a^^ IN MEMORIAM FLORIAN CAJORl Digitized by tine Internet Archive in 2007 witii funding from IVIicrosoft Corporation http://www.arcliive.org/details/collegealgebraOOfiterich COLLEGE ALGEBRA BY WILLIAM BENJAMIN FITE PROFESSOR OF MATHEMATICS IX COLUMBIA UNIVERSITY D. C. HEATH & CO., PUBLISHERS BOSTON NEW YORK CHICAGO Copyright, 19 13, By D. C. Heath & Co. PREFACE An effort has been made to present here the elementary principles of algebra in a simple and direct way, and to give rigorous proofs of the theorems used. It seemed desirable, however, to pass over certain delicate points that are beyond the comprehension of the student, and to omit altogether sev- eral difficult proofs. The treatment is concrete, and graphical methods have been used freely. The introduction to complex numbers that is commonly given seems to me to be arbitrary and unconvincing; and I have sought by putting forward the concrete, geometrical side of the question and by careful attention to the proper sequence of the ideas introduced to present these numbers in a rational and convincing way. Students almost invariably think that the symbol oo repre- sents a number called "infinity," just as the symbol 5 repre- sents a number called " five," and this view is fostered, to some extent at least, by the conventional use of the word " infinity," and by the employment of a symbol to represent it. It seemed worth while, therefore, to depart from this conventional treat- ment and to write the chapter on infinite series without using either the symbol for infinity or the word itself, except as it is implied in the phrase "infinite series." I have selected the problems with a view to convincing the student that algebra is a body of principles by the aid of which certain kinds of important information can be obtained from data that do not give this information explicitly. Many prob- lems have been selected that illustrate the simpler principles of physics, geometry, and analytic geometry. The analytic geometry problems are somewhat of an innovation in a text of iii ivi306040 iv PREFACE this kind, and objection may be made that the difficulties of this subject ought not to be added to those inherent in the algebra. But there is a decided advantage in making the stu- dent feel that mathematics is one. For the sake of adding to the flexibility of the book two proofs of the Binomial Theorem have been given, one in the chapter on Permutations and Combinations, and the other in the chapter on Mathematical Induction. Either chapter may be omitted without the necessity of omitting the Binomial Theorem. Attention is called to the examples given in the text of the chapter on Mathematical Induction to show the necessity of both parts of the proof by induction. Articles 134-138 should be omitted by students who have not studied trigonometry. I am greatly indebted to Professors D. R. Curtis s of North- western University, W. B. Carver of Cornell University, and R. G. D. Richardson and H. P. Manning of Brown University for many suggestions of importance and value. WILLIAM BENJAMIN FITE. Columbia University, New York City. CONTENTS CHAPTEE PA6K I. The Fundamental Operations .... 1 11. Factors and Multiples 19 III. Fractions .27 IV. Linear Equations in One Unknown . . .38 V. Systems of Linear Equations in Two or More Unknowns 47 VI. Fractional and Negative Exponents. Radicals 73 ^ VII. Quadratics 84 VIIL Systems of Equations in Two Unknowns Solv- able BY Means of Quadratics . . . 110 "^ IX. Progressions 123 X. Permutations and Combinations .... 134 XL Mathematical Induction 147 XII. Complex Numbers 153*^^ XTII. Theory of Equations 168 XIV. Determinants 201 XV. Inequalities 221 XVI. Partial Fractions 228 XVII. Logarithms 238 XVIIL Variation 257 XIX. Infinite Series 261 Appendix 278 COLLEGE ALGEBRA CHAPTER I THE FUNDAMENTAL OPERATIONS 1. The operations of addition, subtraction, multiplication, and division are called the fundamental operations of algebra. We shall assume that the student is familiar with the details of performing these operations and shall confine our attention here to a brief consideration of their important properties. 2. The representation of points by numbers. — Since the letters used in algebra represent numbers, it is of great impor- tance for the student to have a clear conception of the nature of numbers. He will acquire this best by thinking of numbers as representatives of points. Consider an indefinite straight line called the axis, a fixed point or origin on this line, and a given length , which we shall call the unit length. ? 4 1 We shall let the point A which is on the line at a distance to the right of the origin equal to this unit length be repre- sented by the number 1. The point B twice this distance to the right of the origin will be represented by the number 2, and similarly any point to the right of the origin on this line is represented by the number which gives its distance from the origin in terms of the given unit of length. The points to the left of the origin on the line are also represented by the num- bers which give their respective distances from the origin. But it is necessary to distinguish these numbers in some way 1 2 COLLEGE ALGEBRA from the representatives of the points to the right of the origin. This is done by prefixing the minus sign to these numbers, and then to make this general scheme more symmetrical, we should prefix a plus sign to the representatives of the points to the right of the origin. Thus, — 5 represents the point five units to the left of the origin, while + 5 represents the point five units to the right of the origin. The two points are the same distance from the origin, but in opposite direction from it. It is this difference that is indicated by the two opposite signs. Definitions. — The numbers that represent the points on the left of the origin are called negative numbers and those that represent the points on the right of the origin are called posi- tive numbers. The plus sign that is a part of the positive numbers is usually not written since its omission causes no ambiguity. These positive and negative numbers, together with the number 0, which represents the origin, are called real numbers for a reason that is explained in Chapter XII. The quotient of two integers is called a rational number. Every integer is a rational number since it is the quotient of itself and 1. The number that represents a point is called the abscissa of this point. The distance of a point from the origin is called the absolute value of the number that represents the point. Thus, two numbers that represent points on opposite sides of the origin but equidistant from it have the same absolute value ; as, for example, the numbers + 7 and — 7. The absolute value of these numbers is 7. 3. Meaning of "greater than" and "less than." — If the point A has the abscissa a and is to the right of the point B which has the abscissa b, a is said to be greater than b, and b less than a. Thus every positive number is greater than and every negative number is less than 0. Moreover every posi- tive number is greater than any negative number. For ex- THE FUNDAMENTAL OPERATIONS 3 ample, 5 is greater than — 6. This relationship is indicated thus ; 5 > — 6, or — 6 < 5. These statements are read " 5 is greater than — 6 " and " — 6 is less than 5 " respectively. We express the fact that the numbers a and h represent dif- ferent points by the symbol a=^b, which is read, " a is not equal to bJ' 4. Definition of sum. — If ^ and B are any points on the axis and if the segment AC equals the segment OB in magni- tude and direction, the abscissa c of the point O is called the Bum of the abscissae a and b of the points A and B respec- tively. f — ? ? — 4 This connection of the numbers a, b, and c is expressed thus : a -f 6 = c. The process of finding the sum of two numbers is called addition. By the sum of three numbers a, b, and c we mean the sum of a -I- & and c. If we represent this sum by a + 6 + c, then a -}- 6 + c = (a + 6) + c. 5. Assumptions concerning real numbers and the operation of addition. — We shall make the following assumptions con- cerning real numbers and the operation of addition. A careful inspection of a figure will convince the student of the appro- priateness of most of these assumptions. I. If a and b are any numbers, there is one arid only one number x such that a-\-b = x. II. Addition is commutative. This means that a -\-h = h + a. III. The sum of three numbers is the same, iri'espective of the way in which they are grouped. Thus, a + 6 + c = (a + 6) -f- c = a + (?> + c). This is the associative law of addition. 4 COLLEGE ALGEBRA IV. Ifa-\-b = a-\-c, then b = c. This is the law of cancellation for addition. V. There is one and only one number' x such that a; + ic = a?. This number is 0. VI. For every number a, there is one and only one number (—a) such that a + (— a) = 0. If a is a negative number, it is clear that the (— «) of this assumption must be a positive number. The familiar statement that if equal numbers be added to equal numbers the sums are equal numbers is equivalent to Assumption I. 6. Consequences of the assumptions. — The following theo- rems are consequences of these assumptions. 1. If a is any number, a + = a. 2. If a and b are any number s, there is one and only one num- ber X such that a=b-\-x. Definition. — The process of finding x when a and b are given is called the process of subtracting b from a. In this process a is called the minuend, b the subtrahend, and x the difference or remainder. Subtraction is indicated by the minus sign. Thus, the relation of «, h, and x just described is expressed by the equation a — h = X. The process of subtracting b from a is equivalent to that of finding the length and direction of the segment BAj A and B being the points whose abscissae are a and b respectively. The statement that if equal numbers be subtracted from equal numbers the remainders are equal numbers is equivalent to Theorem 2. THE FUNDAMENTAL OPERATIONS 5 3. The result of adding ( — b) to any number a is the same as the result of subtracting b from a. That is, a+(-&)=a-6. 4. The result of subtracting (—6) from any number a is equal to the result of adding b to a. That is, a - (— 6) = a + h. Although Theorems 1-4 are consequences of Assumptions I-V, they seem so obvious from the definition of addition and an inspection of a figure that we shall omit the formal proofs. The familiar rule for finding the difference of two numbers, — namely, to change the sign of the subtrahend and add the resulting number to the minuend, — is an immediate conse- quence of Theorems 3 and 4. 7. Parentheses. — An expression like a — b-\-c-\-d indi- cates that (—6) is to be added to a, c added to the resulting sum, and d to this last sum. That is, a — 6 + c + cZ is a sim- plified form for \_{a — b) -\- c'] -{- d. Now [(a _ 6) + c] + d = (a - 6) + (c + d) (III) = a-|-[-6 + (c + -5(/.-2)-f6[6-(5 + 3Z>)+7]. 8. 4a-(-3a-[-4a-J-2a-3a-f i;-10a]). 9. 7 X -{- 2[3 X - \2 X -(y - 2 x)l - 2(x - 8 y)]. 10. a-(-a-S-7a-3[a-3a-4]H-3J-[3a-a + l]). 11. a(x + y)- h{3x-2y). 12. 2a(p? - W) - h [a52 _ 3 (4 a^ - 5 W)']. 13. x-lx-lx-ix-l)]]. 14. 3 m - 2 ?i[l -f- 3 m (1 - m 4- w)]. 15. 2{x + y + z)-3(x-y-\-z)+2{x^-y-z). 16. ^ah-3h\a-\-h-2{3a-2b)\. 17. a(a; 4- ?/) -f h{x - y)- x{a -f 6) -f y{a - b). 12 COLLEGE ALGEBRA 18. (a + 6 — c) + (a — 6 + c) + (— a + 6 4-c). 19. 4m + 2 Jn — 3(m — n)J— 3m — 271. 20. Sm — 2n — 2\n — 3{m — n)\ — 4:m. In the following polynomials inclose all the terms containing x^ in parentheses preceded by the plus sign and inclose the terms containing x in parentheses preceded by the minus sign : 21. 4: x^ -{- ax^ — 3 X -\- bx + 5. 22. x^ + 2ax + mx^ — 2x^3-\-kx. 23. hx + x'^ -{■ ex + \. 24. 2 a'W -2h'^x- aV + x^. 25. — ax -\- a? — hx — 2 X — ^. 26. 7 ic — 6 ic'^ + 4 a — 3 6x — c:i(?. 27. ax^ + hx + c-3ix? -lOx + 4.- hxK 28. 3 + a + ic + £c'^ — mx — na;^ — vic. 29. — 3 a; + 5 — aa^^ + fta;'^ 4- cx^ — ax — bx — ex. 30. 3(ax^ — 2bx-^c)—5(a-\-cx — bx'^). 12. Addition and subtraction of polynomials. — The problem of expressing the sum of the polynomials 3x— 4:y-\-6z — 2, — 5x — 6y-[-z, and 3x-\-2y-{-z-\-5m the simplest possible form is the same as the problem of simplifying the expression (3x-4.y-\-6z-2) + {-5x-6y-\-z)-^(3x-\-2y-\-z-\-5). Now this is equal to Sx — 4:y + 6z-2 — 5x-6y-{-z-{-3x-\-2y + z-h5 = 3x—5x + 3x-4:y—6y-{-2y-^6z-\-z-\-z-2-\-5 = x—Sy + Sz-}-3. The usual rule for finding the sum of these polynomials is merely a more direct way to this same result. And in general the rule for finding the sum of any polynomials is a direct THE FUNDAMENTAL OPERATIONS 13 consequence of Theorem 5 and Assumption II. The rule for simplifying the sum of two or more similar terms is a conse- quence of Assumption XIII. We get the rule for subtracting one polynomial from another in a simi- lar way by using Theorem 6 instead of Theorem 5. EXERCISES 1. Add 7 a3 _ 4 a + 2 a2 - 5, -2 a + 5 a^ -\-l - 2 a% and 4 -I- 3 a + 2 a^ + a'. 2. Add — 5oiy^y — 2 xy^ — o? — y^^ a^ -\- y^, a^ — y^, and a^ + Sa^y-j-Sxy^ + y\ 3. Subtract S a^ -\' S a'b - 5 ab'' + 2b^ from a' - 6 a^ft + 4 aft^ -f- 3 b^, and then add the subtrahend to the result. Can you predict what the final result will be ? 4. Subtract x^ — y^ from a^ -{- y^. 5. Subtract aj^ -f 2/^ from a^ — y^. 6. From the sum of y"^ — S y -\- 2, 7 y - 2 + y"^, and 6 2/^ -f 6y -\- 6 subtract y^ -\- 5 y + 6. 7. From the sum of 4: a^b A- 5 ab% 5 a^ -2 ab^ -hSb% and 2a'-2b^-{-Sa'b-i-7ab^ subtract 7 a^ -{- 7 a^b -{- 10 ab^ -^ b . 8. Add {r-^s-\-t)x+ (r — s + t)y-\-(r-{-s — t)z, (2r — s)x •i- {s -\- 3t)y -\- {r -{■ s -{-t)Zj and {s-\-3t)x-{-(2r-\-4:t)y-h{s-{-3t)z. 9. From the sum of ma^ -{• 3 na^ -\- px — q and nx^ — mx^-^ 2qx+p subtract (m -\-n)a^-\- {m — 3n)x^-{-{p-\-q)x-\- (q—p). 10. Subtract (a + b)x-\- {b-\-c)y + (G + a)z iiom (c-\-a)x-\- {a-{-b)y -^{b-{-c)z. 11. Subtract {c -{- a)x -\- {a -\-b)y -{- (b -{- c)z from (a-{-b)x-\- {b + c)y-{-{c + a)z. 12. Does a consideration of the results in Exs. 10 and 11 suggest to you any general principle ? Consider also Exs. 4 and 5. 14 COLLEGE ALGEBRA 13. Add a~\-b — c, a — b-\-c, and —a-\-b-\-c. (See Ex. 18, §11). 14. Subtract 2(a-\-b — c) from a -f 6 -}- c. 15. From the sum of a^ + 3 a-b + 3ab- + b^ and a^ — S a^b + 3 ab^ — b^ subtract the sum of a^ -f b^ and a^ — b'^. 16. Subtract a^ + b^ from a'^ + 3 a^ft + 3 a^^ + ?)3^ and a^-b^ from a3-3a26 + 3a&2-63, 17. Is there any difference between the result in Ex. 15 and the sura of the results in Ex. 16 ? Could you have answered this question without going through the details of Exs. 15 and 16 ? 18. Subtract ax^-\-bx + c from dx~ -\-ex-\-f. 19. Is the result of subtracting a number from zero the same as that of subtracting zero from this number ? 20. What must be added to 7 a^ — 8 a^ + 4 a 4- 7 in order that the sum shall be 2 a^ + 4 a^ + 6 a - 1 ? 21. What must be added to A afy — 6 xy^ + 8 / in order that the sum shall be 5 x^ -{- 7 xy^ -\- y^ ? 