although the number of moves necessary is indefinitely great, they can all be accomplished in a definite time. XV, 94] PROGRESSIONS 161 94. Repeating Decimals. If we express the fraction | decimally by dividing 12 by 33 in the usual way, we find that the quotient is .363636 , the dots indicating that the divi- sion process never stops (or is never exact) but leads to a never-ending decimal. However, the figures appearing in this decimal are seen to repeat themselves in a regular order, since they are made up of 36 repeated again and again. Such a decimal is called a repeating decimal. More generally, a repeating decimal is one in which the figures repeat them- selves after a certain point. Thus, .12343434 -, 1.653653653 , are repeating decimals. Let us now turn the question around. Thus, suppose that a certain repeating decimal is given, as for example .272727 , and let us ask what fraction when divided out gives this decimal. This kind of question is usually too difficult to answer in arithmetic, but it can be easily answered as follows by use of the formula in 92. Thus the decimal .272727 may be written in the form 100 o o ~i~ ioooooo~f" *" This is an infinite geometric progression in which a = r= T J Tr . The sum of this progression to infinity must be the value of the given decimal. Hence, the desired value is = 27 100 = 27 = 3 A 100 99 99 11 This answer may be checked by dividing 3 by 11, the re- sult being .272727 -, which is the given decimal. NOTE. It is shown in higher mathematics that every rational fraction in its lowest terms (that is, every number of the form a/6, where a and 6 are integers prime to each other) gives rise when divided out to a never-ending, repeating decimal, while every irra- tional number (such as V2) gives rise when expressed decimally to a never-ending non-repeating decimal. M 162 SECOND COURSE IN ALGEBRA [XV, 94 EXERCISES Find the values of the following repeating decimals and check your answer for each of the first six. 1. .414141 . 4. .3414141 -.. 2. .898989 . 3. .543543543 SOLUTION. .3414141 = .3 + .0414141 = .3 +^(.414141-) 6. .6535353 - 6. 1.212121 7. 3.2151515 _ 3 _,_'! ^ 41 ^100 ~10 + 10 X 100 X 99 = A_i_^L = 338 = 169 A 10 990 990 495' 8. 5.032032032 9. 6.008008008 10. 34.5767676 MISCELLANEOUS PROBLEMS I. ARITHMETIC PROGRESSION 1. What will be the cost of digging a 20-foot well if the digging costs 50 cents for the first foot and increases by 25 cents for each succeeding foot ? 2. Fifty-five logs are to be piled so that the top layer shall consist of 1 log, the next layer of 2 logs, the next layer of 3 logs, etc. How many logs will lie on the bottom layer ? 3. In a potato race 30 potatoes are placed at the dis- tances 6 feet, 9 feet, 12 feet, etc., from a basket. A player starts from the basket, picks up the potatoes and carries them, one at a time, to the basket. How far does he go altogether in doing this? XV, 94] PROGRESSIONS 163 4. A row of numbers in arithmetic progression is written down and afterwards all erased except the 7th and the 12th, which are found to be 10 and 15, respectively. What was the 20th number? 5. If your father gives you as many dimes on each of your birthdays as you are years old on that day, how old will you be when the total amount he has given you in this way amounts to $12? 6. How many arithmetic means must be inserted be- tween the numbers 4 and 25 in order that their sum may amount to 87 ? 7. Prove that equal multiples of the terms of an arith- metic progression are in arithmetic progression. 8. Prove that the sum of n consecutive odd integers, be- ginning with 1, is n . 9. The sum of three numbers in arithmetic progression is 30 and the sum of their squares is 462. What are the numbers ? [HINT. The numbers may be represented as x y, x, x+y. Form two equations and solve for x and y.} 10. If a person saves $20 the first month and $10 each month thereafter, how long before his total savings will amount to $1700? 11. Divide 80 into four parts which are in arithmetic progression and which are such that the product of the first and fourth is to the product of the second and third as 2:3. 12. Find the sum of the first 40 terms of an A.P. in which the ninth term is 136 and the sum of the first nineteen terms is 2527. 13. lid = 2, n = 21, and 5 = 147, find a and I 14. Show that if, in any A. P., the values of d, I, and S are given, then the formula for a is 164 SECOND COURSE IN ALGEBRA [XV, 94 II. GEOMETRIC PROGRESSION 15. A wheel in a certain piece of machinery is making 32 revolutions per second when the steam is turned off and the wheel begins to slow down, making one half as many revolutions each second as it did the preceding second. How long before it will be making only 2 revolutions per second? 16. Show that if a principal of $p be invested at r % compound interest, the sum of money accumulating at the ends of successive years will form a geometric progression, while if the investment be made at simple interest, the sums accumulating will form an arithmetic progression. 17. From a cask of vinegar ^ the contents is drawn off and the cask then filled by pouring in water. Show that if this is done 6 times, the cask will then contain more than 90% water. [HINT. Call the original amount of vinegar 1, then express (as a proper fraction) the amount of water in the cask after the first refilling, second refilling, etc.] 18. A set of concentric circles is drawn, each having a radius half that of the circle just outside it. Show that the limit toward which the sum of their circumferences is ap- proaching is equal to twice the circumference of the largest circle. 19. A dipper when hung on a wall often swings back and forth for a time, the swings gradually dying out. If the first swing occupies 1 second, and each succeeding swing takes .9 as long as the one before it, how long before the dipper comes to rest? 20. It is found by experiment that the number of bacteria in a sample of milk doubles every 3 hours. What increase will there be in 24 hours, assuming that all outside conditions remain the same? XV, 94] PROGRESSIONS 165 21. In Fig. 61 a series of ordinates equally spaced from each other has been drawn, the first one being laid off 1 unit long, the second one being laid off equal to the first one increased by % its length, the third being equal to the second in- creased by ^ its length, etc. Show that these ordinates represent the successive terms of the G. P. whose first term is 1 and whose common ratio is 14-. In this sense, the figure may be called the dia- gram for the G. P. in which a = l, r=l-J. 22. Draw the diagram for the G. P. in which (a) a = l, r = H; (6) a = 2, r = l; (c) a = 4, r = J. [HINT. Use 8 ordinates only, spacing them at any convenient but equal distance apart.] 23. Prove that any series of numbers formed by writing down the reciprocals of the successive terms of a geometric progression is itself a geometric progression. 24. Three numbers whose sum is 24 are in arithmetic pro- gression, but if 3, 4, and 7 be added to them respectively, the results form a geometric progression. Find the numbers. 25. If a series of numbers are in G. P., are their squares likewise in G. P. ? Answer the same for their cubes ; also for their square roots and their cube roots. Answer the same questions for an A. P. [HiNT. See that your reasoning is general, that is, do not base it upon an examination of some special cases.] CHAPTER XVI RATIO AND PROPORTION 95. Ratio. The quotient of one number divided by another of the same kind is called their ratio. Thus the ratio of 6 inches to 3 inches is |-, or 2 ; the ratio of 5 Ib. to 3 Ib. is |> etc. Note that in each of these cases the ratio is simply a fraction of the kind studied in arithmetic. The first number, or dividend, is called the antecedent; the second number, or divisor, is called the consequent. Thus, in the ratio -J, the antecedent is 3 and the consequent is 4. EXERCISES 1. What is the ratio of. 10 yards to 2 yards? of 7 yards to 3 yards? 2. State (as a fraction in its simplest form) the value of each of the following ratios. (a) 5 to 25. (c) i to |. (e) 1 to 3. (g) 18 x z to 4 z 2 . (6) 16 to 12. (d) 2toi. (/) 3 a to 6 b. (h) x*-y*tox-y. 3. State which is the antecedent and which the conse- quent in each of the parts of Ex. 2. 4. What is the ratio of 10 inches to 2 feet? [HINT. First reduce the 2 feet to inches so that we may com- pare like numbers, that is numbers measured in the same unit.] 5. The dimensions of a certain grain bin are 3 feet by 6 feet by 7 feet. What is the ratio of its cubical contents to that of a bin whose dimensions are 3 feet 6 inches by 5 feet by 1 yards? 166 XVI, 96] RATIO AND PROPORTION 167 6. If one square has its sides each twice as long as the sides of another square, what is the ratio of the area of the first square to that of the second? [HINT. Let a =a side of the smaller square.] 7. If one cube has its edges each twice as long as the edges of another cube, what is the ratio of the volume of the first cube to that of the second ? 8. Show that if a cylinder and a cone have the same circular base and the same height, the ratio of the volume of the cylinder to that of the cone is 3 : 1. FIG. 62. 9. When we sharpen a lead pencil a cer- tain part of the cylindrical lead is exposed. What part of the exposed lead is cut off when a smooth conical point is made? 96. Proportion. A proportion is an expression of equality between two ratios, or fractions. For example, since j- is the same as f-, we have the proportion ^ = -|. Likewise, we may write f = , f=if> -f = -f, etc. ; hence all these are true proportions. But f = -J- is not a proportion since these two fractions are unequal. Every proportion is thus seen to be an equality of the form a/b = c/d, where a, 6, c, and d are certain numbers. These four numbers are called the terms of the proportion. The first and fourth (that is, a and d) are called the extremes of the proportion, while the second and third (b and c) are called the means. Besides writing a proportion in the form a/6 = c/d, it may be written in the form a : b : : c : d, or also in the form a : d = c:d. In all cases it is read " a is to b as c is to d," and it means that the fraction a/6 equals the fraction c/d. 168 SECOND COURSE IN ALGEBRA [XVI, 96 EXERCISES 1. Using the language of proportion, read each of the following statements. (a) i = f. (c) 2: -1=8: -4. (6) 1:4 = 3:12. (d) :::4:3. 2. State which are. the extremes and which the means in each part of Ex. 1. 3. State such proportions as you can make out of the following four quantities : 3 inches, 6 inches, 12 inches, 24 inches. [HINT. 3 inches is to 6 inches as . Make other proportions also.] 4. State such proportions as you can make out of the following four quantities : 1 inch, 3 inches, 1 foot, 1 yard. [HINT. First express all quantities in inches.] 5. State such proportions as you can make out of the following : 1 pint, 1 quart, 1 gallon, 2 gallons. 6. Do as in Ex. 5 for the following : 2 seconds, 1 minute, 1 hour, a day and a quarter. 7. Do as in Ex. 5 for the following : 1 cent, 1 dollar, 1 centimeter, 1 meter. [HINT. Compare money ratio with distance ratio.] 8. Do as in Ex. 5 for the following : 4 ounces, 1 pound, 1 gallon, 1 quart. 97. Algebraic Proportions. If we consider the algebraic fraction (a 2 6)/(a& 2 ), we see (upon dividing both numerator and denominator by ab) that it reduces to a/6. In other words, we have o 2 6 = a ab 2 b This is an example of an algebraic proportion. Similarly, 2x 2 y_ x 4 xyz 2 z XVI, 98] RATIO AND PROPORTION 169 is an algebraic proportion and may be written if desired in the form 2 x 2 y : 4 xyz = x : 2 z. Likewise, since a 2 -6 2 __ _ we have 98. Fundamental Theorem. Let a/b = c/d be any pro- portion. By multiplying both sides of this equality by bd, we obtain or ad = be. This result may be stated in the following theorem. THEOREM A. In any proportion, the product of the means is equal to the product of the extremes. This theorem is useful in testing the correctness of a proportion. Thus 6:9=14:21 is a correct proportion be- cause the product of the means, which is 9 X 14, is equal to the product of the extremes, which is 6 X21 ; but 6 : 9 = 8 : 15 is not correct because 9 X8 is not equal to 6 Xl5. Similarly, x 3 : x 2 y = x:y, because x 2 y - x = x z y. EXERCISES By means of the theorem of 98, test the correctness of the following proportions. 1. 5:6 = 15:18. 5. 17:19 = 21:23. 2. 3:2 = 5:6. 6. 2 a : ab = W x : 5 bx. 3. 4: -1 = 8: -2. 7. 3 m 2 : (a-b) =6 m : 2 m(a-b). 4. i:| = 8:4. 8. 9. a 2 -6 2 : 170 SECOND COURSE IN ALGEBRA [XVI, 98 By means of the theorem of 98, determine the value which x must have in the following proportions. 10. z:4 =3:2. [HINT. The theorem gives 4 3 = x 2, or 12 = 2 x.] 11. 10:z = 2:5. 13. (z-5) : 4: : 2: 3. 12. 25/32 = 8/z. 14. (x-3)/(x-4) = 5/6. 15. What number bears the same ratio to 4 as 16 does to 6? [HINT. Let x represent the unknown number and form a pro- portion.] 16. Divide 35 into two parts whose ratio shall be f . [HINT. Let x be one part. Then 35 x will be the other part.] 17. Divide 35 into two parts such that the lesser dimin- ished by 4 is to the larger increased by 9 as 1 : 3. 18. A man's income from two investments is $980. The two investments bear interest rates which are in the ratio of 5 to 6. What income does he receive from each? 19. Concrete for sidewalks is a mixture made of two parts sand to one part cement. How much of each is required to make a walk containing 1500 cubic feet? 20. Prove that no four consecutive numbers, as n, ft+1, ft +2, ft +3, can form a proportion in the order given. 99. Application to Similar Figures. When two geometric figures have the same shape, though not necessarily the same size, they are called similar figures. Thus any two circles are similar figures; likewise, any two squares, or any two cubes, or any two spheres. Two triangles may be similar, as , illustrated in Fig. 63. The following facts are shown in / , geometry regarding any two similar figures. XVI, 99] RATIO AND PROPORTION 171 (a) Corresponding lines are proportional. Thus, in the two similar triangles of Fig. 63, if the side AB of the one is twice as long as the corresponding side A'B' of the other, then BC is twice as long as B'C'. That is, AB = BC A'B' B'C'' In the same way, we have also AB CA A'B' C'A' (6) Areas are proportional to the squares of corresponding lines. Thus, if one circle has a radius of length R and another circle has a radius of length r, the area, A, of the first is to the area, a, of the second as R 2 is to r 2 . That is we have the proportion A /a = R 2 /r 2 . (c) Volumes are proportional to the cubes of corresponding lines. Thus, if one sphere has the radius R and another has the radius r, the volumes V and v of the spheres are such that V/v = R 3 /r 3 . EXERCISES ON SIMILAR FIGURES 1. In the two similar triangles shown in 99 suppose AB = 2 feet, A'B' = l foot 4 inches, and BC = 3 feet. How long will B'C' be? 2. If a tree casts a shadow 40 feet long when a post 3 feet high casts a shadow 4 feet long, how high is the tree ? 3. Compare the areas of two city lots of the same shape if a side of the one is three times as long as the corresponding side of the other. Does it matter what the shape of each is ? 4. If a certain bottle holds -J pint, how much will a bottle of the same shape but only half as high hold ? 172 SECOND COURSE IN ALGEBRA [XVI, 100 100. Mean Proportional. If the means of a proportion are equal, either is called the mean proportional between the extremes. Thus 2 is the mean proportional between 1 and 4 because =}. Likewise, 2 x is the mean proportional between x 2 and 4, because x 2 /2x =2 z/4. NOTE. The mean proportional between a and 6 is always equal to Vo6, for we must have a/x=x/b. Hence, clearing of fractions, x 2 = ofc/andjtheref ore x = Vo6. This will be a surd ( 42) unless the product ab is a perfect square. For example, the mean proportional between 2 and 3 is the surd V2 3, or V6 =2.44949 (table). 101. Third and Fourth Proportionals. The third propor- tional to two numbers a and b is that number x such that a: b = b:x. Thus the third proportional to 2 and 3 is the value of x in the equation ^ ^ o = ~- Solving, = = 4 \. Arts. O X The fourth proportional to three numbers a, b, and c is that number x such that a:b = c:x. Thus the fourth proportional to 2, 3, and 4 is the value of x in the equation 2/3 = 4/x. Solving, x = 6. Ans. EXERCISES 1. Find the mean proportional between 8 and 18. [HINT. Let* be the desired mean. Then8/x =x/lS. Solveforz.] Find the mean proportional between each of the follow- ing pairs of numbers. In cases where the answer is a numer- ical surd, use the table to find its approximate value. 2. 9 and 81. 6. | and |f. 3. 6 and 7. 7. 2 and 3^. 4. 5 and 20. 8. 2 x 2 y and 32 xy 2 . a 2 -2a-3 K K A 10 Q - ^ 6. 5 and 19. 9. - and a+1 a 3 XVI, 101] RATIO AND PROPORTION 173 Find the third proportional to each of the following pairs. 10. 3 and 4. 12. 2| and 3^. 14. z 2 -9 and z-3. 11. 18 and 50. 13. 2 x and x. 15. 2 and 6. Find the fourth proportional to each of the following sets. 16. 3, 4, and 5. 19. V2, \/6, and A/12. 17. 5, 4, and 2. 20. 3 a, 2 6, and c. 18. 2, 3^, 4i. 21. x, y, and xy. NOTE. In Exs. 16-21, the numbers must be placed in the pro- portion in the order in which they are given, as in the illustrative examples of 101. 22. In the semicircle ABC suppose DE drawn perpendicular to AB. Then (as shown in geometry) the length of DE will be the mean proportional A between the lengths of AE and EB. Fl <*- 64. If A E = 4 inches and EB = 16 inches, find DE. 23. The figure shows a circle and a point P outside it from which are drawn two lines PS and PT. The first of p these lines (called a secant) cuts through the circle at two points R and S while the second line (called a tangent) just touches the circle at the point T. In all such cases, the tangent length, PT, is the mean pro- portional between the whole secant, PS, and its external segment, PR (as shown in geometry). Find the length of PT if PR = 4 and RS=11. 24. If a, b, c, d are unequal numbers such that a:b = c:d t show that no number x can be found such that a-\-x\ b+x = c-\-x \d-\-x. FIG. 65. 174 SECOND COURSE IN ALGEBRA [XVI, 102 102. Second Fundamental Theorem. THEOREM B. // the product of two numbers is equal to the product of two other numbers, either pair may be made the means of a proportion in which the other two are taken as the extremes. PROOF. Suppose mn = xy. Dividing both members by nx gives m/x = y/n, or m : x = y : n, which is one of the possible proportions mentioned in the theorem. Similarly, if we divide both members of mn = xy by ny we obtain m/y = x/n, or m:y = x:n, which is another of the possible proportions mentioned in the theorem. The other possible proportions are x:m = n:y and n:x = y : m. The proof of these is left to the pupil. For example, the equality 2 9=3 6 gives rise to the propor- tions 2 : 3=6: 9, 2:6 = 3:9, 9:3=6:2, and 9: 6=3: 2. 103. Inversion in a Proportion. THEOREM C. If four quantities are in proportion, they are in proportion by inversion; that is the second term is to the first as the fourth is to the third. PROOF. We are to show that if a/b = c/d, then b/a = d/c. Since a/b = c/d, we have, by Theorem A, ad = bc. Therefore, by Theorem B, we may write b/a = d/c. For example, f = f gives by inversion the new proportion f- = f . 104. Alternation in a Proportion. THEOREM D. // four quantities are in proportion, they are in proportion by alterna- tion; that is the first term is to the third as the second is to the fourth. The proof is left to the pupil. First use Theorem A. See proof of Theorem C. For example, f = f gives by alternation the new proportion =f 105. Composition in a Proportion. THEOREM E. // four quantities are in proportion, they are in proportion by XVI, 107] RATIO AND PROPORTION 175 composition; that is the sum of the first two terms is to the second term as the sum of the last two terms is to the last term. PROOF. We are to show that if a/b = c/d, then (a+b)/b = (c+d)/d. Since a/b = c/d, we may add 1 to each member of this equa- tion, thus giving +!=< + !, which reduces to , b a o a For example, f = f gives by composition the new proportion (2+3)/3 = (6+9)/9, orf=V 5 . 106. Division in a Proportion. THEOREM F. // four quantities are in proportion, they are in proportion by division; that is, the difference between the first two terms is to the second term as the difference between the last two terms is to the last term. PROOF. We are to show that if a/b = c/d, then (a &)/& = (c-d)/d. Since a/b = c/d, -we may subtract 1 from each side of this equation, thus obtaining f- 1=^-1, which reduces to *LZ&=^. b a o d For example, f =f gives by division the new proportion (3-2)/2 = (9-6)/6, or|=f. 107. Composition and Division. THEOREM G. // four quantities are in proportion, they are in proportion by composi- tion and division; that is the sum of the first two terms is to their difference as the sum of the last two terms is to their difference. PROOF. We are to show that if a/b = c/d, then ab cd 176 SECOND COURSE IN ALGEBRA [XVI, 107 By Theorems E and F, we have a-j-b_c-\-d , ab_c d ~l d~' ! ~b 5" By dividing the first of these equations by the second, member by member, we obtain the desired result, namely a b cd For example, f = |- gives by composition and division the-new pro- portion (3 +2) /(3-2) = (9 +6) /(9-6), or f =-^. 108. Several Equal Ratios. THEOREM H. In a series of equal ratios, the sum of the antecedents is to the sum of the conse- quents as any antecedent is to its consequent. PROOF. We are to show that if a/b = c/d = e/f=g/h= , then a+c+e+g-\ _a_^c _e_g _ b+d+f+h+~- b d f h Let k be the value of any one of the equal ratios, so that JL_ _c r._e k _g 6'* d'* / -A' = fc&, c = kd, e = kf, g = kh, . Hence b+d+f+h+- or a+c+e+g-\ _^a_c ' _e_Q [_ b+d+f+h+- b d f h For example, the three equal ratios f = f = f give the new pro- portions 2+4+6^2^.4,6 3+6+9 369* or 12 = 2 = 4 = 6 18 3 6 9* XVI, 108] RATIO AND PROPORTION 177 EXERCISES 1. Given the proportion f =J ^-. Write down the vari- ous proportions to be obtained from this by (1) inversion, (2) alternation, (3) composition, (4) division, (5) composi- tion and division. Note that each new proportion thus ob- tained is a true one. 2. Show that if a/b = c/d, then a 2 /b 2 = c 2 /d 2 ', in other words, if four numbers are in proportion, their squares are also in proportion. 3. Show that if four numbers are in proportion, their cubes are also in proportion; likewise their square roots; likewise their cube roots. 4. Show by means of Theorem A ( 98) that a/b = a 2 /b 2 is not a true proportion, unless a = b. In other words, the ratio of two numbers is not in general the same as the ratio of their squares. Prove similarly that the ratio of two num- bers is not in general the same as the ratio of their cubes, or of their square roots, or of their cube roots. 5. If a : b = c : d, establish the following proportions. (a) a 2 :b 2 c 2 = l:d 2 . SOLUTION. It suffices to show here that the product of the extremes is equal to the product of the means, for if these two products are the same, the proportion in question is true by Theorem B. Thus we are to prove that a z d z = 6 2 c 2 . Now, we know (by hypothesis) that a:b=c:d, or ad = bc. Squaring gives, as desired, a?d 2 = b z c 2 , thus completing the proof. (b) ac:bd = c 2 :d 2 . [HINT. Remember to use the hypothesis, namely that a : b = c : d.] (c) Vad:Vb = Vc:l. (d) a:a+b = a+c: a+b+c+d.. (e) a+b: c+d=Va 2 +b 2 : Vc 2 +d 2 . (/) a+b+c+d: ab+cd=a+bcd: ab c+d. (g) If a : b = c : d, and x:y = z:w, show that ax : cz = by : dw. CHAPTER XVII VARIATION 109. Direct Variation. One quantity is said to vary directly as another when the two are so related that, though the quantities themselves may change, their ratio never changes. Thus the amount of work a man does varies directly as the number of hours he works. For example, if it takes him 4 hours to draw 10 loads of sand, we can say it will take him 8 hours to draw 20 loads. Here the first ratio is ^ and the second is -$ and the two are seen to be equal, though the numbers in the second have been changed from what they were in the first. In general, if the man works twice as long, he will draw twice as much ; if he works three times as long, he will draw three times as much, etc. ; all of which implies that the ratio of the time he works to the amount he draws in that time never changes. EXERCISES Determine which of the following statements are true and which false, giving your reason in each instance. 1. The amount of electricity used in lighting a room varies directly as the number of lights turned on. 2. The amount of water in a cylindrical pail varies directly as the height to which the water stands in the pail. 178 XVII, 110] VARIATION 179 3. The amount of gasoline used by an automobile in any given time (one week, say) varies directly as the amount of driving done. 4. The time it takes to walk from one place to another at any given rate (3 miles an hour, say) varies directly as the distance between the two places. 5. The time it takes to walk any given distance (5 miles, say) varies directly as the rate of walking. 6. The perimeter of a square varies directly as the length of one side. 7. The circumference of a circle varies directly as the length of the radius. 8. The area of a square varies directly as the length of one side. 9. x varies directly as 10 x. 10. x varies directly as 10 x 2 . 110. Inverse Variation One quantity, or number, is said to vary inversely as another when the two are so related that, though the quantities themselves may change, their product never changes. Thus the time occupied in doing any given piece of work varies inversely as the number of men employed to do it. For example, if it takes 2 men 6 days, it will take 4 men only 3 days. The point to be observed here is that the first product, 2X6, equals the second product, 4X3. In general, if twice as many men are employed it will take half as long ; if three times as many men are employed, it will take one third as long, etc. In all these cases, the number of men employed multiplied by the corresponding time required to do the work remains the same. NOTE. The term varies inversely as is due to the fact that in case xy never changes (as required by the above definition), it follows that x + (\/y) never changes, since xy=x + (\/y). That is, x varies directly as the reciprocal, or inverse, of y ( 109). 180 SECOND COURSE IN ALGEBRA [XVII, 110 EXERCISES Determine which of the following statements are true and which false, giving your reason in each instance. 1. The time it takes water to drain off a roof varies in- versely as the number of (equal sized) conductor pipes. 2. The time it takes to walk any given distance (5 miles, say) varies inversely as the rate of walking. (Compare Ex. 5, p. 179.) 3. The weight of a pail of water varies inversely as the amount of water that has been poured out of it. 4. x varies inversely as 10/z. 6. x varies inversely as 10/z 2 . 111. Joint Variation. One quantity, or number, is said to vary jointly as two others when it varies directly as their product. Thus the area of a triangle varies jointly as its base and altitude, for if A be the area of any triangle and b its base and h its altitude, we have A=^bh, which may be written A/bh=%. Whence A varies directly as the product bh ( 109), that is the ratio of A to bh is always the same, namely ^ in this instance. EXERCISES I Determine whether the following statements are true, giving your reason in each instance. 1. The area of a rectangle varies jointly as its two dimensions, that is as its length and breadth. 2. The pay received by a workman varies jointly as his daily wage and the number of days he works. 3. The amount of reading matter in a book varies jointly as the thickness of the book and the distance between the lines of print on the page. XVII, 113] VARIATION 181 4. The interest received in one year from an investment varies jointly as the principal and rate. 5. The volume of a' rectangular parallelepiped (such as an ordinary rectangular shaped box) varies jointly as its length, breadth, and height. [HINT. Here we have one quantity varying jointly as three others. First make a definition yourself of what such variation means.] 112. Variables and Constants. When we say that the amount of work a man does varies directly as the number of hours he works (see 109), we are dealing with two quanti- ties, namely the amount of work done and the time used in doing it. But it is to be observed that these are not being regarded as fixed quantities, but rather as changeable ones, the only essential idea being that their ratio never changes. In general, quantities which are thus changeable throughout any discussion or problem are called variables, while quan- tities which do not change are called constants. 113. The Different Types of Variation Stated as Equa- tions. We may now state very briefly and concisely what is meant by the different types of variation mentioned in 109-111 and certain other important types also. To do this, let us think of x, y, and z as being certain variables and k as being some constant. Then (1) To say that x varies directly as y means (by 109) that T - = fc, or x = ky. y (2) To say that x varies inversely as y means (by 110) that xy k, or x = -. y (3) To say that x varies jointly as y and z means (by 1 11) that 1* = k, or x kyz. yz 182 SECOND COURSE IN ALGEBRA [XVII, 113 Two other important types of variation are described below : (4) To say that x varies directly as the square of y means that - = k, or x = ky 2 . (5) To say that x varies inversely as the square of y means that xy 2 = k, or x = y 2 In all these types of variation it is important to observe that the value which must be given to the constant k depends upon the particular statement or problem in hand. For example, consider the statement that " The area of a rec- tangle varies jointly as its two dimensions." This means (see (3)) that if we let A be the variable area and a and b the variable dimensions, then A = kab. But in this case we know by arithmetic that A = ab, so the value of k here must be 1. On the other hand, consider the statement that " The area of a triangle varies jointly as its base and altitude." Letting A be the variable area and b and h the variable base and altitude, respectively, this means that A = kbh. But here, as we know from arithmetic, k = \. EXERCISES Convert each of the following statements into equations, supplying for each the proper value for the constant k mentioned in 113. 1. The circumference of a circle varies directly as the radius. [HINT. Let C stand for circumference and r for radius.] 2. The circumference of a circle varies directly as the diameter. 3. The area of a circle varies directly as the square of the radius. XVIT, 113] VARIATION 183 4. The area of a circle varies directly as the square of the diameter. 5. The area of a sphere varies directly as the square of the radius. (See 14, (/).) 6. The volume of a rectangular parallelepiped varies jointly as its length, breadth, and height. (See Ex. 5, p. 181.) 7. Interest varies jointly as the principal, rate, and time. 8. The volume of a sphere varies directly as the cube of the radius. [HINT. First supply for yourself the definition of what this type of variation means.] 9. The volume of a circular cone varies jointly as the alti- tude and the square of the radius of the base. (See p. 103.) 10. The distance, measured in feet, through which a body falls if dropped vertically downward from a position of rest (as from a window ledge) varies directly as the square of the number of seconds it has been falling. [HINT. It is found by experiments in physics that the value of the constant A; is in this case 32 (approximately).] 11. The following, like Ex. 10, are statements of well- known physical laws. Convert each into an equation with- out, however, attempting to supply the proper value of k, since to do so requires a study of physics. (a) If a body is tied to a string and swung round and round in a circle (as in swinging a pail of water at arm's length from the shoulder) , the force, F, with which it pulls outward from the center (called centrifugal force) varies directly as the square of the velocity of the motion. (6) The intensity of the illumination due to any small source of light (such as a candle) varies inversely as the square of the distance of the object illuminated from the source of light. 184 SECOND COURSE IN ALGEBRA [XVII, 113 (c) When an elastic string is stretched out, as represented in Fig. 66, the tension (force tending to pull it apart at any point) varies directly as the length to which the string has been stretched (Hooke's Law). FIG. 66. (d) The pressure per square inch which a given amount of gas (such as air, or hydrogen, or oxygen, or illuminating gas) exerts upon the sides of the receptacle which holds the gas (such as a bag) varies inversely as the volume of the receptacle (Boyle's Law). FIG. 67. For example, whenever air is con- fined in a rubber balloon, as in the first drawing in Fig. 67, it exerts a certain pressure upon each square inch of the interior surface. If the balloon be squeezed, as in the second drawing in Fig. 67 (no air being allowed to escape), until its volume is half of what it was before, this pressure will be exactly doubled. 114. Problems in Variation. The problems naturally arising in the study of variation fall into two general classes as follows : (1) Those in which the value of the constant k mentioned in 113 can be determined from the statement of the problem and forms an essential part in the solution. This kind of problem is illustrated by Exs. 1-10 on pages 185-187. The solution given for Ex. 1 should be well understood before the pupil undertakes Exs. 2-10. XVII, 114] VARIATION 185 (2) Those in which it is not necessary to know the value of k. Such problems are illustrated in Exs. 11-20, pp. 187-189. The pupil is advised to work several problems from each group rather than to confine his attention to either. EXERCISES I. ILLUSTRATIONS OF CASE (1) 1. In a fleet of ships all made from the same model (that is, of the same shape, but of different sizes) the area of the deck varies directly as the square of the length of the ship. If the ship whose length is 200 feet has 5000 square feet of deck, how many square feet in the deck of the ship which is 300 feet long? SOLUTION. Let A represent the area of deck on the ship whose length is I. Then the given law of variation, expressed as an equation ( 113), is (1) A=kl z . (k = some constant) Since the ship which is 200 feet long has 5000 square feet of deck, it follows from (1) that we must have = /c(200) 2 . This equation tells us that the value of k in the present problem must be .5000 = 5000 = 1 (200) 2 200X200 8* Placing this value of k in (1), gives us an equation which deter- mines completely the relation between A and I in the present problem, that is (2) A = iZ*. Now the problem asks how many square feet of deck there are in the ship whose length is 300 feet. This can be found by simply placing = 300 in (2) and solving for A. Thus A =4 X(30Q) 2 = 300 * 3Q P:= 11,250 squ are feet. Ans. O O 186 SECOND COURSE IN ALGEBRA [XVII, 114 NOTE. Observe that the first step in the above solution is to express as an equation the law of variation belonging to the problem. Next, the constant k is determined. After this, the first equation is rewritten in its more exact form obtained by assigning to k its value. The answer is then readily obtained. These steps should be followed in working each of the Exs. 2-10. 2. In a fleet of ships all of the same model, the ship whose length is 200 feet contains 6000 square feet in its deck. How long must a similar ship be made if its deck is to contain 13,500 square feet? 3. To make a suit of clothes for a man who is 5 feet 8 inches high requires 6 square yards of cloth. How much cloth will be required to make a suit for a man of similar build, whose height is 6 feet 2 inches? [HINT. The areas of any two similar figures vary directly as the squares of their heights.] 4. If 10 men can do a piece of work in 20 days, how long will it take 25 men to do it? [HINT. The time required varies inversely as the number of men employed.] 5. The horsepower required to propel a ship varies di- rectly as the cube of the speed. If the horsepower is 2000 at a speed of 10 knots, what will it be at a speed of 15 knots? 6. A silver loving-cup (such as is sometimes given as a prize in athletic contests) is to be made, and a model is first prepared out of wood. The model is 8 inches high and weighs 12 ounces. What will the loving-cup cost if made 10 inches high, it being given that silver is 17 times as heavy as wood and costs $2.20 an ounce? [HINT. The volumes and hence the weights of any two similar figures vary directly as the cubes of their heights. See 99 (c).] XVII, 114] VARIATION 187 7. When electricity flows through a wire, the wire offers a certain resistance to its passage. The unit of this resist- ance is called the ohm, and for a given length of wire the resistance varies inversely as the square of the diameter. If .a certain length of wire whose diameter is \ inch offers a resistance of 3 ohms, what will be the resistance of a similar wire (same length and material) ^ of an inch in diameter? 8. Three spheres of lead whose radii are 6 inches, 8 inches, and 10 inches respectively are melted and made into one. What is the radius of the resulting sphere ? 9. On board a ship at sea the distance of the horizon varies directly as the square root of one's height above the water. If, at a height of 20 feet, the horizon is 5.5 miles dis- tant, what is its distance as seen from a light-house 80 feet above sea-level? 10. The horsepower that a shaft can safely transmit varies jointly as its speed in revolutions per minute and the cube of its diameter. A 3-inch steel shaft making 100 revo- lutions per minute can transmit 85 horsepower. How many horsepower can a 4-inch shaft transmit at a speed of 150 revolutions per minute ? II. ILLUSTRATIONS OF CASE (2) 11. Knowing that the force of gravitation due to the earth varies inversely as the square of the distance from the earth's center (Newton's Law of Gravitation), find how far above the earth's surface a body must be taken in order to lose half its weight. SOLUTION. Letting W represent the weight of a given body at the distance d from the earth's center, the law stated above, when ex- pressed as an equation, becomes ( 1 ) W = - - (k = some constant) a* 188 SECOND COURSE IN ALGEBRA [XVII, 114 Now let Wi represent the weight of the body when on the surface. Remembering that the earth's radius is 4000 miles (approximately), equation (1) gives (2) k 4000 2 Next, let x represent the desired distance, namely the distance above the surface at which the same body loses half its weight. At this distance its weight will consequently be %Wi, while its dis- tance from the earth's center is now 4000 +x. So (1) gives (3) IKi = , & 2 (4000 +z) 2 Dividing equation (3) by equation (2), noting the cancelation of W\ on the left and of the (unknown) k on the right, we obtain 1 = 400Q 2 2 (4000 +xY It remains only to solve this equation for x. Clearing of fractions, (4000 +z) 2 = 2 - 4000 2 = 4000 2 2. Extracting the square root of both members, 4000 +x =4000 \/2. Solving, x =4000 V2 -4000 =4000( V2 -1) miles. Ans. To find the approximate value of this answer, we have (see table) V2 = 1.41421 so that x =4000(1.41421 -1) = 4000 X. 41421 = 1656.84 miles. Ans. NOTE. Observe that the first step in the above solution (as also in the preceding exercises) is to. express as an equation the law of variation belonging to the problem. Then write down the two special equations which express the particular conditions given in the problem and divide one of these equations by the other to eliminate the unknown k. The answer is then readily obtained. A similar process should be followed in working the remaining exercises of this list. 12. Show that the earth's attraction at a point on the sur- face is over 5000 times as strong as at the distance of the moon, that is at the (approximate) distance of 280,000 miles. [HINT. Call W\ the weight of a given body on the surface, and let W 2 represent the weight of the same body at the distance of the moon from the earth's center. Then use the law expressed in (1) of the solution of Ex. 11.1 XVII, 114] VARIATION 189 13. A book is being held at a distance of 2 feet from an incandescent lamp. How much nearer must it be brought in order that the illumination on the page shall be doubled? (See Ex. 11 (6), p. 183.) 14. If two like coins (such as quarter dollars) were melted and made into a single coin of the same thickness as the origi- nal, show that its diameter would be \/2 times as great. [HINT. Call D the diameter of the given coins and A the area of each. Note that the area of the new coin will then be 2 A. Use the result stated in Ex. 3, p. 182.] 15. Find the result in Ex. 14 when four equal-sized coins are used. 16. Show that a falling body will pass over the second 3 feet of its descent in about .4 of the time it takes it to pass over the first 3 feet. (See Ex. 10, p. 183.) 17. The time required for a pendulum to make a complete oscillation (swing forward and back) varies directly as the square root of its length. By how much must a 2-foot pendu- lum be shortened in order that its time of complete oscilla- tion may be halved ? 18. If the diameter of a sphere be increased by 10%, by what per cent will the volume be increased ? 19. Show that if a city is receiving its water supply by means of a main (large pipe) from a reservoir, the supply can be increased 25% by increasing the diameter of the main by about 12%. 20. It is desired to build a ship similar in shape to one already in use but having a 40% greater cargo space (or hold). By what per cent must the beam (width of the ship) be in- creased. (See 99 (c).) 190 SECOND COURSE IN ALGEBRA [XVII, 115 115. Variation Geometrically Considered. If a variable y varies directly as another variable x, we know ( 113) that this is equivalent to having the equation y = kx, where k is some constant. If the value of & is 1, this equation takes the definite form y = x, and we may now draw its graph, the result being a certain straight line. If, on the other hand, k = 2, we have y = 2 x } and this again is an equation whose 234 FIG. 68. DIRECT VARIATION. graph may be drawn, leading to a straight line, but a differ- ent one. In general, whatever the value of jfe, the corre- sponding equation has a straight-line graph. The fact that in all cases the graph is a straight line characterizes this type of variation, that is, characterizes the type in which one variable varies directly as another. The figure shows the lines corresponding to several different values of k. XVII, 115] VARIATION 191 In case a variable y varies inversely as another variable x, we know ( 113) that there exists an equation of the form y = k/x, where k is some constant. If we let fc = l, this be- comes y=l/x. By letting x take a series of values and determining the corresponding values of y from this equa- tion (thus forming a table as in 57) we obtain the graph. Similarly, corresponding to the value A; = 2 we have y = 2/x, 1 2 3 4 X FIG. 69. INVERSE VARIATION. and this equation has a definite graph which is different from the one just mentioned. In general, whatever the value of fc, the corresponding equation has a graph, but it is now to be noted that these graphs are not straight lines; they are hyperbolas. (See Ex. 2, 78.) The figure shows the curves corresponding to several different values of k. NOTE. Though these curves differ in form, they have the follow- ing feature in common : Through the origin draw any two straight 192 SECOND COURSE IN ALGEBRA [XVII, 115 lines (dotted in figure). Then the intercepted arcs AB, CD, EF, GH, etc., are similar, that is the smallest arc when simply magnified by the proper amount produces one of the others. EXERCISES Draw diagrams to represent the geometric meaning of each of the following statements. 1. y varies directly as the square of x. 2. y varies inversely as the square of x. 3. y varies as the cube of x. 4. y varies directly as x, and y = 6 when x = 2. [HINT. The diagram here consists of a single line.] 5. y varies inversely as x, and y = 6 when x = 2. 6. The cost of n pounds of butter at 40j per pound is c = 40 n. 7. The amount of the extension, e, of a stretched string is proportional to the tension, t, and e = 2 in. when t = 10 Ib. (See Ex. 11 (c), p. 184.) 8. The pressure, p, of a gas on the walls of a retaining vessel varies inversely as the volume, v ; and p = 40 Ib. per square foot when v = 10 cu. ft. 9. The length, L, of any object in centimeters is propor- tional to its length, I, expressed in inches; and L = 2.54 when 1=1. NEWTON (Sir Isaac Newton, 1642-1727) Discoverer of the law of gravitation and famous in algebra for his discov- ery of the binomial theorem. Inventor of the branch of higher mathematics called the Calculus, wherein rates of motion and other changing, or variable, quantities are extensively studied. CHAPTER XVIII EXPONENTS I. POSITIVE INTEGRAL EXPONENTS 116, Powers. Involution. Just as a 2 = u - a ; a 3 = a -a a ; etc., so we define the nth power of a, where n is any posi- tive integer, as follows : a n = a'Ci'a'a'a--a (n factors) . The process of finding the power of a number, or expression, is called involution. 117. Laws of Exponents. There are five fundamental laws of exponents which are as follows, it being understood that m and n everywhere stand for positive integers : I. MULTIPLICATION LAW. This law for multiplying two powers of the same quantity is Qm . ^n = -^7n+n < PROOF. a m = a a - a a a a (m factors). ( 116) a n a- a- a- a a a (n factors). Hence a m . a n ={a a- a a a (m factors)} {a a a a a (n factors)} = a- a- a - a--- a (m-\-n) f actors = a m+n . ( 116) Therefore a m - a n = a m+n . ILLUSTRATIONS. 194 SECOND COURSE IN ALGEBRA [XVIII, 117 II. DIVISION LAW. The law for dividing one power by another power of the same quantity is a m -^- a n = a m ~ n . PROOF. a m -=- a = = * ' * ' * ' **'" a ' a ' a '" a ( m factor3 ) a n tf tf fa JL (n factors) = a a a a a (mri) factors = a m ~ n . ( 116) Therefore a m -j-o n = a m ~ n . ILLUSTRATIONS. 3 6 ^.32 =3 6-2 =3 4 . ( _ 2 )5 + ( _2)3 = ( -2)2 ; x 8 -r-z 5 =x 3 (a +b) 7 -7- (a +6) 3 = (a +6) 4 . III. LAW FOR THE POWER OF A POWER. The law for raising a power of a quantity to a new power is mn a PROOF. (a m ) n = a m a m a m a m - a (n factors) ( 116) = a m+m+m+m + "- +m (n terms)^ (La\V I) Therefore (a m } n = a mn since m +m -\-m + +m to n terms = ran. ILLUSTRATIONS. ( 4 2)3 =4 2X3 = 46 ; (_2)3}3 = (_ 2 )9; (X 4)5 =:C 20 ; {(a +6)3)4 = (a +6) 12 . IV. LAW FOR THE POWER OF A PRODUCT. The law for raising to a power a product of two quantities is (ab) n = a n b n . PROOF. (aZ>) (afc) (06) (n factors) (116) {a a a a (n factors)} [b b b b b (n factors)} a n b n . Therefore ILLUSTRATIONS. (2 X3) 4 = 2 4 X3 4 ; {( -3) ( -2)} 3 = ( -3) 3 ( -2) 3 ; {(a +6) (c +d)} 3 = (a +6) 3 XVI11, 117] EXPONENTS 195 V. LAW FOR THE POWER OF A QUOTIENT. The law for raising to a power the quotient of two quantities is Jb, PROOF. M n /|j . M . /^ ... ^\ ( n factors) ( 116) a (n factors) a b - b - b b (n factors) b n Therefore ILLUSTRATIONS. (|V~fJ; f-^V \o/ o \ o / o" (x\ 7 _x^. /a -|-b\ 5 _ (q-j-b) 5 W Z/ 7 ' \a b) (a 6) 5 EXERCISES Find the results of the indicated operations in the following cases, using one (or more) of the five laws in 117. 1. 2 5 2 3 . 8. I 2a t a . 15. x l + x 2 . 2( 1 "\3 / 1 "\2 Q ?r 1 /yr+1 1C ->jl2 ^yj6 . ^ i) \ L) . o. Z Z . lu. 7/fr "!~7fl . 4/y.lO .7.2 -11 /7P+g 1 . /^l+r 1 Q nm,^_rA . ju JL , XJ.. ;/ y . xo. ly . y . 5. m 12 -m 13 . 12. 8 3 -^8 2 . 19. i 2a ^-^. 6. y b -y n . 13. (-3) 5 ^(-3) 3 . 20. z^+z*- 1 . 22. w + p - 1 -7-0 1 +'\ 25. S(-8) 3 J 4 . 23. (a+b) 2r +(a+b) r - 1 . 26. (x 6 ) 4 . 28. (m 4 ) 8 . 24. (2 5 ) 3 . 27. (?/ 3 ) 7 . 29. (a 2 6) 3 . [HINT ro Ex. 29. First use Law IV, then Law III.] 30. (z 3 !/ 2 ) 2 . 32. (a 2 6 3 c) 4 . 34. S(a+6) 2 (c+d) 3 | 4 . 31. (a6c) 3 . 33. (ra 2 n 3 w 4 ) 3 . 35. (xY) 2m - 36. (r 2 < o\3 />?5\4 /r n2 37. f ? ) - 38. ^ ) . 39. n J . . V n J \s 196 SECOND COURSE IN ALGEBRA [XVIII, 117 40. f*- a Y- . 42. c^y Y^y. 44. f-^v. V2/y V2/V \ V 43. f-Y-^C- 9 Y- 45. (- W \y*J V 2/ 3w 118. Roots. Evolution. Just as VcT means the number whose square gives a, and Vo means the number whose cube gives a, etc., so we define the nth root of a, Va, to be the number whose nth power gives a, that is we agree that Thus Vx l2 =x 4 , because (z 4 ) 3 =z 12 . ( 117, Law III.) Similarly Va 16 6 8 = a 4 6 2 , because (ct 4 6 2 ) 4 = ct 16 b 8 . NOTE. In case n = 2, we write simply V" instead of v/ . The number n is called the incfex of the root. The number under the sign V~, as a, is called the radicand. The process of finding the root of a number or expression is called evolution. 119. Rule for Finding the nth Root of an Expression. The nth root of an expression may be obtained readily in case the expression itself is an exact nth power. This is illustrated in the following examples. EXAMPLE 1. To find the value of \^m*n 9 . SOLUTION. The expression m e n 9 may be written as an exact cube, namely (m 2 n 3 ) 3 . (Laws IV and III of 117.) Therefore \/m 6 n 9 = V(m 2 n 3 ) 3 = m 2 w 3 . Ans. (118) EXAMPLE 2. To find the value of SOLUTION. The expression x V may be written as an exact 5th z 15 power, namely f ^) - (Laws IV and III of 117.) Therefore A/^- 6 ==A/(^V=^- Ans. (118) XVIII, 119] EXPONENTS 197 Observe that the answer to Example 1 (namely ra 2 n 3 ) is the result of simply dividing each exponent of the radicand (namely w 6 n 9 ) by the index of the desired root (namely 3). Similarly, in Ex. 2 if we simply divide each exponent in f- by 5 we get the answer immediately. Thus, in practice, & we use the following rule. To find the nth root of an exact nth power, divide the expo- nent of each factor of the radicand by n. Thus NOTE. It jwill be recalled ( 39) that, unless otherwise stated, the symbol Va means the positive number whose square is a. Thus V9=+3, the other root, 3, being represented by V9. This agreement is made in order to bring about perfect definiteness in the use of the symbol V . EXERCISES Determine (from the definition in 1 18) the value of : _ a 625. 2. V-27. 3. -v/SL 6. \/A. 8. Determine by means of the Rule in 119 the value of: 10. ^64 a 6 6 6 . [HINT. Write 2 6 for 64.] 23. 24. 11. A/625 a 8 6 4 . 12. \/-27m?n. 13. - 14. - 15. 22. 198 SECOND COURSE IN ALGEBRA [XVIII, 120 II. FRACTIONAL, ZERO, AND NEGATIVE EXPONENTS 120. Introduction of General Exponents. Thus far we have considered only positive integral exponents. Such symbols as a 3/4 and or 2 thus have no meaning for us as yet since there can be no such thing as taking a as a factor three fourths times, or minus two times. However, we shall now see that by extending our definitions we can assign perfectly definite meanings to these symbols as well as to all others wherein fractional, zero, or negative exponents occur. 121. Meaning of a Fractional Exponent. If a 374 is to obey the multiplication law ( 117) then a . a . a . a = ^ That is (a 3/4 ) 4 = a 3 , so that we must have = 3/4+3/4+3/4+3/4 _ Thus we naturally take vV to be the meaning of a 3 / 4 . Similarly (if the multiplication law is to hold true), the meaning of a 273 is v'a 2 , while that of a 475 is x/a 4 , etc. So, in all cases a m/n means the nth root of a m , that is, Qinln _ ^/Q m . Thus 82/3 = ^82 = ^64 = 4. Similarly, _ _ (2^4)8/4 = #( x 8g4)3 = ^24012 = X *y3. ($66 Rule 111 119.) EXERCISES Express with radical sign and find the value of : 1. 8 173 . 6. 27 273 . 11. G/ 10 ) 175 - 2. 4 1/2 . 7. 32 275 . 12. (i/ 10 ) 275 . 3. 9 1/2 . 8. 81 374 . 13. 4. 27 173 . 9. 64 176 . 14. 6. (-8) 173 . 10. (x 6 ) 173 . 15. S(a+6) 3 i 273 . XVIII, 123] EXPONENTS 199 Express with radical signs : 16. 2 2/3 . 18. 4 3/2 . 20. (a 2 ) 4 / 3 . 