13. Multiplication of polynomials. — The rule for forming the product of two polynomials is a direct consequence of Assumption XIII. Consider the product of the two binomials a + 6 and x-^y. By Assumption XIII (a -{-b)(x + y) = (a+6) x + (a+b)y=ax-\-bx+ay+by. The product of any two polynomials can be treated step by step in a similar way. Homogeneous polynomials. — A polynomial all of whose terms are of the same degree is said to be homogeneous. Thus, 5 ic2 -f 7 ?/2 and a^ -f- 3 a^b + 3 a&2 -f 68 are homogeneous polyno- mials of degrees 2 and 3 respectively. THE FUNDAMENTAL OPERATIONS 15 The product of two homogeneous polynomials is a homo- geneous polynomial whose degree is equal to the sum of the degrees of the factors. This fact is useful in checking the work of forming the product of two homogeneous polynomials. EXERCISES Simplify the following indicated products and use the fore- going check whenever it is applicable : 1. {x + 2y){x + Sy). 2. (a + 5) (a -8). 3. {x^ + 2xy + f){x'-2xy + f). 4. (x^-2xy + y''){x'-^x^y-{-^xy^-f). 5. {p? + ab-\-h-){a''-ah + h-). 6. (a + ar + ai-^ + ai-^ -f- ar^) (1 — r). 7. (a - h) {a* + a^b + a%^ + aW + h"). .8. {x^y + zjix + y-z). 9. (a + 6 -f c) (2 a - 6 -f 3 c). • 10. (a2 + 6a + 4)(a3-4a2 + 5a-f 2). 11. {a + h^x — y){a — h — x-\-y). 12. (3 a^ - 5 a^ft + 7 a&2 -I- 4 6^ (4 a^ 4- 10 a6 + 5 62) . 13. (x^ -\- y"^ -^ z^ ^-2 xy -\-2 yz -\-2 zx){x + y -^ z). 14. (2a-36 + 5)(2a-36-4). 15. (wi* + m~n^ -f- n^) (m* — m^n^ + n*) . 16. (c3 + 9(r-f-27c + 27)((r-6c + 9). 17. (16-Sx + 4:i^-23(^ + x')(2-\-x). 18. (r' + 5r~4)(j'2 + 5r-3). 19. (7ft2-(-56^-9c2)(6a2-46'-^ + c2). 20. (aj2 + 3a;-f 9)(x2-3a;-f 9). 16 COLLEGE ALGEBRA 14. Special products. — The following special products should be memorized by the student in order that he may be able to write down products of these types from an inspection of the factors. He should verify each of the formulae by actual multiplication. He should also translate each one into ordinary language. 1. {a + hy = a'-{-2ab-\-h\ 2. (a^hf = a^ + 3 a?h + Z ay + h\ 3. {a-\-h){a-h) = a''-h\ 4. (a + & + c)2 = a^ + 62 4- c^ + 2 a& + 2 6c + 2 ca. EXERCISES Write the results of the following indicated multiplications : 1. {a-Vf, 11. {x^2f. 2. {a-hf. 12. (5 a -4)1 3. {2x + 3yf. 13. (10 + 9 6) (10 -9 6). 4. (2x + ^yf. 14. (a + iy. 5. (2a; + 3?/)(2a;-32/). 15. (J x + 10y)(10 y -7 x). 6. (3 a -6)2. 16. {^m + n + r)\ 7. (a-6-c)2. 17. (a^ + a^ + l). 8. (2 a; -3 2/ + 4 2)2. 18. {^r-sf. 9. (a + 6 + c)(a + 6-c). 19. {s-^rf. 10. (2 m — 5 w) (2 m — 5 w). 20. {x-\-y — z){x — y -^z). 15. Division of poljrnomials, — If ^ and B are two polyno- mials such that the degree of A in some letter is equal to, or greater than, the degree of B in this letter, the process of find- ing two polynomials Q and i? such that A = Bq^R, and R is of lower degree in the given letter than B, is called the process of dividing A by B. In this process A is called the dividend, B the divisor, Q the quotient, and B the re- THE FUNDAMENTAL OPERATIONS 17 mainder. If R=0 we say that A is divisible by B, or that the division is exact. If the divisor is a monomial, it follows immediately from Assumption XIII that the quotient is the sum of the partial quotients obtained by dividing each term of the dividend by the divisor. How the quotient is obtained when the divisor contains more than one term is best shown by an example. Divide 12 x^ — 7 x^y — 14 xy"^ + 5 y^ by 4 x — 5 y. 12 ic3 - 1 x^y- 14 xy^ + 5 y^l ix -by 12 x» - 15 x'^y 1 3x^4- 2x2/ -2/2 8 x'-^y — 14 xy'^ 8 x^y - 10 xy^ — 4 xy2 + 5 2/3 — 4 xy"^ + 5 1/3 We know that the required quotient is Sx"^ + 2xy — y^ and that the remainder is 0, since the products of the successive terms of the quotient and the divisor when subtracted in turn from the dividend give a final remainder of 0. Before beginning this work of finding the successive terms of the quo- tient, the student should see to it that the dividend and the divisor are arranged according to the descending powers of some letter, although when the division is exact they may be arranged according to the ascend- ing powers. The relation A = BQ-\-B, holds true for all values of the letters involved and is for this reason called an identity. (See § 31.) EXERCISES Perform the following indicated divisions : 1. (a^-b'-)^(a-b), 2. (x' -j-aff ^i/) -ir (x' -{-xy ^y"), 3. Q^-f)^(x-y). 5. (a*-|-4a3 + 6a2 + 4a-hl)-^(a2-f 2a-f 1). 18 COLLEGE ALGEBRA 6. (a^4-3a^ + 7a2 + 8ci + 6)-v-(a2 + a-f-3). 8. (r' - s') --- {r -\- s). 9. (a^-h') ^ (a + b). 10. [(a _ by- 5(a - 6) + 6] - [(a - ?>) - 3] . 11. (m^ 4-7^3) -J- (m- 71). 12. (a;2+ 5 aj + 6)-- (.^• + 2). 13. (o5_?>5)h-(«-^6). 14. {a'-h')-r- (a-\-b). 15. (64 + 9?>2 + 81)-(62 + 36+9). 16. (x'' + 5x + 4)^(x'' + Sx + 2). 17. (a^ + 9 a^fe + 27 ab^+ 27 ?>3) -j- (a2 4. e a6 -f 9 62). 18. (a^ + 9 a26 + 27 a^^ _^ 27 6^) _^(« 4. 3 5). 19. (x^-y^)-^(x-y). 20, (aj4_^4)^(^^,^)^ CHAPTER II FACTORS AND MULTIPLES 16. The problem of factoring a polynomial — that is, of find- ing two polynomials whose product is the given polynomial — is, in general, more difficult than that of dividing one polyno- mial by another one, since in factoring both the divisor and the quotient are unknown. There are, however, certain types of polynomials that can be factored readily. The more com- mon of these and their factors are given here. 17. 1. Polynomials with monomial factors. — It follows from the distributive law for multiplication that a factor of all the terms of a polynomial is a factor of the polynomial. For example, Ix^y'^ is a factor of every term of the polynomial 6 x^ip- + 8 x'^y^ — 10 x^y3, and is therefore a factor of the polynomial. 6 ic3y2 + 8 x4«/3 - 10 5C52/8 = 2 xhf^(^ + 4 a-y - 5 xhf). 2. Polynomials that can be factored by grouping their terms. — The following is a typical polynomial of this form : ax -\- ay -{■ az ■\- hx ■\- hy -\- hz —{ax + ay + az) + (&x + 6y + hz) = «(» + ?/ + 2;) + &(a- + 2/ + iS!) = (x + ?/ f ^)(a + &). 3. The difference of two squares. a2-62^ (a+6)(a-6). Many polynomials that are more complicated in appearance than this one are in reality of this form. For example, a2 -f 2 a6 + 62 _ a;2 _|. 2 a:?/ - y2 = (a2 + 2 ah + l)^)-ix?- -2xy + y'^) = (a-\-hy-(x-yy = (a-{-b + x — y){a-\-b — x + y). 19 20 COLLEGE ALGEBRA 4. The sum of two cubes. a^ + b^ = (a-i- 6)(a2- ab-^i^). 5. The difference of two cubes. 6. Trinomial squares. a'^ + 2ab-hb^=(a-h by. a2 - 2 a& + ft2 = (a - by. 7. The quadratic trinomial ax"^ -i-bx+ c. If four numbers k, I, m, and n can be found such that km = a, Im + kn = b, and Z?i = c, then aa;^ + 5^ + =(A:x + l)(mx + n). 8. Polynomials of four terms that are the cubes of bi- nomials. a^-\-Sa''b-[-Sat^+b^ = (a+ by. a3 _ 3 flSft + 3 a62 _ iy3 =(a _ 5)3. 9. The sum of two like odd powers. If n is an odd positive interger a** + 6" is divisible by a + 6. 10. The difference of two like powers. For every positive integral value of n, a" — 6** is divisible by a — 6. This statement and the one under 9 can be proved by means of the factor theorem (§ 142, Cor.) 18. Whether a given polynomial can be factored or not depends somewhat upon our point of view. Thus, the polyno- mial a^ — 2 has the factors x + V2 and x — V2 although it has none with rational numerical coefficients. Also x^ — y has the factors X + 'Vy and x — ^y, but none that are free from radicals. In this chapter we shall confine our attention to polynomials whose numerical coeflBcients are integers, and when we speak of the factors of a polynomial we shall mean factors that are free from radicals and that have integral numerical coefficients. FACTORS AND MULTIPLES 21 Definition. — A polynomial that has no factors of this kind is said to be prime. A polynomial whose numerical coefficients are integers with a common factor greater than 1 is not prime since this numeri- cal factor is a factor of the kind just described. The work of factoring can be checked by making use of the fact that the product of all the factors found should equal the given polynomial. EXERCISES Factor the following polynomials into their prime factors : 1. 12 a'b'-\- IS a'b'- 21 a'b'. 16. a2-5a + 4. 2. (a + 6)2-1. 17. m^ — 25 m2 — m + 5. 3. Sx^-\-12x^-j-6x-{-l. 18. -2x^-\-Sxy + 27y\ 4. 9a2 + 24a6c+16 6V. 19. x* -\- 34: xY -^ 225 y*. 5. 27-64a^. 20. a^ +4.a^ + 2 a-\-S. 6. ^9-x^-\-4:xy-4:f. 21. 54: a^x^ - 16 b^x^. 7. a2-f4a-77. 22. 6x'^-x-15. 8. a^ — b^ — a + b. 23. a^ — x*. 9. a^ — 2a3 + l. 24. 7n^ -\- 2 xy — x'^ — y\ 10. x^ -\- x"^ -\- X -\- 1. 25. x^ — y\ 11. a'b — aV. 26. x^ + y^. 12. 2xy-x^~y\ 27. 15 a" ^- 4:1 ab + 2S b\ 13. Q?-3x'^y + 3x7f-y^-z\ 28. {x-{-yf-21. 14. 12o}-{-13ab-35b\ 29. Qx^ + 3 x" -\-2 x-^1, 15. 6a^2_,_7^2/-20 2/2. 30. a^ + 6 a^ + 12 a + 8. 19. Highest common factor. — A polynomial that is a factor of each of two or more polynomials is called a common factor of these polynomials. A given set of polynomials may have several common factors of the same degree. Thus, 2 a;2 _ 4 0^ 4- 2 2^2 and 4 x^ — 4 y^ have the common factors x — y^ — x + y, 2x — 2y, and — 2 x + 2 y of the first degree. 22 COLLEGE ALGEBRA Among all the common factors of the highest degree of two or more polynomials there is always one the greatest common divisor of whose numerical coefficients is equal to the greatest common divisor of all the numerical coefficients of the given polynomials. We call this a highest common factor of the polynomials. Thus, 2x — 2y isa, highest common factor of the polynomials given in the preceding illustration. But — 2 ic + 2 ?/ is also a highest common factor of these polynomials ; and in general, any set of polynomials has two highest common factors which differ only in sign. In most cases it makes no difference which of these two highest com- mon factors of the given polynomials we take, and it is there- fore customary to speak of " the highest common factor," as if there were but one. If the given polynomials can be factored into their prime factors, their H. C. F. (highest common factor) can be deter- mined immediately by inspecting these prime factors. 20. Lowest common multiple. — A polynomial that is divisi- ble by each of two or more polynomials is called a common multiple of these polynomials. A given set of polynomials has many common multiples of the same degree. Thus, 2 x^ — 4: xy -\- 2 y"^ and 4 x^ — 4 y2 have the common multiples 4 x^— 4 x^y — 4 xy^ + 4 2/^, 8 x^ — 8 x^y — 8 xy'^ + 8 y^, and many others which are of the third degree. Among all the common multiples of the lowest degree of two or more polynomials there is always one the greatest common divisor of whose numerical coefficients is equal to the least common multiple of the greatest common divisors of the numerical coefficients of the respective polynomials. We call this a lowest common multiple of the polynomials. Thus, 4x^-4x^y-4xy^ + 4 two polynomials given in the preceding illustration. FACTORS AND MULTIPLES 23 Every set of polynomials has two lowest common multiples which differ only in sign. The expression "the lowest com- mon multiple " refers to either one. If the given polynomials can be factored, their L. C. M. (lowest common multiple) can readily be formed. EXERCISES Find the H. C. F. and the L. C. M. of the following poly- nomials : 1. 25,30,40. 2. '3? —]^^x^— y^. 3. a2 + 2a6-hZ>2, a^ + 3 a6 -j- 2 61 4. h' + f, W-f. 5. x^ + 2/2, y^— i/, x^ — y\ 6. a^-243, a;2-6a; + 9, a;2-9. 7. ax — hx — ay -\- by, a^-\-ab — 2 b\ 8. a2 4-3 a- 28, Sa^- 20 a. 9. a^ — x^, a^ — a^, a^ — x"^. 10. x''-\-x-6, 2x'^-\-2x —12. 11. a''-¥, ¥-a\ 12. x^ + l x + 12, aj2 4_ 5 X -I- 6. 13. 6 a2 - 13 a5 + 6 b\ 10 a^ _ 13 aft - 3 b\ 14. x^ + 2/^ ^^ ~~ y^- 15. 24 m2 -14. mn- 20 n^, 24 m^ + 6 mw - 45 ?^2. 16. 16a4-81ft^ 8a3-27 6^ 17. aj2 _|_ 6 icy -}- 9 ?/^ a^ + 9 a;2^ + 27 a^?/^ + 27 f. 18. ao^ + 2 aa;2 -j- 3 aa; + 6 a, 4 aa; + 8 a. 19. 15 a2 + 11 a6 + 2 62, 5 a^ -f- 7 a6 + 2 b\ 20. a-\-b, a-\-2b, a + 3 6. 24 COLLEGE ALGEBRA 21. Euclid's Method for finding the highest common factor of two or more polynomials. — If we cannot find the factors of the given polynomials, the preceding method cannot be used. In such a case, if the polynomials contain only one letter, we can fall back on the following method, which is called Euclid's Method. Let A and B be any two polynomials in x whose H. C. F. we wish to find. Arrange them according to the descending powers of X, and if the degree of B is not greater than the degree of A divide A by B. Denote the quotient by Qi and the remainder by Bi. Then (§15) A = BQ,-{-B„ or A-BQi = R^. Now any factor of J5 is a factor of BQi. Hence any common factor of A and J5 is a factor of ^ — BQi, which is the same as Ri ; and is therefore a common factor of B and i?i. Con- versely, any common factor of B and J?i is a factor ot BQi-\- Bi, which is the same as A ; and is therefore a common factor of A and B. The highest common factor of A and B must accord- ingly be the same as the highest common factor of B and J?i. We can then shift the problem to the finding of the highest common factor of B and B^. Now R^ is of lower degree in x than B. We therefore proceed as before and divide B by R^. If the quotient is Q2 and the remainder R2, we have B=R,Q2 + R2, or B-RiQ2 = R2. Reasoning similar to that used before shows that the highest common factor of B and Ri is the highest common factor of Ri and i?2- By continuing in this way we can establish the following series of identities : FACTORS AND MULTIPLES 25 El = i^aQs + Bsf Bk — B,,^iQk+2 + B^^2i in which Bk+2 tloes not contain x. This is possible since each B is of lower degree in x than the preceding one, and if the process is continued far enough we must come to an B that does not contain x. It is clear from the method of formation of these equations that the highest common factor of B^ and B^+i is also the high- est common factor of A and B. If now Bk+2 = 0, B,,+i is the highest common factor of Bj, and i?t+i and therefore also of A and B. But if Bk+2 ^ 0, ^ and B have no common factor con- taining x. Example. — Find the H. C. F. of x^ + 2 a;2 _ 13 ^ + 10 and x^ + x:^ - 10 X + 8. These two polynomials are of the same degree and either may there- fore be taken as the divisor. We select the second one. x^-^2x^-13x + 10 \ a^ + x2 - 10 a; -H 8 x» + x^-10x-\- S 1 x2- Sx + ~Y\ x^+ a;2-10x + 8 x + 4 x3-3x2+2x 4x2-12x + 8 4a;2_i2a;+8 Here i?2 = and therefore i?i = a;2 - 3 x + 2 is the H. C. F. In applying this process the given polynomials should first be divided by any monomial factors they may have and account taken of these factors in the final result. Fractional coefficients should be avoided by introducing or removing numerical factors w^henever necessary at any stage of the process. This can always be done in such a way as not to affect the final result. 26 COLLEGE ALGEBRA 22. Lowest common multiple of two polynomials. — If F is the highest common factor of the polynomials A and B, and A = A'F, B — B'Fj then A' and J3' have no common factor. Now = AB^ — A'B. Hence is a common multiple of A and B. Suppose now that M is any common multiple of A and B. Then we should have M= AQ = A'FQ. But M must also be divisible by B, which is the same as B'F. Hence Q must be divisible by B'. In order that M be the lowest com- mon multiple, Q must be the same as B'. Hence the L. C. M. AB of A and B is A'FB' = AB' = . This is equivalent to say- ing that The lowest common multiple oftivo polynomials is equal to their product divided by their highest common factor. EXERCISES Find the H. C. F. and the L. C. M. of the following poly- nomials : 1. a^ - a;2 + 3 a; -f 5, and y^+^x^-^x-h. 2. a;^-f-4a^ + 5aj2 + 4aj + l and a;^ + 5 a;^ -f 3 x^ - 10 a; - 4. 3. 6 a:^ _ 22 a;2 + 15 a; - 2 and 6 a;4 - a-3 + 4 a;2 - 13 aj + 6. 4. 2 a:^ _j_ 12 a;2 + 19 a; + 3 and 2 a^ -f 13 aj2 _^ 17 a! + 3. 5. a;^ -f 3 a^ 4- 6 a;2 + 3 a? - 5 and a;4 + a^ - 11 a;2 - 10 a; + 10. 6. a^ -f- 6 a;2 + 11 a; -f 6 and a.-3 + 8 a)2 + 11 a; - 20. 7. 2 a^ + 4 a3 - 2a2 - 4 a and a^ 4- 2 a2 - 3. 8. ?/4 + 3 2/3 + 4 2/2 - 3 2/ - 5 and ?/^ + 3 ?/3 4- 6 / + 3 2/ + 5. 9. a' + 3 a2 -f- 8 a -f 6 and a^ 4- 3 a^ -f 9 a2 + 8 a + 6. 10. 62 _ e 5 4. 2 and W -'lb'' -22h + 8. 11. a;5 _|. 4 ^ ^ 3.2 _ 5 ^ _^ 1 and a;^ 4- a;^ ^ 5 ^j^ + 5 a; 4- 5. CHAPTER III FRACTIONS 23. Definition and principles. — A fraction is the indicated quotient of two numbers. Thus, 2 -3 _L4 V2^ 3' 5 ' -.5 3 The principles upon which the usual operations with fractions rest can be readily derived from the assumptions of Chapter I. These principles are : 1. Tlie value of a fraction is not changed by multiplying the numerator and the denominator by the same number. Consider the fraction - . It follows from the definition of b division that cl a = b • -. b Hence, ak = bk --. ' b But this is only another way of saying that ak_a bk~b' This holds for all values of k except 0, and since division by a number is equivalent to multiplication by the reciprocal of this number, we have also proved that the value of a fraction is not changed by dividing the nu- merator and the denominator by the same number. 2. Tlie sign of a fraction is changed if the sign of either the numerator or the denominator is changed. This is an immediate consequence of the law of signs for division, which is derived from the assumptions and the definition of division. 3. The sum of two fractions loith a common denominator is a fraction whose numerator is the sum of the numerators of the given fractions and whose denominator is the common denominator. 27 28 COLLEGE ALGEBRA Let the given fractions be - and - . c c ^ow a = C' -, c and b = c ' . (Definition of division.) c Hence, a + ?> = c- 4- c- c c = c ( - + - ). (Distributive law for multiplication.) \c cj From this it follows that ^^±^ = - + - . (Definition of division.) c c c In a similar way we see that a—h _a b c c c In view of what has just been proved we can say that divi- sion is distributive as to addition. This includes the statement that division is distributive as to subtrac- tion since subtraction is included under addition. (See 3, Chapter I.) a ' c 4. The sum and the difference of any two fractions, - a7id -, a uiK eyaut/ vu — bd bd "^"^"^ For by 1, a _ad b^bd' and c_bc d~bd' Hence by 3, a b' c ad-\-bc d ~ bd '■ and a b c ad — be d bd FRACTIONS 29 5. The product of two fractions is a fraction whose numerator is the product of their numerators and whose denominator is the product of their denominators. -ci j^ a c c For, ^' ^'-^ = ^'-J^ h d d (Definition of division, and IX, Chapter I.) and hd ' -^ . ~ =b • - • - • d= a* - > d = ac. b d b d d (Definition of division, VIII, and IX, Chapter I.) Hence, - • - = — . (Definition of division.) ' b d bd ^ ^ 6. Tlie quotient resulting from dividing one fraction by another is a fraction luhose numerator is the product of the numerator of the dividend and the denominator of the divisor and whose de- nominator is the product of the denominator of the dividend and the numerator of the divisor. If the dividend is - and the divisor - , we have, from the b d definition of division, ^ a_b c b^c' d d a Hence, ad = -' be, d a ad^h be c* d and When we are dealing with more than two fractions the principles to be employed are obvious extensions of those just developed. 30 COLLEGE ALGEBRA The following reductions are applicable to rational fractions ; that is, fractions which are equivalent to indicated quotients of polynomials. 24. Reduction of certain fractions to mixed expressions. — If the degree of the numerator of a fraction in a certain letter is equal to, or greater than, the degree of the denominator in this letter, we can arrange both numerator and denominator according to the descending powers of this letter and divide the numerator by the denominator. Then, if the given frac- tion is -, we have the identity a ,-r 6 = ^ + 6' where g is a quotient, and r the remainder, of this division. Here g is a polynomial and - is a fraction whose numerator is of lower degree than its denominator. Hence, if the degree of the numerator of a fraction in any letter is equal to, or greater than, the degree of the denominator in this letter, the fraction is equal to the sum of a polynomial and a fraction whose 7iumerator is of lower degree in this letter than its denominator. 25. Reduction of a fraction to its lowest terms. — A fraction is said to be in its lowest terms when its numerator and de- nominator have no common factor. A fraction can be reduced to its lowest terms by dividing the numerator and denominator by their highest common factor (§ 23, 1). 26. Reduction of several fractions to equivalent fractions with a common denominator of lowest possible degree. — Find the L. C. M. of the denominators, which is called the lowest common denominator of the fractions, and divide this by the denominator of each fraction. Then multiply the numerator and the denominator of each fraction by the corresponding quotient. The resulting fractions are equivalent to the orig- FRACTIONS 31 inal ones (§ 23, 1) and have the required common denominator. It is unnecessary actually to perform the multiplication of each denominator by the corresponding quotient, since we know beforehand that the product will be the lowest common denominator. 27. Addition and subtraction of fractions. — In order to add or subtract fractions, first reduce them to equivalent fractions with a common denominator and then proceed in accordance with § 23, 3. In order to add a polynomial to a fraction, consider the polynomial as a fraction with the denominator 1 and then pro- ceed as directed. 28. Multiplication and division of fractions. — The rules for multiplying and dividing fractions are given in § 23, 5, 6. The given fractions should be reduced to their lowest terms before applying these rules. EXERCISES Reduce the following fractions to mixed expressions : a + b a^-1 ar^ + 2a^- a; + 4 a^-a^-^2 X-4: x'-hl x'-l 4:X ar — 1 3. -J^. 6. ^ a — 1 x-^ 1 Reduce the following fractions to their lowest terms: ^ 18 xy ^^ m^ + 5m^4-2m + 10 12x2/"* * m2-f7m + 10 8 ^^-^ 11 ^-(y-h^y x''-6x-{-9 ' (x + yf-z^ ^ a;5_y5 ^^ a^ + 4aj2-f-8a; + 5 y^-a^ a^-Sx-'+Sx-^-T 32 COLLEGE ALGEBRA Perform the following indicated operations and bring youi results to their simplest forms : 13. ^- + ^_. 15. ^ ^'^ K + l a-l 0^-20 + 1 a^ — 1 14. -i ^ + -?- 16. '" + ^ . + ^ + 3 K + l » — 1 s + S a;^+5x + e ii?+l x+12 (x -!)(«- 2) {x-2){x-Z) (»-3)(x-l) 18. m^-m + 1— !^. 21. 2aW.10a^ m + 1 5 3?y* 3 6^ 19. *+-^- o„ n^yw 25. a bj\a bj\ a — hj\ a — b w} + mn + mr + **** . w? — t^ m"^ — mn — mr + nr m^ — n^ 10i» + 25 - (>+Mh^ 3x2 ym^ + 71^ m'^ — n^ 28.1 i+l^ 1 \ ^ /m -\- n .m — n\ J \m — n m + nj 1 r^jT^-r} 30. ^~^ . ^^y . 3 a; a;2 -|- 3 a;^/ — 4 2/2 FRACTIONS 33 Simplify the following : 31. ^i^. 34. "^ m N_ ^ R 35. H 1 + 1 + 5508 29. Ratio. — The fraction - is called the ratio of a to h. The h numerator a is called the antecedent of the ratio and the de- nominator h the consequent. Two numbers a and 6 are said to be commensurable if there is a number r such that - and - are integers. If we call these integers x and y respectively, that is, if - = a; and - = y, then r r - = -• Hence the ratio of the two commensurable numbers is h y a rational fraction. Not all numbers are commensurable. For example, V2 and 1 are incommensurable, since if they were com- mensurable we should have ^^ = - or 2 = — • Hence, 2 y2 = ^2. But 1 y y-^ ever}'' prime factor of x^ and of y^ occurs to an even power and this rela- tion requires that 2 occur as a factor of x^ to a power one higher than the power to which it occurs in y^. 30. Proportion. — The statement that two fractions are equal is called a proportion. The numerators and denominators of the fractions are called the terms of the proportion. Thus, «, b, c, and d are the terms of the proportion - = -• This propor- b d tion is sometimes read "a is to 6 as c is to d" The first and the last terms are called the extremes, and the other two the means, of the proportion. 34 COLLEGE ALGEBRA If we clear the equation - = - b d of fractions, we get ad — he. Hence, in any proportion the product of the means is equal to the product of the extremes. The student can verify that if a_c b~d' then ^ = i c d and - = - . a G The former of these two is said to have been obtained from the original one by alternation and the latter by inversion. Suppose that -==- = - = r. ^^ h d f Then a = hr, c =dr, and e =fr. Hence, a + c -^ e = br -\- dr -\- fr = r (b -{- d -\- f), -I a + c + e ace and , , ^ = r = " = - = -,. b + d^f b d f If we had started with more than three equal ratios, it is easy to see that an analogous procedure would have led to an analogous result. We conclude therefore that 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. If - = -,/ is called the fourth proportional to a and b and c. If — =— , m is called a mean proportional between a and b. m b If - = -, ^ is called the third proportional to a and b. b t FRACTIONS 35 EXERCISES AND PROBLEMS 1. In the diatonic music scale the notes O, D, E, F, G, A, and B are produced respectively by c, d, e, /, g, a, and b vibra- tions of the air per second, where - = -^ =•<-- = - and - = — = e b a 5 g 2 d — = - • Find the number of vibrations per second that pro- 2c 6 f ^ duce each of these notes, it being assumed that C is produced by 256 vibrations per second. 2. If the terms of a ratio are positive and the same positive number is added to each, in what way is the value of the ratio changed? Let. the ratio be - and let x be the positive number. Then com- pare the values of - and ^-±-?. b b + X Theorem from Geometry In a right triangle the perpendicular dropped from the vertex of the right angle to the hypotenuse is a mean proportional between the segments into which it divides the hypotenuse. 3. The hypotenuse of a right triangle is 20 inches long and the perpendicular from the vertex of the right angle upon the hypotenuse is 8 inches long. Where is the foot of this perpendicular ? 4. Prove that if -= -, then b d (a) (&) (c) a-\-b b c-\-d d a-b b c-d d a-{-b c-\-d a—b c—d Proportion (a) is said to have been derived from the given proportion by composition, (6) by division, and (c) by composition and division. 36 COLLEGE ALGEBRA 5. Find the mean proportionals between 16 and 36. 6. Find the third proportional to 15 and 28. 7. Find the fourth proportional to 13, 28, and 36. Theorem from Geometry. — A perpendicular dropped from any point in the circumference of a circle to a diameter is a mean proportional between the segments into which it divides the diameter. 8. A perpendicular drawn from a point in the circumference of a circle of radius 6 inches to a diameter divides this diameter into two parts, the ratio of whose lengths is equal to ^. Find the length of this perpendicular. 9. In the triangle ABC, AB = S, BC = 17, and CA = 20. Through what point of BC must a line DE parallel to AB be drawn in order that DE = ^AB? DE^EC AB BC' Hint, ±^ = A property of similar triangles. — The areas of similar triangles are proportional to the squares of two corresponding sides. Thus, if S\ and 82 are the areas respectively of the similar triangles ABC and DEF, then Si^AB^ ^BC^ ^CJ^. S2 DE^ EF^ Flf 10. If AB = 9 inches, what must be the length of DE in order that the triangle DEF be twice as large as the triangle ^B(7? 11. If Si = 20 square inches, xS'2 = 25 square inches, and JBC=4 inches, how long is EF? FRACTIONS 37 12. In the triangle ABC the line DE is to be drawn parallel to AB in such a way that the tri- angle cut off shall be one third of ABC. At what point in ^4 (7 must Z) be ? Formula from Physics. — If two falling bodies pass over the distances Sj feet and Sg feet in ti seconds and ^2 s t^ seconds respectively, then — = -^. , ^2 ^2 13. What is the ratio of the distances passed over by the two bodies if the second one falls three times as long as the first one ? 14. If a body falls 64 feet in 2 seconds, how far will a body fall in 5 seconds ? Theorem from Geometry. — If a plane be drawn parallel to the base of a pyramid V-ABC, cutting the pyramid in the sec- tion DEF, and if we denote the areas of ABC and DEF by JSi and S2 respec- tively, then where VH and VK are the distances of the base and the cutting plane respec- tively from the vertex. 15. In a pyramid whose altitude is 10 inches, how far from the vertex must a cutting plane be in order that the section DEF be three fourths as large as the base ? 16. If the area of the base of the pyramid is 14 square inches, and the altitude is 9 inches, what is the area of the section midway between the base and the vertex ? CHAPTER IV LINEAR EQUATIONS IN ONE UNKNOWN 31. Definition of equation. — The statement that two alge- braic expressions are equal is called an equation. If the statement contains only one letter whose value is unknown, it may be considered as a description of a number. A number answers this description if the two expressions have the same value when this number is substituted for the letter whose value is unknown. An identity is a description that fits every number. Thus, the two members of the identity, (cc — 1)2 = a:'2— 2 x + 1, are equal to one another regardless of what value is substituted for x. An identity is sometimes indicated by the symbol = . Thus, (x - 1)2 = a;2 - 2 X + 1. On the other hand, the two members of the equation ^x -\-l = X — 1 do not have the same value for most values of X. When x = 2, for example, the left member equals 7 and the right member 1. As a matter of fact, — 1 is the only value of x that makes the two members of this equation equal ; that is, —1 is the only number that answers the description given by this equation. An equation that does not fit every number is called an equation of condition, because it imposes a condition on the numbers that satisfy it. If each of its members is a polynomial in the letter repre- senting the number described, the equation is called a rational integral equation. If neither of these polynomials is of degree greater than 1 in the letter representing the number described, the equation is said to be of the first degree, or linear. 