22. 2 x l/ \ 24. m 2/3 n 3 / 4 . 17. 3 2/3 . 19. a 473 . 21. (3z) 3/4 . 23. (2 a&) 3/4 . 25. (z+t/) 5/6 . Express with fractional exponents : 26. V^ 4 . 29. 2v / z*~. 32. Sv^n 2 ". 35. V(a+6) 3 . 27. V^ 5 . 30. ^(-a) 2 . 33. 28. v^S. 31. Vabc. 34. 122. Meaning of a Zero Exponent. If a is to obey the multiplication law (117), then a m -a = a m +, that is a m a = a m . Dividing both members of the last equality by a m gives a = a -f- a m = 1 . That is, This means that the zero power of any number a (except 0) must always be taken equal to 1. Thus 3 = 1; (-32) = l;()o = l; x o = i ; ( mn ) = l; (o+6)=l; etc. 123. Meaning of Negative Exponents. If a~ m is to obey the multiplication law (117), then a m a~ m = a m-n = a, that is a m a~ m =l ( 122). Dividing both members of the last equation by a m gives This means that a negative power of any number a must al- ways be taken equal to 1 divided by the corresponding positive power of a. 2' 8' (-4)' 1 Similarly, (+!>)-"* = L =. 200 SECOND COURSE IN ALGEBRA [XVIII, 123 EXERCISES Express with positive exponents and find the values of each of the following expressions. 1. S- 2 . 5. 2- 1 3- 2 . 9. 8 2 4~*. 13. 81~ 1/4 . 2. 4~ 2 . 6. 4 - S- 3 . 10. 8~ 1/3 . 14. 64~ 1/6 . 3. 2~ 4 . 7. 7-4-4. 11. (-8)- 1/3 . 15. (-125)~ 1/3 . 4. 8. 8. 2~ 3 -8 4-. 12. 27~ 1/3 . 16. (-32)- 1 / 5 . Write with positive exponents each of the following ex- pressions. 17. x 3 y~*. 20. (2 a)" 3 ?) 3 . 23. ~ l m i n~*. 18. x~ 2 y 2 (Law II) cr 3 3. (a- 372 ) 475 = a'- 372 ) ' 4/5 = a- 675 . (Law III) 4. (a-^6 2 )- 1/4 = a(- 3 )(- 174 ) 6 2 <- 1/4 > = a 374 &- 172 . (Law IV) 202 SECOND COURSE IN ALGEBRA [XVIII, 125 HISTORICAL NOTE. The idea of using exponents to mark the power to which a quantity is raised is due to the French mathe- matician and philosopher Descartes (1596-1650) ; see the picture fac- ing p. 41), but he used only positive integral exponents, as in a 1 , a 2 , a 3 , a 4 , . The English mathematician Wallis (1616-1703) en- couraged the use of fractional and negative exponents and caused them to be brought into general use. EXERCISES In the following exercises, assume that the laws of 117 hold true for all rational exponents. LAW I Multiply 1. a 2 by a- 1 . 8. a 1/2 6 1/3 by a 1/2 6 2/3 , 2. a 3 by a~ 2 . 9. ra 1/2 rc 1/3 by m 3/2 rc 2 / 3 . 3. a 3 by or 3 . 10. p~ 1/4 q by 4 p 5/4 . 4. a by cr 4 . 11. n 2 by 6n~ 3 . 5. cr 2 by a- 3 . 12. x m ~ n by x m+n . 6. a 173 by a 2/3 . 13. z (m+n)/2 by x (m ~ n)/2 . 7. ar 1/2 byz 3/2 . 14. a 1/2 +6 1/2 by a 1/2 6 1/2 . 16. a 2/3 +a 1/3 6 1/3 +& 2/3 by a 1/3 -6 1/3 . [HINT. Follow the rule for multiplying one polynomial by another, as given in 8.] 16. m 1/3 +m 1/6 n 1/6 +n 1/3 by m 1/3 -m 1/6 n 1/6 +n 1/3 . Carry out the following indicated operations. 17. (a 1/2 +6 1/2 )(a 1/2 -6 1/2 ). 20. 18. (x 2/3 +2/ 2/3 )(x 2/3 -i/ 2/3 ). 21. 19. (a 1/2 +6 1/2 ) 2 . 22. 23. (z 2 24. (x 2 25. (2z 2/3 -3z 1/3 +4)(2+3ar 1/3 ). 26. x 2 - XVIII, 125] EXPONENTS 203 LAW II Divide 27. a 4 by a 5 . 29. z 2 by or 2 . 31. (ran) 2 / 3 by (mnY /s 28. a 3 bya. 30. z 372 by z~ 1/2 . 32. x l/ Y /2 by y~ l/2 . 33. x*-\-x 2 y 2 -\-y* by z 2 ?/ 2 . SOLUTION. x ~*~ x ^ +3r = _g |_?J. _| ^_=^_. X 2^2 X 2^2 X 2^2 X 2^2 ^/2 34. a 3 +a 2 +abya 4 . 35. a~ 4 +a~ 2 6+6 2 by a~ 2 b. 36. x 4 -|-2 az 3 -|-5 z" 1 ?/ Q?]r^~}~y* by x 1 2/~ 2 . 37. a 6 by a 1/2 +6 1/2 . SOLUTION, a 61 a+a 1/2 b 1/2 a 1/2 -6 1/2 . a 1/2 6 1/2 6 38. a-6bya 1/2 -6 1/2 . 41. z-1 by z 2/3 +z 1/3 +l. 39. a-f6 by a 1/3 +6 1/3 . 42. x - y by z 174 - 1/ 1/4 . 40. z 2 +2/ 2 by z 2/3 +2/ 2/3 . 43. m 2 -n 3 by w 1/3 +n 1/2 . LAW III Simplify 44. (a 1 / 2 ) 3 . SOLUTION. As in Illustration 3 of 125, the answer is a 1/2X3 , or a 3 / 2 . 45. (a 1 / 2 ) 2 . 46. (or 173 ) 6 . 47. (or 5 ) 2 . 48."(8- 1/3 ) 2 . 49. (16~ 1/2 ) 3 . 50. Var 2 . [HINT. V^ = ( a ~ 2 ) 1 / 2 by 121.] 61. VaF" 2 . 52. v^z 372 . 204 SECOND COURSE IN ALGEBRA [XVIII, 125 LAWS IV AND V Simplify 63. SOLUTION. As in Illustration 4 of 125, we have (a~ 4 6 2/3 ) -1 / 2 = 4)(-l/2) . 52/3(-l/2) = a 25-l/3. 64. (a l73 &- 172 ) 6 . 65. (z 473 2T 3 )- 173 . 66. vV 172 6- 3 . 57. Vz 473 ?/- 3 . 58. 59. 60. 61. [HINT to Ex. 62. See Illustration 5 in 125.] 63. . 64. * MISCELLANEOUS EXERCISES Expand, by use of Formulas VI and VII of 10. 1. (a: 172 -i/ 172 ) 2 . 3. (a 173 +6 173 ) 2 . 2. (a^+Zr 1 ) 2 . 4. (1+2 x 172 ) 2 . Simplify, expressing results with positive exponents. 6. -VWxVa-* z y ^T^/W" 10. Solve forz and check each result in the following equations. 11. z 3/4 = 8. [HiNT. Write z 3 ' 4 in the form Or 1 / 4 ) 3 .] 12. z 275 = 9. 14. z 372 = 72. 16. 25ar 273 = l. 13. z 473 = 16. 16. la; 273 = 25. 17. or 372 -27 = 0. CHAPTER XIX RADICALS 126. Important Formulas. In 3 we defined VcTas mean- ing that number or expression which, when raised to the nth power, would give a ; that is (1) (W = a. Unless a is an exact nth power of some number or expres- sion, we agreed ( 41) to call Va a radical of the nth order. Thus V5, V23, Vj, V.05, Vx+y, Vm 2 +n 2 are radicals of the second order; #5~, v^ ^f' v/x+y are radicals of the third order, etc. Moreover, we saw ( 44) that there exist two general formulas as follows : (2) _ Va (3) And, in connection with the study of fractional exponents, we havejseen ( 121) that the meaning of a 1/n must be taken to be Va, that is we have the formula (4) a 1/n =Va. The four formulas (1), (2), (3), (4) contain all that is essential in the study of radicals. In fact, we have already seen in Chapter IX how (1), (2), and (3) are thus used. In the present chapter we shall review and extend those studies, making use now of (4) also. 205 206 SECOND COURSE IN ALGEBRA [XIX, 127 127. Simplification of Radicals. EXAMPLE 1. Simplify V75. SOLUTION. V75 = V25X3 = V25 X Vjj = 5 Vs. Ans. (Formula (2), 126) Ay-J A fit *T EXAMPLE 2. Simplify SOLUTION. Ans. EXERCISES Before undertaking the following exercises, review 44, including the note on page 72. These resemble the exercises on pages 73, 74, but in some instances are more difficult because they refer to radicals of as high orders as the fifth, sixth, and seventh. Simplify each of the following expressions. 3/3. ^3 \* 9 29. 30. A 31. 27 32. 33. 34. 8 / 4 \125 [HINT. See Note in 45.] 35. 81 37. XIX, 128] RADICALS 207 EXERCISES Write each of the following in a form having no coefficient outside the radical sign. First review the similar exercises on page 74. 1. 3^2. SOLUTION. 3^2 = ^27X^2 = v / 27><2 = ^54. Ans. 2. 2\/3. 7. 4oV2a. 12. 3. 2\/2. 8. GsVjfi?. 2r 13 ^ 4. 3A/3. 9. 5. 2"V / f. 10. mv^n. 2x 6. 3^6. 11. 2a^. T 2 128. Reduction of Radicals of Different Orders to Equiva- lent Radicals of the Same Order. EXAMPLE. Reduce V% v^, and ^5 to equivalent radi- cals of the same order. SOLUTION. By use of Formula (4), 126, and the laws of expo- nents ( 117) we may write V2"= 2 1/2 = 2 6/12 = v^ = V64, #3= 3 1 / 3 = 3 4 / 12 = v^3* = v/81, ^5 =51/4 =53/12 = ^5?= 1^ NOTE. As now expressed, the given radicals may be compared as to their magnitudes. Thus we see V5 is the greatest of the three since (the orders of the radicals being now the same) it has the largest radicand, namely, 125. An examination of the process just followed leads to the following rule. To reduce radicals to equivalent radicals of the same order : 1. Express the radicals with fractional exponents. 2. Reduce the exponents to a common denominator. 3. Rewrite the results thus obtained in radical form. 208 SECOND COURSE IN ALGEBRA [XIX, 128 EXERCISES Reduce each of the following groups to equivalent radicals of the same order. 1. V3andV?. 3. Vjfand V/3. 3. 7 +4 Vs. 5. 6-V2a 7. 7-V|a 9. 20-6VIT. 10. Establish Formula (3) of 135 by a process similar to that used in establishing Formula (2). CHAPTER XX LOGARITHMS I. GENERAL CONSIDERATIONS f 136. Definition of Logarithms. If we ask what power of 10 must be used to give a result of 100, the answer is 2 because 10 2 =100. Another common way of stating this is to say that " the logarithm of 100 is 2." In the same way, the power of 10 needed to give 1000 is 3 because 10 3 = 1000, and this is briefly stated by saying that " the logarithm of 1000 is 3." Similarly, the power of 10 that gives .1 is 1 because 10~ 1 = 1 1 j y , or .1 ( 123), and this is equivalent to saying that " the logarithm of .1 is 1." Likewise, the loga- rithm of .01 is -2. Why? From these illustrations we readily see what is meant by the logarithm of a number. It may be denned as follows : The logarithm of a number is the power of 10 required to give that number. NOTE. A more general definition will be given in 151, but this is the one commonly used in practice. The fact that the logarithm of 100 is 2 is written log 100 = 2. Similarly, we have log 1000 = 3, log .1 = - 1, log .01 = -2, etc. t Parts I and II give definitions and essential theorems which should be well understood before Part III, which describes the im- portant applications, is taken up. 216 XX, 137] LOGARITHMS 217 EXERCISES 1. What is the meaning of log 10000 ? What is its value f 2. What is the value of log .001? Why? 3. What is the value of log .00001 ? Why? 4. What is the value of log 10? 5. What is the value of log 1 ? (See 122.) 6. As a number increases from 100 to 1000 how does its logarithm change? 7. As a number decreases from .1 to .01 how does its logarithm change? Answer the same as the number goes from .01 to .001 ; from 1 to 10 ; from 1 to 1000. 8. Explain why the following are true statements : (a) log 100000 = 5. (6) log .0001 =-4. (c) logv^0 = i. [HINT. Remember VlO = 10 1/2 . (d) log >^IO = i. (e) log ^100 = 1. [HINT. Remember \/100 = v/10 2 = 10 2 ' 3 . ( 121.)] 137. Logarithm of Any Number. Suppose we ask what the value is of log 236. What we are asking for (see defini- tion in 136) is that value which, when used as an exponent to 10, will give 236 ; that is we wish the value of x which will satisfy the equation 10 X = 236. This question resembles those in 136, but is different because we cannot immediately arrive at the desired value of x by mere inspection. All we can say here at the beginning is that x must lie somewhere between 2 and 3, because 10 2 =100 and 10 3 =1000, and 236 lies between these two numbers. In order to find z to a finer degree of accuracy, it is now natural to try for it such values 218 SECOND COURSE IN ALGEBRA [XX, 137 as 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, and 2.9, all of which lie between 2 and 3. The result (which for brevity we shall here state without proof) is that when x = 2.3 the value of 10* is slightly less than our given number 236, while if we take x= 2.4 the value of 10* is slightly greater than 236. Thus x lies somewhere between 2.3 and 2.4. In other words, the value of log 236 correct to the first decimal place (compare 37) is 2.3. It is now natural, if we wish to obtain x to still greater accuracy, to try for it such values as 2.31, 2.32, 2.33, 2.34, 2.35, 2.36, 2.37, 2.38, and 2.39, all of which lie between 2.3 and 2.4. The result (which again is here stated without proof) is that when x = 2.37 the value of 10 X is slightly less than our number 236, while if we take # = 2.38 the value of 10* is slightly greater than 236. This means that the second figure of the decimal is 7, after which we may say that the value of log 236 correct to two places of decimals is 2.37. Proceeding farther in the same manner, it can be shown that when x = 2.372 the value of 10* is slightly less than 236, while for x = 2.373 the value of 10* is slightly greater than 236. Thus the value of log 236 correct, to three places of decimals is 2.372. Similarly, it can be shown that the number in the fourth decimal place is 9, and this is as far as it is necessary to carry out the process, since the result is then sufficiently accurate for all ordinary purposes. In summary, then, we have log 236 = 2.3729, this value being correct to four places of decimals. NOTE. It thus appears that logarithms do not in general come out exact, though they do so for such exceptional numbers as 100, 1000, 10,000, .1, .01, etc. (Compare 37.) They can be ex- pressed only approximately, yet as accurately as one pleases by carrying out the decimal far enough. In this respect they resemble such numbers as V2. \/2, V3, etc. XX, 138] LOGARITHMS 219 Other examples of logarithms are given below. Note especially the decimal part of each, which is correct to four places. log 283 = 2.4518 log 196 = 2.2923 log 17 = 1.2304 log 6 = 0.7782 log 3.410 = 0.5328 log 5.75 = 0.7597 138. Characteristic. Mantissa. We have seen that the logarithm of a number consists (in general) of an integral part and a decimal part. Thus log 236 =2.3729. Here the integral part is 2 and the deci- mal part is .3729. Similarly, in log 6 = 0.7782 the integral part is 0, while the decimal part is .7782. These two parts of every logarithm are given special names as follows : The integral part of a logarithm is called the characteristic of the logarithm. The decimal part of a logarithm is called the mantissa of the logarithm. Thus the characteristic of log 236 is 2, while its mantissa is .3729. (See above.) Similarly, the characteristic of log 6 is 0, while its mantissa is .7782. EXERCISES 1. What is the characteristic of log 100? What the mantissa? Answer the same questions for log 1000, log 10, and log 1. 2. What is the characteristic of log 156 ? [HINT. Note that 156 lies between 10 2 and 10 3 .] 3. What is the characteristic of log 276? of log 1376? of log 97? of log 18? of log 5? of log 11? of log 14798-? 4. For what kind of number can one tell by inspection both the characteristic and the mantissa of its logarithm? (See 136.) 220 SECOND COURSE IN ALGEBRA [XX, 1I19 139. Further Study of Characteristic and Mantissa. We have seen ( 138) that log 236 = 2.3729, which is the same as saying that (1) 10 2 - 3729 = 236. Let us now multiply both members of (1) by 10. The left side becomes 10 2 - 3729+1 or 10 3 - 3729 ( 117, Law I) while the right side becomes 2360. That is, we have 10 3 - 3729 = 2360, which is the same as saying that log 2360 = 3.3729 If, instead of multiplying both sides of (1) by 10, we divide both by 10, we obtain in like manner lO 2 - 3729 " 1 = 23.6 ( 117, Law II). That is, we have 10 1 - 3729 = 23.6, which is the same as saying that log 23.6 = 1.3729 Finally, if we divide both sides of (1) by 10 2 , or 100, we obtain 10 2 - 3729 ~ 2 = 2.36. That is, we have 10- 3729 = 2.36 which is the same as saying that log 2.36 = 0.3729 What we now wish to do is to compare the results which we have just been obtaining, and for this purpose they are ar- ranged side by side in a column below. log 2360 = 3.3729 log 236 = 2.3729 log 23.6=1.3729 log 2.36 = 0.3729 Note that the mantissas here appearing on the right are all the same, namely .3729, while the numbers appearing on the left (that is, 2360, 236, 23.6, and 2.36) are alike except for the position of the decimal point, that is they contain the same significant figures. This illustrates the following important rule. (2) XX, 141] LOGARITHMS 221 RULE I. // two or more numbers have the same significant figures (that is, differ only in the location of the decimal point) , their logarithms will have the same mantissas, that is their logarithms can differ only in their characteristics. Thus log 243, log 2430, log 24.3, log 2.43, log .243, and log .0243 all have the same mantissas. It is only their characteristics that can be different. EXERCISES Apply Rule I, 139, to tell which of the following loga- rithms have the same mantissas. log .167 log 8100 log 16.7 log 81 log .0072 log .081 log 7.2 log 720 log 1670 log 16700 II. To DETERMINE THE LOGARITHM OF ANY NUMBER 140. Purpose of This Part. When we wish to determine the value of a logarithm, as, for example, to find log 236, we can work out the characteristic and mantissa as explained in 137, but this requires considerable time. What we do in practice is to use certain simple rules for determining the characteristic, and we determine the mantissa directly from certain tables which have been carefully prepared for the purpose. We shall now state these rules ( 141-143) and explain the tables and how to use them ( 144-146). 141. Characteristics for Numbers Greater than 1. If we look again at the results in (2) of 139, we see that the characteristic of log 2360 is 3. Thus the characteristic is 1 less than the number of figures to the left of the decimal point. NOTE. 2360 is the same as 2360., so that there are four figures here to the left of the decimal point. SECOND COURSE IN ALGEBRA [XX, 141 Again, we see from (2) of 139 that the characteristic of log 236 is 2 and this, as in the case already examined, is 1 less than the number of figures to the left of the decimal point. NOTE. 236 is the same as 236., so there are three figures here to the left of the decimal point. Similarly, since the characteristic of log 23.6 is 1 (see (2) of 139) this again obeys the same law as just observed in the other two cases, that is, the characteristic is 1 less than the number of figures to the left of the decimal point. Finally, since the characteristic of log 2.36 is 0, the same law is again present here. Explain. The law which we have just observed can be shown in like manner to hold good for the characteristic of the loga- rithm of any number greater than 1 ; hence we may state the following general rule. RULE II. The characteristic of the logarithm of a number greater than 1 is one less than the number of figures to the left of the decimal point. Thus the characteristic of log 385.9 is 2 ; that of log 8.679 is 0. EXERCISES State, by Rule II, 141, the characteristic of the logarithm of each of the following numbers. 1. 476.5 5. 89.65 9. 500.005 2. 325. 6. 105,000. 10. 3076.8 3. 8976. 7. 17.694 11. 41. 4. 1.6 8. 2.0815 12. 3.25679 State how many figures precede the decimal point of a number if the characteristic of its logarithm is 13. 3. 14. 2. 15. 0. 16. 1. 17. 4. 18. 5. XX, 142] LOGARITHMS 223 142. Characteristics for Positive Numbers Less Than 1. We have seen (see (2) in 139) that log 2.36 = 0.3729, which is the same as saying that (1) 10- 3729 = 2.36 Let us now divide both members of this relation by 10. We thus obtain ( 117, Law II) 10 o.372 9 -i = .236 (or 10- 1+0 - 3729 = .236) , which gives us (by 136) log .236 =-1+0.3729 Observe that 1 +0^3729 is really a negative quantity, being equal to -(1-0.3729) which reduces to -0.6271. However, it is more convenient for our present purposes to keep the longer form 1+0.3729. Note that this cannot be written as 1.3729 be- cause this last is equal to -1-0.3729 instead of -1+0.3729. If, instead of dividing both members of (1) by 10, we divide both by 10 2 , or 100, we obtain JQO.3729-2 = <0 23 6 (or lQ-2+0.3729 = .Q236), which means that log .0236 =-2+ 0.3729 Similarly, by dividing (1) by 10 3 , or 1000, we find that log .00236= -3+0.3729 Finally, if we divide (1) by 10 4 , or 10000, we find that log .000236= -4+0.3729 Let us now compare the four results just obtained. Be- ginning with the last result, we see that in the number .000236 there are three zeros immediately to the right of the decimal point, that is, between the decimal point and the first significant figure. Corresponding to this, the charac- teristic on the right is minus four. Hence the characteristic is negative and 1 more numerically than the number of zeros between the decimal point and the first significant figure. 224 SECOND COURSE IN ALGEBRA [XX, 142 Similarly, in the number .00236 there are two zeros between the decimal point and the first significant figure, and corre- sponding to this there is a characteristic on the right of minus three. Hence, as before, the characteristic here is negative and numerically 1 more than the number of zeros between the decimal point and the first significant figure. This statement, which is true in all cases mentioned above, can be proved for the characteristic of the logarithm of any positive number less than 1. Hence we have the following rule. RULE III. The characteristic of the logarithm of a (positive) number less than 1, is negative, and is numerically 1 greater than the number of zeros between the decimal point and the first significant figure. Thus the characteristic of log .0076 is -3 ; that of log .28 is -1. NOTE. The logarithm of a negative number is an imaginary quantity (as shown in higher mathematics), and hence we shall con- sider here the logarithms of positive numbers only. 143. Usual Method of Writing a Negative Characteristic. In 142 we saw that log .236= -1+0.3729. If we add 10 to this quantity and at the same time subtract 10 from it, we do not change its value, but we give it the new form 9+0.3729-10, which is the same as 9.3729-10. That is, we may write. log .236 = 9.3729 -10. This is the form used in practice. Likewise, instead of writing log .0236= -2+0.3729 (see 142) we write in practice log .0236 = 8.3729-10, and similarly we write log .00236 = 7.3729 -10. XX, 144] LOGARITHMS 225 Thus the usual method of expressing the characteristic of 1 is to write 910 for it ; if it is 2, we write 8 10 for it ; if it is 3, we write 7 10 for it, etc. For example, log .0076 has the characteristic 7 10. EXERCISES State, by Rule III, 142, the value of the characteristic of the logarithm of each of the following ; state how it would be written if expressed in the usual form described in 143. 1. .06 -2, or 8-10. Ans. 2. .0087 5. .0835 8. .00978 3. .75 6. .835 9. .12345 4. .00067 7. .33764 How many zeros lie between the decimal point and the first significant figure of a number when the characteristic of its logarithm is 10. -3. 11. 9-10. 12. -5. 13. 8-10. 14. 7-10. 144. Determination of Mantissas. Use of Tables. Sup- pose we wish to determine completely the value of log 187. By Rule II, 141, we know that the characteristic is 2. To find the mantissa, we turn to the tables (p. 226) and look in the column headed N for the first two figures of the given number, that is, for 18. The desired mantissa is then to be found on the horizontal line with these two figures and in the column headed by the third figure of the given num- ber, that is, in the column headed by 7. Thus in the present case the mantissa is found to be .2718. NOTE. For brevity, the decimal point preceding each mantissa is omitted from the tables. It must be supplied as soon as the mantissa is used. The complete value (correct to four decimal places) of log 187 is therefore 2.2718. 226 SECOND COURSE IN ALGEBRA N O 1 2 3 4 5 6 7 8 9 1O 11 12 13 14 0000 0414 0792 1139 1461 0043 0453 0828 1173 1492 0086 0492 0864 1206 1523 0128 0531 0899 1239 1553 0170 0569 0934 1271 1584 0212 0607 0969 1303 1614 0253 0645 1004 1335 1644 0294 0682 1038 1367 1673 0334 0719 1072 1399 1703 0374 0755 1106 1430 1732 15 16 17 18 19 1761 2041 2304 2553 2788 1790 2068 2330 2577 2810 1818 2095 2355 2601 2833 1847 2122 2380 2625 2856 1875 2148 2405 2648 2878 1903 2175 2430 2672 2900 1931 2201 2455 2695 2923 1959 2227 2480 2718 2945 1987 2253 2504 2742 2967 2014 2279 2529 2765 2989 2O 21 22 23 24 3010 3222 3424 3617 3802 3032 3243 3444 3636 3820 3054 3263 3464 3655 3838 3075 3284 3483 3674 3856 3096 3304 3502 .3692 3874 3118 3324 3522 3711 3892 3139 3345 3541 3729 3909 3160 3365 3560 3747 3927 3181 3385 3579 3766 3945 3201 3404 3598 3784 3962 25 26 27 28 29 3979 4150 4314 4472 4624 3997 4166 4330 4487 4639 4014 4183 4346 4502 4654 4031 4200 4362 4518 4669 4048 4216 4378 4533 4683 4065 4232 4393 4548 4698 4082 4249 4409 4564 4713 4099 4265 4425 4579 4728 4116 4281 4440 4594 4742 4133 4298 4456 4609 4757 30 31 32 33 34 4771 4914 5051 5185 5315 4786 4928 5065 5198 5328 4800 4942 5079 5211 5340 4814 4955 5092 5224 5353 4829 4969 5105 5237 5366 4843 4983 5119 5250 5378 4857 4997 5132 5263 5391 4871 5011 5145 5276 5403 4886 5024 5159 5289 5416 4900 5038 5172 5302 5428 35 36 37 38 39 5441 5563 5682 5798 5911 5453 5575 5694 5809 5922 5465 5587 5705 5821 5933 5478 5599 5717 5832 5944 5490 5611 5729 5843 5955 5502 5623 5740 5855 5966 5514 5635 5752 5866 5977 5527 5647 5763 5877 5988 5539 5658 5775 5888 5999 5551 5670 5786 5899 6010 40 41 42 43 44 6021 6128 6232 6335 6435 6031 6138 6243 6345 6444 6042 6149 6253 6355 6454 6053 6160 6263 6365 6464 6064 6170 6274 6375 6474 6075 6180 6284 6385 6484 6085 6191 6294 6395 6493 6096 6201 6304 6405 6503 6107 6212 6314 6415 6513 6117 6222 6325 6425 6522 45 46 47 48 49 6532 6628 6721 6812 6902 6542 6637 6730 6821 6911 6551 6646 6739 6830 6920 6561 6656 6749 6839 6928 6571 6665 6758 6848 6937 6580 6675 6767 6857 6946 6590 6684 6776 6866 6955 6599 6693 6785 6875 6964 6609 6702 6794 6884 6972 6618 6712 6803 6893 6981 50 51 52 53 54 6990 7076 7160 7243 7324 6998 7084 7168 7251 7332 7007 7093 7177 7259 7340 7016 7101 7185 7267 7348 7024 7110 7193 7275 7356 7033 7118 7202 7284 7364 7042 7126 7210 7292 7372 7050 7135 7218 7300 7380 7059 7143 7226 7308 7388 7067 7152 71':;.-, 7316 7396 LOGARITHMS 227 N 1 2. 