38 LINEAR EQUATIONS IN ONE UNKNOWN 39 A number that answers the description given by the equation is said to satisfy the equation, and is called a root of the equa- tion. The process of finding the number or numbers described by an equation is called the process of solving the equation. At the opposite extreme from identities stand descriptions (equations) that do not fit any number. For example, 2x + l = 5a; + 4 — 3a;, 32. Equivalent equations. — Two equations that describe the same numbers are said to be equivalent. The equations (a; — 1) (a; - 2)= and {x - \Y{x - 2) = each de- scribe the numbers 1 and 2 ; but they are not considered equivalent since the latter in a sense describes the number 1 twice. (See § 148.) The following principles concerning equivalent equations are of frequent application in the solution of equations. 1. If we add the same number to each member of an equation and equate the sums, the equation thus formed is equivalent to the original one. If we represent the left member of the equation by L and the right member by R, we have L = R. (1) Either L, or R, or both of them contain x. Let the new equation be L + 8 = R + S. (2) Now any number described by equation (1), when put in place of cc, makes L=R, and, therefore, L-\-S=R-\-S (I, Chap- ter I). Moreover, any number described by equation (2), when put in place of £c, makes L-\-S — R-\- 8, and, therefore, L== R (IV, Chapter I). That is, the two equations describe the same number and are therefore equivalent. The case of the subtraction of the same number from each member of the equation is included in this proof since the subtraction of any number /S'is equivalent to the addition of (— S) (3, Chapter 1). 40 COLLEGE ALGEBRA 2. If we multiply or divide both members of an equation by the same number and equate the results, the equatio7i thus formed is equivalent to the original one, provided that the number by which we multiply or divide is not zero and is not expressed in the terms of the uyiknown number of the equation. Let the original equation be L = R, (1) and the new one LM= RM. (2) Now any number described by equation (1) when put in place of ic, makes L = Ry and therefore LM— RM (VII, Chapter I). Moreover, any number described by equation (2), when put in place of X, makes LM=RM, and therefore L=R (X, Chapter I). That is, the two equations describe the same numbers and are therefore equivalent. The argument just given covers the case of dividing both members by the same number, since to divide a number by M is equivalent to multi- plying it by J-. M If the M of the demonstration were 0, equation (2) would be = 0. But this is an identity and is not equivalent to (1). The reason for saying that the number by which we multiply or divide the two members of the equation shall not be ex- pressed in terms of the unknown of the equation will be clear from the following examples : Consider the equation 2,x + S = x + 2. If we multiply each member by x — 3 and equate the products, we get (2 5c + 3)(a; - 3) = (x + 2)(x - 3). Now the student can readily verify that - 1 is the only number that answers the first description, whereas both — 1 and 3 answer the second one. Hence the two equations are not equivalent — the second one has a root which is not a root of the first one. Consider now the equation x^-9 = x + S. LINEAR EQUATIONS IN ONE UNKNOWN 41 If we divide both members by a- 4- 3 and equate the quotients, we get The student can verify that 4 and — 3 answer the first description and that 4 is tlie only number answering the second one. Hence the two equations are not equivalent. 33. From Assumption I it is clear that we can transpose any term from one member of the equation to the other pro- vided we change the sign before it. We can find the number described by any linear equation by proceeding in the following way : 1. If any of the numerical coefficients are fractions^ multiply both members of the equation by the least common multiple of the denominators of these coefficients and equate the products. 2. Remove any parentheses that may occur in the resulting equation. 3. Transpose all the terms that involve the unknown number to the left member, and all the other terms to the right member. 4. Simplify each member of the resulting equation by com- bining like terms. 5. Divide both members of the last equation by the coefficient of the u7iknown number in the left member. This process gives a series of equivalent equations, and there- fore the number given by the last equation must be the number described by the original equation. It is not necessary that these steps be taken in this order. 34. Example. Solve the equation I^ _ 1 = i£. (1) 3 6 4 Multiplying each member by 12, 28x-2 = 15a;. (2) Transposing, 28 x - 15 a; = 2. (3) Combining like terms, 13 a; = 2. (4) Dividing each member by the coefficient of a;, 42 COLLEGE ALGEBRA This work can be checked by substituting this value of x for x in equa- tion (1). The two members of the equation should then have the same value. 7 • i^TT 1^14 1 ^ 28 - 13 ^ 5 3 6 39 6 78 26 4 52 ~ 26 * Since the two members of this equation have the same value for this value of oj, we conclude that equation (1) has been correctly solved. EXERCISES Solve each of the following equations or show that it is an identity, and check your results : 1. 2a;+5=5x-f2. 3. ^^-|(3 + 2x) = 6. • 2. ^ + ^ = ^-0.. 4. ^-30.4-1 = ^- 2 3 4 2 2 5. (a; + l)' + (.'c + 2)2 = (a;-3)2-}-(aj-4)2. 6. (r_l)2 4.(r-2)2=(r-3)2 4-(r-4)2. 2 4 8. a^ 4- 63 _ 3 rah = 0. (Solve for r) 9. 5a; + 72/ + 3 = 0. (Solve for ic.) 10. 5 a; + T 2/ + 3 = 0. (Solve for 2/.) 11. aic + &2/ + c = 0. (Solve for ?/.) 12. ax 4- &2/ + c = 0. (Solve for £c.) 13. M = M + pk- (Solve for p^>j 14. s — So = ^ g^i^ — v4,. (Solve for Wo-) 15. 2/^ = 2 a (« + c). (Solve for a;.) 16. x^ {y — pa?) = yp^. (Solve for 2/.) 17. 5-4(a;-l)+| = 3-4a;. 18. p{p-y)= x{x + 2/). (Solve for 2/.) LINEAR EQUATIONS IN ONE UNKNOWN 43 19 _^4._^=:44--5I^. ' 410 330 6765 ^ 27 21 .770 8 . 770 21. V =VQ + gt. (Solve for t.) 22. a:(m + ?i — 5) + 3 a; — 8 = 9(7+ a; — m). 23. 3(5a;-4)-i(6 + a')=2. 25. 23-8=(^-2)(22 + 22 + 4). 26. 2/ — 2 = ??i (x — 1). (Solve for x.) 27. ma5 = nTF — nx. (Solve for x.) 28. 2.5a;-6.75 = 2.45 + .75a;. 29. From each of the following equations form a new one by equating the products obtained by multiplying each mem- ber by the expression at the right. Do any of the new equa- tions possess roots not possessed by the corresponding original equations ? ^ (a) a;- 2 = 5. x-1. (h) 3x + 4 = 7a;-3. 5. (c) a; + 6 = 0. X. 30. From each of the following equations form a new one by equating the quotients obtained by dividing each member by the expression at the right. Do any of the new equations possess all of the roots of the corresponding original equations? (a) a;2-l = 0. a; + l. (6) x'-lx^O. X. (c) 2 a; + 10=0. 2. 44 COLLEGE ALGEBRA \M N V' PROBLEMS ^" V '^\ 1. The sum of three consecutive integers is 24. What are they ? 2. The sum of three consecutive integers is a. What are they ? Are there any restrictions on a implied by the condi- tions of this problem ? 3. The sum of four consecutive odd integers is a. AVhat are they ? What restrictions are imposed on a by the condi- tions of this problem ? 4. At what time between four and five o'clock are the hands of a watch opposite each other? 5. Can the hands of a watch be opposite each other twice within an hour ? Count the number of times the hands of a watch are oppo- site each other between noon and midnight. 6. How far apart must the centers of two pulleys be if each has a diameter of 16 inches, and the connecting belt is 25 feet long? Volume of an anchor ring. — When the circle C is revolved about the line MN a solid is generated that is called an anchor ring. It is shown in Integral Calculus that the volume of this anchor ring is equal to the area of the circle multi- plied by the* length of the path de- ' scribed by the center of the circle. 7. If the radius of the circle is 3 inches, how far must the center of the circle be from the axis in order that the volume of the anchor ring shall be 1425 cubic inches ? 8. What must be the outer radius of a circular ring that is 3 centimeters wide and contains 240 square centimeters ? 9. What must be the outer radius of a circular ring that is 3 inches wide and contains a square inches ? What restrictions on a are implied in this problem ? LINEAR EQUATIONS IN ONE UNKNOWN 45 10. A runs around a circular track in 30 seconds and B in 35 seconds. Two seconds after B starts, A starts from the same place in the same direction. AVhen will they meet ? 11. A runs around a circular track in a seconds, and B in h seconds. Two seconds after B starts, A starts from the same place in the same direction. When will they meet ? Describe what happens when a = h. 12. If in Problem 11 A starts 2 seconds before B, when will they meet ? 13. If A runs around a circular track in 40 seconds, how fast must B go in order that they may meet every 18 seconds when going in opposite directions ? For the principles of the lever involved in Problems 14-18 see the Ap- pendix. 14. A uniform beam 12 feet long weighs 14 pounds per linear foot, and has a weight of 8 pounds attached to it 5 feet from one end. When it is lying horizontal, how many pounds must a man lift in order to pick up this end of it ? 15. A uniform beam 30 feet long balances on a fulcrum 1 foot 8 inches from its center when a man weighing 150 pounds stands on one end. How much does the rail weigh per linear foot ? 16. The roadway of a bridge is 25 feet long, and the weight of the bridge is 5 tons. What pressure is borne by each sup- port at the ends when a wagon weighing 2 tons is one fourth of the way across ? The bridge may be considered a lever with the fulcrum at one of the supports and the power at the other support. The measure of the power is the reaction of the support on the bridge, and is therefore equal to the pressure on this support. If X = no. of tons pressure due to the wagon on the support nearer the wagon, then, neglecting the weight of the bridge, we have 25 X = 2 . ^. Hence, ^i/^^"^-^^ 46 COLLEGE ALGEBRA The pressure due to the wagon on the other support is then 2 — x = ^. But the bridge itself exerts a pressure of 2^ tons on each support. Hence the pressures on the supports are 4 tons and 3 tons respectively. 17. A uniform beam 30 feet long and weighing 175 pounds to the foot rests with its ends on two supports. What is the pressure on each of these supports when the beam carries a load of 4 tons 6 feet from one end ? 18. A beam is supported at its ends and a weight of TF pounds is placed on it m feet from one end and n feet from the other. Find the pressure on each of the supports due to the weights. 19. A freight train whose rate is 18 miles per hour passes a certain station h hours and m minutes before a passenger train. What must be the rate of the passenger in order that it may overtake the freight in t hours ? CHAPTER V SYSTEMS OF LINEAR EQUATIONS IN TWO OR MORE UNKNOWNS 35. We have seen that a conditional equation of the first degree in one unknown is a description of a number and that there is only one number that answers such a description. In this chapter we shall see that an equation of the first degree in two unknowns is a description of a relation between two numbers, but that this description differs from the one given by an equation in one unknown in that there are a great many num- bers related to each other in the way described by the equation. Consider, for example, the equation 5x + 2y = 4t. The student can readily verify that any pair of values of x and y in the following table are related to each other in the way described by this equation : X -8 2 1 -3 -f 1 6 — — — — — y 22 2 -3 -'i- — — — — -7 -1 -1 5 6 The student should complete the table by finding the values of y that are related to the second four values of x in the given way, and finding the values of x that are related to the last six values of y in the given way. 36. Graphical representation. — A description of the rela- tion between two numbers by means of an equation can also be given by means of a graphical representation. The following explanations and illustrations are given with a view to enabling the student to become familiar with this method of representing these relations. In Chapter I we looked upon the real numbers as the repre- sentatives of the points of a straight line with reference to a 47 X-L •A J — J 48 COLLEGE ALGEBRA given origin and a given unit length. In a similar way pairs of real numbers can be looked upon as the representatives of the points of a plane. We take two perpendicular straight lines in this plane and their point of intersec- tion as the origin on each line. If X and y are a pair -• X of numbers we lay off x on the horizontal line, in the way just recalled, and y in a similar way on the verti- cal line, counting distances above the origin as positive and those below as nega- tive. Then through the point representing each of these num- bers we draw a line perpendicular to the line on which the point lies. The point of intersection of these two perpen- diculars is the point represented by the pair of numbers x and y. In the figure given above, the points A, B, C, and D are represented by the pairs of numbers (2, 1) ; (- f, 5) ; (- 1, — |) ; and (2, - -J), respectively. We can also say that these four points represent these four pairs of numbers respectively. 37. Definitions. — If the numbers x and y represent the point A in the way just described, x is called the abscissa of Aj and y its ordinate. The two numbers together are called the coordinates of the point. The abscissa of a point is always given first. Thus, the point (2, 3) is the point whose abscissa is 2 and ordinate 3. The two perpendicular lines are called the axes of coordi- nates, or coordinate axes. The axis on which the abscissae are laid off is called the X-axis, and the one on which the ordinates are laid off is called the j^-axis. The ic-axis is usually horizontal. EQUATIONS IN TWO OR MORE UNKNOWNS 49 38. Every pair of values of x and y is represented by a point, and every point represents a pair of values of x and ?/, which are its coordinates. These can be found by drawing perpendiculars from the point to the axes and measuring the distances from the origin to the feet of these perpendiculars. f), (1, 1), (4, 4). 4). EXERCISES Draw two axes and locate the following points, using \ inch or 1 centimeter as the unit distance : 1. (2,3), (5, -2), (0,1). 2. (-2, -2), (-3,-3), (- 3- (-f,|),(0,0), (3, -3), (i, -^; 4. (7, 0), (5, 0), (4, 0), (1, 0), (-3, 0), (- 8, 0). 5. (0, -10), (0, - 8), (0, - 7), (0, -4), (0, 1), (0, 2), (0, 3). 6. Determine by accu- rate measurements the coordinates of the points A, B, C, D, E, and F in the accompanying figure. 