3 4 5 6 7 8 9 55 56 57 58 59 7404 7482 7559 7634 7709 7412 7490 7566 7642 7716 7419 7497 7574 7649 7723 7427 7505 7582 7657 7731 7435 7513 7589 7664 7738 7443 7520 7597 7672 7745 7451 7528 7604 7679 7752 7459 7536 7612 7686 7760 7466 7543 7619 7694 7767 7474 7551 7627 7701 7774 6O 61 62 63 64 7782 7853 7924 7993 8062 7789 7860 7931 8000 8069 7796 7868 7938 8007 8075 7803 7875 7945 8014 8082 7810 7882 7952 8021 8089 7818 7889 7959 8028 8096 7825 7896 7966 8035 8102 7832 7903 7973 8041 8109 7839 7910 7980 8048 8116 7846 7917 7987 8055 8122 65 66 67 68 69 8129 8195 8261 8325 8388 8136 8202 8267 8331 8395 8142 8209 8274 8338 8401 8149 8215 8280 8344 8407 8156 8222 8287 8351 8414 8162 8228 8293 8357 8420 8169 8235 8299 8363 8426 8176 8241 8306 8370 8432 8182 8248 8312 8376 8439 8189 8254 8319 8382 8445 70 71 72 73 74 8451 8513 8573 8633 8692 8457 8519 8579 8639 8698 8463 8525 8585 8645 8704 8470 8531 8591 8651 8710 8476 8537 8597 8657 8716 8482 8543 8603 8663 8722 8488 8549 8609 8669 8727 8494 8555 8615 8675 8733 8500 8561 8621 8681 8739 8506 8567 8627 8686 8745 75 76 77 78 79 8751 8808 8865 8921 8976 8756 8814 8871 8927 8982 8762 8820 8876 8932 8987 8768 8825 8882 8938 8993 8774 8831 8887 8943 8998 8779 8837 8893 8949 9004 8785 8842 8899 8954 9009 8791 8848 8904 8960 9015 8797 8854 8910 8965 9020 8802 8859 8915 8971 9025 8O 81 82 83 84 9031 9085 9138 9191 9243 9036 9090 9143 9196 9248 9042 9096 9149 9201 9253 9047 9101 9154 9206 9258 9053 9106 9159 9212 9263 9058 9112 9165 9217 9269 9063 9117 9170 9222 9274 9069 9122 9175 9227 9279 9074 9128 9180 9232 9284 9079 9133 9186 9238 9289 85 86 87 88 89 9294 9345 9395 9445 9494 9299 9350 9400 9450 9499 9304 9355 9405 9455 9504 9309 9360 9410 9460 9509 9315 9365 9415 9465 9513 9320 9370 9420 9469 9518 9325 9375 9425 9474 9523 9330 9380 9430 9479 9528 9335 9385 9435 9484 9533 9340 9390 9440 9489 9538 90 91 92 93 94 9542 9590 9638 9685 9731 9547 9595 9643 9689 9736 9552 9600 9647 9694 9741 9557 9605 9652 9699 9745 9562 9609 9657 9703 9750 9566 9614 9661 9708 9754 9571 9619 9666 9713 9759 9576 9624 9671 9717 9763 9581 9628 9675 9722 9768 9586 9633 9680 9727 9773 95 96 97 98 99 9777 9823 9868 9912 9956 9782 9827 9872 9917 9961 9786 9832 9877 9921 9965 9791 9836 9881 9926 9969 9795 9841 9886 9930 9974 9800 9845 9890 9934 9978 9805 9850 9894 9939 9983 9809 9854 9899 9943 9987 9814 9859 9903 9948 9991 9818 9863 9908 9952 9996 228 SECOND COURSE IN ALGEBRA [XX, 144 Again, suppose we wish to determine log 27.6. The char- acteristic (by 141) is 1. The mantissa, by Rule I, 139, is the same as that of log 276 ; and the latter, as given in the tables, is .4409. Therefore, log 27.6 = 1.4409. Ans. As a last example, suppose we wish to determine log .0173. The characteristic (by 142) is 2, or 8-10. The mantissa, by the rule in 139, is the same as that of log 173 and the latter, as obtained from the tables, is .2380. Therefore, log .0173 = 8.2380 -10. Ans. These examples illustrate how the tables together with Rules II and III, 141, 142, enable us to determine com- pletely the logarithm of any number provided it contains no more than three significant figures. We may now sum- marize our results in the following rule. RULE IV. To find the logarithm of a number of three signifi- cant figures : 1. Look in the column headed N for the first two figures of the given number. The mantissa will then be found on the hori- zontal line opposite these two figures and in the column headed by the third figure of the given number. 2. Prefix the characteristic according to Rules II and III, ' 141, 142. EXERCISES Determine the logarithm of each of the following numbers, expressing all negative characteristics as explained in 143. 1. 451. 2. 318. 3. 861. 4. 900. 5. 72.5 [HINT. Note how log 27.6 was obtained in 144.] 6. 7.25 7. 93. [HINT. Write as 93.0] 8. 9. [HINT. Write as 9.00] 9. .0136 10. .936 11. .0036 [HINT. Write as .00360] XX, 145] LOGARITHMS 229 12. 8560. 15. .45 18. .000235 13. .081 16. 61.7 19. i 14. .8 17. 23,500. 20. f. 145. To Find the Logarithm of a Number of More Than Three Significant Figures. Stippose we wish to determine log 286.7. Here we have four significant figures while our tables only tell us the mantissas of numbers having three (or less) significant figures (as in 144 and in the preceding ex- ercises). In such cases we proceed as follows : From the tables log 286 = 2.4564 | log 286.7 - ? Difference = 2.4597 - 2.4564 = .0033 log 287 = 2.4597 J Since 286.7 lies between 286 and 287, its logarithm must lie between their logarithms. Now, an increase of one unit in the number (in going from 286 to 287) produces an increase of .0033 in the mantissa. It is therefore assumed that an increase of .7 in the number (in going from 286 to 286.7) pro- duces an increase of .7 of .0033, or .00231, in the mantissa. Therefore log 286.7 = 2.4564+.7 of .0033 = 2.4564+.00231 = 2.45871, so that log 286.7 = 2.4587 (approximately). Ans. In practice the answer is quickly obtained as follows : The difference between any mantissa and the next higher one in the table (neglecting the decimal point) is called a tabular difference. The tabular difference in this example is 4597^564, or 33. Taking .7 of this gives 23.1, which (keeping only the first two figures) we call 23, and adding this to 4564 gives 4587. This, therefore, is the required mantissa of log 286.7, so that log 286.7 = 2.4587 (approximately) . Ans. 230 SECOND COURSE IN ALGEBRA [XX, 145 Similarly, in finding log 286.75 the tabular difference (as before) is 33. Taking .75 of 33 gives 24.75, which (keeping only two figures) has the approximate value 25. Hence the mantissa of log 286.75 is 4564+25 =4589. Therefore log 286.75 =2.4589, Ans. Below are two examples further illustrating how the above processes are quickly carried out in practice. The student should form the habit of writing the work in this form. EXAMPLE 1. Determine the value of log 48. 731 SOLUTION. Mantissa of log 487 = 6875 1m,, -, . . Mantissa of log 488 =6884 ) Tabular dlff ^nce =9 .31 X9 = 2.79 = 3 (approximately). Hence mantissa of log 48.731 =6875+3 = 6878. Therefore log 48.731 = 1.6878 Ans. EXAMPLE 2. Determine the value of log .013403 SOLUTION. Mantissa of 134 = 1271 Mantissa of 135 = 1303 ( Tabular difference =32. .03 X32 = .096 = 1 (approximately). Hence mantissa of log .013403 = 1271+1 = 1272. Therefore log .013403= -2 +.1272 = 8.1272 -10. Ans. NOTE. The process which we have employed for determining a mantissa when it does not actually occur in the tables is called interpolation. When examined carefully, it will be seen that the pro- cess is based upon the assumption that if a number is increased by any fractional amount of itself, the logarithm of the number will like- wise be increased by the same fractional amount of itself. Thus, in finding the mantissa of log 286.7 at the middle of p. 229, we assumed that the increase of .7 in going from 286 to 286.7 would be accom- panied by like increase of .7 in the logarithm. Such an assumption, though not exactly correct, is very nearly so in most cases and is therefore sufficiently accurate for all ordinary purposes. XX, 146] LOGARITHMS 231 Tables of logarithms much more extensive than those on pages 226, 227 have been prepared and are commonly used. See, for ex- ample, The Macmillan Tables. By means of these, any desired mantissa may usually be obtained as accurately as is necessary, directly, that is without interpolation. EXERCISES Obtain the logarithm of each of the following n'umbers. 1. 678.4 8. 4.806 15. 62.856 2. 231.3 9. 1.508 16. 541.07 3. 785.4 10. 3.276 17. 6.3478 4. 492.6 11. .4567 18. 3.1416 6. 856.8 12. .08346 19- 1.7096 6. 42.17 13. 856.34 20. .15786 7. 9.567 14. 243.47 21. .085679 146. To Find the Number Corresponding to a Given Loga- rithm. Thus far we have considered how to determine the logarithm of a given number, but frequently the problem is reversed, that is, it is the logarithm that is given and we wish to find the number having that logarithm. The method of doing this is the reverse of the method of 144-145, and is illustrated in the following examples. EXAMPLE 1. Find the number whose logarithm is 1.9547 SOLUTION. Locate 9547 among the mantissas in the table. Having done so, we find in the column N on the line with 9547 the figures 90. These form the first two figures of the desired number. At the head of the column containing 9547 is 1, which is therefore the third figure of the desired number. Hence the number sought is made up of the figures 901. The given characteristic being 1, the number just found must be pointed off so as to have two figures to the left of its decimal point (Rule II, 141). Therefore the number is 90.1. Ans. 232 SECOND COURSE IN ALGEBRA [XX, 146 EXAMPLE 2. Find the number whose logarithm is 0.6341 SOLUTION. As in Example 1, we look among the mantissas of the table to find 6341. In this case we do not find exactly this man- tissa, but we see that the next less mantissa appearing is 6335, while the one next greater is 6345. The numbers corresponding to these last two mantissas are seen to be 430 and 431 respectively. Whence, if x represents the num- ber sought, we have Mantissa of log 430 = 6335 j D - ff =6 1 Mantissa of log x = 6341 J > Tabular difference = 10. Mantissa of log 431 =6345 Since an increase of 10 in the mantissa produces an increase of 1 in the number, we assume that an increase of 6 in the mantissa will produce an increase of ^, or .6, in the number. Hence the number sought has the figures 4306. Since the given characteristic is 0, the number must be 4.306 (141). Ans. NOTE 1. The pupil will observe that in Example 1 the given mantissa actually occurs in the tables, while in Example 2 it does not, thus making it necessary in this last case to interpolate. (See the Note in 145.) NOTE 2. The number whose logarithm is a given quantity is called the antilogarithm of that quantity. Thus 100 is the anti- logarithm of 2, 1000 is the antilogarithm of 3, etc. EXERCISES Find the numbers whose logarithms are given below. 1. 1.8751 9. 1.4893 2. 2.9405 10. 2.8588 3. 0.3856 11. 3.7430 4. 3.5866 12. 0.5240 6. 9.6955-10 13. 0.6970 6. 8.7152-10 14. 9.7400-10 7. 7.4900-10 15. 8.3090-10 8. 6.8519-10 16. 7.5308-10 NAPIER (John Napier, 1550-1617) Famous as the inventor of logarithms and first to show the advantage of using them in reducing the labor of ordinary computations. Interested and active also in the political and religious controversies of his day. XX, 147] LOGARITHMS 233 III. THE USE OF LOGAKITHMS IN COMPUTATION 147. To Find the Product of Several Numbers. The pro- cesses of multiplication, division, raising to powers, and ex- traction of roots, as carried out in arithmetic, may be greatly shortened by the use of logarithms, as we shall now show. Let us take any two numbers, for example 25 and 37, and determine their logarithms. We find that log 25 = 1.3979 and log 37 = 1.5682. T^iis means ( 136) that 25 = 10 1 - 3979 and 37 = 10 1 - 5682 Multiplying, we thus have 25X37 = l0 1 -+i.B682 ( 117) Law j) The last equality means ( 136) that log (25X37) = 1.3979+1.5682, or log (25X37) =log 25+log 37. Similarly, if we start with the three numbers 25, 37, and 18 we can show that log (25X37X18)= log 25+log 37+log 18. Thus we arrive at the following important rule. RULE V. The logarithm of a product is equal to the sum of the logarithms of its factors. Thus log (13 X. 0156X99.8) =log 13+log .0156+log 99.8. The way in which this rule is used to find the value of the product of several numbers is shown below. EXAMPLE 1. To find the value of 13 X. 0156X99.8 SOLUTION. log 13= 1.1139 log .0156= 8.1931-10 log 99.8= 1.9991 Adding, 11.3061 -10, or 1.3061 Hence, by Rule V, the logarithm of the desired product is 1.3061. It follows that the product itself is the number whose logarithm is 1.3061. When we look up this number (as in 146) we find it to be 20.23. Hence 13 X. 0156X99.8 =20.23 (approximately). Arcs. 234 SECOND COURSE IN ALGEBRA [XX, 147 EXAMPLE 2. To find the value of 8.45X.678X.0015X956X.111 SOLUTION. log 8.45= 0.9269 log .678= 9.8312-10 log .0015= 7.1761-10 log 956= 2.9805 log .111= 9.0453-10 /Adding, 29.9600-30=9.9600-10. f Hence, by Rule V, the logarithm of the* desired product is seen to be 9.9600 -10. Therefore the product itself is found (as in 146) to be -912 (approximately). Ans. These examples lead to the following rule. RULE VI. To multiply several numbers : 1. Add the logarithms of the several factors. 2. The sum thus obtained is the logarithm of the product. 3. The product itself can then be determined as in 146. NOTE. It may happen (as in Example 2) that the sum of several logarithms is negative. In such cases it is best to write the sum in such a form that it will end with 10, thus conforming always to 143. EXERCISES Find, by Rule V, 147, the value of each of the following logarithms. 1. log (35.1X7.29). 3. log (145.7 X 8.35 X. 00456). 2. log (5X3.17X.0016). 4. log (3.456 X. 001798 XI. 456). Find (by Rule VI, 147) the value of 5. 56.8X3.47 X. 735 Check your answer by multiplying out the long way as in arith- metic. Compare the two results and see how great was the error committed by following the short (logarithmic) method. Compare also the time required for the two methods. XX, 148] LOGARITHMS 235 6. .975X42.8X3.72 7. 896X40.8X3.75X.00489 8. 34.56X18.16X.0157 [HINT. See 145.] 9. 576.8X43.25X3.576X.0576 10. 60.573X8.087X.008915X1.2387 11. 23X23X23X23X23X23X23, (or 23 7 ) 12. 1.2X2.3X3.4X4.5X5.6X6.7X7.8 13. .31X5.198X6.831X2.584X.00312X.07568 14. Since 25 X 15 = 375 we know by Rule V, 147, that the logarithm of 25 added to the logarithm of 15 is equal to the logarithm of 375. Show that the values given in the tables for log 25, log 15, and log 375 confirm this result. Invent and try out several other similar problems for yourself. 148. To Find the Quotient of Two Numbers. Let us take any two numbers, for example 41 and 29, and look up their logarithms. We find log 41 = 1.6128 log 29=1.4624 These mean that 41==10 1.6128 and 29 = 10 1 - 4624 Whence, dividing the first of these equalities by the second, we obtain 101 ' 6128 ~ M624 ( 117 > Law n ) The last equality means that log (41^-29) = 1.6128-1.4624 = log 41-log 29. 236 SECOND COURSE IN ALGEBRA [XX, 148 This result illustrates the following general rule. RULE VII. The logarithm of a quotient is equal to the logarithm of the dividend minus the logarithm of the divisor. Thus log (467.3 -J-.00149) =log 467.3 -log .00149 The way in which this rule is used to find the value of the quotient of two numbers is shown below. EXAMPLE 1. To find the value of 236 -i-4. 15 SOLUTION. log 236=2.3729 log 4.15 =0.6180 Subtracting, 1.7549 Hence the logarithm of the desired quotient is 1.7549 (Rule VII) The number whose logarithm is 1.7549 is found (as in 146) to be 56.875 Therefore 236-^4.15=56.875 (approximately). Ans. EXAMPLE 2. To find the value of 1 . 46 -^ .00576 SOLUTION. log 1.46 =0.1644 = 10.1644 - 10 (See Note below.) log .00576 = 7.7619-10 Subtracting, 2.4025 The number whose logarithm is 2.4025 is found to be 252.64 Therefore 1 .46 -T- .00576 = 252. 64 (approximately) . Ans. Thus we have the following rule. RULE VIII. To find the quotient of two numbers : 1. Subtract the logarithm of the divisor from the logarithm of the dividend. 2. The difference thus obtained is the logarithm of the quo- tient. 3. The quotient itself can then be determined as in 146. NOTE. To subtract a negative logarithm from a positive one, or to subtract a greater logarithm from a less, increase the charac- teristic of the minuend by 10, writing 10 after the mantissa to compensate. Thus, in Example 2, we wished to subtract the nega- tive logarithm 7.7619-10 from the positive one 0.1644. There- fore 6.1644 was written in the form 10.1644-10, after which the subtraction was easily performed. XX, 149] LOGARITHMS 237 EXERCISES Find, by Rule VII, 148, the value of each of the follow- ing logarithms. 1. log(13-f-9). 3. log (38.76 -J-. 0017). 2. log (217-5-8.16). 4. log (8.764 -f- 114.3). Find, by Rule VIII, 148, the value of each of the follow- ing quotients. 5. 246 -i- 15.7 Check your answer by dividing out the long way as in arith- metic. Compare the two results and see how great was the error committed by following the short (logarithmic) method. 6. 34.7^-5.34 8. 45.67^38.01 7. 389.7^-4.353 9. 3.25 -f-. 00876 [HINT. See 145.] [HINT. See Note in 148.] 10. 49.6 -i- 87.3 n 40.3X6.35 3.72 [HINT. Find the logarithm of the numerator by Rule V, 147.] 12 .0036X2.36 24.3 X. 695 X. 0831 .0084 8.40 X. 216 14. Since 27-:-9 = 3 we know, by Rule VII, 147, that the logarithm of 9 subtracted from the logarithm of 27 is equal to the logarithm of 3. Show that the values given in the tables for log 9, log 27, and log 3 confirm this result. Invent and try out several other similar problems for yourself. 149. To Raise a Number to a Power. Let us take any number, for example 25, and raise it to any power, say the fourth. We then have 25 4 , which means 25X25X25X25. Hence, by Rule V, 147, we must have log 25 4 = log 25-flog 25+log 25+log 25, or log 25 4 = 4 log 25. 238 SECOND COURSE IN ALGEBRA [XX, 149 This illustrates the following rule. RULE IX. The logarithm of any power of a number is equal to the logarithm of the number multiplied by the exponent indicating the power. Thus log 3.17 10 = 10 log 3.17 ; similarly, log. 00174 6 = 6 log .00174. The way in which this principle is used to raise a number to a power is shown below. EXAMPLE 1. To find the value of 2.37* SOLUTION. log 2.37= 0.3747 4 Multiplying, 1.4988 Hence Iog2.37 4 = 1.4988 (Rule IX) The number whose logarithm is 1.4988 is found to be 31.535 Therefore 2.37 4 =31.525 (approximately). Ans. EXAMPLE 2. To find the value of .856 5 SOLUTION. log .856 = 9.9325 - 10 Multiplying, 49.6625-50 = 9.6625 - 10 The number whose logarithm is 9.6625-10 is .4597 ( 146) Therefore .8565= . 4597 (approximately). Ans. Thus we have the following rule. RULE X. To raise a number to a power : 1. Multiply the logarithm of the number by the exponent indicating the power. 2. The result thus obtained is the logarithm of the answer. 3. The answer itself can then be determined as in 146. EXERCISES Find, by Rule IX, 149, the value of each of the following logarithms. 1. log 16 6 2. log 3.12 3 3. log .0176 2 4. log 36.64 4 XX, 150] LOGARITHMS 239 Find, by Rule X, 149, the value of each of the following expressions. 5. 8.82 3 Check your answer by raising 8.82 to the third power as in arithmetic. Compare the two results and see how great was the error committed by following the short (logarithmic) method. 6. 4'.12 4 7. 4.123 4 8. .175 5 [HINT. See Ex. 2 in 149.] 9. 81 3 X.015 2 [HINT. Combine the rules of 147 and 149.] 10. 43X8.9 2 X.075 3 8.76X53.9X4.5 3 ' 2.3 2 X3.15X5.14 3 [HINT. Use Rules VI, VIII, X.] 12. Since 9 3 = 729 we know, by Rule IX, 149, that three times the logarithm of 9 is equal to the logarithm of 729. Show that the values given in the tables for log 9 and log 729 confirm this result. Invent and try out several other similar problems for yourself. 150. To Extract Any Root of a Number. Let us take any number, for example 36, and consider any root of it, say the fifth, that is, let us consider \ / 36l Supposing x to be the value of the desired root, we have z 5 = 36. ( 118) Now the logarithm of the first member of this equality is equal to 5 log x by Rule IX. Hence 5 log x = \og 36, or log x = ^ log 36. This illustrates the following rule. RULE XI. The logarithm of the root of a number is equal to the logarithm of the radicand divided by the index of the root. Thus log ^73=^ log 2.73 ; similarly, log -v/.01685=| log .01685. The way in which this principle is used to extract the roots of numbers in arithmetic will now be shown. 240 SECOND COURSE IN ALGEBRA [XX, 150 EXAMPLE 1. To find the value of 'V / 85.2 SOLUTION. log 85.2 = 1 .9304, so that -J- of log 85.2=0.4826. Hence log ^85^2" =0.4826. (Rule XI) The number whose logarithm is 0.4826 is 3.038 ( 146) Therefore ^85.2=3.038 (approximately). Ans. EXAMPLE 2. To find the value of v^.0875 SOLUTION. log .0875 = 8.9420 - 10, so that i of log .0875 = ^8.9420 - 10) = i(48.9420 -50) = 9.7884 - 10. (See Note below.) The number whose logarithm is 9.7884-10 is .6143 ( 146) Therefore ^.0875 = . 6143 (approximately). Ans. These examples lead to the following rule. RULE XII. To find any root of any number. 1. Divide the logarithm of the number by the index of the root. 2. The quotient thus obtained is the logarithm of the desired root. 3. The root itself can then be determined as in 146. NOTE. To divide a negative logarithm, write it in a form where the negative part of the characteristic may be divided exactly by the divisor giving 10 as quotient. Thus, in Example 2, we wrote 8.9420-10 in the form 48.9420-50 after which the division by 5 was easily done and resulted in a form ending in 10. EXERCISES Find, by Rule XI, 150, the value of each of the following logarithms. 1. log vT6 2. log v^332 3. log ^0175 4. log v/38^6" Find, by Rule XII, 150, the value of each of the following. 5. Check your answer by extracting the square root of 315 (correct to three decimal places) as in arithmetic. Compare the two results and see how great was the error committed by following the short (logarithmic) method. XX, 150] LOGARITHMS 241 8. [HINT. See Example 2 in 150.] 9. -^8.76 X. 0153 [HINT. Use Rules IX and XI.] _ 1576 X 9. 13 2 11. 3.8X5.32 3 12. Since A/49 = 7 we know, by Rule XI, 150, that one half the logarithm of 49 is equal to the logarithm of 7. Show that the values given in the tables for log 49 and log 7 con- firm this result. Invent and try out several other similar problems for yourself. APPLIED PROBLEMS Solve the following exercises by logarithms. 1. How many cubic feet of air are there in a schoolroom whose dimensions are 50.5 ft. by 25.3 ft. by 10.4 ft. ? 2. How many gallons will a rectangular tank hold whose dimensions are 8 ft. 10 in. by 9 ft. 3 in. by 10 ft 1 in. ? 3. How much wheat will a cylindrical bin hold if the diameter of the base is 9 ft. 5 in. and the height is 40 ft. 4 in. ? 4. How much would a sphere of solid cork weigh if its diameter was 4 ft. 3 in., it being known that the specific gravity of cork is .24? (See Example 14 (e), page 6.) [HINT. To say that the specific gravity of cork is .24 means that any volume of cork weighs .24 times as much as an equal volume of water. Water weighs 62.5 pounds per cubic foot.] 5. The diameter d in inches of a wrought-iron shaft re- quired to transmit h horse power at a speed of n revolutions per minute is given by the formula d = \ Find the * n diameter required when 135 horse power is to be transmitted at a speed of 130 revolutions per minute. 242 SECOND COURSE IN ALGEBRA [XX, 150 6. The amount to which P dollars will accumulate at r% compound interest in n years is given by the formula Find A if P = $500, r = 5, and n = W. Find A if P = $100, r = 3.5, and n=15. 7. By means of Formula 3 of 65, find the area of the triangle whose sides are 3.15 in., 4.87 in., and 2.68 in. 8. The height H of a mountain in feet is given by the formula H = 49,00(/^ r Vl +\ \R+rJ\ 900 / where R, r are the observed heights of the barometer in inches at the foot and at the summit of the mountain, and where T, t are the observed Fahrenheit temperatures at the foot and summit. Find the height of a mountain if the height of the barom- eter at the foot is 29.6 inches and at the summit 25.35 inches, while the temperature at the foot is 67 and at the summit 32. 9. By means of the formula in Ex. 6 answer the follow- ing question : How long will it take a sum of money to double itself if placed at compound interest at 5 % ? 14.2 years. Ans. GENERAL LOGARITHMS *161. Logarithms to Any Base. In 136 we defined the loga- rithm of a number as the power to which 10 must be raised to obtain that number. Thus, from such equalities as 10 2 = 100, 10 3 = 1000, etc., we had log 100 = 2, log 1000 = 3, etc. Strictly speaking, this defines the logarithm of a number to the base 10, or, as it is usually called, a common logarithm. We may and frequently do use some other base than 10. For example, since 3 2 =9, 3 3 =27, 3 4 =81, etc., we can say that the logarithm of 9 to the base 3 is 2, the logarithm of 27 to the base 8 is 3, XX, 154] LOGARITHMS 243 the logarithm of 81 to the base 3 is 4, etc. The usual way of denot- ing this is to write Iog 3 9 =2, Iog 3 27 =3, Iog 3 81 =4, etc. Observe that the number being used as the base is thus placed to the right and just below the symbol log. Similarly, we have Iog 2 16=4, Iog 8 64 = 2, Iog 6 125=3, etc. Thus we have the following general definition. The logarithm of any number x to a given base a is the power of a required to give x. It is written \og a x. Any positive number except 1 may be used as the base. NOTE. When the base a is taken equal to 10 (that is, in the usual case) we write simply log x instead of EXERCISES State first the meaning and then the value of 1. Iog 2 4. 2. Iog 2 8. 3. Iog 4 16. 4. logsf 6. Iog 2 i. 6. log^. 7. Iog 5 .2 8. Iog 8 32. *152. Logarithm of a Product. We can now show that Rule V, 147, holds true whatever the base. That is, if M and N are any two numbers, and a the base, then log a MN=log a M+log a N. PROOF. Let x = log a M and y = log a N. Then a x = M and a" = N (151). Hence a*-a = MN, or a x + = MN (117, Law I). But the last equality means that log a MN=x+y=\og a M+log a N. ( 151) *153. Logarithm of a Quotient. Rule VII, 148, holds true whatever the base. That is, if M and N are any two numbers, then lQg.(M+N) = \Og a M-log a N. PROOF. Let x = log M and y = \og a N. Then a x = M and a = N. ( 151). Hence, a* -=- a" = M -=- N, OTa x ~v = M + N ( 117, Law II). But the last equality means that log a (M + N)=x-y=log a M-log a N. ( 151) *164. Logarithm of a Power of a Number. Rule IX, 149, holds true whatever the base. That is, if M is any number and n any (posi- tive integral) power, then log M re = n logaM. PROOF. Letz = log a Af. Then a* = M ( 151) and hence a nx = M n ( 117, Law III). But the last equality means that log M n = nx=n log M. ( 151 ) 244 SECOND COURSE IN ALGEBRA [XX, 155 *155. Logarithm of a Root of a Number. Rule XI, | 150, holds true whatever the base. That is, if M is any number and n any (posi- tive integral) root, then = log M. n PROOF. Let x=log a M Then a x = M (151) and hence ( *)i/ = M 1 / n , ora* /n = VM" ( 121). But the last equality means that n n *166. Summary. From the results established in 151-155 it appears that Rules V-XII, 147-150, are not only true when the base is 10 (as was there taken) but they are true for any base. Complete tables have been worked out for various bases other than 10, but we shall not consider them further here. NOTE. The reason why 1 cannot be used as a base is that 1 to any power is equal to 1, that is, we cannot get different numbers by raising 1 to different powers. *157. Historical Note. Logarithms were first introduced and employed for shortening computation by JOHN NAPIER (1550-1617), a Scotchman. (See the picture facing p. 233.) However, he did not use the base 10, this being first done by the English mathemati- cian BRIGGS (1556-1631), who computed the first table of common logarithms and did much to bring logarithms into general use. *158. Calculating Machines. The Slide-Rule. Machines have been invented and are now coming into very general use, especially by engineers, by which the processes of multiplication, division, involution, and evolution can be immediately performed. The construction of these machines depends upon the principles of logarithms, but to describe the machines and their methods of working would take us beyond the scope of this text. The simplest machine of this kind is the slide rule, the use of which is easily understood. A simple slide rule with directions is inexpensive and may ordinarily be secured from booksellers. I A A I c (( l,I|p!.l ^111! i""l" ^ ! ^ D FIG. 70. THE SLIDE RULE. PAET III. SUPPLEMENTARY TOPICS CHAPTER XXI FUNCTIONS 159. The Function Idea. In ordinary speech we make such statements as the folio whig : 1. The area of a circle depends upon the length of its radius. 