7. Draw a line parallel ^ to the a;-axis and 3 units above it, and measure the coordinates of 5 points on it. 8. Draw a line parallel to the 2/-axis and 2 units to the left of it, and measure the coordinates of 5 points on it. 9. Lay off the points whose coordinates are the successive pairs of values of x and y in the table of § 35. Can a straight line be drawn through these points ? ?Z) 50 COLLEGE ALGEBRA x^ 39. Locus of an equation. — For a given equation in x and y there is a certain line, straight or curved, such that the coordi- nates of every point on the line satisfy the equation, and that every pair of values of x and y that satisfy the equation are the coordinates of a point on the line. This line is called the locus of the equation. It exhibits graphically the relation between the unknowns (or variables) that is described by the equa- tion in the sense that those numbers which are the coordi- nates of points of this line, and only those numbers, are related to each other in the way described by the equation. For example, the locus of the equation 2 X - 3 .V = 6 is the straight line of the accompanying figure. EXERCISES Lay off carefully 10 points on the locus of each of the following equations : 1. x = y. 2. x — y = 5. 3. 2x = 3y. 5. -5»+62/+7 = 0. 6. a; + 2/ + l = 0. 7. x = 0. 8. y = 0. 9. x + y = 0. 14. -^-•^ = 1. 5 8 10. y = Sx. 15. 4a; -f 5 = 0. 11. M=^- 16. 17. a;-4 = 0. 2/ + 5 = 0. 12. -M=^- 18. Sx-^5y = 2. 13. i-.=i. 19. 20. y-l = 3{x-2). y+S = -5(x-l) EQUATIONS IN TWO OR MORE UNKNOWNS 51 40. Graphical solution. — It is proved in Analytic Geometry that the locus of every equation of the first degree in two variables is a straight line. It is because of this fact that equations of the first degree are called linear equations. Any pair of values of the unknowns that satisfy each of two linear equations in two unknowns must be the coordinates of a point on the locus of each equation ; that is, must be the coordinates of the point of intersection of these loci. Hence, since two straight lines intersect in not more than one point, two linear equations in two unknowns cannot be satisfied by more than one pair of values of the unknowns unless they are dependent. (See § 42.) Thus we see that although a single linear equation in two unknowns does not describe two par- ticular numbers, two such equations taken together do give us such a description, provided that they are not dependent and not inconsistent. (See § 42.) The values of the unknowns that satisfy two linear equa- tions can be found by plotting the loci of these equations and measuring the coordinates of their point of intersection ; but results obtained in this way are only approximate since the locus of an equation cannot be drawn with perfect accuracy and since the coordinates of a point of intersection of two loci cannot be measured with perfect accuracy. 41. Algebraic solution. — These values of the unknowns can also be found without reference to the loci of the equations. We shall illustrate two ways of doing this : 1. Solution by the method of substitution. rind the numbers described by the equations 3a;-4y^7, (1) 7x + 5y=9. (2) Solve (1) for x, x = '^-^^ . (3) 3 52 COLLEGE ALGEBRA Substitute this value of x for x in (2), 3 Solve (4) for y, 49 + 28 ?/ + 15 ?/ = 27, 43 2/ =-22, ^ 43 Substitute this value of ?/, for ?/ in (3), gg "^"43 ^= ^ = _71 43 (4) Check. — From (1), 3.71-4f_22U§01 = 7. 43 \ 43; 43 From (2), 7 • ^ + 5 f - ^^^ = §^ = 9. ^ ^' 43 V 43; 43 From a study of this example, the student should formulate a general rule of procedure. 2. Solution by the method of addition or subtraction. Multiply each member of equation (1) by 5 and each member of (2) by 4. 15a;-202/ = 35. (6) 28 a; + 20 ?/ = 36. (6) The sum of the left members of (5) and (6) equals the sum of the right members. Hence, _ "^-43- Substitute this value of a; for x in (1), 4b ^ 213 - 172 ?/ = 301, 172 2/ =-88, The check is the same as in the other method. From a study of this example the student should formulate a general rule of procedure. EQUATIONS IN TWO OR MORE UNKNOWNS 53 42. Inconsistent and dependent equations. — Sometimes two linear equations in two unknowns have no common solution. This is the case when the loci of the two equations are parallel lines. The student should verify that 3 aj _|_ 4 2/ = 1, and S X -^ iy = 5 are two such equations. Such equations are said to be inconsistent. Sometimes two equations convey the same information con- cerning the relations of the unknowns. Such equations are said to be dependent. Two dependent equations have the same locus. The equations 3 a; -j- 4 2/ = 1, and 6x-{-8y = 2 are dependent equations. The second one gives no information concerning x and y that is not given by the first one. EXERCISES Solve the following pairs of equations graphically and also by one of the other methods. See that the results agree approximately. 1. 2x-y = l, 3x-\-2y = 12. 2. x-^y-\-S = 0, 4,x-5y + 3 = 0. 3. 5x + 2y=2, 7x-5y = S4:. 7. 2x-Sy-\-4t = 0, 5. X 3 -y=- 5 -1, 4^2 = 1. a 3; x+4:y + 5 = 0, 6; x+8y = — 10. 3y-2x-2=0. 3-4:X = y, 2 ' 4 . ' Sx + 2y = 6, 4. 2x-^3y = 6y «_l_^^l^ 8. 3-4:X = y, 54 COLLEGE ALGEBRA x — 5 y-4: = -8" 12. 3x-\-4.y-l=z 0, 12x — 5y-^6 = = 0. 13. 5x-2y + ll = = 0, e-i.=y^. 14. a b ax -\-by = c. 10. ?i±i_^jz^ 6 -9 '. 11. x + y + l = (), The lines of Ex. 14 cannot be plotted until definite values are assigned to a, h, and c. 17. ^+y=i, m n _E ^ = 1. 4 m 5 n In solving the following equations do not clear of fractions, but solve for - and -. These equations are not linear and their X y loci are not straight lines. Do not try to solve them graphically. 1. ax -\- by + c = 0, 16. ^ + 1 = 1, Ix 4- my + 76 = 0. a b ^-^ = 1 a b 18. X y 19. r 9 „ 20. « + ^ = e, X y 12+4 = 1. a; 2^ - + - =/• a; 2/ 43. Solution by formulae. — Consider the two equations a^x + hy = Ci, (1) ttgic -{- b^ = Cg. (2) The student should verify that if «i&2 — «2&i ^^ the numbers described C162 — C2&1 0162 — 0,2^1 by these equations are h — r h and 2/ = ^^"^ - ^^^^ ai&2 — Cf20l EQUATIONS IN TWO OR MORE UNKNOWNS 55 These formulae can be easily remembered by means of the following device : The symbol ^' ^' the cross products 0^)2 and a^b^. That is, will be used to represent the difference of CI2 62 tti^a — 0t2^i' For example, 12 51 U 3 6 _ 20 = - 14, and = i-(-56) = 56i. This symbol is called a determinant of the second order. The values of x and y which were obtained as the solution of the preceding equations can now be expressed as follows : X = Cl 61 ai Cl C2 62 , y = a^ c-z ai bi «i &i a2 62 a2 h2 44. We observe that the two denominators are the same determinant. The numbers in the first row of this deter- minant are the coefficients of x and y respectively in the first equation, and the numbers in the second row are the coefficients of x and y in the second equation. The numer- ator of the fraction that gives the value of a; is the same as the denominator with the exception that the coeffi- cients of X (the numbers in the first column) have been replaced by the constant terms of the respective equa^ tions ; and in the fraction that gives the value of y the numerator is the same as the denominator with the exception that the coefficients of y (the numbers in the second column) have been replaced by the constant terms of the respective equations. By means of these formulae we can write down immediately the solutions of any pair of consistent, independent, linear equations in two unknowns. 56 COLLEGE ALGEBRA For example, the values of x and y that satisfy the equations 7x+ 12?/ = 3 and a; - 2 w = 5 X = and 3 12 6 -2 _ -6-00 - 14 - 12 33 7 12 "13' 1 -2 7 3 1 5 85- 3 - 14 - 12 16 7 12 13 1 -2 The student should observe that these formulae apply only when each equation is in the form ax -\- hy = c. If it is not in this form originally, it must first be reduced to this form. If one of the letters does not appear in an equation, it should be regarded as present with the coefficient zero. (See Ex. 11.) In solving Equations (1) and (2) of § 43 we get and {afii — ajb^)x = C162 — C261 The common coefficient of x and y in these equations is the de- terminant in the denominators of the formulae, while the right members of the equations are the determinants in the numera- tors of the formulae. Hence if in these formulae the determinant in the denomina- tors is zero, while those in the numerators are different from zero, the equations are inconsistent. If all the determinants are zero, the equations are dependent. EQUATIONS IN TWO OR MORE UNKNOWNS 57 EXERCISES Solve the following pairs of equations by means of the formulae : 10. Sx — 5y = 2, 11. 5 05 + 19 2/ = 14, 3 a; = 2. 12. kx -\- ly -\- m = Of 2x — Sy = 4:m. 13. a(x — y)-\-b(x-\-y)=b + c, a{x -\-y)— b(x — y)=:b — c. 14. ^- + f = 1, a b ^ + ^ = 1. 2a 46 15. (a-\-b)x-\-(a — b)y = Sab, (a — b)x + (a -f b)y — ab. 16. ax -{- by -\- c = Of ax — by — c = 0. 17. Ix + my = n, 2 Za; + 3 my = n. 18. 3 « + 1 = 5, x-\-y = 2. 19. a(aj + 2/) + 6(aJ - y) = a, (a 4- 6)a? + (a — 6)?/ = 6. c a 1. 9x-\-y = 5, 3x-\-ny = ll. 2. 5 + 7"^^ ^ - -^ = 1. 3 4 3. 2/ - 5 = 4 a;, 2/-3=:7(a: + l). 4. .v-|=H^ + i). y-i = H^-i)' 5. y-T a; + 10 5 "- 3 ' a; _ 4 y-S 2^3* 6. 4a54-3y = 2, 8aj + 6?/ = l. 7. 5^2 ' 3^7 8. - + ^ = 1, 4^2 a; + 2 2/ = 4. 9, a; _ 1 _ ^ _ 2 3 -8 ' . + 1=1. 58 COLLEGE ALGEBRA 45. Two descriptions of the relations connecting three un- known numbers are not sufficient to identify these numbers. Eor example, there are many sets of values of x, y, and z that answer the two following descriptions : 2a;+32/ + 2 = 4, (1) -4a;4-5y + 62; = 7. (2) The student should make a table exhibiting at least five sets of values of X, y, and z that answer these descriptions. How many more such sets of values are there ? There is however only one set of values of the unknowns that answer three descriptions of the first degree of the rela- tions connecting them. Consider, for example, the description x+y+z^Z, (3) in connection with (1) and (2). If we multiply each member of (1) by 2 and add each mem- ber of the resulting equation to the corresponding member of (2), we get 11 2, + 8. = 15. (4) Then if we multiply each member of (3) by 2 and subtract the products from (1), we get y-z = -2, (5) Now every set of values of x, y, and z that satisfy equations (1), (2), and (3) must also satisfy equations (4) and (5). But the only values of y and z that satisfy (4) and (5) are Moreover if y and z have these values, we must have a? = -f^ in order that (1), (2), and (3) be satisfied. Hence there is only one set of values of a?, y, and z that satisfy equations (1), (2), and (3). It was not necessary to proceed in exactly this way in order to get the solution of the given equations. We might have EQUATIONS m TWO OR MORE UNKNOWNS 59 chosen a different unknown for the first elimination and ac- cordingly have combined the original equations differently. But the same variable must always be eliminated in the first two eliminations. The work of solving a system of three linear equations in three unknowns can be checked, as in the case of equations with two unknowns, by substituting in the equations the values found for the unknowns. 46. Inconsistent and dependent equations. — It may be that there are no values of ic, y^ and z that satisfy all three equa- tions. Consider, for example, the following equations : 2x+3y+5r = 4, (1) x-\.y + z = ^, (2) 4 X + 6 y + 2 = 5. (3) If we multiply each member of (1) by 2 and subtract the products from the corresponding members of (3), we get = -3. This obvious untruth was obtained on the supposition that there is a set of values of cc, y^ and z that satisfy both (1) and (3). Hence the sup- position must be wrong and (1) and (3) are inconsistent. Exercise. — Explain in detail where the existence of a set of values of oj, y, and z satisfying (1) and (3) was assumed in getting the result =— 3. Sometimes the three equations are such that one of them gives no more information concerning the relations connecting the unknowns than is given by the other two equations. The equations 2 x -|- 3 y + 4 2; = 5, (1) - « + 2 y - 3 ;? = 2, (2) x^hy + z = 'J, (3) form such a set, inasmuch as the two members of (3) are the sums of the corresponding members of (1) and (2), and any set of values of a:, y, and z satisfying (1) and (2) must also satisfy (3). The student should give to z in (1) and (2) any convenient value and solve the resulting equations for x and y. He will then find that these values of a:, y, and z satisfy (3) . 60 COLLEGE ALGEBRA Such equations are said to be dependent. If each of the equations contains information not contained in the others, the equations are said to be independent. Note, — If the student has studied solid geometry, he maybe inter- ested in the fact that an equation of the first degree in three variables represents a plane, just as an equation of the first degree in two variables represents a straight line. The loci of two inconsistent equations are parallel planes. The loci of three dependent equations are three planes through the same line. EXERCISES ' Solve the following sets of equations and check your results : 1. 2x-^Sy-z = 5, 8. A.B.O o 3 + 2+^-^' 2. 4:X + 2y + 2z = U. 2x-4:y-6z = -6, 2^5-^3-^' Sx-\-5y-\-2z=lS, — x-{-2y-\-z = 0. ABC 63 5 3 2 5 * 3. Tx-\-5y-\-Sz = S, 1 1 , — 6x-{-2y-^oz = 6j 9. X y 14 a; + 10 2/ + 16 ^ = 4. 4. a; + 2/ + l = 0, y z y + z = 4, 1 1 z-{-x=l. Z X 6. -x-y + 2z=:12, 4:X-Sy-2z = 10, Solve for -, -, and i X y z 2x + 4:y + Sz = 3. 6. ax-^by = 1, cy-\-dz = 1, ez-\-fx = l. 10. ^ + B+0 = 3, A-\-B-C=2, A-B+C=5. 7. 4.A-5B + 6C=23, 11. A-{-B==0, 2A-\-7 B-{-3C=A0, B+C=l, 9^ + 2^ + 8(7=61. C + A = 3. EQUATIONS IN TWO OR MORE UNKNOWNS 61 12. 14^ + 20/+c = -149, 17. 3^1 + 2 5 + 4 0+4=0, 16 g + 18/+ c = - 145, 2A-5B-G = 20, 65^ + 4/- c = 13. ^ + 5- (7=0. 13. c = 0, 18. Ax-3y-4.z = 0, 2/+c + l = 0, 5a; + 62^ + 102; = 7, 2gr + c + l = 0. x-\-2z=^l. 14. 5 + 2^ + 4/+c = 0, ^^- 12x + 5y + Sz^-U, 5 + 4^ + 2/+ 0=0, Ta; + 2/ + 2;+| = 0, 15. 16. 13+6^-4/+c = 0. ^ + 2^ + 2. = -3. 20. X + 2/ + 2;=3, X -2y + z=:0, 3 x-\-4:y — z = 6. X 2 + ! + . = !, X -M=^' - ■i-^+l=^- tt 6 c a c a c -1. aiX-\-biy + c^z = = ^1, a^ +60^ + ^22: = = ^2, «3aJ + &32/ + C32; = = t^3. 21. 47. Solution by formulae. — The equations in Ex. 21 of the preceding article are typical of every set of three simultaneous linear equations in three unknowns, since any such set can be obtained from these by assigning proper values to ai, b^ Ci, c?i, <^2> ^2? — 026103 , . _ did^cz + g2 Q V A / N PQ = V(5 - 2)2 + (4 - 2)2 = V13. In general, if P represents the point (x^, y^) and Q the point (^2j 2/2)? RQ = X2 — Xi and RP = 2/1 — y^- Hence, PQ = ^(x, - a^O' + (2/1 - 2/2)= = V {X, - x,y + (2/2 - 2/1)'. If we represent the distance PQ by d, we have the formula This important formula enables us to find the distance be- tween two points when the coordinates of these points are given. 66 COLLEGE ALGEBRA EXERCISES Find the distances between the following pairs of points : 1. (1,3), (2,5). 3. (-2, -7), (-3,6). 2. (0,0), (4, -3). 4. (10, 0), (0, 2). 6. (8, -3), (-3,8). 6. Show that the quadrilateral whose vertices are the points (^> 2), (—4, 2), (—4, —5), and (3, —5) respectively has equal sides. 52. The equation of a circle. — Every point in the circum- ference of a circle whose radius is 5 and whose center is at the point (2, 3) is 5 units distant from the center. Hence if x smd y are the coordinates of such a point, we have (x-2y + (y-Sy = 25. If X and y are the coordinates of a point inside the circle, ^^^"^ (a^-2)2 + (2/-3)2<25; and if they are the coordinates of a point outside the circle, ^^^"^ (x-2y + {y-Sy>25. The points in the circumference of this circle are therefore the only points whose coordinates answer the description given by this equation. We say accordingly that this equation is the equation of the circle and that the circle is the locus of this equation. EXERCISES Write the equations of the following circles, and draw the circles : 1. Radius 3 and center at (—2, 1). 