2. The time it takes to go from one place to another de- pends upon the distance between them. 3. The power which an engine can exert depends upon the pressure per square inch of the steam in the boiler. Another way of stating these facts is as follows : 1. The area of a circle is a function of the length of its radius. 2. The time it takes to go from one place to another is a function of the distance between them. 3. The power which an engine can exert is a function of the pressure per square inch of the steam in the boiler. The idea thus conveyed by the word function is that we have one magnitude whose value is determined as soon as we know the value of some other one (or more) magnitudes upon which the first one depends. This idea is at once seen to be universal in everyday experience and for that reason it be- comes of great importance in mathematics, f In the present t The extended formal study of the function idea enters into that branch of mathematics known as the Calculus. 245 246 SECOND COURSE IN ALGEBRA [XXI, 159 chapter we shall indicate briefly how it is related to some of the subjects treated in the preceding chapters, noting es- pecially the significance of the idea when considered graphi- cally. 160. Types of Algebraic Functions. An expression of the form (1) wr+fli, where the coefficients a Q and ai have any given values (except a must not be 0) is called a linear function of x. Observe that every such expression depends for its value upon the value assigned to #, and is determined as soon as x is known. Hence it is a function of x in the sense explained in 159. It is called a linear function since it is of the first degree in x. (Compare 26.) For example, 2 x +3 is a linear function of x. Here we have the form (1) inwhicha = 2 and ai=3. Similarly, 3x 2, x 4,-z+i and 3x are linear functions of x. (Why?) Likewise, 3^+2 is a linear function of t, while r+5 is a linear function of r, etc. As an example of a linear function in everyday experience, sup- pose that in Fig. 71 a person starts from the point P and moves to the right at the rate of 15 miles per hour, and let Q be the point 10 S E h 10 >! FIG. 71. miles to the left of P. Then we may say that the distance of the traveler from Q is a linear function of the time he has been traveling, for if t represent the number of hours he has been traveling, his dis- tance from P is 15 1 (see 62) and hence his distance from Q is 15 + 10. This is seen to be a linear function of t, being of the form (1) in which a = 15 and 01 = 10. . Likewise, the interest which a given principal, P, will yield in one year is a linear function of the rate, for, if r be the rate, the interest XXI, 160] FUNCTIONS 247 in question is given by the formula P X-^:,or r, and this is seen , -LvJO J.LHJ p to be of the form (1) in which o = rrr:, and ai=0. J_ \j\J .___ - An expression of the form (2) a x 2 +aix+a 2 , where the coefficients a , i, and a 2 have any given values (except that a must not be 0) is called a quadratic function of x. For example, 2rr 2 +3 x 1 is a quadratic function of x because it is of the form (2) in which a = 2, a\ = 3, 02 = 1. Likewise, x 2 +-|- x ; x 2 +i; z 2 +3 x ; 5 x 2 ; z 2 are quadratic functions of x. (Why?) Again, we may say that the area of a square is a quadratic func- tion of the length of one side, for if x be the length of side, the area is x z and this is of the form (2) in which a = 1, a\ = a 2 =0. Similarly, the area of a circle is a quadratic function of the radius. (Why?) An expression of the form (3) flo* 3 +ai* 2 +a 2 Jt+a 3 , where the coefficients a , i, 2 and a 3 have any given values (except that a must not be 0) is called a cubic function of x. For example, 3 x 3 -x*+x-l ; 4z 3 -z; z 3 -2x 2 + l; 5x 3 ; x 3 , etc. (Why?) Again, we may say that the volume of a cube is a cubic function of the length of one edge. (Why?) ! Also, the volume of a sphere is a cubic function of the radius. (Why?) It may now be observed that the expressions (1), (2), and (3) are but special forms of the more general expression (4) a x n +ai^ n ~ 1 +fl2X n - 2 H \-a n _iX+a n where it is understood that n can be any positive integer, while the coefficients OQ, ai, a 2 , a,, have any given values (except that a must not be 0). This is called the general 248 SECOND COURSE IN ALGEBRA [XXI, 160 integral rational function of x, or, more simply, a polynomial in x. It reduces to the linear function (1) when n=l; to the quadratic function (2) when n = 2 ; etc. Expressions such as Vx, and all others composed merely of powers or roots (or both) of x are classed under the name of algebraic functions. Since all functions of the form (4) are composed of integral powers only of x, they are but special cases of the algebraic functions just mentioned. EXERCISES 1. Show that the thickness of a book is a linear function of the number of its pages. [HINT. Let x be the number of pages, d be the thickness of each page, and D the thickness of each cover. Now build up the formula for the thickness of the book and note which of the functional types in 160 is present.] 2. The supply of gasoline in a tank was very low, its depth being but 1 inch all over the bottom, when it was re- plenished from a pipe which delivered 3 gallons per minute. Show that the amount in the tank at any moment during the filling was a linear function of the time since the filling began. 3. Show that the force which a steam engine has at any moment at its cylinder is a linear function of the area of the piston ; also that it is a linear function of the boiler pressure of the steam per square inch. 4. A certain room contains a number of 16-candle-power electric lights and a number of Welsbach gas-burners. Show that the amount of illumination at any time is a linear function of the number of electric lights turned on. Is this true regardless of the number of gas-burners already lighted ? XXI, 160] FUNCTIONS 249 5. Show that the perimeter of a square is a linear func- tion of the length of one side ; also that the circumference of a circle is a linear function of its radius. 6. Show that if each side of a square be increased by x, the corresponding increase in the area will be a quadratic function of x. [HINT. Let a = the length of one side of the original square. Then the area is a 2 and the area of the new square is (a +z) 2 . Now formulate the expression for the increase in area.] 7. Show that if the radius of a circle be increased by x, the corresponding increase in area will be a quadratic func- tion of x. 8. Show that if the edge of a cube be increased by x the corresponding increase in volume will be a cubic function of x. State and prove the corresponding statement for a sphere. 9. Show that if y varies directly as x (see 113), then y is a linear function of x. Is the converse of this statement necessarily true, namely if y is a linear function of x, then y varies directly as x ? 10. When y varies as the square of x, to which one of the functional types mentioned in 160 does y belong? Answer the same question when y varies inversely as x ; when y varies inversely as the square of x. 11. A certain linear function of x takes the value 5 when x = l and takes the value 8 when x = 2. Determine com- pletely the form of the function. SOLUTION. Since the function is linear, it is of the form aox +i. Since this expression must (by hypothesis) be equal to 5 when x = 1, we have a l+ai=5. Likewise, placing x=2, gives a 2+ai=8. Solving these two equations for a and ai we obtain oo=3, ai=2. The desired function is therefore 3 x +2. Ans. 250 SECOND COURSE IN ALGEBRA [XXI, 160 12. A certain linear function of x takes the value 14 when x = 3, and takes the value 6 when x= 1. Deter- mine completely the form of the function. 13. A certain quadratic function takes the value when x = 1, and the value 1 when x = 2, and the value 4 when x = 3. Determine completely the form of the function. 14. Show that the area of any triangle is an algebraic function of the sum of its three sides. (See Formula 3 in 65.) 161. Functions Considered Graphically. By the graph of a function is meant the line or curve which results when some letter, as y, is placed equal to the function and the graph is drawn of the equation thus obtained. The purpose of the graph is to bring out clearly and quickly to the eye the relation between the given function and the quantity (vari- able) upon which it depends for its values. The method of drawing such graphs is precisely the same as that given in 29, p. 43 for equations of the first degree, and in 57, p. 90, for quadratic equations. Thus, in order to obtain the graph of the function x*, we place y=*x z and proceed to draw the graph of this equation in the way explained in 29, that is, we assign various values to x and compute (from this equation) the corresponding values of y, then we plot each point thus obtained and finally draw the smooth curve passing through all such points. Below is a table of several values of x and y thus computed; and the graph is shown in Fig. 72. When x = -2 -1 1 2 3 4 then y = -8 -1 1 8 27 64 The portion of the curve lying to the right of the y-a,xis extends up- ward indefinitely, while the portion to the left of the same axis ex- tends downward indefinitely. Note that, from the way this curve has XXI, 161] FUNCTIONS 251 been drawn, it at once brings out to the eye the value of the given function x 3 for any value of the letter x upon which this func- tion depends, the function values being the ordinates ( 28) of the points on the curve. For example, at x = 2 the corresponding ordinate measures 8, which is the function value then present. This curve may be used as a graphical table of cubes of numbers. Thus, if x = 1.5, 2/ = 3.4, approximately, etc. Likewise, if y is given first, the curve shows the cube root of y; for example, if y=4, x is about 1.6. The figure may be drawn by the student on a much larger scale ; the values of x and y can be read much more accurately from such a figure than from the small figure on , this page. Another means of improving the accu- racy of the figure is to take a longer dis- tance on the horizontal line to represent one unit than is taken to represent one unit on the vertical scale. FIG. 72. The graph of every linear function is a straight line. The graph of every other algebraic function is a curved line. -B FIG. 73. For example, in considering the graph of the linear function \x 5, we place y=\ x 5. But this is an equation of the first degree between x and y and hence ( 29) its graph is a straight line. Fig. 73 shows the result. Note that the graph cuts the z-axis in one point. The abscissa of this par- ticular point is 4, which indicates that 4 is the root, or solution, of the equation f x 5=0, for it is this value of x that makes y=Q. 252 SECOND COURSE IN ALGEBRA [XXI, 161 The graph of every quadratic function belongs to the class of curves known as parabolas. A parabola resembles in form an oval, open at one end. It never cuts the z-axis in more than two points. Fig. 74 shows the graph of the quadratic function x z +x-2. Note that the curve cuts the z-axis at two points whose abscis- sas are 2 and 1, respectively. This indicates that 2 and 1 are the roots of the quadratic equation x 2 +x 2=0. 1-0 FIG. 74. FIG. 75. The general form of the graph of a cubic function is that of an indefinitely long smooth curve which cuts the x-axis in no more than three points. Fig. 75 shows the graph of the cubic function x 3 3 z 2 x +3. It cuts the rr-axis at three points whose abscissas are respectively 1, 1, and 3. These values, therefore, are the roots of the cubic equa- tion z 8 -3z 2 -z+3=0. XXI, 161] FUNCTIONS 253 Similarly, the general form of the graph of the rational integral function of the fourth degree is that of an indefinitely long smooth curve which cuts the x-axis in no more than four points. And it may be said likewise that the graph of the general integral function of degree n (see (4), 160) is an indefinitely long smooth curve which cuts the z-axis in no more than n points. Fig. 76 shows, for example, the graph of 2 x 4 5 x 3 +5 x - 2, this being a function of the fourth degree. The four points where the curve cuts the z-axis have abscissas which are equal respectively to 1, -1-, 1, and 2. These values, therefore, are the roots of the equation 2 x 4 5 z 3 +5 x 2 = 0. FIG. 76. FIG. 77. Fractional expressions give rise to more complex graphs, which may have more than one piece. Fig. 77 shows, for example, the graph of 1/rc. If we let y = l/x, y varies inversely as x ( 110). The -curve is therefore similar to those drawn in 115, Fig. 69. The graph consists of two branches and belongs to the class of curves known as hyperbolas. These we have already met in 78. 254 SECOND COURSE IN ALGEBRA [XXI, 161 EXERCISES Draw the graphs of the following functions by plotting several points on each and drawing the curve through them. Try to plot enough points so that the form and location of the various waves, or arches, of the curve will be brought out clearly, as in the figures of 161. Note how many times the curve cuts the x-axis and make such inferences as you can regarding the roots of the corresponding equation. [HINT. When the graph of a quadratic function fails to cut the x-axis, this indicates that the roots of the corresponding quadratic equation are imaginary. (See 57, 60.) Similarly, when the graph of a cubic function cuts the z-axis in but one point, this indi- cates that there is but one real root to the corresponding equation, the other two roots being imaginary. In general, the number of times the graph cuts the x-axis indicates the number of real roots of the corresponding equation, the number of imaginary roots being the degree of the equation minus the number of real roots.] 1. 3x+4. 2. x. 3. x z -x-2. 4. z 2 -4. 5. x 2 +l. 6. x*-3x*-x+3. 7. CHAPTER XXII MATHEMATICAL INDUCTION BINOMIAL THEOREM 162. Mathematical Induction. The three following purely arithmetic relations are easily seen to be true : l+2+3 = We might at once infer from these that if n be any positive integer, there exists the algebraic relation (1) 1+2+3+4+ +n = (n+l), the dots indicating that the addition of the terms on the right continues up to and including the number n. For example, if n = 8, this would mean that l+2+3+4+5+6+7+8=f(8 + l). - Again, if n = 10, it would mean that 1+2+3+4+5+6+7+8+9 + 10=^(10 + 1). That these are indeed true relations is discovered as soon as we simplify them. Let the pupil convince himself on this point. It is now to be carefully observed that the inference just made, namely that (1) is true- for any n, is not yet justified, strictly speaking, from anything we have done, for we have only shown that (1) holds good for certain special values of n, and we could never hope to do more than this however long we continued to try out the formula in this way. Something more than a knowledge of special cases must always be known before any perfectly certain general inference can be made. For example, the fact that Saturday was cloudy for 38 weeks in suc- cession gives no certain information that it will be so on the 39th week. 255 256 SECOND COURSE IN ALGEBRA [XXII, 162 We shall now show how the general formula (1) may be established free from all objection, that is in a way that leaves no possible question as to its truth in all cases. Let r represent any one of the special values of n for which we know (1) to be true. Then (2) 1+2+3+4+ ."+r=|(r+l). Let us add (r+1) to both sides. The result is 1+2+3+4+.- H-r+(r+l) = (r+l) + (r+l). 2i In the second member of the last equation we may write while the first member has the same meaning as Thus, (2) being given us, it follows that we may write (3) But (3) is seen to be precisely the same as (2) except that r+1 now replaces r throughout. Stated in words, this re- sult means that if (1) is true when n = r, as we have supposed, then it holds true necessarily for the next greater value of n, which is r+1. The original fact which we wished to establish (namely, that (1) is true for any n) now follows without difficulty. In fact, we know (see beginning of this section) that (1) is true when n = 4, from which it now follows that it must be true also when n = 5. Being true when n = 5, the same reasoning says it must be true also when n = 6. Being true when n = 6, it must be likewise true when n = 7. Proceeding in this way, we may reach any integer n we may mention, however large it may be. Hence (1) is true [for any such value of n. XXII, 162] BINOMIAL THEOREM 257 This method of reasoning illustrates what is termed mathematical induction. Another example of the process will now be given, the steps being arranged, however, in a more condensed form. EXAMPLE. Prove by mathematical induction that (1) 1+3+5+7'H ----- \-(2n l)=n 2 . (n = any positive integer) SOLUTION. When n = 1, the formula gives 1 = I 2 ; when n = 2, it gives l+3=2 2 ; when n=3, it gives l+3+5=3 2 , all of which arithmetical relations are seen to be correct. Let r represent any value of n for which the formula has been proved. Then (2) 1+3+5+7+.. . +( 2 r -l) = r 2 . Adding (2 r+1) to each member gives (3) 1+3+5+7+.- -+(2 r+1) =r 2 + (2 r + l)=r 2 +2 rfl = (r + l) 2 . But (3) is the same as (2) except that r has been replaced through- out by r + 1. Hence, if (1) is true for any value of n, such as r, it is necessarily true also for that value of n increased by 1. Now, we know (1) to be true when n=3. (See above.) Hence it must be true when n = 4. Being true when n = 4, it must be true when w=5, etc., and in this way we now know that (1) is true for any value (positive integral) of n whatever. EXERCISES Prove the correctness of each of the following formulas by mathematical induction, n always being understood to be any positive integer. [HINT. First try out f orn = 1, n = 2, and n = 3. Let r represent a number for which the formula holds. Add 2(r + l) to both mem- bers of the resulting equation and compare results.] 2. 3+6+9+12+ +3 rc= (n+1). 3. 4. 2 2 +4 2 +6 2 +---+(2n) 2 =n(n+l)(2ri 5. s 258 SECOND COURSE IN ALGEBRA [XXII, 162 6 _1 ,_1 i__l I [_ 1 = n 1-2 2-3 3-4 n(w+l) n+l' 7. 2+2 2 +2 3 +2 4 H r-2 = 2(2-l). 8. Prove that if n is any positive integer, a n b n is divisi- ble by a b. [HINT. Since o r+1 -6 r+1 =a(a r -b r )+b r (a-b), it follows that a r+1 b r+l will be divisible by a 6 whenever a r b r is divisible by 0-6.] 9. Prove that a 2n b 2n is divisible by a +b. 163. The Binomial Theorem. If we raise the binomial (a -\-x) to the second power, that is find (a+x) 2 , the result is a 2 -}- 2 ax+x 2 (10). Similarly, by repeated multiplica- tion of (a+x) into itself, we can find the expanded forms for (a+z) 3 , (a+x) 4 , (a+x) 5 , etc. The results which we find in this way have been placed for reference in a table below : > = a 2 +2ax+x 2 . ' = a 3 +3 a 2 z+3 az 2 +z 3 . (ct-hx) 4 = tt 4 -|-4 a 3 x-\^6 a 2 x 2 -f-4 ax s -}-x*. (a-f-x) 5 = tt 5 -|-6 a 4 x-j-10 a 3 x 2 -\-10 a 2 x s -\-Q ax*-\-x 5 , etc. Upon comparing these, we see that the expansion of (a+x) n , where n is any positive integer, has the following properties : 1. The exponent of a in the first term is n, and it decreases by 1 in each succeeding term. The last term, or x n , maybe regarded as ax n . (See 122.) 2. The first term does not contain x. The exponent of x in the second term is 1 and it increases by 1 in each succeeding term until it becomes n in the last term. 3. The coefficient of the first term is 1 ; that of the second term is n. 4. // the coefficient of any term be multiplied by the exponent of a in that term, and the product be divided by the number of the term, the quotient is the coefficient of the next term. XXII, 163] BINOMIAL THEOREM 259 For example, the term 6 o 2 x 2 , which is the third term in the ex- pansion of (o+z) 4 (see p. 258) has a coefficient, namely 6, which may be derived by multiplying the coefficient of the preceding term (which is 4) by the exponent of a in that term (which is 3) and dividing the product thus obtained by the number of that term (which is 2). 5. The total number of terms in the expansion is n-\- 1. The results just observed regarding the expansion of (a-\-x) n , where n is any positive integer, may be summarized and condensed int.o a single formula as follows : (a+x) n = 1 2 1-2-3 the dots indicating that the terms are to be supplied in the manner indicated up to the last one, or (w+l)st. This formula is called the binomial theorem. By means of it, one may write down at once the expansion of any binomial raised to any positive integral power. That the formula is true in all cases, when n is a positive integer, will be proved in detail in 165. We assume its truth here for those small values of n for which its correctness is easily tested. NOTE. The formula is generally attributed to Sir Isaac Newton (1642-1727) ; see the picture facing p. 193. EXAMPLE 1. Expand (a+z) 6 . SOLUTION. Here n = 6, so the formula gives ''' .6-5-4-3-2 5 .6-5-4-3.2- 1 x6 ^l- 2- 3-4- 5 ^1.2.3-4.5.6 Simplifying the various coefficients by performing the possible cancelations in each, we obtain 5 -Fz 6 . Ans. 260 SECOND COURSE IN ALGEBRA [XXII, 163 NOTE. It may be observed that the coefficients of the first and last terms turn out to be the same ; likewise the coefficients of the second and next to the last terms are the same, and so on symmetri- cally as we read the expansion from its two ends. This feature is true of the expansion of (a +#) to any power. (Note the expansions of (o+x) 2 , (a-f-z) 3 , (a+rc) 4 , etc., as given at the beginning of 163.) EXAMPLE 2. Expand (2 m) 5 . SOLUTION. Here a =2, x=m, and n=5. The formula thus gives S' *' j* 2 2 (-m) 3 1 ^ o Simplifying the coefficients (as in Example 1) this becomes (2-m) 5 =2 5 +5- 2 4 (-m)+10- 2 3 (-m) 2 + 10- 2 2 (-ra) 3 Making further simplifications, we obtain (2-m) 5 = 32-80m+80ra 2 -40ra 3 + 10ra 4 -m 5 . Ans. NOTE. The result for (2 x) 5 is the same as that for (2+x) 6 except that the signs of the terms are alternately positive and negative instead of all positive. A similar remark applies to the expansion of every binomial of the form (a x) n as compared to that of (a+x) n . EXERCISES Expand each of the following powers. 1. (x+y)*. 9. (a 2 -z 2 ) 4 . 1? (I , IV. 2. (a+6) 4 . 10. (2a+l) 4 . (x yj 3. (x vY. 11. (x 3 vY. 40 / u, a/ 4. (a-6) 4 . 12. (1+x 2 ) 6 . 5. (2+r) 5 . 13. (l-o:) 8 . 6. (a+x) 7 . 14. (x ^-) 5 . 7. (fif-3) 6 . 15. (3 a 2 - 1) 4 . 2a 8. (a 2 +x) 6 . 16. (a+x) ' XXII, 1653 BINOMIAL THEOREM 261 *164. The General Term of (a+x) n . The third term in the expansion of (a+x) n , as given by the formula in 163, is (n-l Observe that the exponent of x is 1 less than the number of the term ; the exponent of a is n minus the exponent of x ; the last factor of the denominator equals the exponent of x ; in the numerator there are as many factors as in the denominator. Precisely the same statements can be made as regards the fourth term, or 1 2 3 In the same way, it appears that the above statements can be made of any term, such as the rth, so that the formula for the rth term is rth L ' 2i ' O (r 1) EXAMPLE. Find the 7th term of (2 6 -c) 10 . SOLUTION. Here a = 2 b, x = (c), n = 10, and r = 7. Therefore (using the formula), the desired 7th term is *! 2- 3- L 5 6 6 5 ' ( 26 ) 4 (- c ) 6 = 210 ( 2& ) 4 (- c ) 6 = 3360^c. Ans. EXERCISES Find each of the following indicated terms. 1. 5th term of (+*)-. ? . 6th term of t x +Vf l . 2. 6th term of (x-y}*. 3. 7th term of (2+ z) 9 . 8. 9th term of ^- 4. 10th term of (w-n) 14 . /2 2 x 12 6. 6th term of (*-&.) 9 ' 5th term of (f -f J '_ 6. 20th term of (1 +x) 24 . 10. 4th term of (2 \/2 - v/3) 6 . 165. Proof of the Binomial Theorem. The way in which the binomial formula was established in 163 is, strictly speaking, open to objection because we there made sure of its correctness only for certain special values of n, such as ft = 2, n = 3, n = 4, and n==5. Though the formula holds 262 SECOND COURSE IN ALGEBRA [XXII, 165 true, as we saw, in these cases, it does not follow necessarily that it is true in every case, that is for eveiy positive inte- gral value of n. We can now establish this fact, however, by the process of mathematical induction, when n is a positive integer. Let ra represent any special value of n for which the formula has been established (as, for example, 2, 3, 4, or 5). Then we have (1) m(m-l) - (m-r+2) 1-2-3- (r-1) Let us now multiply both members of this equation by a+z. On the left we obtain (a+x) m+l . On the right we shall have the sum of the two results obtained by multiply- ing the right side of (1) first by a and then by x, that is we shall have the sum of the two following expressions : m(m-l) ... ( m - r+2 ) 1-2.3-. (r-1) and Adding these, and making the natural simplifications in the resulting coefficients of a m x, a m ~ l x 2 , etc., and equating the final result to its equal on the left (namely (a+x) m+1 , as noted above) gives (2) 1 & o (T i) XXII, 167] BINOMIAL THEOREM 263 But (2) is precisely (1) except for the substitution of ra+1 for m throughout. Hence, if the binomial formula holds for any special value of n, as m, it necessarily holds for the next larger value, namely m-j-1. But we have already ob- served that it holds when n = 5. It must, therefore, hold when n = 5+l, or 6. But if it holds when n = Q, it must likewise hold when n = 6+l, or 7. Thus we may proceed until we arrive at any chosen value of n whatever. That is, the for- mula must be true for any positive integral value of n. *166. The Binomial Formula for Fractional and Negative Ex- ponents. In case the exponent n is not a positive integer but is fractional or negative, we may still write the expansion of (a-\-x) n by the formula of 163, but it will now contain indefinitely many terms instead of coming to an end at some definite point, that is we meet with an infinite series. (Compare 92.) For example, the formula gives 1-2 1 2 3 = a 1 / 2 +4- a- 1 / 2 .? +n a~ 3/2 x 2 -f~~ a - 5/2 x 3 + 1-2 1-2-3 = a^+a-^x-a-*< Here we have written only the first four terms of the expansion, but we could obtain the 5th term in the same way and as many others in their order as might be desired. *167. Application. If in (a+x} n the value of x is small in com- parison to that of a (more exactly, if the numerical value of x/a is less than 1) then the first few terms of the expansion furnish a close approximation to the value of (a+z) n . This fact is often used to find approximate values for the roots of numbers in the manner illustrated below. EXAMPLE. Find the approximate value of VlO. SOLUTION. Write VlO = V9 + l = V(3 2 + l) and expand this last form by the binomial formula. Thus (using the final result 264 SECOND COURSE IN ALGEBRA [XXII, 167 in the worked example of 166), we have 1 / 2 - 1-- J-(3 2 )- 3/2 - I 2 o ,_L_ 1.1 2-3 8-3 3 16-3 5 = 3+.166666-.004629+.000257 =3.162288 (approximately). Observe that the value of VlO as given in the tables is 3.16228, thus agreeing with that just found so far as the first five places of decimals are concerned. Whenever extracting roots by this process we use the following general rule. Separate the given number into two parts, the first of which is the nearest perfect power of the same degree as the required root, and ex- pand the result by the binomial theorem. *EXERCISES Write the first four terms in the expansion of each of the following expressions. 1. (o+aO/. 6. (2 o+6)3/4. 2. (a+z)- 2 . 6. (a 3 -z 2 )-3/4. 3. (l+x)V3. 7. 4. (2- a;)- 1 / 4 . 8. 9. Find the 6th term in the expansion of [HINT. Use the formula in 164, with n =-J and r = 6.] Find the 10. 5th term of (a+rr) 1 / 2 . 13. 9th term of (a - 11. 7th term of (a+x)~ 2 / 3 . 14. 10th term of 12. 8th term of (1+z) 1 / 3 . 15. 6th term of \/2a+6. Find the approximate values of the following to six decimal places and compare your results for the first three examples with those given in the tables. 16. VT7. 17. V27. 18. \/9 19. ^11. 20. [HINT. Write 14 = 16-2=24-2.] CHAPTER XXIII THE SOLUTION OF EQUATIONS BY DETERMINANTS 168. Definitions. The symbol a b c d is called a determinant of the second order, and is defined as follows : a b Thus 8 3 2 4 7 2 -10 Q = adbc. 8 -4-2 -3 = 32-6 = 26. = 7 -4-(-2) -3 = 28+6 = 34. 6] = 4[-50+18] = 4(-32) = -128. The numbers a, b, c, and d are called the elements of the determinant. The elements a and d (which lie along the diagonal through the upper left-hand corner of the determinant) form the principal diagonal The letters b and c (which lie along the diagonal through the upper right-hand corner) form the minor diagonal. 265 266 SECOND COURSE IN ALGEBRA [XXIII, 168 From these definitions, we have the following rule. To evaluate any determinant of the second order, subtract the product of the elements in the minor diagonal from the product of the elements in the principal diagonal. EXERCISES Evaluate each of the following determinants. 1. 3. 7 1 4 -8 8 7 -2 3 -2 6 7 -3 6. 5 7. f 8. 2a 3a 7a 36 56 66 x+y 3 x y 4 169. Solution of Two Linear Equations. Let us consider a system of two linear equations between two unknown letters, x and y. Any such system is of the form (1) (2) where ai, 61, Ci, etc., represent known numbers (coefficients). This system may be solved for x and y by elimination, as in Chapter VII. Thus, multiplying (1) by 6 2 and (2) by 61 and then subtracting the resulting equations from each other, the letter y is eliminated and we reach the equation (i6 2 0^61)0; = 6 2 Ci bic%. Therefore (3) x Likewise, we may eliminate x by multiplying (1) by 02 and XXIII, 169] DETERMINANTS 267 (2) by i and subtracting the resulting equations from each other. This gives (ai& 2 0261) y = aid Therefore (4) y It is now clear, by 168, that the numerators and denomi- nators in (3) and (4) are all determinants of the second order ; and by the definition of 168, (3) and (4) may be written respectively in the forms 61 (5) x = y= These forms for the solution of (1) and (2) are easily re- membered. In particular, observe that : 1. The determinant for the denominator is the same for both x and y. 2. The determinant for the numerator of the z-value is the same as that for the denominator except that the numbers Ci and c% replace the ai and a^ which occur in the first column of the denominator determinant. 3. The determinant for the numerator of the ^/-value is the same as that for the denominator except that the num- bers Ci and 02 replace the bi and 6 2 which occur in the second column of the denominator determinant. The usefulness of the forms (5) lies in the fact that they express the solution of a system of two linear equations in condensed form, enabling us to write down the desired values of x and y immediately, without the usual process of elimina- tion. This will now be illustrated. 268 SECOND COURSE IN ALGEBRA [XXIII, 169 EXAMPLE. Solve by determinants the system (6) (7) x-7y=-S. SOLUTION. Using the forms (5), we have at once x = 18 3 -8 -7 _1 8 . (_7)-(-8)- 3 -126+24 -102 R 2 3 2(-7)-l 3 -14-3 -17 1 -7 2 18 1 8 2- (-8)-!- 18 -16-18 -34 o 2 3 2( 7) 1 3 -14-3 -17 1 -7 The solution desired is therefore (x = 6, y=2). Ans. CHECK. Substituting 6 'for # and 2 for y in (6) and (7) gives 12 +6 = 18 and 6 - 14 = -8, which are true results. EXERCISES Solve each of the following pairs of equations by determi- nants, checking your answers for each of the first three. 1. 2. 3. 4. Sx+3y=-7. 3x-7 y=-&, = 14. 7. 8. I 2x-y=lS. I ax-\-by = r, \bx ay = s. j .2a4-.56 = 30, 5 * \. 4 a-. 8 6= -16. XXIII, 170] DETERMINANTS 269 170. Determinants of the Third Order. The symbol (1) is called a determinant of the third order. Its value is defined as follows : This expression, as we shall see presently, is important in the study of equations. The expression (2) is called the expanded form of the determinant (1). It is important to observe that this expanded form may be written down at once as follows. Write the determinant with the first two columns repeated at the right and first note the three diagonals which then run down from left to right (marked +). The product of the elements in the first of these diagonals is ai 6 2 c 3 , and this is seen to be the first term of the expanded form (2). Similarly, the product of the elements in the second of these diagonals is 6iC 2 3 , which forms the second term of (2) ; and likewise the third diagonal furnishes at once the third term of (2). Next consider the three diagonals which run up from left to right (marked with dotted lines) . The product of the elements in the first of these is o 3 6 2 Ci, and this is the fourth term of (2), provided it be taken negatively, that is preceded by the sign . Similarly, the other two dotted diagonals of (3) furnish the last two terms of (2), provided they be taken negatively. NOTE. Every determinant of the third order when expanded contains a total of six terms. EXAMPLE. Expand and find the value of the determinant 379 214 632 270 SECOND COURSE IN ALGEBRA [XXIII, 170 SOLUTION. Repeating the first and second columns at the right, we have 379 214 632 3 7 2 1 6 3 The diagonals running down from left to right give the three products 3-1-2, 7-4-6, 9-2-3, which form the first three terms of the expansion. The diagonals running up from left to right give the products 6-1-9, 3-4-3, 2-2-7, which, when taken negatively, form the three remaining terms of the determinant. The complete expanded form of (3) is, therefore, 3- 1- 2+7-4- 6+9-2- 3-6- -1-9-3-4. 3-2- 2- 7, which reduces to 6+168+54-54-36-28 = 110. Ans. * EXERCISES Expand and find the value of the following determinants. 3. 4. 13 7 24 6 35-4 -7 3 2 2 -4 6 8 -5 -3 8 2 3 6 5 . 3 7 2a 3 66 3 a 2 -56 a. -26 5. 6. 7. 8. x 7 1 23-4 4 2 1 a 6 2 -453 210 a 6 c d e f x y z 1 x-y x+y XXIII, 171] DETERMINANTS 271 *171. Solution of Three Linear Equations. Let us consider a sys- tem of three linear equations between three unknown letters, such as x, y, and z. Any such system is of the form (1) where 01, 61, ci, di, a 2 , b 2 , etc., represent known numbers (coefficients). This system may be solved for x, y, and z by elimination, as in 35, but the process is long. We shall here state merely the results, which are as follows (compare with (3) and (4) of 169) : _dibzC3-\-d 2 b 3 c\-}-dzbiCz dsbzCi _ gib-ids -\-dzb 3 di -\-d 3 bidzd 3 bzdi dib 3 dza 3 bidz w (2) It is clear by 170 that in these values for x, y, and z, each nu- merator and denominator is the expanded form of a determinant of the third order. In fact, it appears from the definition in 170, that we may now express these values of x, y, and z in the following condensed (determinant) forms : i 61 ( (3) x = 61 b 3 61 d 3 The importance of these expressions for x, y, and z lies in the fact that they give at once the solution of any system such as (1) in very compact and easily remembered forms. The following features should be especially noted : 1. The denominator determinant is the same in all three cases. (Compare statement 1 of 169.) 2. The determinant for the numerator of the z-value is the same as that for the denominator determinant except that the numbers di, dz, dz replace the 01, a 2 , 03 which occur in the first column of the denominator determinant. 272 SECOND COURSE IN ALGEBRA [XXIII, 171 3. Similarly, the numerator of the y-v&lue is formed from that of the denominator determinant by replacing the second column by the elements di, d 2 , d 3 ; while the numerator of the 2-value is formed from that of the denominator determinant by replacing the third column by the elements d if d 2 , d 3 . (Compare statements 2 and 3 of 169.) The readiness with which (3) may be used in practice to solve a system of three linear equations is illustrated by the following EXAMPLE. Solve the system 2x-y+3 = 35, z+37/-15=-2z, SOLUTION. Arranging the equations as in (1) of 171, the given system is 2 x-y +3 z =35, Therefore, using (3) of 171, we have at once 0+180-2-9-280-0_-lll_ : + 12-6-27-16-0 ~~^ 35 -1 3 15 3 2 1 4 2 -1 3 1 3 2 3 4 2 35 3 1 15 2 3 1 2 -1 3 1 3 2 3 4 2 -1 35 1 3 15 3 4 1 2 -1 3 1 3 2 3 4 _o , R ~ 3 ' _0+3+210-135-4-Q^ 74 -37 -37 = -2, _6 + 140-45-315-120+l^-333 -37 " -37 = 9. XXIII, 172] DETERMINANTS 273 The desired solution is, therefore, (3=3, 0= 2, 2 = 9). Ans. CHECK. With x = 3, y = 2, z = 9, it is readily seen that the three given equations are satisfied. EXERCISES Solve each of the following systems by determinants. 3-20+2=5, f 3+20=0, 1. { 33+60-42=3, 6. 3+2= -3, 83-100+32=34. [40+3-22 = 4. f 23-3 0-42 = 25, 3. 3 +0-2 =-4, 4. -9 f 3+0 = 3 a, 33+20-2 = 2, 7. <|z+2=56, 23-0+2=^. * 172. Determinants of Higher Order. Determinants of the fourth order exist and are studied in higher algebra, as are deter- minants of the fifth order, sixth order, etc. Moreover, determi- nants of the fourth order bear a similar relation to the solving of four linear equations between four unknown letters, as determinants of the third order bear to the solving of three linear equations be- tween three unknown letters ; and a similar remark may be made regarding determinants of the fifth order, sixth order, etc. In all cases, the solutions of such systems of equations can be expressed very simply by means of determinants. APPENDIX TABLE OF POWERS AND ROOTS EXPLANATION 1. Square Roots. The way to find square roots from the Table is best understood from an example. Thus, suppose we wish to find Vl.48. To do this we first locate 1.48 in the column headed by the letter n. We find it near the bottom of this column (next to the last number). Now we go across on that level until we get into the column headed by Vn. We find at that place the number 1.21655. This is .our answer. That is, Vl.48 =1.21655 (approxi- mately) . If we had wanted V14.8 instead of Vl.48 the work would have been the same except that we would have gone over into the column headed VlO n (because 14.8=10X1.48). The number thus located is seen to be 3.84708, which is, therefore, the desired value of V14.8. Again, if we had wished to find Vl48 the work would take us back again to the column headed Vn, but now instead of the answer being 1.21655 it would be 12.1655. In other words, the order of the digits in Vl48 is the same as for Vl.48, but the decimal point in the answer is one place farther to the right. Similarly, if we desired V1480 the work would be the same as before except that we must now use the column headed VlO n and move the decimal point there occurring one place farther to the right. This is seen to give 38.4708. Thus we see how to get the square root of 1.48 or any power of 10 times that number. 275 276 APPENDIX In the same way, if we wish to find V.148, or V.0148, or V.00148, or the square root of any number obtained by dividing 1.48 by any power of 10 we can get the answers from the column headed Vn or VlOn by merely placing the decimal point properly. Thus, we find that V.148 = .384708, V^148 = .121655, VML48 = .0384708, etc. What we have seen in regard to the square root of 1.48 or of that number multiplied or divided by any power of 10 holds true in a similar way for any number that occurs in the column headed n, so that the tables thus give us the square roots of a great many numbers. 2. Cube Roots. Cube roots are located in the tables in much the same way as that just described for square roots, but we have here three columns to select from instead of two, namely the columns headed Vn, VlO n, VlOO n. Illustration. v/1.48 occurs in the column headed %/n and is seen to be 1.13960. -v/14.8 occurs in the column headed -^10 n and is seen to be 2.4552. \/148 occurs in the column headed v'lOO n and is seen to be 5.28957. To get vO48 we observe that .148 = ^M^ = ^jjj: = ^ ^148. * .LvJ JLLHJU -L\J Thus, we look up \/148 and divide it by 10. The result is instantly seen to be .528957. Similarly, to get \/.0148 we observe that ^0148 = -vS| = A/J^ = i 3/Ul8. Thus, we look up \/14^Fand 10U 1UUU 10 divide it by 10, giving the result .24552. To get ^.00148 we observe that V/.00148 = A/T?H = A V ^8' so -LvJv/vJ JLU that we must divide Vl.48 by 10. This gives .11396. Similarly the cube root of any number occurring in the column headed n may be found, as well as the cube root of any number obtained by multiplying or dividing such a number by any power of 10. TABLE OF POWERS AND ROOTS 277 3. Squares and Cubes. To find the square of 1.48 we naturally look at the proper level in the column headed n 2 . Here we find 2.1904, which is the answer. If we wished the square of 14.8 the result would be the same except that the decimal point must be moved two places to the right, giving 219.04 as the answer. Similarly the value of (148) 2 is 21904.0 etc. On the other hand, the value of (.148) 2 is found by moving the decimal place two places to the left, thus giving .021904. Similarly, (.0148) 2 = .00021904, etc. To find (1.48) 3 we look at the proper level in the column headed n 3 where we find 3.24179. The value of (14.8) 3 is the same except that we must move the decimal point three places to the right, giving 3241.79. Similarly, in finding (.148) 3 we must move the decimal place three places to the left, giving .00324179. Further illustrations of the way to use the tables will be found in 43. EXERCISES Read off from the tables the values of each of the following ex- pressions. 1. \/4l 4. v'GTO 7. V93/7 10. v/,00154 2. V8.9 5. V^9 8. V93.7 11. V.000143 3. >/67 Q. VjOlG 9. V.00154 12. v^.000143 278 Table I Powers and Roots n n 2 V^ vio n* * VWn ^ioow 1.00 1.0000 1.00000 3.16228 1.00000 1.00000 2.15443 4.64159 1.01 1.02 1.03 1.0201 1.0404 1.0609 1.00499 1.00995 1.01489 3.17805 3.19374 3.20936 1.03030 1.06121 1.09273 1.00332 1.00662 1.00990 2.16159 2.16870 2.17577 4.65701 4.67233 4.68755 1.04 1.05 1.06 1.0816 1.1025 1.1236 1.01980 1.02470 1.02956 3.22490 3.24037 3.25576 1.12486 1.15762 1.19102 1.01316 1.01640 1.01961 2.18279 2.18976 2.19669 4.70267 4.71769 4.73262 1.07 1.08 1.09 1.1449 1.1664 1.1881 1.03441 1.03923 1.04403 3.27109 3.28634 3.30151 1.22504 1.25971 1.29503 1.02281 1.02599 1.02914 2.20358 2.21042 2.21722 4.74746 4.76220 4.77686 1.10 1.2100 1.04881 3.31662 1.33100 1.03228 2.22398 4.79142 1.11 1.12 1.13 1.2321 1.2544 1.2769 1.05357 1.05830 1.06301 3.33167 3.34664 3.36155 1.36763 1.40493 1.44290 1.03540 1.03850 1.04158 2.23070 2.23738 2.24402 4.80590 4.82028 4.83459 1.14 .15 .16 1.2996 1.3225 1.3456 1.06771 1.07238 1.07703 3.37639 3.39116 3.40588 1.48154 1.52088 1.56090 1.04464 1.04769 1.05072 2.25062 2.25718 2.26370 4.84881 4.86294 4.87700 .17 .18 .19 1.3689 1.3924 1.4161 1.08167 1.08628 1.09087 3.42053 3.43511 3.44964 1.60161 1.64303 1.68516 1.05373 1.05672 1.05970 2.27019 2.27664 2.28305 4.89097 4.90487 4.91868 1.20 1.4400 1.09545 3.46410 1.72800 1.06266 2.28943 4.93242 1.21 1.22 1.23 1.4641 1.4884 1.5129 1.10000 1.10454 1.10905 3.47851 3.49285 3.50714 1.77156 1.81585 1.86087 1.06560 1.06853 1.07144 2.29577 2.30208 2.30835 4.94609 4.95968 4.97319 1.24 1.25 1.26 1.5376 1.5625 1.5876 1.11355 1.11803 1.12250 3.52136 3.53553 3.54965 1.90662 1.95312 2.00038 1.07434 1.07722 1.08008 2.31459 2.32079 2.32697 4.98663 5.00000 5.01330 1.27 1.28 1.29 1.6129 1.6384 1.6641 1.12694 1.13137 1.13578 3.56371 3.57771 3.59166 2.04838 2.09715 2.14669 1.08293 1.08577 1.08859 2.33311 2.33921 2.34529 5.02653 5.03968 5.05277 1.30 1.6900 1.14018 3.60555 2.19700 1.09139 2.35133 5.06580 1.31 1.32 1.33 1.7161 1.7424 1.7689 1.14455 1.14891 1.15326 3.61939 3.63318 3.64692 2.24809 2.29997 2.35264 1.09418 1.09696 1.09972 2.35735 2.36333 2.36928 5.07875 5.09164 5.10447 1.34 1.35 1.36 1.7956 1.8225 1.8496 1.15758 1.16190 1.16619 3.66060 3.67423 3.68782 2.40610 2.46038 2.51546 1.10247 1.10521 1.10793 2.37521 2.38110 2.38697 5.11723 5.12993 5.14256 1.37 1.38 1.39 1.8769 1.9044 1.9321 1.17047 1.17473 1.17898 3.70135 3.71484 3.72827 2.57135 2.62807 2.68562 1.11064 1.11334 1.11602 2.39280 2.39861 2.40439 5.15514 5.16765 5.18010 1.40 1.9600 1.18322 3.74166 2.74400 1.11869 2.41014 5.19249 1.41 1.42 1.43 1.9881 2.0164 2.0449 1.18743 1.19164 1.19583 3.75500 3.76829 3.78153 2.80322 2.86329 2.92421 1.12135 1.12399 1.12662 2.41587 2.42156 2.42724 5.20483 5.21710 5.22932 1.44 1.45 1.46 2.0736 2.1025 2.1316 1.20000 1.20416 1.20830 3.79473 3.80789 3.82099 2.98598 3.04862 3.11214 1.12924 1.13185 1.13445 2.43288 2.43850 2.44409 6.24148 5.26359 6.26564 1.47 1.48 1.49 2.1609 2.1904 2.2201 1.21244 1.21655 1.22066 3.83406 3.84708 3.86005 3.17652 3.24179 3.30795 1.13703 1.13960 1.14216 2.44966 2.46620 111071! 6.27763 5.28957 6.30146 Powers and Roots 279 n n z Vn VlOw n B Vn ^10 n VWQn 1.50 2.2500 1.22474 3.87298 3.37500 1.14471 2.46621 5.31329 1.51 1.52 1.53 2.2801 2.3104 2.3409 1.22882 1.23288 1.23693 3.88587 3.89872 3.91152 3.44295 3.51181 3.58158 1.14725 1.14978 1.15230 2.47168 2.47712 2.48255 5.32507 5.33680 5.34848 1.54 1.55 1.56 2.3716 2.4025 2.4336 1.24097 1.24499 1.24900 3.92428 3.93700 3.94968 3.65226 3.72388 3.79642 1.15480 1.15729 1.15978 2.48794 2.49332 2.49867 5.36011 5.37169 5.38321 1.57 ; 1.58 1.59 2.4649 2.4964 2.5281 1.25300 1.25698 1.26095 3.96232 3.97492 3.98748 3.86989 3.94431 4.01968 1.16225 1.16471 1.16717 2.50399 2.50930 2.51458 5.39469 5.40612 5.41750 1.60 2.5600 1.26491 4.00000 4.09600 1.16961 2.51984 5.42884 1.61 1.62 1.63 2.5921 2.6244 2.6569 1.26886 1.27279 1.27671 4.01248 4.02492 4.03733 4.17328 4.25153 4.33075 1.17204 1.17446 1.17687 2.52508 2.53030 2.53549 5.44012 5.45136 5.46256 1.64 1.65 1.66 2.6896 2.7225 2.7556 1.28062 1.28452 1.28841 4.04969 4.06202 4.07431 4.41094 4.49212 4.57430 1.17927 1.18167 1.18405 2.54067 2.54582 2.55095 5.47370 5.48481 5.49586 1.67 1.68 1.69 2.7889 2.8224 2.8561 1.29228 1.29615 1.30000 4.08656 4.09878 4.11096 4.65746 4.74163 4.82681 1.18642 1.18878 1.19114 2.55607 2.56116 2.56623 5.50688 5.51785 5.52877 1.70 2.8900 1.30384 4.12311 4.91300 1.19348 2.57128 5.53966 1.71 1.72 1.73 2.9241 2.9584 2.9929 1.30767 1.31149 1.31529 4.13521 4.14729 4.15933 5.00021 5.08845 5.17772 1.19582 1.19815 1.20046 2.57631 2.58133 2.58632 5.55050 5.56130 5.57205 1.74 1.75 1.76 3.0276 3.0625 3.0976 1.31909 1.32288 1.32665 4.17133 4.18330 4.19524 5.26802 5.35938 5.45178 1.20277 1.20507 1.20736 2.59129 2.59625 2.60118 5.58277 5.59344 5.60408 1.77 1.78 1.79 3.1329 3.1684 3.2041 1.33041 1.33417 1.33791 4.20714 4.21900 4.23084 5.54523 5.63975 5.73534 1.20964 1.21192 1.21418 2.60610 2.61100 2.61588 5.61467 5.62523 5.63574 1.80 3.2400 1.34164 4.24264 5.83200 1.21644 2.62074 5.64622 1.81 1.82 1.83 3.2761 3.3124 3.3489 1.34536 1.34907 1.35277 4.25441 4.26615 4.27785 5.92974 6.02857 6.12849 1.21869 1.22093 1.22316 2.62559 2.63041 2.63522 5.65665 5.66705 5.67741 1.84 1.85 1.86 3.3856 3.4225 3.4596 1.35647 1.36015 1.36382 4.28952 4.30116 4.31277 6.22950 6.33162 6.43486 1.22539 1.22760 1.22981 2.64001 2.64479 2.64954 5.68773 5.69802 5.70827 1.87 1.88 1.89 3.4969 3.5344 3.5721 1.36748 1.37113 1.37477 4.32435 4.33590 4.34741 6.53920 6.64467 6.75127 1.23201 1.23420 1.23639 2.65428 2.65901 2.66371 5.71848 5.72865 5.73879 1.90 3.6100 1.37840 4.35890 6.85900 1.23856 2.66840 5.74890 1.91 1.92 1.93 3.6481 3.6864 3.7249 1.38203 1.38564 1.38924 4.37035 4.38178 4.39318 6.96787 7.07789 7.18906 1.24073 1.24289 1.24505 2.67307 2.67773 2.68237 5.75897 5.76900 5.77900 1.94 1.95 1.96 3.7636 3.8025 3.8416 1.39284 1.39642 1.40000 4.40454 4.41588 4.42719 7.30138 7.41488 7.52954 1.24719 1.24933 1.25146 2.68700 2.69161 2.69620 5.78896 5.79889 5.80879 1.97 1.98 1.99 3.8809 3.9204 3.9601 1.40357 1.40712 1.41067 4.43847 4.44972 4.46094 7.64537 7.76239 7.88060 1.25359 1.25571 1.25782 2.70078 2.70534 2.70989 5.81865 5.82848 5.83827 280 Powers and Roots n n 2 v^ vio n 8 tfi ^Wn VWOn 2.00 4.0000 1.41421 4.47214 8.00000 1.25992 2.71442 5.84804 2.01 2.02 2.03 4.0401 4.0804 4.1209 1.41774 1.42127 1.42478 448330 4.49444 4.50555 8.12060 8.24241 8.36543 1.26202 1.26411 1.26619 2.71893 2.72344 2.72792 5.85777 5.86746 5.87713 2.04 2.05 2.06 4.1616 4.2025 4.2436 1.42829 1.43178 1.43527 4.51664 4.52769 4.53872 8.48966 8.61512 8.74182 1.26827 1.27033 1.27240 2.73239 2.73685 2.74129 5.88677 6.89637 5.90594 2.07 2.08 2.09 4.2849 4.3264 4.3681 1.43875 1.44222 1.44568 4.54973 4.56070 4.57165 8.86974 8.99891 9.12933 1.27445 1.27650 1.27854 2.74572 2.75014 2.75454 5.91548 5.92499 5.93447 2.10 4.4100 1.44914 4.58258 9.26100 1.28058 2.75892 5.94392 2.11 2.12 2.13 4.4521 4.4944 4.5369 1.45258 1.45602 1.45945 4.59347 4.60435 4.61519 9.39393 9.52813 9.66360 1.28261 1.28463 1.28665 2.76330 2.76766 2.77200 5.95334 5.96273 5.97209 2.14 2.15 2.16 4.5796 4.6225 4.6656 1.46287 1.46629 1.46969 4.62601 4.63681 4.64758 9.80034 9.93838 10.0777 1.28866 1.29066 1.29266 2.77633 2.78065 2.78495 5.98142 5.99073 6.00000 2.17 2.18 2.19 4.7089 4.7524 4.7961 1.47309 1.47648 1.47986 4.65833 4.66905 4.67974 10.2183 10.3602 10.5035 1.29465 1.29664 1.29862 2.78924 2.79352 2.79779 6.00925 6.01846 6.02765 2.20 4.8400 1.48324 4.69042 10.6480 1.30059 2.80204 6.03681 2.21 2.22 2.23 4.8841 4.9284 4.9729 1.48661 1.48997 1.49332 4.70106 4.71169 4.72229 10.7939 10.9410 11.0896 1.30256 1.30452 1.30648 2.80628 2.81050 2.81472 6.04594 6.05505 6.06413 2.24 2.25 2.26 5.0176 5.0625 5.1076 1.49666 1.50000 1.50333 4.73286 4.74342 4.75395 11.2394 11.3906 11.5432 1.30843 1.31037 1.31231 2.81892 2.82311 2.82728 6.07318 6.08220 6.09120 2.27 2.28 2.29 5.1529 5.1984 5.2441 1.50665 1.50997 1.51327 4.76445 4.77493 4.78539 11.6971 11.8524 12.0090 1.31424 1.31617 1.31809 2.83145 2.83560 2.83974 6.10017 6.10911 6.11803 2.30 5.2900 1.51658 4.79583 12.1670 1.32001 2.84387 6.12693 2.31 2.32 2.33 5.33G1 5.3824 5.4289 1.51987 1.52315 1.52643 4.80625 4.81664 4.82701 12.3264 12.4872 12.6493 1.32192 1.32382 1.32572 2.84798 2.85209 2.85618 6.13579 6.14463 6.15345 2.34 2.35 2.36 5.4756 5.5225 5.5696 1.52971 1.53297 1.53623 4.83735 4.84768 4.85798 12.8129 12.9779 13.1443 1.32761 1.32950 1.33139 2.86026 2.86433 2.86838 6.16224 6.17101 6.17975 2.37 2.38 2.39 5.6169 5.6644 5.7121 1.53948 1.54272 1.54596 4.86826 4.87852 4.88876 13.3121 13.4813 13.6519 1.33326 1.33514 1.33700 2.87243 2.87646 2.88049 6.18846 6.19715 6.20582 2.40 5.7600 1.54919 4.89898 13.8240 1.33887 2.88450 6.21447 2.41 2.42 2.43 5.8081 5.8564 5.9049 1.55242 1.55563 1.55885 4.90918 4.91935 4.92950 13.9975 14.1725 14.3489 1.34072 1.34257 1.34442 2.88850 2.89249 2.89647 6.22308 6.23168 6.24025 2.44 2.45 2.46 5.9536 6.0025 6.0516 1.56205 1.56525 1.56844 4.93964 4.94975 4.95984 14.5268 14.7061 14.8869 .34626 .34810 1.34993 2.90044 2.90439 2.90834 6.24880 6.25732 6.26583 2.47 2.48 2.49 6.1009 6.1504 6.2001 1.57162 1.57480 1.57797 4.96991 4.97996 4.98999 15.0692 15.2530 15.4:382 1.35176 1.35358 1.35540 2.91227 2.91620 2.92011 6.27431 6.28276 6.29119 Powers and Roots 281 n W 2 Vn VlOw n s Vn VlOn ^ioo^ 2.50 6.2500 1.58114 5.00000 15.6250 1.35721 2.92402 6.29961 2.51 2.52 2.53 6.3001 6.3504 6.4009 1.58430 1.58745 1.59060 5.00999 5.01996 5.02991 15.8133 16.0030 16.1943 1.35902 1.36082 1.36262 2.92791 2.93179 2.93567 6.30799 6.31636 6.32470 2.54 2.55 2.56 6.4516 6.5025 6.5536 1.59374 1.59687 1.60000 5.03984 5.04975 5.05964 16.3871 16.5814 16.7772 1.36441 1.36620 1.36798 2.93953 2.94338 2.94723 6.33303 6.34133 6.34960 2.57 2.58 2.59 6.6049 6.6564 6.7081 1.60312 1.60624 1.60935 5.06952 5.07937 5.08920 16.9746 17.1735 17.3740 1.36976 1.37153 1.37330 2.95106 2.95488 2.95869 6.35786 6.36610 6.37431 2.60 6.7600 1.61245 5.09902 17.5760 1.37507 2.96250 6.38250 2.61 2.62 2.63 6.8121 6.8644 6.9169 1.61555 1.61864 1.62173 5.10882 5.11859 5.12835 17.7796 17.9847 18.1914 1.37683 1.37859 1.38034 2.96629 2.97007 2.97385 6.39068 6.39883 6.40696 2.64 2.65 2.66 6.9696 7.0225 7.0756 1.62481 1.62788 1.63095 5.13809 