2. Radius 1 and center at (0, 0). 3. Radius 4 and center at (4, 0). 4. Radius 6 and center at (0, 6). EQUATIONS IN TWO OR MORE UNKNOWNS 67 5. Radius 2 and center at (2, 2). 6. Radius 5 and center at (— 3, — 4). 7. Radius 3 and center at (5, — 2). 8. Radius 4 and center at (0, 0). 9. Radius 7 and center at (5, 3). 10. Radius 3|- and center at (7, — 6). 53. The equation (x - hf + (y- ky = r^, which is the equation of the circle with the radius r and the . center at the point {h, k), can be written in the form ic2 + 2/' - 2 /ix - 2 % + 7i2 -f A;2 - ?^ = 0. This is equivalent to the form x^ + y' + 2gx + 2fy-\-c = 0, if we let 2 gr stand for — 2 /i, 2/ stand for —2 k, and c stand for h^-\-k''- r". The equation of every circle can be written in this form. We can take advantage of this fact to find the equation of the circle that passes through three given points not on the same straight line. For example, suppose that we want to find the equation of the circle that passes through the points (7, 10), (8, 9), and (—3, — 2). The equation must be in the form x'^-\-y2 + 2gx-\-2fy + c = 0, and it must be satisfied by the coordinates of each of these points. Hence, 49 + 100 + 14^^ + 20/+ c = 0, 64 + 81 + 16 gr + 18/4-0 = 0, 9 + 4-6sr-4/+c = 0. Or 14 5r + 20/+c=-149, 16p + 18/+c=-145, 65r + 4/- c = 13. 68 COLLEGE ALGEBRA This is a system of three linear equations in the three unknowns, gr, /, and c. If we solve these equations by either of the methods already considered, we get j, = _ 2, / = _ 4, c = - 41. Hence the equation of the circle sought is The student should verify directly that this equation is satisfied by the coordinates of the given points. PROBLEMS 1. On the Centigrade thermometer the temperature of melt- ing ice is marked 0° and that of boiling water 100°. Write down the equation giving the relation between the reading of the Centigrade and the Fahrenheit thermometers, and give a graphical means of changing from one of these readings to the other by drawing the locus of this equation. 2. The following temperatures (Fahrenheit) were reported by the New York Swi for Jan. 16, 1912. Degrees Degbeks 8 A.M 8 9a.m 2 10 A.M 6 11 A.M 10 Noon 12 Ip.M 11 2 p.M 13 3p,m 13 4 p.M 13 From the figure of the preceding problem measure off the corresponding temperatures on the Centigrade thermometer. 3. A train passes a station at the rate of 10 miles an hour. Half an hour afterwards another train follows it at the rate of 12 miles an hour. Determine graphically how long it will take the latter to overtake the former. Let y = number of miles traveled by the first train in x hours. Then y = 10 X. The locus of this equation is the line marked (1) in the figure. 1 A.M .... 10 2 A.M .... 8 3 A.M . ... 8 4 A.M .... 6 5 A.M .... 4 6 A.M .... 4 7 A.M . ... 3 7.30 A.M . . . . 2 EQUATIONS IN TWO OR MORE UNKNOWNS 69 It represents graphically the rela- tion between the distance trav- eled and the time of traveling of the first train. When the first train has traveled x hours, the second one has traveled x — ^ hours. If now we let y = no. of miles traveled by the second train, y = 12(x — I). The locus of this equation is the line marked (2) in the figure. The abscissa of the point of intersec- tion of these two loci represents the number of hours from the time of departure of the first train to the time of meeting. 4. Determine graphically how long it will take an automo- bile going 25 miles an hour to overtake another one that had 15 miles start and is going at the rate of 16 miles an hour. 5. Determine graphically where two trains will meet if one of them starts from Detroit for Chicago at the rate of 30 miles an hour at the same time the other one starts from Chicago for Detroit at the rate of 40 miles an hour. The distance from Detroit to Chicago is 285 miles. 6. A man rows 9 miles downstream in 2 hours and returns in 4 hours. How fast does he row in still water, and what is the rate of the current ? In solving this problem assume that the rate downstream is equal to the rate of rowing in still water increased by the rate of the stream, and that the rate upstream is equal to the difference of these two rates. 7. A and B working together do one fifth of a piece of work in a day. After they have worked together three days, B stops and A finishes in three days more. How long would it have taken each one to do the work alone ? 8. A glass of water weighs 14 ounces. When full of sulphuric acid of specific gravity 1.35, the glass weighs 70 COLLEGE ALGEBRA 17.5 ounces. Eind the weight of the glass when empty and the number of ounces of water it holds. See the Appendix for an explanation of specific gravity. 9. A certain vessel weighs p pounds when full of a sub- stance of specific gravity Si, and it weighs q pounds when full of a substance of specific gravity So. What is the weight of the vessel, and how much of each substance does it hold ? 10. Two weights of 25 pounds and 30 pounds respectively balance when resting on a lever at unknown distances from the fulcrum. If 5 pounds are added to the first weight, the other one must be moved 2 feet further from the fulcrum to main- tain the balance. What was the original distance from the fulcrum to each of the weights ? 11. Two weights of w^ and Wg pounds respectively balance when resting on a lever at unknown distances from the ful- crum. If p pounds are added to the first weight, the other one must be moved / feet further from the fulcrum to maintain the balance. What was the original distance from the fulcrum to each of the weights ? 12. A weight of 20 pounds is placed at random on a beam of unknown length that is supported at its ends. It is found that this produces a pressure of 15 pounds on the support at one end in addition to that caused by the weight of the beam. It is also found that the same pressure is produced on this support by a weight of 25 pounds placed 2 feet further from this end. How long is the beam ? 13. A weight of p pounds is placed at random on a beam of unknown length that is supported at its ends. It is found that this produces a pressure of a pounds on the support at one end in addition to that caused by the weight of the beam. It is also found that the same pressure is produced on this support by a weight oi p-\-n pounds placed m feet further from this end. How long is the beam ? EQUATIONS IN TWO OR MORE UNKNOWNS 71 14. Represent graphically the change in pressure borne by each support of the bridge described in Problem 16, p. 45, as the wagon moves across the bridge. Let X = the number of tons pres- sure due to the wagon on the first support. Let y = the number of feet from the first support to the wagon. Then 26 x =2(25 -y), or 25x+2y=50. The point on the locus of this equa- tion whose ordinate is the distance of the wagon at any moment from the first support has for abscissa the pressure in tons on this support at this moment. 15. A beam 18 feet long is supported at its ends. Repre- sent graphically the change in the pressure on each support as a weight of 150 pounds moves across the beam. 16. A locomotive weighing 600 tons moves across a bridge 70 feet long that is supported by stone abutments at the ends. Represent graphically the change in the pressure on the two supports. Find the equation of the circle that passes through each of the following sets of three points. Draw each circle. 17. (3,2), (1,4), (-2,1). 18. (0, 0), (3, 0), (0, 5). 20. (-5, -5), (3, 3), (4,0). 21. (-3, -6), (2,7), (4,8). 19. (-2,1), (3,4), (4, -2). 22. Find the distances from the vertices to the points of contact of the circle inscribed in the triangle whose sides are 9, 14, and 18 respectively. 72 COLLEGE ALGEBRA 23. Find the distances from the vertices to the points of contact of the circle inscribed in the triangle whose vertices are(-21, -22), (11,2), (-14,2). 24. Three circles are drawn with centers at the points (_26, -39), (46, -9), and (-10,24) respectively and such that they are tangent to one another externally. What are the equations of these circles ? 25. Find the equations of the three circles that are tangent to one another externally and have their centers at the points (-16, -6|), (5, 2^), and (-5, 12). 26. In the triangle ABC, AB — 5 inches, BC = 8 inches, and CA = 10 inches. What must be the radii of three circles with centers at A, B, and C respec- tively in order that those whose centers are at B, C shall be tan- gent to each other externally and shall each be tangent internally to the one whose center is at ^ ? 27. In the triangle ABC, AB = c, BC=a, CA=b. What must be the radii of three circles with centers at A, B, and C respectively in order that those whose centers are at B £tnd C shall be tangent to each other externally and shall each be tangent internally to the one whose center is at -4 ? CHAPTER VI FRACTIONAL AND NEGATIVE EXPONENTS. RADICALS 54. The fundamental laws of exponents. — The funda- mental laws in the use of positive integral exponents are ex- pressed by the formulae : I. «"• . a" = a*"+". II. a"^ -i-a'' = a'"~". (When n is less than m.) III. {a^^Y = a"*". IV. (ab)"' = a'^b"'. \bj 6- 55. Fractional and negative exponents. — Now heretofore we have considered only positive integral exponents, since the purpose of an exponent was to indicate how many times a number was to be taken as a factor. It turns out however to be possible to use fractional and negative exponents in a simple and natural way that is also consistent with these laws. For example, if we agree that a^ shall be a symbol for a num- ber such that 1 1 a^ . a^ = a, we shall have a definition that is consistent with the first of these laws. But this is equivalent to saying that a^ is one of two equal factors of a. We accordingly agree that this symbol shall represent the same number as Va. More generally, if p and q are any two positive integers and £ we agree that a* shall be a symbol for a number such that p p a« • a* ••• (to 5 factors) = a^, 73 74 COLLEGE ALGEBRA we shall have a definition that is consistent with the first law. p But this makes a^ one of q equal factors of a^, or a gth root p of aF. We agree then that a' shall be a symbol for a number such that p _ It would be in accordance with the fundamental laws of exponents to agree that 92 =— 3. But in view of the fact that a positive number has only one positive even root and that any real number has only one real odd root it is agreed to call the positive even root of a positive number and the real odd root of a negative number principal roots. Radical signs and fractional exponents are used to indicate these principal roots. Thus, we say that 9^= V9 = 3, and - 9^ = - V9 = - 3. It should be noted that these remarks do not apply to even roots of negative numbers. (See § 133) . 56. Suppose now that x = -Va, and y = ^b. Then af = a, and ]f = b. Hence, x^y^ = ab, and (xyy = ab. (Law IV.) Therefore, xy = \Jab^ or ■V~a'^b = ^'ab, 11 1 or a^'¥ = {ciby. A direct application of this relation shows that 1 1 _ :? {cfiy - (iapy = VcP' = a*. p 1 Hence a' may be looked upon as the pth power of a' as well as the qth. root of a^. For example, 9^ = V9 = 3, 2^ = v/23 = y/S, (a - 6)^ = \/(a-by = ( ^a^^)*. Note. — The symbol ai is read " a to the power p divided by g," or *' a exponent p divided by g." FRACTIONAL AND NEGATIVE EXPONENTS 75 If a and b are negative and q is even, care must be taken 1 1 in forming the product a* • ¥. For example, we would natu- rally say that (- 1)^ • (- 1)^ = (- 1 • - 1)^ = (1)^ = 1. But, as we shall see in § 133, we interpret the symbol (— 1)^ in such a way that (— 1)^ • (— 1)^ = -1. 57. If we agree that a^ shall be a symbol for a number such that that is, if we agree that this symbol shall obey the first law of exponents, we must say that a i a" = — = 1. It is agreed that this relation shall hold for all values of a except 0. We do not give any meaning to the symbol 0^. Also, in accordance with the first law of exponents, we agree that a~'"{m a positive number) shall be a symbol for a number such that a"*" . «"» = a^ == 1. This is equivalent to saying that a number with a negative exponent or, as we shall say, raised to a negative power, shall be equal to the reciprocal of this number raised to the corre- sponding positive power, «-"' = — . a"* With this agreement, the value of a fraction will not be affected if any factor is transferred from the numerator to the denominator, or from the denominator to the numerator, pro- vided the sign of its exponent is changed. Thus, a.2 ^ a;2?/-3 ^ 1 . {x + 2 yY(x - 3 y)^ ^ (x + 2 yY{x -3 y)5(a-6)-3 ^3^4 z^ x-^yH^ ' (a - 6)3(a + 4 6)2 (a + 4 6)2 = (x + 2 yy^x - 3 yy(a - b)-^{a + 4 6)-2. Vi 76 COLLEGE ALGEBRA The student is cautioned that this applies only to factors of the numerator and of the denominator, and not to separate terms. Thus, ?!-+l! is not equal to -^ or to ^* 3^'^ y-a + z^ 58. We have agreed upon a use of fractional, zero, and negative exponents that is consistent with the first law of ex- ponents. It can be shown that this use of these exponents is consistent with all five of the fundamental laws of exponents. It is even possible to use irrational exponents in a way con- sistent with these laws. Hence we can consider these laws, as given in § 54, as applicable to all real exponents. This use of fractional, zero, and negative exponents merely gives us new ways for indicating operations with which we were already familiar. If we were to use these exponents in a way not consistent with the fundamental laws of exponents, the usefulness of algebra as a means for getting information would be seriously impaired. EXERCISES Write each of the following expressions without the use of the radical sign : 1. Vl9. 4. ^■^J{x^ryf. 2. v'24. 5. \V^fz-\ 3. V3l 6. ^{x-Zyf{^x^^y). 8. i#. 36\62 9 4/ (4 a + 3 6)^(2 g-r^y^ • \ {x-Vyfi?x-2yf 10. vn. FRACTIONAL AND NEGATIVE EXPONENTS 77 Write each of the following expressions without the use of fractional or negative exponents : 11 A. 14. 8~i ■ ^* X5. (-*)-f. 12. x^y-i 16. S^V^-l 13. 3^xi 17. G\)i 18. (o^-{-2x'y-5xy^-{-3f)i 19. r i._V^ . • 20. (2a;-3 2/)^(3a;-2 2/)^(2a; + 3 2/)'^. Write each of the following expressions without negative exponents and in as simple a form as possible : 21. (ff)^. 22. (.64)-l 23. it)-'- 24. 8456". 25. a + a-i a — tt~^ 26. 2 — 2-2 2 + 2-2 27. 1 X-' + 2/"' Multiply : 28. 7x-*. 29. (7x)-\ 30. (64 a-'b'c^yi 31. 2 32. 33. 1 , (e^-e-y ^ ' 4 34. a^ + a^6^ + 5 by ci^ _ ^i. 35. Va^ + Vi/* by a;^ — 2/^. 36. m^ + m^n^ + n^ by m^ — n^. 78 COLLEGE ALGEBRA Divide : 37. a^-\-b^hj a^ + bK 38. 16 m2 - 81 n^ by 2 m^ - 3 riK 39. 