5.14782 5.15752 18.3997 18.6096 18.8211 1.38208 1.38383 1.38557 2.97761 2.98137 2.98511 6.41507 6.42316 6.43123 2.67 2.68 2.69 7.1289 7.1824 7.2361 1.63401 1.63707 1.64012 5.16720 5.17687 5.18652 19.0342 19.2488 19.4651 1.38730 1.38903 1.39076 2.98885 2.99257 2.99629 6.43928 6.44731 6.45531 2.70 7.2900 1.64317 5.19615 19.6830 1.39248 3.00000 6.46330 2.71 2.72 2.73 7.3441 7.3984 7.4529 1.64621 1.64924 1.65227 5.20577 5.21536 5.22494 19.9025 20.1236 20.3464 1.39419 1.39591 1.39761 3.00370 3.00739 3.01107 6.47127 6.47922 6.48715 2.74 2.75 2.76 7.5076 7.5625 7.6176 1.65529 1.65831 1.66132 5.23450 5.24404 5.25357 20.5708 20.7969 21.0246 1.39932 1.40102 1.40272 3.01474 3.01841 3.02206 6.49507 6.50296 6.51083 2.77 2.78 2.79 7.6729 7.7284 7.7841 1.66433 1.66733 1.67033 5.26308 5.27257 5.28205 21.2539 21.4850 21.7176 1.40441 1.40610 1.40778 3.02570 3.02934 3.03297 6.51868 6.52(552 6.53434 2.80 7.8400 1.67332 5.29150 21.9520 1.40946 3.03659 6.54213 2.81 2.82 2.83 7.89G1 7.9524 8.0089 1.67631 1.67929 1.68226 5.30094 5.31037 5.31977 22.1880 22.4258 22.6652 1.41114 1.41281 1.41448 3.04020 3.04380 3.04740 6.54991 6.55767 6.56541 2.84 2.85 2.86 8.0656 8.1225 8.1796 1.68523 1.68819 1.69115 5.32917 5.33854 5.34790 22.9063 23.1491 23.3937 1.41614 1.41780 1.41946 3.05098 3.05456 3.05813 6.57314 6.58084 6.58853 2.87 2.88 2.89 8.2369 8.2944 8.3521 1.69411 1.69706 1.70000 5.35724 5.36656 5.37587 23.6399 23.8879 24.1376 1.42111 1.42276 1.42440 3.06169 3.06524 3.06878 6.59620 6.60385 6.61149 2.90 8.4100 1.70294 5.38516 24.3890 1.42604 3.07232 6.61911 2.91 2.92 2.93 8.4681 8.5264 8.5849 1.70587 1.70880 1.71172 5.39444 5.40370 5.41295 24.6422 24.8971 25.1538 1.42768 1.42931 1.43094 3.07584 3.07936 3.08287 6.62671 6.63429 6.64185 2.94 2.95 2.96 8.6436 8.7025 8.7616 1.71464 1.71756 1.72047 5.42218 5.43139 5.44059 25.4122 25.6724 25.9343 1.43257 1.43*19 1.43581 3.08638 3.08987 3.09336 6.64940 6.65693 6.66444 2.97 2.98 2.99 8.8209 8.8804 8.9401 1.72337 1.72627 1.72916 5.44977 5.45894 5.46809 26.1981 26.4636 26.7309 1.43743 1.43904 1.44065 3.09684 3.10031 3.10378 6.67194 6.67942 6.68688 282 Powers and Roots ft n 2 V^ VWn n* Vn VWn VlMn 3.00 9.0000 1.73205 5.47723 27.0000 1.44225 3.10723 6.69433 3.01 3.02 3.03 9.0601 9.1204 9.1809 1.73494 1.73781 1.74069 5.486:35 5.49545 5.50454 27.2709 27.5436 27.8181 1.44385 1.44545 1.44704 3.11068 3.11412 3.11756 6.70176 6.70917 6.71657 3.04 3.05 3.06 9.2416 9.3025 9.3636 1.74356 1.74642 1.74929 5.51362 5.52268 5.53173 28.0945 28.3726 28.6526 1.44863 1.45022 1.45180 3.12098 3.12440 3.12781 6.72395 6.73132 6.73866 3.07 3.08 3.09 9.4249 9.4864 9.5481 1.75214 1.75499 1.75784 5.54076 5.54977 5.55878 28.9344 29.2181 29.5036 1.45338 1.45496 1.45653 3.13121 3.13461 3.13800 6.74600 6.75331 6.76061 3.10 9.6100 1.76068 5.56776 29.7910 1.45810 3.14138 6.76790 3.11 3.12 3.13 9.6721 9.7344 9.7969 1.76352 1.76635 1.76918 5.57674 5.58570 5.59464 30.0802 30.3713 30.6643 1.45967 1.46123 1.46279 3.14475 3.14812 3.15148 6.77517 6.78242 6.78966 3.14 3.15 3.16 9.8596 9.9225 9.9856 1.77200 1.77482 1.77764 5.60357 5.61249 5.62139 30.9591 31.2559 31.5545 1.46434 1.46590 1.46745 3.15483 3.15818 3.16152 6.79688 6.80409 6.81128 3.17 3.18 3.19 10.0489 10.1124 10.1761 1.78045 1.78326 1.78606 5.63028 5.63915 5.64801 31.8550 32.1574 32.4618 1.46899 1.47054 1.47208 3.16485 3.16817 3.17149 6.81846 6.82562 6.83277 3.20 10.2400 1.78885 5.65685 32.7680 1.47361 3.17480 6.83990 3.21 3.22 3.23 10.3041 10.3684 10.4329 1.79165 1.79444 1.79722 5.66569 5.67450 5.68331 33.0762 33.3862 33.6983 1.47515 1.47668 1.47820 3.17811 3.18140 3.18469 6.84702 6.85412 6.86121 3.24 3.25 3.26 10.4976 10.5625 10.6276 1.80000 1.80278 1.80555 5.69210 5.70088 5.70964 34.0122 34.3281 34.6460 1.47973 1.48125 1.48277 3.18798 3.19125 3.19452 6.86829 6.87534 6.88239 3.27 3.28 3.29 10.6929 10.7584 10.8241 1.80831 1.81108 1.81384 5.71839 5.72713 5.73585 34.9658 35.2876 35.6113 1.48428 1.48579 1.48730 3.19778 3.20104 3.20429 6.88942 6.89643 6.90344 3.30 10.8900 1.81659 5.74456 35.9370 1.48881 3.20753 6.91042 3.31 3.32 3.33 10.9561 11.0224 11.0889 1.81934 1.82209 1.82483 5.75326 5.76194 5.77062 36.2647 36.5944 36.9260 1.49031 1.49181 1.49330 3.21077 3.21400 3.21722 6.91740 6.92436 6.93130 3.34 3.35 3.36 11.1556 11.2225 11.2896 1.82757 1.83030 1.83303 5.77927 5.78792 5.79655 37.2597 37.5954 37.9331 1.49480 1.49629 1.49777 3.22044 3.22365 3.22686 6.93823 6.94515 6.95205 3.37 3.38 3.39 11.3569 11.4244 11.4921 1.83576 1.83848 1.84120 5.80517 5.81378 5.82237 38.2728 38.6145 38.9582 1.49926 1.50074 1.50222 3.23006 3.23325 3.23643 6.95894 6.96582 6.97268 3.40 11.5600 1.84391 5.83095 39.3040 1.50369 3.23961 6.97953 3.41 3.42 3.43 11.6281 11.6964 11.7649 1.84662 1.84932 1.85203 5.83952 5.84808 6.85662 39.6518 40.0017 40.3536 1.50517 1.50664 1.50810 3.24278 3.24595 3.24911 6.98637 6.99319 7.00000 3.44 3.45 3.46 11.8336 11.9025 11.9716 1.85472 1.8.7742 1.86011 5.86515 5.87367 5.88218 40.7076 41.0636 41.4217 1.50957 1.51103 1.51249 3.25227 3.25542 3.25856 7.00680 7.01358 7.02035 3.47 3.48 3.49 12.0409 12.1104 12.1801 1.86279 1.86548 1.86815 5.89067 5.89915 5.90762 41.7819 42.1442 42.5085 1.51394 1.51540 1.51685 3.26169 3.26482 3.26795 7.02711 7.03385 7.04058 Powers and Roots 283 n n? Vn vio n* ^n VWH ^lOOw 3.50 12.2500 1.87083 5.91608 42.8750 1.51829 3.27107 7.04730 3.51 3.52 3.53 12.3201 12.3904 12.4609 1.87350 1.87617 1.87883 5.92453 5.93296 5.94138 43.2436 43.6142 43.9870 1.51974 1.52118 1.52262 3.27418 3.27729 3.28039 7.05400 7.06070 7.06738 3.54 3.55 3.56 12.5316 12.6025 12.6736 1.88149 1.88414 1.88680 5.94979 5.95819 5.96657 44.3619 44.7389 45.1180 1.52406 1.52549 1.52692 3.28348 3.28657 3.28965 7.07404 7.08070 7.08734 3.57 3.58 3.59 12.7449 12.8164 12.8881 1.88944 1.89209 1.89473 5.97495 5.98331 5.99166 45.4993 45.8827 46.2683 1.52835 1.52978 1.53120 3.29273 3.29580 3.29887 7.09397 7.10059 7.10719 3.60 12.9600 1.89737 6.00000 46.6560 1.53262 3.30193 7.11379 3.61 3.62 3.63 13.0321 13.1044 13.1769 1.90000 1.90263 1.90526 6.00833 6.01664 6.02495 47.0459 47.4379 47.8321 1.53404 1.53545 1.53686 3.30498 3.30803 3.31107 7.12037 7.12694 7.13349 3.64 3.65 3.66 13.2496 13.3225 13.3956 1.90788 1.91050 1.91311 6.03324 6.04152 6.04979 48.2285 48.6271 49.0279 1.53827 1.53968 1.54109 3.31411 3.31714 3.32017 7.14004 7.14657 7.15309 3.67 3.68 3.69 13.4689 13.5424 13.6161 1.91572 1.91833 1.92094 6.05805 6.06630 6.07454 49.4309 49.8360 50.2434 1.54249 1.54389 1.54529 3.32319 3.32621 3.32922 7.15960 7.16610 7.17258 3.70 13.6900 1.92354 6.08276 50.6530 1.54668 3.33222 7.17905 3.71 3.72 3.73 13.7641 13.8384 13.9129 1.92614 1.92873 1.93132 6.09098 6.09918 6.10737 51.0648 51.4788 51.8951 1.54807 1.54946 1.55085 3.33522 3.33822 3.34120 7.18552 7.19197 7.19840 3.74 3.75 3.76 13.9876 14.0625 14.1376 1.93391 1.93649 1.93907 6.11555 6.12372 6.13188 52.3136 52.7344 53.1574 1.55223 1.55362 1.55500 3.34419 3.34716 3.35014 7.20483 7.21125 7.21765 3.77 3.78 3.79 14.2129 14.2884 14.3641 1.94165 1.94422 1.94679 6.14003 6.14817 6.15630 53.5826 54.0102 54.4399 1.55637 1.55775 1.55912 3.35310 3.35607 3.35902 7.22405 7.23043 7.23680 3.80 14.4400 1.94936 6.16441 54.8720 1.56049 3.36198 7.24316 3.81 3.82 3.83 14.5161 14.5924 14.6689 1.95192 1.95448 1.95704 6.17252 6.18061 6.18870 55.3063 55.7430 56.1819 1.56186 1.56322 1.56459 3.36492 3.36786 3.37080 7.24950 7.25584 7.26217 3.84 3.85 3.86 14.7456 14.8225 14.8996 1.95959 1.96214 1.96469 6.19677 6.20484 6.21289 56.6231 57.0666 57.5125 1.56595 1.56731 1.56866 3.37373 3.37666 3.37958 7.26848 7.27479 7.28108 3.87 3.88 3.89 14.9769 15.0544 15.1321 1.96723 1.96977 1.97231 6.22093 6.22896 6.23699 57.9606 58.4111 58.8639 1.57001 1.57137 1.57271 3.38249 3.38540 3.38831 7.28736 7.29363 7.29989 3.90 15.2100 1.97484 6.24500 59.3190 1.57406 3.39121 7.30614 3.91 3.92 3.93 15.2881 15.3664 15.4449 1.97737 1.97990 1.98242 6.25300 6.26099 6.26897 59.7765 60.2363 60.6985 1.57541 1.57675 1.57809 3.39411 3.39700 3.39988 7.31238 7.31861 7.32483 3.94 3.95 3.96 15.5236 15.6025 15.6816 1.98494 1.98746 1.98997 6.27694 6.28490 6.29285 61.1630 61.6299 62.0991 1.57942 1.58076 1.58209 3.40277 3.40564 3.40851 7.33104 7.33723 7.34342 3.97 3.98 3.99 15.7609 15.8404 15.9201 1.99249 1.99499 1.99750 6.30079 6.30872 6.31664 62.5708 63.0448 63.5212 1.58342 1.58475 1.58608 3.41138 3.41424 3.41710 7.34960 7.35576 7.36192 284 Powers and Roots n n 2 V^ VWn n? Vn i/Wn VIMn 4.00 16.0000 2.00000 6.32456 64.0000 1.58740 3.41995 7.36806 4.01 4.02 4.03 16.0801 16.1604 16.2409 2.00250 2.00499 2.00749 6.33246 6.34035 6.34823 64.4812 64.9648 65.4508 1.58872 1.59004 1.59136 3.42280 3.42564 3.42848 7.37420 7.38032 7.38644 4.04 4.05 4.06 16.3216 16.4025 16.4836 2.00998 2.01246 2.01494 6.35610 6.36396 6.37181 65.9393 66.4301 66.9234 1.59267 1.59399 1.59530 3.43131 3.43414 3.43697 7.39254 7.39864 7.40472 4.07 4.08 4.09 16.5649 16.6464 16.7281 2.01742 2.01990 2.02237 6.37966 6.38749 6.39531 67.4191 67.9173 68.4179 1.59661 1.59791 1.59922 3.43979 3.44260 3.44541 7.41080 7.41686 7.42291 4.10 16.8100 2.02485 6.40312 68.9210 1.60052 3.44822 7.42896 4.11 4.12 4.13 16.8921 16.9744 17.0569 2.02731 2.02978 2.03224 6.41093 6.41872 6.42651 69.4265 69.9345 70.4450 1.60182 1.60312 1.60441 3.45102 3.45382 3.45661 7.43499 7.44102 7.44703 4.14 4.15 4.16 17.1396 17.2225 17.3056 2.03470 2.03715 2.03961 6.43428 6.44205 6.44981 70.9579 71.4734 71.9913 1.60571 1.60700 1.60829 3.45939 3.46218 3.46496 7.45304 7.45904 7.46502 4.17 4.18 4.19 17.3889 17.4724 17.5561 2.04206 2.04450 2.04695 6.45755 6.46529 6.47302 72.5117 73.0346 73.5601 1.60958 1.61086 1.61215 3.46773 3.47050 3.47327 7.47100 7.47697 7.48292 4.20 17.6400 2.04939 6.48074 74.0880 1.61343 3.47603 7.48887 4.21 4.22 4.23 17.7241 17.8084 17.8929 2.05183 2.05426 2.05670 6.48845 6.49615 6.50384 74.6185 75.1514 75.6870 1.61471 1.61599 1.61726 3.47878 3.48154 3.48428 7.49481 7.50074 7.50666 4.24 4.25 4.26 17.9776 18.0625 18.1476 2.05913 2.06155 2.06398 6.51153 6.51920 6.52687 76.2250 76.7656 77.3088 1.61853 1.61981 1.62108 3.48703 3.48977 3.49250 7.51257 7.51847 7.52437 4.27 4.28 4.29 18.2329 18.3184 18.4041 2.06640 2.06882 2.07123 6.53452 6.54217 6.54981 77.8545 78.4028 78.9536 1.62234 1.62361 1.62487 3.49523 3.49796 3.50068 7.53025 7.53612 7.54199 4.30 18.4900 2.07364 6.55744 79.5070 1.62613 3.50340 7.54784 4.31 4.32 4.33 18.5761 18.6624 18.7489 2.07605 2.07846 2.08087 6.56506 6.57267 6.58027 80.0630 80.6216 81.1827 1.62739 1.62865 1.62991 3.50611 3.50882 3.51153 7.55369 7.55953 7.56535 4.34 4.35 4.36 18.8356 18.9225 19.0096 2.08327 2.08567 2.08806 6.58787 6.59545 6.60303 81.7465 82.3129 82.8819 1.63116 1.63241 1.63366 3.51423 3.51692 3.51962 7.57117 7.57698 7.58279 4.37 4.38 4.39 19.0969 19.1844 19.2721 2.09045 2.09284 2.09523 6.61060 6.61816 6.62571 83.4535 84.0277 84.6045 1.63491 1.63619 1.63740 3.52231 3.52499 3.52767 7.58858 7.59436 7.60014 4.40 19.3600 2.09762 6.63325 85.1840 1.63864 3.53035 7.60590 4.41 4.42 4.43 19.4481 19.5364 19.6249 2.10000 2.10238 2.10476 6.64078 6.64831 6.65582 85.7661 86.3509 86.9383 1.63988 1.64112 1.64236 3.53302 3.53569 3.53835 7.61166 7.61741 7.62316 4.44 4.45 4.46 19.7136 19.8025 19.8916 2.10713 2.10950 2.11187 6.66333 6.67083 6.67832 87.5284 88.1211 88.7165 1.64359 1.64483 1.64606 3.54101 3.54367 3.54632 7.62888 7.63461 7.64032 4.47 4.48 4.49 19.9809 20.0704 20.1601 2.11424 2.11660 2.11896 6.68581 6.69328 6.70075 89.3146 89.9154 90.5188 1.64729 1.64851 1.64974 3.54897 3.55162 3.55426 7.64603 7.65172 7.65741 Powers and Roots 285 n n 2 Vw VlOw n 8 Vn ^10 n \/100w 4.50 20.2500 2.12132 6.70820 91.1250 1.65096 3.55689 7.66309 4.51 4.52 4.53 20.3401 20.4304 20.5209 2.12368 2.12603 2.12838 6.71565 6.72309 6.73053 91.7339 92.3454 92.9597 1.65219 1.65341 1.65462 3.55953 3.56215 3.56478 7.66877 7.67443 7,68009 4.54 4.55 4.56 20.6116 20.7025 20.7936 2.13073 2.13307 2.13542 6.73795 6.74537 6.75278 93.5767 94.1964 94.8188 1.65584 1.65706 1.65827 3.56740 3.57002 3.57263 7.68573 7.69137 7.69700 4.57 4.58 4.59 20.8849 20.9764 21.0681 2.13776 2.14009 2.14243 6.76018 6.76757 6.77495 95.4440 96.0719 96.7026 1.65948 1.66069 1.66190 3.57524 3.57785 3.58045 7.70262 7.70824 7.71384 4.60 21.1600 2.14476 6.78233 97.3360 1.66310 3.58305 7.71944 4.61 4.62 4.63 21.2521 21.3444 21.4369 2.14709 2.14942 2.15174 6.78970 6.79706 6.80441 97.9722 98.6111 99.2528 1.66431 1.66551 1.66671 3.58564 3.58823 3.59082 7.72503 7.73061 7.73619 4.64 4.65 4.66 21.5296 21.6225 21.7156 2.15407 2.15639 2.15870 6.81175 6.81909 6.82642 99.8973 100.545 101.195 1.66791 1.66911 1.67030 3.59340 3.59598 3.59856 7.74175 7.74731 7.75286 4.67 4.68 4.69 21.8089 21.9024 21.9961 2.16102 2.16333 2.16564 6.83374 6.84105 6.84836 101.848 102.503 103.162 1.67150 1.67269 1.67388 3.60113 3.60370 3.60626 7.75840 7.76394 7.76946 4.70 22.0900 2.16795 6.85565 103.823 1.67507 3.60883 7.77498 4.7? 4.72 4.73 22.1841 22.2784 22.3729 2.17025 2.17256 2.17486 6.86294 6.87023 6.87750 104.487 105.154 105.824 1.67626 1.67744 1.67863 3.61138 3.61394 3.61649 7.78049 7.78599 7.79149 4.74 4.75 4.76 22.4676 22.5625 22.6576 2.17715 2.17945 2.18174 6.88477 6.89202 6.89928 106.496 107.172 107.850 1.67981 1.68099 1.68217 3.61903 3.62158 3.62412 7.79697 7.80245 7.80793 4.77 4.78 4.79 22.7529 22.8484 22.9441 2.18403 2.18632 2.18861 6.90652 6.91375 6.92098 108.531 109.215 109.902 1.68334 1.68452 1.68569 3.62665 3.62919 3.63172 7.81339 7.81885 7.82429 4.80 23.0400 2.19089 6.92820 110.592 1.68687 3.63424 7.82974 4.81 4.82 4.83 23.1361 23.2324 23.3289 2.19317 2.19545 2.19773 6.93542 6.94262 6.94982 111.285 111.980 112.679 1.68804 1.68920 1.69037 3.63676 3.63928 3.64180 7.83517 7.84059 7.84601 4.84 4.85 4.86 23.4256 23.5225 23.6196 2.20000 2.20227 2.20454 6.95701 6.96419 6.97137 113.380 114.084 114.791 1.69154 1.69270 1.69386 3.64431 3.64682 3.64932 7.85142 7.85683 7.86222 4.87 4.88 4.89 23.7169 23.8144 23.9121 2.20681 2.20907 2.21133 6.97854 6.98570 6.99285 115.501 116.214 116.930 1.69503 1.69619 1.69734 3.65182 3.65432 3.65681 7.86761 7.87299 7.87837 4.90 24.0100 2.21359 7.00000 117.649 1.69850 3.65931 7.88374 4.91 4.92 4.93 24.1081 24.2064 24.3049 2.21585 2.21811 2.22036 7.00714 7.01427 7.02140 118.371 119.095 119.823 1.69965 1.70081 1.70196 3.66179 3.66428 3.66676 7.88909 7.89445 7.89979 4.94 4.95 4.96 24.4036 24.5025 24.6016 2.22261 2.22486 2.22711 7.02851 7.03562 7.04273 120.554 121.287 122.024 1.70311 1.70426 1.70540 3.66924 3.67171 3.67418 7.90513 7.91046 7.91578 4.97 4.98 4.99 24.7009 24.8004 24.9001 2.22935 2.23159 2.23383 7.04982 7.05691 7.06399 122.763 123.506 124.251 1.70655 1.70769 1.70884 3.67665 3.67911 3.68157 7.92110 7.92641 7.93171 286 Powers and Roots n W 2 Vn VlOw W 8 Vn vwn ^iooli 5.00 25.0000 2.23607 7.07107 125.000 1.70998 3.68403 7.93701 5.01 5.02 5.03 25.1001 25.2004 25.3009 2.23830 2.24054 2.24277 7.07814 7.08520 7.09225 125.752 126.506 127.264 1.71112 1.71225 1.71339 3.68649 3.68894 3.69138 7.94229 7.94757 7.95285 5.04 5.05 5.06 25.4016 25.5025 25.6036 2.24499 2.24722 2.24944 7.09930 7.10634 7.11337 128.024 128.788 129.554 1.71452 1.71566 1.71679 3.69383 3.69627 3.69871 7.95811 7.96337 7.96863 5.07 5.08 5.09 25.7049 25.8064 25.9081 2.25167 2.25389 2.25610 7.12039 7.12741 7.13442 130.324 131.097 131.872 1.71792 1.71905 1.72017 3.70114 3.70357 3.70600 7.97387 7.97911 7.98434 5.10 26.0100 2.25832 7.14143 132.651 1.72130 3.70843 7.98957 5.11 5.12 5.13 26.1121 26.2144 26.3169 2.26053 2.20274 2.26495 7.14843 7.15542 7.16240 133.433 134.218 135.006 1.72242 1,72355 1.72467 3.71085 3.71327 3.71569 7.99479 8.00000 8.00520 5.14 5.15 5.16 26.4196 26.5225 26.6256 2.26716 2.26936 2.27156 7.16938 7.17635 7.18331 135.797 136.591 137.388 1.72579 1.72691 1.72802 3.71810 3.72051 3.72292 8.01040 8.01559 8.02078 5.1T 5.18 5.19 26.7289 26.8324 26.9361 2.27376 2.27596 2.27816 7.19027 7.19722 7.20417 138.188 138.992 139.798 1.72914 1.73025 1.73137 3.72532 3,72772 3.73012 8.02596 8.03113 8.03629 5.20 27.0400 2.28035 7.21110 140.608 1.73248 3.73251 8.04145 5.21 5.22 5.23 27.1441 27.2484 27.3529 2.28254 2.28473 2.28692 7.21803 7.22496 7.23187 141.421 142.237 143.056 1.73359 1.73470 1.73580 3.73490 3.73729 3.73968 8.04660 8.05175 8.05689 5.24 5.25 5.26 27.4576 27.5625 27.6676 2.28910 2.29129 2.29347 7.23878 7.24569 7.25259 143.878 144.703 145.532 1.73691 1.73801 1.73912 3.74206 3.74443 3.74681 8.06202 8.06714 8.07226 5.27 5.28 5.29 27.7729 27.8784 27.9841 2.29565 2.29783 2.30000 7.25948 7.26636 7.27324 146.363 147.198 148.036 1.74022 1.74132 1.74242 3.74918 3.75155 3.75392 8.07737 8.08248 8.08758 5.30 28.0900 2.30217 7.28011 148.877 1.74351 3.75629 8.09267 5.31 5.32 5.33 28.1961 28.3024 28.4089 2.30434 2.30651 2.30868 7.28697 7.29383 7.30068 149.721 150.569 151.419 1.74461 1.74570 1.74680 3.75865 3.76101 3.76336 8.09776 8.10284 8.10791 5.34 5.35 5.36 28.5156 28.6225 28.7296 2.31084 2.31301 2.31517 7.30753 7.31437 7.32120 152.273 153.130 153.991 1.74789 1.74898 1.75007 3.76571 3.76806 3.77041 8.11298 8.11804 8.12310 5.37 5.38 5.39 28.8369 28.9444 29.0521 2.31733 2.31948 2.32164 7.32803 7.33485 7.34166 154.854 155.721 156.591 1.75116 1.75224 1.75333 3.77275 3.77509 3.77743 8.12814 8.13319 8.13822 5.40 29.1600 2.32379 7.34847 157.464 1.75441 3.77976 8.14325 5.41 5.42 5.43 29.2681 29.3764 29.4849 2.32594 2.32809 2.33024 7.35527 7.36206 7.36885 158.340 159.220 160.103 1.75549 1.75657 1.75765 3.78209 3.78442 3.78675 8.14828 8.15329 8.15831 5.44 5.45 5.46 29.5936 29.7025 29.8116 2.33238 2.33452 2.33666 7.37564 7.38241 7.38918 160.989 161.879 162.771 1.75873 1.75981 1.76088 3.78907 3.79139 3.79371 8.16331 8.16831 8.17330 5.47 5.48 5.49 29.9209 30.0304 30.1401 2.33880 2.34094 2.34307 7.39594 7.40270 7.40945 163.667 164.567 165.469 1.76196 1.76303 1.76410 3.79603 3.79834 3.80065 8.17829 8.18327 8.18824 Powers and Roots 287 ft n* Vn vio n s #n VWn ^100^ 5.50 30.2500 2.34521 7.41620 166.375 1.76517 3.80295 8.19321 5.51 5.52 5.53 30.3601 30.4704 30.5809 2.34734 2.34947 2.35160 7.42294 7.42967 7.43640 167.284 168.197 169.112 1.76624 1.76731 1.76838 3.80526 3.80756 3.80985 8.19818 8.20313 8.20808 5.54 5.55 5.56 30.6916 30.8025 30.9136 2.35372 2.35584 2.35797 7.44312 7.44983 7.45654 170.031 170.954 171.880 1.76944 1.77051 1.77157 3.81215 3.81444 3.81673 8.21303 8.21797 8.22290 5.57 5.58 5.59 31.0249 31.1364 31.2481 2.36008 2.36220 2.36432 7.46324 7.46994 7.47663 172.809 173.741 174.677 1.77263 1.77369 1.77475 3.81902 3.82130 3.82358 8.22783 8.23275 8.23766 5.60 31.3600 2.36643 7.48331 175.616 1.77581 3.82586 8.24257 5.61 5.62 5.63 31.4721 31.5844 31.6969 2.36854 2.37065 2.37276 7.48999 7.49667 7.50333 176.558 177.504 178.454 1.77686 1.77792 1.77897 3.82814 3.83041 3.83268 8.24747 8.25237 8.25726 5.64 5.65 5.66 31.8096 31.9225 32.0356 2.37487 2.37697 2.37908 7.50999 7.51665 7.52330 179.406 180.362 181.321 1.78003 1.78108 1.78213 3.83495 3.83722 3.83948 8.26215 8.26703 8.27190 5.67 5.68 5.69 32.1489 32.2624 32.3761 2.38118 2.38328 2.38537 7.52994 7.53658 7.54321 182.284 183.250 184.220 1.78318 1.78422 1.78527 3.84174 3.84399 3.84625 8.27677 8.28164 8.28649 5.70 32.4900 2.38747 7.54983 185.193 1.78632 3.84850 8.29134 5.71 5.72 5.73 32.6041 32.7184 32.8329 2.38956 2.39165 2.39374 7.55645 7.56307 7.56968 186.169 187.149 188.133 1.78736 1.78840 1.78944 3.85075 3.85300 3.85524 8.29619 8.30103 8.30587 5.74 5.75 5.76 32.9476 33.0625 33.1776 2.39583 2.39792 2.40000 7.57628 7.58288 7.58947 189.119 190.109 191.103 1.79048 1.79152 1.79256 3.85748 3.85972 3.86196 8.31069 8.31552 8.32034 5.77 5.78 5.79 33.2929 33.4084 33.5241 2.40208 2.40416 2.40624 7.59605 7.60263 7.60920 192.100 193.101 194.105 1.79360 1.79463 1.79567 3.86419 3.86642 3.86865 8.32515 8.32995 8.33476 5.80 33.6400 2.40832 7.61577 195.112 1.79670 3.87088 8.33955 5.81 5.82 5.83 33.7561 33.8724 33.9889 2.41039 2.41247 2.41454 7.62234 7.62889 7.63544 196.123 197.137 198.155 1.79773 1.79876 1.79979 3.87310 3.87532 3.87754 8.34434 8.34913 8.35390 5.84 5.85 5.86 34.1056 34.2225 34.3396 2.41661 2.41868 2.42074 7.64199 7.64853 7.65506 199.177 200.202 201.230 1.80082 1.80185 1.80288 3.87975 3.88197 3.88418 8.35868 8.36345 8.36821 5.87 5.88 5.89 34.4569 34.5744 34.6921 2.42281 2.42487 2.42693 7.66159 7.66812 7.67463 202.262 203.297 204.336 1.80390 1.80492 1.80595 3.88639 3.88859 3.89080 8.37297 8.37772 8.38247 5.90 34.8100 2.42899 7.68115 205.379 1.80697 3.89300 8.38721 5.91 5.92 5.93 34.9281 35.0464 35.1649 2.43105 2.43311 2.43516 7.68765 7.69415 7.70065 206.425 207.475 208.528 1.80799 1.80901 1.81003 3.89519 3.89739 3.89958 8.39194 8.39667 8.40140 5.94 5.95 5.96 35.2836 35.4025 35.5216 2.43721 2.48926 2.44131 7.70714 7.71362 7.72010 209.585 210.645 211.709 1.81104 1.81206 1.81307 3.90177 3.90396 3.90615 8.40612 8.41083 8.41554 5.97 5.98 5.99 35.6409 35.7604 35.8801 2.44336 2.44540 2.44745 7.72658 7.73305 7.73951 212.776 213.847 214.922 1.81409 1.81510 1.81611 3.90833 3.91051 3.91269 8.42025 8.42494 8.42964 288 Powers and Boots n w 2 Vw VWJi W 3 va VWn ^100 6.00 36.0000 2.44949 7.74597 216.000 1.81712 3.91487 8.43433 6.01 36.1201 2.45153 7.75242 217.082 1.81813 3.91704 8.43901 6.02 36.2404 2.45357 7.75887 218.167 1.81914 3.91921 8.44369 6.03 36.3609 2.45561 7.76531 219.256 1.82014 3.92138 8.44836 6.04 36.4816 2.45764 7.77174 220.349 1.82115 3.92355 8.45303 6.05 86.6025 2.45967 7.77817 221.445 1.82215 3.92571 8.45769 6.06 36.7236 2.46171 7.78460 222.545 1.82316 3.92787 8.46235 6.07 36.8449 2.46374 7.79102 223.649 1.82416 3.93003 8.46700 6.08 36.9664 2.46577 7.79744 224.756 1.82516 3.93219 8.47165 6.09 37.0881 2.46779 7.80385 225.867 1.82616 3.93434 8.47629 6.10 37.2100 2.46982 7.81025 226.981 1.82716 3.93650 8.48093 6.11 37.3321 2.47184 7.81665 228.099 1.82816 3.93865 8.48556 6.12 37.4544 2.47386 7.82304 229.221 1.82915 3.94079 8.49018 6.13 37.5769 2.47588 7.82943 230.346 1.83015 3.94294 8.49481 6.14 37.6996 2.47790 7.83582 231.476 1.83115 3.94508 8.49942 6.15 37.8225 2.47992 7.84219 232.608 1.83214 3.94722 8.50403 6.16 37.9456 2.48193 7.84857 233.745 1.83313 3.94936 8.50864 6.17 38.0689 2.48395 7.85493 234.885 1.83412 3.95150 8.51324 6.18 38.1924 2.48596 7.86130 236.029 1.83511 3.95363 8.51784 6.19 38.3161 2.48797 7.86766 237.177 1.83610 3.95576 8.52243 6.20 38.4400 2.48998 7.87401 238.328 1.83709 3.95789 8.52702 6.21 38.5641 2.49199 7.88036 239.483 1.83808 3.96002 8.53160 6.22 38.6884 2.49399 7.88670 240.642 1.83906 3.96214 8.53618 6.23 38.8129 2.49600 7.89303 241.804 1.84005 3.96427 8.54075 6.24 38.9376 2.49800 7.89937 242.971 1.84103 3.96638 8.54532 6.25 39.0625 2.50000 7.90569 244.141 1.84202 3.96850 8.54988 6.26 39.1876 2.50200 7.91202 245.314 1.84300 3.97062 8.55444 6.27 39.3129 2.50400 7.91833 246.492 1.84398 3.97273 8.55899 6.28 39.4384 2.50599 7.92465 247.673 1.84496 3.97484 8.56354 6.29 39.5641 2.50799 7.93095 248.858 1.84594 3.97695 8.56808 6.30 39.6900 2.50998 7.93725 250.047 1.84691 3.97906 8.57262 6.31 39.8161 2.51197 7.94355 251.240 1.84789 3.98116 8.57715 6.32 39.9424 2.51396 7.94984 252.436 1.84887 3.98326 8.58168 6.33 40.0689 2.51595 7.95613 253.636 1.84984 3.98536 8.58620 6.34 40.1956 2.51794 7.96241 254.840 1.85082 3.98746 8.59072 6.35 40.3225 2.51992 7.96869 256.048 1.85179 3.98956 8.59524 6.36 40.4496 2.52190 7.97496 257.259 1.85276 3.99165 8.59975 6.37 40.5769 2.52389 7.98123 258.475 1.85373 3.99374 8.60425 6.38 40.7044 2.52587 7.98749 259.694 1.85470 3.99583 8.60875 6.39 40.8321 2.52784 7.99375 260.917 1.85567 3.99792 8.61325 6.40 40.9600 2.52982 8.00000 262.144 1.85664 4.00000 8.61774 6.41 41.0881 2.53180 8.00625 263.375 1.85760 4.00208 8.62222 6.42 41.2164 2.53377 8.01249 264.609 1.85857 4.00416 8.<2671 6.43 41.3449 2.53574 8.01873 265.848 1.85953 4.00624 8.63118 6.44 41.4736 2.53772 8.02496 267.090 1.86050 4.00832 8.63566 6.45 41.6025 2.53969 8.03119 268.336 1.86146 4.01039 8.64012 6.46 41.7316 2.54165 8.03741 269.586 1.86242 4.01246 8.64459 6.47 41.8609 2.54362 8.04363 270.840 1.86338 4.01453 8.64904 6.48 41.9904 2.r>4. r >r,,s 8.04984 272.098 1.86434 4.01660 8. 65350 6.49 42.1201 2.54755 8.05605 273.359 1.8651*0 4.01866 s.<;r,7<)5 Powers and Roots 289 n n 2 ^n vio n 8 Vn vlO n ^100^ 6.50 42.2500 2.54951 8.06226 274.625 1.86626 4.02073 8.66239 6.51 6.52 6.53 42.3801 42.5104 42.6409 2.55147 2.55343 2.55539 8.06846 8.07465 8.08084 275.894 277.168 278.445 1.86721 1.86817 1.86912 4.02279 4.02485 4.02690 8.66683 8.67127 8.67570 6.54 6.55 6.56 42.7716 42.9025 43.0336 2.55734 2.55930 2.56125 8.08703 8.09321 8.09938 279.726 281.011 282.300 1.87008 1.87103 1.87198 4.02896 4.03101 4.03306 8.68012 8.68455 8.68896 6.57 6.58 6.59 43.1649 43.2964 43.4281 2.56320 2.56515 2.56710 8.10555 8.11172 8.11788 283.593 284.890 286.191 1.87293 1.87388 1.87483 4.03511 4.03715 4.03920 8.69338 8.69778 8.70219 6.60 43.5600 2.56905 8.12404 287.496 1.87578 4.04124 8.70659 6.61 6.62 6.63 43.6921 43.8244 43.9569 2.57099 2.57294 2.57488 8.13019 8.13634 8.14248 288.805 290.118 291.434 1.87672 1.87767 1.87862 4.04328 4.04532 4.04735 8.71098 8.71537 8.71976 6.64 6.65 6.66 44.0896 44.2225 44.3556 2.57682 2.57876 2.58070 8.14862 8.15475 8.16088 292.755 294.080 295.408 1.87956 1.88050 1.88144 4.04939 4.05142 4.05345 8.72414 8.72852 8.73289 6.67 6.68 6.69 44.4889 44.6224 44.7561 2.58263 2.58457 2.58650 8.16701 8.17313 8.17924 296.741 298.078 299.418 1.88239 1.88333 1.88427 4.05548 4.05750 4.05953 8.73726 8.74162 8.74598 6.70 44.8900 2.58844 8.18535 300.763 1.88520 4.06155 8.75034 6.71 6.72 6.73 45.0241 45.1584 45.2929 2.59037 2.59230 2.59422 8.19146 8.19756 8.20366 302.112 303.464 304.821 1.88614 1.88708 1.88801 4.06357 4.06559 4.06760 8.75469 8.75904 8.76338 6.74 6.75 6.76 45.4276 45.5625 45.6976 2.59615 2.59808 2.60000 8.20975 8.21584 8.22192 306.182 307.547 308.916 1.88895 1.88988 1.89081 4.06961 4.07163 4.07364 8.76772 8.77205 8.77638 6.77 6.78 6.79 45.8329 45.9684 46.1041 2.60192 2.60384 2.60576 8.22800 8.23408 8.24015 310.289 311.666 313.047 1.89175 1.89268 1.89361 4.07564 4.07765 4.07965 8.78071 8.78503 8.78935 6.80 46.2400 2.60768 8.24621 314.432 1.89454 4.08166 8.79366 6.81 6.82 6.83 46.3761 46.5124 46.6489 2.60960 2.61151 2.61343 8.25227 8.25833 8.26438 315.821 317.215 318.612 1.89546 1.89639 1.89732 4.08365 4.08565 4.08765 8.79797 8.80227 8.80657 6.84 6.85 6.86 46.7856 46.9225 47.0596 2.61534 2.61725 2.61916 8.27043 8.27647 8.28251 320.014 321.419 322.829 1.89824 1.89917 1.90009 4.08964 4.09163 4.09362 8.81087 8.81516 8.81945 6.87 6.88 6.89 47.1969 47.3344 47.4721 2.62107 2.62298 2.62488 8.28855 8.29458 8.30060 324.243 325.661 327.083 1.90102 1.90194 1.90286 4.09561 4.09760 4.09958 8.82373 8.82801 8.83228 6.90 47.6100 2.62679 8.30662 328.509 1.90378 4.10157 8.83656 6.91 6.92 6.93 47.7481 47.8864 48.0249 2.62869 2.63059 2.63249 8.31264 8.31865 8.32466 329.939 331.374 332.813 1.90470 1.90562 1.90653 4.10355 4.10552 4.10750 8.84082 8.84509 8.84934 6.94 6.95 6.96 48.1636 48.3025 48.4416 2.63439 2.63629 2.63818 8.33067 8.33667 8.34266 334.255 335.702 337.154 1.90745 1.90837 1.90928 4.10948 4.11145 4.11342 8.85360 8.85785 8.86210 6.97 6.98 6.99 48.5809 48.7204 48.8601 2.64008 2.64197 2.64386 8.34865 8.35464 8.36062 338.609 340.068 341.532 1.91019 1.91111 1.91202 4.11539 4.11736 4.11932 8.86634 8.87058 8.87481 290 Powers and Boots n> W 2 V^ VWn n* Vn VWn ^100^ 7.00 49.0000 2.64575 8.36660 343.000 1.91293 4.12129 8.87904 7.01 7.02 7.03 49.1401 49.2804 49.4209 2.64764 2.64953 2.65141 8.37257 8.37854 8.38451 344.472 345.948 347.429 1.91384 1.91475 1.91566 4.12325 4.12521 4.12716 8.88327 8.88749 8.89171 7.04 7.05 7.06 49.5616 49.7025 49.8436 2.65330 2.65518 2.65707 8.39047 8.39643 8.40238 348.914 350.403 351.896 1.91657 1.91747 1.91838 4.12912 4.13107 4.13303 8.89592 8.90013 8.90434 7.07 7.08 7.09 49.9849 50.1264 50.2681 2.65895 2.66083 2.66271 8.40833 8.41427 8.42021 353.393 354.895 356.401 1.91929 1.92019 1.92109 4.13498 4.13693 4.13887 8.90854 8.91274 8.91693 7.10 50.4100 2.66458 8.42615 357.911 1.92200 4.14082 8.92112 7.11 7.12 7.13 50.5521 50.6944 50.8369 2.66646 2.66833 2.67021 8.43208 8.43801 8.44393 359.425 360.944 362.467 1.92290 1.92380 1.92470 4.14276 4.14470 4.14664 8.92531 8.92949 8.93367 7.14 7.15 7.16 50.9796 51.1225 51.2656 2.67208 2.67395 2.67582 8.44985 8.45577 8.46168 363.994 365.526 367.062 1.92560 1.92650 1.92740 4.14858 4.15052 4.15245 8.93784 8.94201 8.94618 7.17 7.18 7.19 51.4089 51.5524 51.6961 2.67769 2.67955 2.68142 8.46759 8.47349 8.47939 368.602 370.146 371.695 1.92829 1.92919 1.93008 4.15438 4.15631 4.15824 8.95034 8.95450 8.95866 7.20 51.8400 2.68328 8.48528 373.248 1.93098 4.16017 8.96281 7.21 7.22 7.23 51.9841 52.1284 52.2729 2.68514 2.68701 2.68887 8.49117 8.49706 8.50294 374.805 376.367 377.933 1.93187 1.93277 1.93366 4.16209 4.16402 4.16594 8.96696 8.97110 8.97524 7.24 7.25 7.26 52.4176 52.5625 52.7076 2.69072 2.69258 2.69444 8.50882 8.51469 8.52056 379.503 381.078 382.657 1.93455 1.93544 1.93633 4.16786 4.16978 4.17169 8.97938 8.98351 8.98764 7.27 7.28 7.29 62.8529 52.9984 53.1441 2.69629 2.69815 2.70000 8.52643 8.53229 8.53815 384.241 385.828 387.420 1.93722 1.93810 1.93899 4.17361 4.17552 4.17743 8.99176 8.99588 9.00000 7.30 53.2900 2.70185 8.54400 389.017 1.93988 4.17934 9.00411 7.31 7.32 7.33 53.4361 53.5824 53.7289 2.70370 2.70555 2.70740 8.54985 8.55570 8.56154 390.618 392.223 393.833 1.94076 1.94165 1.94253 4.18125 4.18315 4.18506 9.00822 9.01233 9.01643 7.34 7.35 7.36 53.8756 54.0225 54.1696 2.70924 2.71109 2.71293 8.56738 8.57321 8.57904 395.447 397.065 398.688 1.94341 1.94430 1.94518 4.18696 4.18886 4.19076 9.02053 9.02462 9.02871 7.37 7.38 7.39 54.3169 54.4644 54.6121 2.71477 2.71662 2.71846 8.58487 8.59069 8.59651 400.316 401.947 403.583 1.94606 1.94694 1.94782 4.19266 4.19455 4.19644 9.03280 9.03689 9.04097 7.40 54.7600 2.72029 8.60233 405.224 1.94870 4.19834 9.04504 7.41 7.42 7.43 54.9081 55.0564 55.2049 2.72213 2.72397 2.72580 8.60814 8.61394 8.61974 406.869 408.518 410.172 1.94957 1.95045 1.95132 4.20023 4.20212 4.20400 9.04911 9.05318 9.05725 7.44 7.45 7.46 55.3536 55.5025 55.6516 2.72764 2.72947 2.73130 8.62554 8.63134 8.63713 411.831 413.494 415.161 1.95220 1.95307 1.95395 4.20589 4.20777 4.20965 9.06131 9.06537 9.06942 7.47 7.48 7.49 55.8009 55.9504 56.1001 2.73313 2.73496 2.73679 8.64292 8.64870 8.65448 416.833 418.509 420.190 1.95482 1.95569 1. 95656 4.21153 4.21341 4.21529 9.07347 9.07752 9.08156 Powers and Boots 291 n n 2 Vn VlOw n 8 Vn ^10 n VlWn 7.50 56.2500 2.73861 8.66025 421.875 1.95743 4.21716 9.08560 7.51 7.52 7.53 56.4001 56.5504 56.7009 2.74044 2.74226 2.74408 8.66603 8.67179 8.67756 423.565 425.259 426.958 1.95830 1.95917 1.96004 4.21904 422091 4.22278 9.08964 9.09367 9.09770 7.54 7.55 7.56 56.8516 57.0025 57.1536 2.74591 2.74773 2.74955 8.68332 8.68907 8.69483 428.661 430.369 432.081 1.96091 1.96177 1.96264 4.22465 4.22651 4.22838 9.10173 9.10575 9.10977 7.57 7.58 7.59 57.3049 57.4564 57.6081 2.75136 2.75318 2.75500 . 8.70057 8.70632 8.71206 433.798 435.520 437.245 1.96350 1.96437 1.96523 4.23024 4.23210 4.23396 9.11378 9.11779 9.12180 7.60 57.7600 2.75681 8.71780 438.976 1.96610 4.23582 9.12581 7.61 7.62 7.63 57.9121 58.0644 58.2169 2.75862 2.76043 2.76225 8.72353 8.72926 8.73499 440.711 442.451 444.195 1.96696 1.96782 1.96868 4.23768 4.23954 4.24139 9.12981 9.13380 9.13780 7.64 7.65 7.66 58.3696 58.5225 58.6756 2.76405 2.76586 2.76767 8.74071 8.74643 8.75214 445.944 447.697 449.455 1.96954 1.97040 1.97126 4.24324 4.24509 4.24694 9.14179 9.14577 9.14976 7.67 7.68 7.69 58.8289 58.9824 59.1361 2.76948 2.77128 2.77308 8.75785 8.76356 8.76926 451.218 452.985 454.757 1.97211 1.97297 1.97383 4.24879 4.25063 4.25248 9.15374 9.15771 9.16169 7.70 59.2900 2.77489 8.77496 456.533 1.97468 4.25432 9.16566 7.71 7.72 7.73 59.4441 59.5984 59.7529 2.77669 2.77849 2.78029 8.78066 8.78635 8.79204 458.314 460.100 461.890 1.97554 1.97639 1.97724 4.25616 4.25800 4.25984 9.16962 9.17359 9.17754 7.74 7.75 7.76 59.9076 60.0625 60.2176 2.78209 2.78388 2.78568 8.79773 8.80341 8.80909 463.685 465.484 467.289 1.97809 1.97895 1.97980 4.26167 4.26351 4.26534 9.18150 9.18545 9.18940 7.77 7.78 7.79 60.3729 60.5284 60.6841 2.78747 2.78927 2.79106 8.81476 8.82043 8.82610 469.097 470.911 472.729 1.98065 1.98150 1.98234 4.26717 4.26900 4.27083 9.19335 9.19729 9.20123 7.80 60.8400 2.79285 8.83176 474.552 1.98319 4.27266 9.20516 7.81 7.82 7.83 60.9961 61.1524 61.3089 2.79464 2.79643 2.79821 8.83742 8.84308 8.84873 476.380 478.212 480.049 1,98404 1.98489 1.98573 4.27448 4.27631 4.27813 9.20910 9.21302 9.21695 7.84 7.85 7.86 61.4656 61.6225 61.7796 2.80000 2.80179 2.80357 8.85438 8.86002 8.86566 481.890 483.737 485.588 1.98658 1.98742 1.98826 4.27995 4.28177 4.28359 9.22087 9.22479 9.22871 7.87 7.88 7.89 61.9369 62.0944 62.2521 2.80535 2.80713 2.80891 8.87130 8.87694 8.88257 487.443 489.304 491.169 1.98911 1.98995 1.99079 4.28540 4.28722 4.28903 9.23262 9.23653 9.24043 7.90 62.4100 2.81069 8.88819 493.039 1.99163 4.29084 9.24434 7.91 7.92 7.93 62.5681 62.7264 62.8849 2.81247 2.81425 2.81603 8.89382 8.89944 8.90505 494.914 496.793 498.677 1.99247 1.99331 1.99415 4.29265 4.29446 4.29627 9.24823 9.25213 9.25602 7.94 7.95 7.96 63.0436 63.2025 63.3616 2.81780 2.81957 2.82135 8.91067 8.91628 8.92188 500.566 502.460 504.358 1.99499 1.99582 1.99666 4.29807 4.29987 4.30168 9.25991 9.26380 9.26768 7.97 7.98 7.99 63.5209 63.6804 63.8401 2.82312 2.82489 2.82666 8.92749 8.93308 8.93868 506.262 508.170 510.082 1.99750 1.99833 1.99917 4.30348 4.30528 4.30707 9.27156 9.27544 9.27931 292 Powers and Roots n W 2 Vii VlOn W 3 Vn VWn ^lOOn 8.00 64.0000 2.82843 8.94427 512.000 2.00000 4.30887 9.28318 8.01 8.02 8.03 64.1601 64.3204 64.4809 2.83019 2.83196 2.83373 8.94986 8.95545 8.96103 513.922 515.850 517.782 2.00083 2.00167 2.00250 4.31066 4.31246 4.31425 9.28704 9.29091 9.29477 8.04 8.05 8.06 64.6416 64.8025 64.9636 2.83549 2.83725 2.83901 8.96660 8.97218 8.97775 519.718 521.660 523.607 2.00333 2.00416 2.00499 4.31604 4.31783 4.31961 9.29862 9.30248 9.30633 8.07 8.08 8.09 65.1249 65.2864 65.4481 2.84077 2.84253 2.84429 8.98332 8.98888 8.99444 525.558 527.514 529.475 2.00582 2.00664 2.00747 4.32140 4.32318 4.32497 9.31018 9.31402 9.31786 8.10 65.6100 2.84605 9.00000 531.441 2.00830 4.32675 9.32170 8.11 8.12 8.13 65.7721 65.9344 66.0969 2-84781 2.84956 2.85132 9.00555 9.01110 9.01665 533.412 535.387 537.368 2.00912 2.00995 2.01078 4.32853 4.33031 4.33208 9.32553 9.32936 9.33319 8.14 8.15 8.16 66.2596 66.4225 66.5856 2.85307 2.85482 2.85657 9.02219 9.02774 9.03327 539.353 541.343 543.338 2.01160 2.01242 2.01325 4.33386 4.33563 4.33741 9.33702 9.34084 9.34466 8.17 8.18 8.19 66.7489 66.9124 67.0761 2.85832 2.86007 2.86182 9.03881 9.04434 9.04986 545.339 547.343 549.353 2.01407 2.01489 2.01571 4.33918 4.34095 4.34271 9.34847 9.35229 9.35610 8.20 67.2400 2.86356 9.05539 551.368 2.01653 4.34448 9.35990 8.21 8.22 8.23 67.4041 67.5684 67.7329 2.86531 2.86705 2.86880 9.06091 9.06642 9.07193 553.388 555.412 557.442 2.01735 2.01817 2.01899 4.34625 4.34801 4.34977 9.36370 9.36751 9.37130 8.24 8.25 8.26 67.8976 68.0625 68.2276 2.87054 2.87228 2.87402 9.07744 9.08295 9.08845 559.476 561.516 563.560 2.01980 2.02062 2.02144 4.35153 4.35329 4.35505 9.37510 9.37889 9.38268 8.27 8.28 8.29 68.3929 68.5584 68.7241 2.87576 2.87750 2.87924 9.09395 9.09945 9.10494 565.609 567.664 569.723 2.02225 2.02307 2.02388 4.35681 4.35856 4.36032 9.38646 9.39024 9.39402 8.30 68.8900 2.88097 9.11043 571.787 2.02469 4.36207 9.39780 8.31 8.32 8.33 69.0561 69.2224 69.3889 2.88271 2.88444 2.88617 9.11592 9.12140 9.12688 573.856 575.930 578.010 2.02551 2.02632 2.02713 4.36382 4.36557 4.36732 9.40157 9.40534 9.40911 8.34 8.35 8.36 69.5556 69.7225 69.8896 2.88791 2.88964 2.89137 9.13236 9.13783 9.14330 580.094 582.183 584.277 2.02794 2.02875 2.02956 4.36907 4.37081 4.37256 9.41287 9.41663 9.42039 8.37 8.38 8.39 70.0569 70.2244 70.3921 2.89310 2.89482 2.89655 9.14877 9.15423 9.15969 586.376 588.480 590.590 2.03037 2.03118 2.03199 4.37430 4.37604 4.37778 9.42414 0.42789 9.43164 8.40 70.5600 2.89828 9.16515 592.704 2.03279 4.37952 9.43539 8.41 8.42 8.43 70.7281 70.8964 71.0649 2.90000 2.90172 2.90345 9.17061 9.17606 9.18150 594.823 596.948 599.077 2.03360 2.03440 2.03521 4.38126 4.38299 4.38473 9.43913 9.44287 9.44661 8.44 8.45 8.46 71.2336 71.4025 71.5716 2.90517 2.90689 2.90861 9.18695 9.19239 9.19783 601.212 603.351 605.496 2.03601 2.03(582 2.03762 4.38646 4.38819 4.38992 9.45034 9.45407 9.45780 8.47 8.48 8.49 71.7409 71.9104 72.0801 2.91033 2.91204 2.91376 9.20326 9.20869 9.21412 607.645 609.800 611.960 2.03842 2.039L':i 2.04003 4.30166 4.39338 4.: s'.>r> 10 9.46152 9.46525 9.46897 Powers and Boots 293 n w 2 Vti- VWn n 8 3/n VWn ^100 n 8.50 72.2500 2.91548 9.21954 614.125 2.04083 4.39683 9.47268 8.51 8.52 8.53 72.4201 72.5904 72.7609 2.91719 2.91890 2.92062 9.22497 9.23038 9.23580 616.295 618.470 620.650 2.04163 2.04243 2.04323 4.39855 4.40028 4.40200 9.47640 9.48011 9.48381 8.54 8.55 8.56 72.9316 73.1025 73.2736 2.92233 2.92404 2.92575 9.24121 9.24662 9.25203 622.836 625.026 627.222 2.04402 2.04482 2.04562 4.40372 4.40543 4.40715 9.48752 9.49122 9.49492 .8.57 8.58 8.59 73.4449 73.6164 73.7881 2.92746 2.92916 2.93087 9.25743 9.26283 9.26823 629.423 631.629 633.840 2.04641 2.04721 2.04801 4.40887 4.41058 4.41229 9.498(51 9.50231 9.50600 860 73.9600 2.93258 9.27362 636.056 2.04880 4.41400 9.50969 8.61 8.62 8.63 74.1321 74.3044 74.4769 2.93428 2.93598 2.93769 9.27901 9.28440 9.28978 638.277 640.504 642.736 2.04959 2.03039 2.05118 4.41571 4.41742 4.41913 9.51337 9.51705 9.52073 8.G4 8.65 8.66 74.6496 74.8225 74.9956 2.93939 2.94109 2.94279 9.29516 9.30054 9.30591 644.973 647.215 649.462 2.05197 2.05276 2.05355 4.42084 4.42254 4.42425 9.52441 9.52808 9.53175 8.67 8.68 8.69 75.1689 75.3424 75.5161 2.94449 2.94618 2.94788 9.31128 9.31665 9.32202 651.714 653.972 656.235 2.05434 2.05513 2.05592 4.42595 4.42765 4.42935 9.53542 9.53908 9.54274 8.70 75.0900 2.<)4958 9.32738 658.503 2.05671 4.43105 9.54640 8.71 8.72 8.73 75.8641 76.0384 76.2129 2.95127 2.95296 2.9546(5 9.33274 9.33H09 9.34345 660.776 663.055 665.339 2.05750 2.05828 2.05907 4.43274 4.43444 4.43613 9.55006 9.55371 9.55736 8.74 8.75 8.76 76.3876 76.5625 76.7376 2.95635 2.95804 2.95973 9.34880 9.35414 9.35949 667.628 669.922 672.221 2.05986 2.06064 2.06143 4.43783 4.43952 4.44121 9.56101 9.56466 9.56830 8.77 8.78 8.79 76.9129 77.0884 77.2641 2.96142 2.96311 2.96479 9.36483 9.37017 9.37550 674.526 676.836 679.151 2.06221 2.06299 2.06378 4.44290 4.44459 4.44627 9.57194 9.57557 9.57921 8 80 77.4400 2.96648 9.38083 681.472 2.06456 4.44796 9.58284 8.81 8.82 8.83 77.6161 77.7924 77.9689 2.96816 2.96985 2.97153 9.38616 9.39149 9.39681 683.798 686.129 688.465 2.06534 2.06612 2.06690 4.44964 4.45133 4.45301 9.58647 9.59009 9.59372 8.84 8.85 8.86 78.1456 78.3225 78.4996 2.97321 2.97489 2.97658 9.40213 9.40744 9.41276 690.807 693.154 695.506 2.06768 2.06846 2.06924 4.45469 4.45637 4.45805 9.59734 9.60095 9.60457 8.87 8.88 8.89 78.6769 78.8544 79.0321 2.97825 2.97993 2.98161 9.41807 9.42338 9.42868 697.864 700.227 702.595 2.07002 2.07080 2.07157 4.45972 4.46140 4.46307 9.60818 9.61179 9.6154C 890 79.2100 2.98329 9.43398 704.969 2.07235 4.46475 9.61900 8.91 8.92 8.93 79.3881 79.5664 79.7449 2.98496 2.98664 2.98831 9.43928 9.44458 9.44987 707.348 709.732 712.122 2.07313 2.07390 2.07468 4.46642 4.46809 4.46976 9.62260 9.62620 9.62980 8.94 8.95 8.96 79.9236 80.1025 80.2816 2.98998 2.99166 2.99333 9.45516 9.46044 9.46573 714.517 716.917 719.323 2.07545 2.07622 2.07700 4.47142 4.47309 4.47476 9.63339 9.63698 9.64057 8.97 8.98 8.99 80.4609 80.6404 80.8201 2.99500 2.99666 2.99833 9.47101 9.47629 9.48156 721.734 724.151 726.573 2.07777 2.07854 2.07931 4.47642 4.47808 4.47974 9.64415 9.64774 9.65132 294 Powers and Boots n n* Vti VWn n* V* VWn VJMn 9.00 81.0000 3.00000 9.48683 729.000 2.08008 4.48140 9.65489 9.01 9.02 9.03 81.1801 81.3604 81.5409 3.00167 3.00333 3.00500 9.49210 9.49737 9.50263 731.433 733.871 736.314 2.08085 2.08162 2.08239 4.48306 4.48472 4.48638 9.65847 9.66204 9.66561 9.04 9.05 9.06 81.7216 81.9025 82.0836 3.00666 3.00832 3.00998 9.50789 9.51315 9.51840 738.763 741.218 743.677 2.08316 2.08393 2.08470 4.48803 4.48969 4.49134 9.66918 9.67274 9.67630 9.07 9.08 9.09 82.2649 82.4464 82.6281 3.01164 3.01330 3.01496 9.523(35 9.52890 9.53415 746.143 748.613 751.089 2.08546 2.08623 2.08699 4.49299 4.49464 4.49629 9.67986 9.68342 9.68697 9.10 82.8100 3.01662 9.53939 753.571 2.08776 4.49794 9.69052 9.11 9.12 9.13 82.9921 83.1744 83.3569 3.01828 3.01993 3.02159 9.54463 9.54987 9.55510 756.058 758.551 761.048 2.08852 2.08929 2.09005 4.49959 4.50123 4.50288 9.69407 9.69762 9.70116 9.14 9.15 9.16 83.5396 83.7225 83.9056 3.02324 3.02490 3.02655 9.56033 9.56556 9.57079 763.552 766.061 768.575 2.09081 2.09158 2.09234 4.50452 4.50616 4.50781 9.70470 9.70824 9.71177 9.17 8.18 9.19 84.0889 84.2724 84.4561 3.02820 3.02985 3.03150 9.57601 9.58123 9.58645 771.095 773.621 776.152 2.09310 2.09386 2.09462 4.50945 4.51108 4.51272 9.71531 9.71884 9.72236 9.20 84.6400 3.03315 9.59166 778.688 2.09538 4.51436 9.72589 9.21 9.22 9.23 84.8241 85.0084 85.1929 3.03480 3.03645 3.03809 9.59687 9.60208 9.60729 781.230 783.777 786.330 2.09614 2.09690 2.09765 4.51599 4.51763 4.51926 9.72941 9.73293 9.73645 9.24 9.25 9.26 85.3776 85.5625 85.7476 3.03974 3.04138 3.04302 9.61249 9.61769 9.62289 788.889 791.453 794.023 2.09841 2.09917 2.09992 4.52089 4.52252 4.52415 9.73996 9.74348 9.74699 9.27 9.28 9.29 85.9329 86.1184 86.3041 3.04467 3.04631 3.04795 9.62808 9.63328 9.63846 796.598 799.179 801.765 2.10068 2.10144 2.10219 4.52578 4.52740 4.52903 9.75049 9.75400 9.75750 9.30 86.4900 3.04959 9.64365 804.357 2.10294 4.53065 9.76100 9.31 9.32 9.33 86.6761 86.8624 87.0489 3.05123 3.05287 3.05450 9.64883 9.65401 9.65919 806.954 809.558 812.166 2.10370 2.10445 2.10520 4.53228 4.53390 4.53552 9.76450 9.76799 9.77148 9.34 9.35 9.36 87.2356 87.4225 87.6096 3.05614 3.05778 3.05941 9.66437 9.66954 9.67471 814.781 817.400 820.026 2.10595 2.10671 2.10746 4.53714 4.53876 4.54038 9.77497 9.77846 9.78195 9.37 9.38 9.39 87.7969 87.9844 88.1721 3.06105 3.06268 3.06431 9.67988 9.68504 9.69020 822.657 825.294 827.936 2.10821 2.10896 2.10971 4.54199 4.54361 4.54522 9.78543 9.78891 9.79239 9.40 88.3600 3.06594 9.69536 830.584 2.11045 4.54684 9.79586 9.41 9.42 9.43 88.5481 88.7364 88.9249 3.06757 3.06920 3.07083 9.70052 9.70567 9.71082 833.238 835.897 838.562 2.11120 2.11195 2.11270 4.54845 4.55006 4.55167 9.79933 9.80280 9.80627 9.44 9.45 9.46 89.1136 89.3025 89.4916 3.07246 3.07409 3.07571 9.71597 9.72111 9.72625 841.232 843.909 846.591 2.11344 2.11419 2.11494 4.55328 4.55488 4.55649 9.80974 9.81320 9.81666 9.47 9.48 9.49 89.6809 89.8704 90.0601 3.07734 3.07896 3.08058 9.73139 9.73653 9.74166 849.278 851.971 854.670 2.11568 2.11642 2.11717 4.55809 4.55970 4.56130 9.82012 9.82357 9.82703 Powers and Roots 295 n n 2 Vn VWn n 8 ^n ^10 n ^100 n 9.50 90.2500 3.08221 9.74679 857.375 2.11791 4.56290 9.83048 9.51 9.52 9.53 90.4401 90.6304 90.8209 3.08383 3.08545 3.08707 9.75192 9.75705 9.76217 860.085 862.801 865.523 2.11865 2.11940 2.12014 4.56450 4.56610 4.56770 9.83392 9.83737 9.84081 9.54 9.55 9.56 91.0116 91.2025 91.3936 3.08869 3.09031 3.09192 9.76729 9.77241 9.77753 868.251 870.984 873.723 2.12088 2.12162 2.12236 4.56930 4.57089 4.57249 9.84425 9.84769 9.85113 9.57 9.58 9.59 91.5849 91.7764 91.9681 3.09354 3.09516 3.09677 9.78264 9.78775 9.79285 876.467 879.218 881.974 2.12310 2.12384 2.12458 4.57408 4.57567 4.57727 9.85456 9.85799 9.86142 9.60 92.1600 3.09839 9.79796 884.736 2.12532 4.57886 9.86485 9.61 9.62 9.63 92.3521 92.5444 92.7369 3.10000 3.10161 3.10322 9.80306 9.80816 9.81326 887.504 890.277 893.056 2.12605 2.12679 2.12753 4.58045 4.58204 4.58362 9.86827 9.87169 9.87511 9.64 9.65 9.66 92.9296 93.1225 93.3156 3.10483 3.10644 3.10805 9.81835 9.82344 9.82853 895.841 898.632 901.429 2.12826 2.12900 2.12974 4.58521 4.58679 4.58838 9.87853 9.88195 9:88536 9.67 9.68 9.69 93.5089 93.7024 93.8961 3.10966 3.11127 3.11288 9.83362 9.83870 9.84378 904.231 907.039 909.853 2.13047 2.13120 2.13194 4.58996 4.59154 4.59312 9.88877 9.89217 9.89558 9.70 94.0900 3.11448 9.84886 912.673 2.13267 4.59470 9.89898 9.71 9.72 9.73 94.2841 94.4784 94.6729 3.11609 3.11769 3.11929 9.85393 9.85901 9.86408 915.499 918.330 921.167 2.13340 2.13414 2.13487 4.59628 4.59786 4.59943 9.90238 9.90578 9.90918 9.74 9.75 9.76 94.8676 95.0625 95.2576 3.12090 3.12250 3.12410 9.86914 9.87421 9.87927 924.010 926.859 929.714 2.13560 2.13633 2.13706 4.60101 4.60258 4.60416 9.91257 9.91596 9.91935 9.77 9.78 9.79 95.4529 95.6484 95.8441 3.12570 3.12730 3.12890 9.88433 9.88939 9.89444 932.575 935.441 938.314 2.13779 2.13852 2.13925 4.60573 4.60730 4.60887 9.92274 9.92612 9.92950 9.80 96.0400 3.13050 9.89949 941.192 2.13997 4.61044 9.93288 9.81 9.82 9.83 96.2361 96.4324 96.6289 3.13209 3.13369 3.13528 9.90454 9.90959 9.91464 944.076 946.966 949.862 2.14070 2.14143 2.14216 4.61200 4.61357 4.61514 9.93626 9.93964 9.94301 9.84 9.85 9.86 96.8256 97.0225 97.2196 3.13688 3.13847 3.14006 9.91968 9.92472 9.92975 952.764 955.672 958.585 2.14288 2.14361 2.14433 4.61670 4.61826 4.61983 9.94638 9.94975 9.95311 9.87 9.88 9.89 97.4169 97. (5144 97.8121 3.14166 3.14325 3.14484 9.93479 9.93982 9.94485 961.505 964.430 967.362 2.14506 2.14578 2.14651 4.62139 4.62295 4.62451 9.95648 9.95984 9.96320 9.90 98.0100 3.14643 9.94987 970.299 2.14723 4.62607 9.96655 9.91 9.92 9.93 98.2081 98.4064 98.6049 3.14802 3.14960 3.15119 9.95490 9.95992 9.96494 973.242 976.191 979.147 2.14795 2.14867 2.14940 4.62762 4.62918 4.63073 9.96991 9.97326 9.97661 9.94 9.95 9.96 98.8036 99.0025 99.2016 3.15278 3.15436 3.15595 9.96995 9.97497 9.97998 982.108 985.075 988.048 2.15012 2.15084 2.15156 4.63229 4.63384 4.63539 9.97996 9.98331 9.98665 9.97 9.98 9.99 99.4009 99.6004 99.8001 3.15753 3.15911 3.16070 9.98499 9.98999 9.99500 991.027 994.012 997.003 2.15228 2.15300 2.15372 4.63694 4.63849 4.64004 9.98999 9.99333 9.99667 TABLE II IMPORTANT NUMBERS A. Units of Length ENGLISH UNITS METRIC UNITS 12 inches (in.) = 1 foot (ft.) 10 millimeters = 1 centimeter (cm.) 3 feet = 1 yard (yd.) (mm.) 6 yards = 1 rod (rd.) 10 centimeters = 1 decimeter (dm.) 320 rods = 1 mile (mi.) 10 decimeters = 1 meter (m.) 10 meters = 1 dekameter (Dm.) 1000 meters = 1 kilometer (Km.) ENGLISH TO METRIC METRIC TO ENGLISH 1 in. = 2.5400 cm. 1 cm. = 0.3937 in. 1 ft. = 30.480 cm. 1m. = 39.37 in. = 3.2808 ft 1 mi. = 1.6093 Km. 1 Km. = 0.6214 mi. B. Units of Area or Surface 1 square yard = 9 square feet = 1296 square inches 1 acre (A.) = 160 square rods = 4840 square yards 1 square mile = 640 acres = 102400 square rods C. Units of Measurement of Capacity DRY MEASURE LIQUID MEASURE 2 pints (pt.) = 1 quart (qt.) 4 gills (gi.) = 1 pint (pt.) 8 quarts = 1 peck (pk.) 2 pints = 1 quart (qt.) 4 pecks = 1 bushel (bu.) 4 quarts = 1 gallon (gal.) 1 gallon = 231 cu. in. D. Metric Units to English Units 1 liter = 1000 cu. cm. = 61.02 cu. in. = 1.0567 liquid quarts 1 quart = .94636 liter = 946.36 cu. cm. 1000 grams = 1 kilogram (Kg.) = 2.2046 pounds (Ib.) 1 pound = .453593 kilogram = 453.59 grams E. Other Numbers T = ratio of circumference to diameter of a circle = 3.14159265 1 radian = angle subtended by an arc equal to the radius = 57 17' 44".8 = 67.2957795 = 180/ir 1 degree = 0.01745329 radian, or w/lSQ radians Weight of 1 cu. ft. of water = 62.425 Jb. 296 INDEX Abscissa, 41. Absolute value, 2. Addition, of expressions, 9; of frac- tions, 32 ; of radicals, 74. Antecedent, 166. Arithmetic progression, 141 ; means, 146. Ascending powers, 8. Axes, coordinate, 41. Base, logarithm to any, 242. Brace, 9. Bracket, 9. Binomial, 8; theorem, 258; proof of theorem, 261. Calculating machines, 244. Change of signs in fractions, 29. Characteristic, 219, 221, 223. Coefficient, 8. Common difference, 141. Common logarithm, 242. Complex numbers, 93. Consequent, 166. Constants, 181. Coordinates of a point, 41. Cube root, 5. Decimals, repeating, 161. Denominator, 29. Descending powers, 8. Determinant, of the second order, 264; of the third order, 269; of higher order, 273. Difference, tabular, 230. Discriminant, 110. Division, formulas and rules, 14. Elements of a determinant, 265. Elimination by substitution, 50; by addition or subtraction, 51. Ellipse, 125. Equation, simple, 36 ; linear, 36 ; of the first degree, 36; solution of, 36; root of, 36; principles useful in the solution of, 37, 38; containing radicals, 65; literal, 95; quadratic, see Quadratic equation. Equations, simultaneous, 47; in- consistent, 48; simultaneous in three unknowns, 54. Evolution, 196. Exponent, 4; fractional, 198; zero, 199 ; fundamental laws for any rational exponent, 201. Exponents, laws of, 193 ; introduc- tion of general, 198; negative in fractions, 200. Extremes of a proportion, 167. Factor, prime, 26. Factoring, type forms of, 19, 23. Factors, common, 26 ; highest com- mon, 26. Formulas, 97, 102, 205. Fractions, definition, 28 ; equiva- lent, 29 ; change of signs in, 29 ; reduction to lowest terms, 30; reduction to lowest common de- nominator, 31 ; addition and sub- traction of, 32 ; multiplication and division of, 36. Functions, idea of, 245 ; types of alge- braic functions, 246; considered graphically, 250. Gear wheel law, 101. Geometric progression, 150; means, 154; infinite, 156. Graph, of an equation, 43 ; deter- mined from two points, 44 ; of a quadratic, 90 ; of a function, 250. 297 298 INDEX Hyperbola, 126. Imaginary numbers, 116; pure, 93; unit, 116; addition and sub- traction of, 118; multiplication of, 119; division of, 120; geo- metric representation of, 121. Inconsistent equations, 47. Index, 5, 68, 196. Involution, 193. Irrational numbers, 68. Letters, use of in algebra, 4. Lever, law of, 100. Limit, variable, 159. Linear equation, 36 ; graph of, 43, 44. Logarithm, 217 ; of any number, 217 ; number corresponding to, 231; of a power, 235; general, 242; common, 242; of a root, 244; tables, 291. Mantissa, 219; determination of, 225. Mathematical induction, 255. Mean proportional, 172. Means of a proportion, 167. Monomial, 8 ; square root of, 62. Multiple, common, 27 ; lowest com- mon, 28. Multiplication, formulas and rules, 11. Negative exponents, 199. Negative numbers, 1 ; operations with, 2. Numbers, negative, 1 ; positive, 1 ; operations with, 2; rational, 68, 108; irrational, 68; complex, 93; real, 93, 108; summary, 109; imaginary, see Imaginary number. Numerator, 29. Order of a determinant, 265, 269. Ordinate, 41. Origin, 41. Parenthesis, 9. Perfect square trinomial, 20. Periods in square root, 60. Polynomial, 8; arranging a, 8; square root of, 62. Power, 4; tables, 274. Powers, 195. Prime factor, 27. Progression, arithmetic, 141 ; geo- metric, 150. Proportion, 167; terms of a, 167; extremes of a, 167 ; means of a, 167; algebraic, 168; fundamental principles of, 168, 174; inversion in a, 174 ; alternation in a, 174 ; composition in a, 174 ; division in a, 175; composition and divi- sion in a, 175 ; several equal ratios in a, 176. Proportional, mean, 172; third and fourth, 172. Quadratic equation, 78; pure, 78; affected, 78; solution of pure, 78 ; solution of affected by fac- toring, 81 ; solution by complet- ing the square, 83, 85; solution by the Hindu method, 86 ; solu- tion by formula, 87 ; graphical solution of, 90, 122 ; having imag- inary solutions, 92; character of the roots of, 109 ; character of roots considered geometrically, 111; for- mation of from given solutions, 114; solution by elimination, 128; simultaneous, 137. Radical, or quadratic radical, 67; of the nth order, 68 ; value of, 68. Radicals, simplification! of, 71, 206; similar, 74; addition and sub- traction of, 74; multiplication of, 75; division of, 77. Radicand, 68, 196. Ratio, 166; of geometric progres- sion, 150. Rational number, 68. Rationalizing the denominator, 2J.1. Root, square, 5; nth, 5, 196. Root of an equation, 36. INDEX 299 Roots, 196; imaginary of a quad- ratic, 108; character of, 111 Simultaneous equations, 47. Slide rule, 244. Solution of an equation, 44. Special products, formulas of, 17. Square root, 5 ; of a number, 59 ; in arithmetic, 59 ; in algebra, 62 ; of trinomials, 62 ; double sign of, 64 ; tables, 274. Subtraction, 8 ; of expressions, 9 ; of fractions, 32; of radicals, 74. Surd, 68; binomial, 214. Table, use of, 68. Term, 8. Terms, like, 8 ; of a proportion, 167. Theorem, binomial, 258. Trinomial, 8; perfect square, 20; square root of, 62. Unit, imaginary, 116. Variables, 181. Variation, direct, 178; inverse, 179; joint, 180; problems in, 185; geometrically considered, 190. Vinculum, 9. Printed in the United States of America. UNIVERSITY OF CALIFORNIA LIBRARY This book is DUE on the last date stamped below. 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