15a^-ah^-QbhyS-^a-2Vb. Expand : 40. (a^ — x^y, 41. (a^ — x^^y. 59. Radicals. — A radical is any algebraic expression of the form -y/a, or b-y/a, where n is a positive integer. In such expressions a is called the radicand, ri the index, and b the coefficient. A radical is said to be in its simplest form when : (a) The radicand is an integer, or a polynomial with integral coefficients if it contains any literal terms. (b) The radicand contains no factors of the type indicated in (a) raised to powers equal to, or greater than, the index of the radical. ^^ ^ -"■ (c) The radicand is not a power whose index has a factor in common with the index of the radical. Thus the radicals V|, V(a — a)*, and Va^ are not in their simplest form. 60. Simplification of radicals. — Certain radicals can be sim- plified by means of one or more of the following reductions. But not every radical can be made to assume the simple form described in the preceding article. (1) Reduction of a fractional radicand to the integral form. This reduction can always be performed as follows : if the radicand is not a single fraction in its lowest terms, put it into this form. Then, if the radical is of index p, make the denominator of the radicand a perfect pth power by multiply- ing the numerator and the denominator of the fraction by a properly chosen expression. The original radical is equal to FRACTIONAL AND NEGATIVE EXPONENTS 79 the pth root of the resulting numerator divided by the pth. root of the resulting denominator, which is rational. Thus, J^ = Jl = ^, ' \3 ^9 3 and in Pla P labP-^ K/abP-^ general, ^'^ ^ ^^^^^ = voo^^ (Fundajnental Law V.) ^b ^ bp b (2) The removal of factors from the radicand. This reduction can be made only when the radicand con- tains factors to powers equal to, or greater than, the index of the radical. The following examples illustrate the method of procedure and are based upon Fundamental Law IV. Examples. VS = \/4T2 = Vi • V2 = 2 V2. ^9. 18. Which is the greater, VlO or -^28 ? V3 or -v^6 ? Vl9 or -y/eE? Perform the following indicated multiplications : 19. 3V45.4V72. 21. -^^ • ^f 20. 6ax^'5ax~K 22. V(a - 6)^ Va + 6)1 FRACTIONAL AND NEGATIVE EXPONENTS 83 23. (Va-Va^)^ 24. (10 - V5) (5 + VlO). 25. ( V2 + V3 + V5) ( V 2 - V3 - V5). 26. { /281 (-11±V281)(- -19±^ /281) _196T12V281 490 --f 30V281 _2(98qF6>/281) _ 6(98=F6V281) .2 5* 90 COLLEGE ALGEBRA EXERCISES Solve the following quadratics by means of the formula: 1. x'-{-Sx-^l = 0. 2. 2y^-5y-3 = 0. x + 1 x' — C) 4. 3 4 5. Ix^ 4- wx + n = 0. 6. mx"^ — 3 a; 4- c = 0. 7. x'^ + ax + b = 0._ 10. 9. 42/2_5?/ = 0. 10. 7n~ — 6 m 4- 5 = 0. n. (8m-6)2 = 256m2. 12. A:a;2 _ 3(A: + l)a^ + 5 = 0. 13. 9(A; + 1)2 = 20 A:. 14. (^-1)2 = 3(2 77-3). 15. 64m2-256(l4-m2)=:0. 8. Sx-\-6x 7 -\-x ~2" ' i -\- X 16. {mx-\-ky = 6x. 18 (x±5y (S^-J^^^ 9 ^ 16 + 3(p2^0. 17. x2 4- 4. 27. 28. 16 19. 17y + 4: = Sy\ 20. a;2-16 = 0. 21. (o2/ + 2)2 + 2/2 = |. 22. bx'^ + bx = l. 23. (aj-2)(7a;-l) = 5. ;£_-2)(mi»-l)=5. 30. x'^ — Sx- m{x -{-2x'^-i-^) = 5x^-{-3. 31. Show that aa^ -\-bx-{-c = a( X 25. cp2_^irtV + 8maj + 64 = 9. 26. mV + 8ma;+64 = 6a;. - (^^±^ = 1. = 1. -' 16 9 x^ (3X + 5y 9 29. 6a;2_|_(^2^_' = i2. 6 4_V62-4 2a — b — -Vb'-^ — 4 ac 2 a 71. Nature of the roots of a quadratic. — Certain things about the roots of a quadratic equation can be determined without solving the equation. If the equation is in the form aa;2 4- &x + c = 0, QUADRATICS 91 the roots are —b± -\/¥ — 4:ac 2 a Hence, when a, b, and c are rational numbers, I. lfb'^ — 4: ac = 0, there is oyily one root ; namely, — In this case it is customary to say that there are two equal roots. II. Ifb^ — 4: ac is the square of a rational number different from zero, the roots are real, unequal, and rational. III. If b^ — 4: ac is positive and not the square of a rational number, the roots are real, unequal, and irrational. IV. If b"^ — Aac is negative, the roots are imaginary and un- equal. The expression 6^ — 4 ac is called the discriminant of the equation. Thus the roots of the equation are real, distinct, and irrational, since its discriminant equals 13. If we call the two roots of the equation ax^ + 6.7; H- c = a?! and x^ respectively, so that 0^1=- -b-^-Vb'- 2a 4 «c , ^ _-b- Wb^- 4 ac 2a i x^ 4- x^ = = -^^, and a .. ^ _ ^" - (&' - 4 ac) _ c 4a^ a nee, the sum of the roots of the equation ax' + bx-^c = is equal to minus the coefficient of x divided by the coefficient of x^, and the product of the roots is equal to the constant term divided by the coefficient of x^. (See § 150.) For example, the sum of the roots of the equation is — I and the product is ^. 92 COLLEGE ALGEBRA The student should verify this statement by solving the equation and forming the sum and the product of the roots. If the equation is not in the standard form, it must be reduced to this form before this rule is applied. Consider, for example, the equation 3 cc2 + 1 z= 5 a; - 4 ic2. Transposing, 7 x^ — 6 x -{- 1 = 0. Hence, the «um of the roots is ^ and the product is ^. EXERCISES Determine without solving the equations the nature of the roots of each of the following equations. Also find the sum and the product of the roots of each equation : 1. 2aj2H-7x4-l = 0. 8. x' = 10x-25. 9 10. x' + 6x-\-9 = 0. 3. x''-x-\-2 = 0. 11. (^x + iy = 5(x-\-2). 4. x''-x-2 = 0. 12. {2 + xy = {l + 2xy. 5. (Sx-\-5y = 6x. 13. 4m2-6m + 5 = 0. 1*4 9 * 15. m2 + 7 = 0. [ 7. 5x^-\-ox = l. 16. x(—2x-^S) = 5. 17. x'-\-2x{2x-l) + {2x-iy-{-x-^{2x-l) + l = 0, 18. {-x-^2y + 6(-x-{-2) + 9 = 7{x-S). 19. aj2 + 3a; + 9 = 0. 20. a;^- 3 a; + 9 = 0. For what values of k will the roots of each of the following equations be equal ? 21. x^-4:kx-\-4: = 0. Here o = 1, 6 = - 4 A:, and c = 4. Hence, b^ - 4tac = IQ k'^ - 16. In order that the roots be equal, we must have 16A;2-16 = 0. QUADRATICS 93 Solving this equation for k we get A; = ± 1. That is, when k in this equation has the vahie 1 , or — 1 , the equation has equal roots. We can readily verify this result. When A: = 1, the equation is a:-4x + 4 = and 2 is the only root ; when A: = — 1, the equation is a; + 4a; + 4 = and — 2 is the only root. 22. x'-Q,kx + 12 = 0. 27. 4a;2-|-4A:aj+4A;2=3(a;-l). 23. ^-kx-^l = 0. 28. k''x'-\-2kx + l = 2x, 24. 10a^ + 6;i^. + ^^-4 = 0. ^'- (^- + 3)^ = 4(. + 2). 30. {2x + ky = 'd{x-^). 25. x^ + 2kx-\-k'^ = ^x. ,,/ , nxo 2 31 ^ (y + i) I ?/^i 26. 9a;2-f 6A:a; + A;2 = 10a;. ' 9 16 32. 3 aj2 4- 5 A;a;(a; - 5) - 2 A;2(aj - 5)^ + 5 x + 3 A:(a; - 5) +4 = 0. 72. Graphical solution of the quadratic equation. — Suppose we draw the locus or graph of the equation 2/ = a;2. The first thing to do is to find pairs of values of x and y that satisfy this equation. This can be done by taking values of x at will and computing the corresponding values of y. Some of the results are given in the following table : X -5 -4 -3 -2 - 1 -i 1 2 1 3 4 5 y 25 16 9 4 1 I 1 4 H 9 16 25 If we plot the points whose coordinates are these corresponding pairs of values of x and y and connect these points by a smooth curve, we shall have an approximate representation of the locus. If now we draw the locus of the equation y = x-^2 with reference to the same set of axes, the two loci will intersect in two points. The coordinates of these two points must satisfy both equations. 94 COLLEGE ALGEBRA or and hence the abscissae are values of x that make the right members of these equations equal ; that is, for values of x equal to these abscissae a;2 = X + 2, or x2 - x - 2 = 0. These abscissae are therefore the roots of this last equation. These considerations suggest a gen- eral method for solving a quadratic equation graphically. If the equa- tion is 9_ h c we draw the loci of the tvro equations and y = x and measure the abscissae of their points of intersection. These abscissae are the roots of the given equation. The locus of the first equation is called a parabola. The locus of the second equation is evidently a straight line. If the line and the parabola have two points in common, the quadratic has two real roots. If the line and the parabola have only one point in common, the quadratic has only one root. If the line and the parabola have no points in common, the roots of the quadratic are imagiiiary. In this case this method gives us no information concerning the roots beyond the fact that they are imaginary. 73. Find the roots of the equation a;2_3aj-f-7 = 0. QUADRATICS 95 We have the locus of the equation already drawn. If we draw the locus of y = Sx-7, we find that the two loci have no points in common and we conclude that the roots of the quadratic are imaginary. That this conclusion is correct can be verified directly by consideration of the value of the discriminant of the given equation. Find the roots of the equation a;"'* + 4 a? + 4 0. Using the same figure for the locus of the equation y = ^\ and drawing the locus of the equation y = — 4 X - 4, we find that the two loci have only one point in common. The abscissa of this point is — 2. Hence — 2 is the only root of the given equation. The student must not expect to get exact solutions by graphical methods. 74. Second graphical method. — There is another graphical method for the solution of a quadratic that is in common use. V It is sufficiently explaiued by the following example. Suppose that we wish to find the roots of the equation Draw the locus of the equation y = y:^ — X — 2. In order to do this we take values of X at will and compute from the equation the corresponding values of y. Then the smooth curve drawn through the points whose coordinates are these corresponding pairs of values of x and y is the approximate locus of the equation. 96 COLLEGE ALGEBRA K this locus has a point in common with the x-axis, the abscissa of such a point is a value of x that makes a;2 - ic - 2 = 0, that is, is a root of this equation, since the ordinate of every point on the ar-axis is 0. In general, in order to sol^re the equation by this method, plot the locus of the equation y = ax^ -\-hx-\- c, and measure the abscissae of the points this locus has in com- mon with the ic-axis. These abscissae are the roots of the given equation. EXERCISES Solve each of the following equations graphically. Do not use the same method for all of the equations. 1. x^-^x-l = 0. 11. iB2-8a; + 16 = 0. 2. ar^_a;-l = 0. 12. ic^ _ g a; + 20 = 0. \'^^ / 3. 2x' + x-2 = 0. 13. a;2_8a; + 6 = 0. ^4. ar + 3a;-|-l = 0. 14. 3a;2 = 0. 5. 5a^ + 7a:-2 = 0. 15. a:^^! 6. 5x'-^-lx-^2 = 0. 16. a.-2 + 5a; + l=0. 7. 2a52 + 8a; + 8 = 0. 17. a;2 + 3a; + 2 = 0. 8. iB2 + 6a? + 10 = 0. 18. («4-4)2=:9. 9. 3a^ + 8a; + 5 = 0. 19. (3a;-8)2 = 0. 10 8a^ + 3a;-5 = 0. 20. 3 aj^-f- 2 » + 1 =0. 75. Equations involving fractions. — When we clear an equation of fractions by multiplying each member by the lowest common denominator of the fractions, the resulting QUADRATICS 97 equation may have roots that are not roots of the original equation. In other words, the new equation may not describe the same numbers as are described by the original equation. Consider, for example, the equation- 2x 1 + 4. x2 _ 9 X- Clearing of fractions, 2 a; = x + 3 + 4(x2 - 9), (x_3)-4(rK2-9) = 0, (a;_3)(l-4a;-12) = 0. One of the roots of this equation is 3. But this evidently cannot be a root of the original equation, since for aj = 3 the left member becomes ^, and this has no numerical value. 76. In many cases the roots of the new equation — that is, the equation resulting from clearing the original equation of fractions — are the same as the roots of the original equation. Some of the circumstances under which this can occur may be inferred from the following theorem : Theorem. — Any root of the new equation that is not a root of the original equation must when substituted for x make some denominator of the original equation zero. Suppose that the equation has been brought to the form by transposing all the terms of the right member to the left member. Here P represents some fractional expression. If D is the lowest common denominator of the fractions oc- curring in P, the new equation is PZ) = 0, and PD is a polynomial in x. If now any value of x makes PD equal to zero and D different from zero, it must make P equal to zero. That is, it must be a root of the original equation. Hence any root of the new equation that is not a root of the original equation must be a root of the equation 7> = 98 COLLEGE ALGEBRA But any root of this last equation must, when substituted for X, make some denominator of the original equation zero. (See 10, Chap. I.) It follows from this that when no denominator of the origi- nal equation contains the unknown, the roots of the new equa- tion must be the same as those of the original equation ; that is, the two equations must be equivalent. 77. Hence, after having solved the new equation, the student should determine by trial whether any of its roots, when sub- stituted for X, makes any denominator of the original equation equal to zero. No root of this kind is a root of the original equation. But all the other roots of the new equation are roots of the latter, and every root of the original equation is a root of the new equation. If we clear the equation of fractions by multiplying each member by a polynomial that is a common denominator not of the lowest degree, the two equations will certainly not be equivalent. EXERCISES Solve each of the following equations : 2. 3. 11. .02 aj2- 3.64 a; -2.5 = 0. 12. .01a;(.3a; + 4.5) = .34a;2_5^ 1 .1.1 2(a;-3)~a;-2 " x-k ^^ QUADRATICS 99 14. M±.-^+l = l. 16. l + ^_ + J_ = «. ar^— 9 a; — 3 a;i» — la? — 2a; 15. ^-^ I ^-H-g^ct 17 _3 ^ = 4. a;4-a a;— a 6 a; + l x — 1 18. ^^ + ^=1. x'-Sx + 'J x-2 78. Equations involving radicals. — Certain equations involv- ing radicals can be solved by the methods described in the following illustrative examples : It is understood that in these problems we are to attach to every radical its principal value. Example 1. Solve the equation V:c'2 - 9 - 4 = 0. Transposing — 4 to the right member, Squaring both members, Check. V25 -9-4 = 4-4 = 0. Vx'- -9 = 4. x^- -9 = 16, X2 = 25, X = ±5. In dealing with rational integral equations we checked our results merely for the sake of guarding against error. But in solving an equation with radicals we sometimes get an ap- parent root that is in reality not a root even when we have proceeded in the regular way. This also happened in the solution of fractional equations. Hence the operation of checking is a necessary part of the solution of such equations. This is illustrated in the next example. Example 2. Solve the equation Vx^=^ + 4 = 0. Transposing, Vx — 6 =—4. Squaring both members, cc — 6 = 16, X = 22. 100 COLLEGE ALGEBRA Check. V22-6 + 4 = Vl6 + 4 = 4+4=^ 0. Hence 22 is not a root of the equation. As a matter ot fact, the equa- tion has no root. The explanation of the appearance of these false roots is as follows : If the original equation is L = Ej the resulting equation is 1/ = R^, or L^-li^ = 0. The roots of this last equation are the roots of the equations L-Ii = and L-\-E = 0. And if the equation L-\- R = has a root, this is also a root of the equation that is not a root of the equation L = R, provided that it does not make L = and /? = 0. In example (1) L = Vx^ — 9 and jf2 = 4. Here the equation iy + i2 = has no root, and therefore the equation U = R^ is equivalent to the equation L=R. But in example (2) 7^ = Vic — 6 and i2 = — 4, and therefore the equation i + i? = has a root ; namely, 22. This is not a root of the original equation, since in this case R^O for any value of X. Example 3. Solve the equation Vx+l + Vx^^ = 5. Transposing y/x — 3 to the right member, Vx+ l=5-\/a;-3. Squaring both members, a; + 1 = 25 — lOVa; - 3 + a; — 3. QUADRATICS 101 Transposing, lOVx — 3 = 21. Squaring both members, 100(x - 3) = 441, x = 7.41. Check. V7.41 -f 1 + \/7.41 - 3 = V8:41 + ViAl = 2.9 + 2.1 = 5. Example 4. Solve the equation Vx2 - 10 X + 41 - Vx2 -f 10 X + 41 = 8. Transposing — Vx^ + 10 x + 41 to the right member, Vx2 _ 10 X 4- 41 = 8 + Vx-2 + 10 X + 41. Squaring both members, x2 _ 10 X + 41 = 64 + 16 Vx'^ + 10 X + 41 4- a;2 + 10 X + 41, -20x-64 = 16Vx2 + lOx + 41, 5x + 16 =-4Vx2 + 10x4-41. Squaring both members, 25 x2 + 160 X + 256 = 16 x-2 + 160 x + 656, • 9 x2 = 400, x = ±^. Check for x= \^-. ^y±o(i. _ ioo. + 41 _ Vioo _,. 20 _(. 41 ^ Vi|^_ VI^^ J/ - -V- =-8. Hence, ^^- is not a root of the equation. Check for x = — ^^. V^lM^ipTri - V^^ _ Aoo _,. 41 = 5^ - J/ = 8. Hence, — %" is a root of the equation. 79. The preceding equations contain no radicals except square roots. The most common irrational equations are of this kind and the method of solution illustrated in these ex- amples is usually applicable in such cases. It may be formu- lated as follows : If necessary, so transpose the terms that one radical stands alone in one member of the equation. 102 COLLEGE ALGEBRA Then form a new equation by equating the squares of the menv- hers of the last equation. If the resulting equation contains radicals, repeat this process until an equation is obtained that is free from radicals. Solve this last equation and test each of its roots by substitut- ing it in the original equation. In simplifying the radicals in order to determine whether a number satisfies the equation, the student must not resort to squaring both members of the equation, since the question he is trying to settle is whether the equation formed in this way is, or is not, equivalent to the original equation. EXERCISES Solve the following equations and check your results : 1. Va? -f • 2 — .'» = 0. ^ Va;^-|- 9 a;-5 _17 3. V4 a; + 13 + a; = 2. 5. ■\/x'' ■\-^x-\-2 = x + l. ^ Va^2 + 16^a;-9 t>- -, 1 z, — =\ K^ \n 7. V.t2 -f 10 a; + 41 - Vif2 - 10 a; + 41 = 8. , , \ 8. Vcc2 + 5i» + 22-Va;2-5a;4-22==2. ^ ^ \ S,' 10. V(a; + 2)2 + 3+V(a;-l)2 + 3 = 5. '^ 11. V8x + 1 = vV + 6 a; — 2. ^12. V(aJ-3)2 + 7+V(aj-|-7)2 + 7=8. '^ 13. V5a;-6-Va;-2 = 2. 14. V3a;4-10 = Va;-l+V2a;-l. QUADRATICS 103 15. 1125 = 1093 VI + .003665 t 16. 1? = 1093 VI + .003665^. Solve for t. This formula gives the velocity of sound in air in feet per second at a temperature of t° Centigrade. The formula in Ex. 15 is a particular case of this. 17. V5 + (6 - xy = 5. 18. \/9 + (4 + a?)2 = V(3 + xf + 16. 19. x — l—Wl + x = 0. 20. 2x + 3 = 4-V^. 21. ■\/{x _ 3)2 +7 - ^{x + 7)2 + 7 = 8. 22. v/2/2-2/ + l+V2/2-22/+l = l. 80. Equations in the form of quadratics. — We can frequently apply the methods for solving quadratics to the solution of equations that involve the unknown only in a certain expres- sion and in the square of this expression. Example 1. Solve the equation 3 ic* - 14 a;2 4- 8 = 0. This can be written 3 {x^Y - 14 x2 + g = 0. Hence, _ 14 ± \/l96 - 96 = 4, orf, 6 ' 3 X = ± 2, or ± Vf. The student should check these results. Example 2. Solve the equation 3 ic - 5 - 4 \/3x-5 + 3 = 0. This can be written CV3a:-5)2 _4V3a:-5 + 3 = 0. Hence, V3a: - 6 = 3, or 1, 3 X — 5 = 9, or 1, X = ^5*, or 2. The student should check these results. 104 COLLEGE ALGEBRA Example 3. Solve the equation x2 - s X -\- 5 - S y/x^ - S X + S = - 5. This can be changed into a quadratic in Vx^ — 3 x + 8 by adding 3 to each member. Thus, x2 - S X + 8 - S Vx^ - S X + 8 =- 2. Hence, \/x2 - 3 a; + 8 = 2, or 1, a;2__3a;4-8 = 4, or 1. 3±VHI,,,3±VEI2 2 ' 2 The student should check these results. EXERCISES Solve the following equations : 4. -6Va;2+l + a;2 + lO = 0. (7;) («2 _ 1)24. 3(a?2 _ i) _^ 2 = 0. 9. a^ + 26a^-27 = 0. 10. a;*- 2x2 + 1 = 0. 11. 2/' -13 2/^^ + 36 = 0. 12. 7Va;2 + 3a; + 2x24-6a;-4 = 0. 13. a;2 + 8a; + l + 3Va;2 + 8a; + 2 = 9. {< . -.15. (z'-j-Szy-2(z^ + 3z)-S = 0. 16. 2:2^-1122 + 12 = 0. 18. (2/2-5)2 = 16. 17. a;« + 7ic3-8 = 0. 19. -6 Va; + l + a; + 10 = 0. 14. lx^±]'-5fx-{--] + 6=:0. X. ''■ {h-'J-ih-'^'^'- QUADRATICS 105 81. The conditions of many problems require that the un- known number be real. If such a problem leads to a quadratic equation and if one of the known numbers is represented by a letter, we can frequently find limitations on the possible values of this known number. The method of procedure is illustrated in the following example. Find two positive numbers such that their product is 36 and their sum s. Let X = one of the numbers. Then s — a: = the other one. Hence x(s-x)= 36, or x^ - sx + 36 = 0. In order for the roots of this equation to be real we must have s^ — 144 ^ 0. Hence the least possible value for s is 12. What are the required numbers for this value of s ? PROBLEMS In solving a quadratic it is generally best to use the formula, unless the method of factoring can be applied readily. Formula from physics. — The number n of complete vibrations per second made by a stretched wire is given by the formula 1 / = 21^1 9S0M where I is the length in centi- j^ 2^ meters between the bridges, M I the weight in grams of the ^ stretching weight, and m the weight in grams of the wire per centimeter of length. 1. What must be the stretching weight on a wire 58.6 centimeters long whose weight per centimeter is .0098 gram in order that it may make 256 complete vibrations per second ? 106 COLLEGE ALGEBRA Formula from physics. — The velocity v of a projectile at any moment is given by the formula V = W — 64 2/ where Vq is the initial velocity and y is the height of the pro- jectile at this moment above the level of the starting point. 2. A projectile is fired from the ground with an initial velocity of 225 feet per second. How high will it be when it has a velocity of 150 feet per second ? 3. Let P be a point on a semicircle whose diameter is ABj and let PB be the perpendicular from P to AB. Being given ^ -^^^ that AB = 10 inches, find the length of ^^ ^\ EP Yfhen AR-^PP =7^ mohes. (PP I \ is a mean proportional between AR ^lI 1^ and RB.) 4. Given the points P and Q with the coordinates (V3, —2) and (V3, 1) respectively. Find the point A on the 2/-axis such that PA-{-AQ = 6. 5. Find the coordinates of all the points on the line y = V2, the sum of whose distances from the points (2, 0) and (—2, 0) is equal to 5. 6. Find the coordinates of the points on the line x = 3 that are 10 units from the point (9, 2). 7. A box is to be 3 feet long and 1 foot wide. How deep must it be in order to have a diagonal of 4 feet ? 8. The points A and B on the avaxis have the abscissae 4 and 12 respectively. What must be the abscissa of the point C on the aj-axis in order that the area of the circle with C as a center and passing through A shall be twice that of the circle with (7 as a center and passing through B? Explain geomet- rically why there should be two answers. 9. What must be the abscissa of C in order that the area of the first circle of Problem 8 be one fourth that of the other one ? QUADRATICS 107 10. The points A and B on the a;-axis have the abscissae — 3 and 8 respectively. What must be the radii of the two circles with centers at A and jB respectively, and tangent to each other in order that the area of the former shall be five times that of the latter? Explain geometrically why there should be two answers. 11. The points A and B on the a>axis have the abscissae a and b respectively. What must be the radii of two circles with centers at A and B respectively, and tangent to each other in order that the area of the former shall be k times that of the latter ? 12. Find the abscissae of the points in which the circle (x-2y + (y-hSy = 25 cuts the £P-axis. Draw the figure. 13. Eind the abscissae of the points in which the circle (x-f 2)2 +(2/ 4- 3)2 =9 cuts the a>axis. Draw the figure. 14. Where does the circle (,, + 5)2 + (2/ -6)2 = 16 cut the 2/-axis ? Draw the figure. 15. What must be the dimensions of a rectangular field that is to contain 3 acres and have a perimeter of 100 rods ? 16. What is the least possible perimeter of a rectangular field that is to contain 10 acres ? 17. What is the least possible perimeter of a rectangular field that is to contain 20 acres ? 18. What is the greatest possible area of a rectangular field inclosed by 200 rods of fence ? 19. What is the area of the greatest rectangle that can be inclosed by a rope 60 feet long ? 20. A man is in a boat 3 miles from the nearest point of a straight beach. If he rows at the rate of 4 miles an hour 108 COLLEGE ALGEBRA and walks at the rate of 5 miles an hour, where ought he to land in order to reach a point on the beach 5 miles from the point directly opposite him in one hour and twenty-seven minutes ? Would it be possible for him to reach his destination in one hour? 21. An open box with a square base and a volume of 392 cubic inches is to be made by cutting 2 inch squares from the corners of a piece of tin and turning up the sides. How large a piece of tin should be used ? 22. A man and a boy working together do a piece of work in 8 days. If we assume that the man could do it alone in 12 days less than the boy could do it, how long would it take each to do it ? 23. The sides of two rectangles, one of which is within the other, are parallel and equal distances apart. How far apart are the sides if the outer rectangle is 10 centimeters by 14 centimeters and is twice as large as the inner one ? Note. — The equation obtained in the solution of a problem should ex- press all the conditions of the problem if possible. But in some prob- lems there are conditions that cannot be expressed in the equation. In Problem 23, for example, the implied condition that the number sought must be less than one half the smaller dimension of the outer rectangle cannot be expressed in the equation. Hence we must apply this condition to the roots after the equation has been solved and reject any that do not satisfy it. 24. Two men rowed 12 miles upstream and back in 7 hours and 30 minutes. The current was running at the rate of 3 miles an hour. How fast would they have gone in still water ? Ex- plain the two solutions. 25. A lever is to be cut from a bar weighing 3J pounds to the foot. How long must it be in order that it may balance about a point 2 feet from one end when a weight of 16 pounds is attached to this end ? Explain the two solutions. QUADRATICS 109 26. A stone is dropped over the edge of a cliff and 12 sec- onds later the sound of its striking the ground below is heard on the cliff. How high is the cliff if the velocity of sound is taken as 1120 feet per second ? 27. Find the outer radius of a hollow spherical shell an inch thick whose volume is 125 cubic inches. 28. Find the outer edge of a hollow cubical shell an inch thick whose volume is 56 cubic inches. 29. The two legs of a right triangle are 3 inches and 4 inches long respectively. If the shorter one remain unchanged, by how much will the longer one need to be lengthened in order to lengthen the hypotenuse 2 inches ? 30. If the longer leg of the right triangle of Ex. 29 remains unchanged, by how much will the shorter one need to be lengthened in order to lengthen the hypotenuse 2 inches ? Center of gravity. — Let ^ ^ ^ A and B be two bodies of ' C ^ ' ^ a , ? „- weights Wi and W2 re- ^ spectively, and let the distances of their centers of gravity from the point in a line with A and 5 be a and b respec- tively. Then, as explained in physics, the distance from of the center of gravity of the two together is 31. A tin can open at the top has a diameter of 4 inches and a height of 6 inches. It weighs 4 ounces and its center of gravity is 2^ inches above the center of its base. How much water must be poured into it in order to bring the center of gravity down to 21 inches. A cubic inch of water weighs f J of an ounce. Explain the two solutions. CHAPTER VIII SYSTEMS OF EQUATIONS IN TWO UNKNOWNS SOLVABLE BY MEANS OF QUADRATICS 82. Case I. One of the equations is linear and the other one is a quadratic; that is, of the second degree. In this case we can proceed as illustrated in the following example. Solve the equations x^ + y^ = 16, (1) x + y = l. (2) From (2) we get y = l — x. Substituting this value of y for y in (1), we get x2^(l-x) (3) 2x2 Then from (3) Hence the pairs of values of x and y that satisfy both equations are .. i-VsI y = The student is familiar with the fact that any linear equation in x and 2/ has for its locus a straight line. He also knows that the locus of equation (1) is a circle with its center at the origin and radius equal to 4. Since the pairs of values of x and y just found satisfy both equa- 110 SYSTEMS OF EQUATIONS IN TWO UNKNOWNS 111 tions, they are the coordinates of the points of intersection of this circle and the straight line represented by equation (2). If only one pair of values of x and y satisfy both equations, the straight line is tangent to the curve. If the values of x and y that satisfy both equations are imaginary, the line does not meet the curve. Conversely, if the line does not meet the curve, we know that the values of x and y which satisfy both equations are imaginary, but we do not know from this fact their particular values. If the equations are satisfied by real values of x and i/, these values can be determined graphically by drawing the locus of each equation and measuring the coordinates of their common points. EXERCISES Solve each of the following pairs of equations. Check your work by means of graphs. The loci of the first equations in Exs. 11 and 12 are circles whose centers are not at the origin. Do not attempt to draw these loci. r 1. aj2 + 2/2 = 9, -^ 7. ic2 + 2/' = 40, \ 2. 3?-\-y'^=^, \l 8. x + y = 0, ^ 2 a; + 2/ = 5. a^ + 2/^ = 35. "3. ic2 + jy' = 16, 9. 2x = by, Sx-2y = 0. x' + y^ = 12. 4. a;2 -1-2/2 = 1, 10. x^ = 10-y% 3x-\-4:y-\-l = 0. y = x-10. 5. x''+2f = 10, 11. a;2-}-2/2-4aj-6 2/ + 9=0, y = x-l. y=x-^2. 6. a;2 + 2/2 = 25, 12. ar^ + Z- 6 a;-f 5 2/ = 2, 2x-{-4:y = S. 3a; + 22/ = l. 112 COLLEGE ALGEBRA 13. 2/2 = 6-a;2, 14. i^-9 = -y^, 2x-\-y + l = 0, 15. a;2 + 2/2 = 1, x-\-y = l. 16. a;2_|_2/2_9 = 0, y = 2. 17. ^2 + 2/2- 9 = 0, 18. i»2 + ?/ = 9, a; = 4. 19. 2/-l = 3(a; + l), iK2 _|_ 2/2 = 3. 20. 2/ + 2=4(a: + 3), 83. Equation of an ellipse. — Not all equations of the second degree in x and y represent circles. We have already met with equations of this kind that represent parabolas (see § 72), and there are equations of this kind that represent other curves. Consider, for example, the equation The student should verify that the following pairs of values of x and y satisfy this equation : a? 3-8 1 1 2 2 -1 -1 -2 -2 y 2 -2 %V2 -§\/2 §V5 -§\/5 §V2 -§\/2 §V5 -SV5 Moreover, x = ± |-n/4 — y^ and y = ± | V9 — o;^. Hence if the abso- lute value of y exceeds 2, a; is imaginary ; and if the absolute value of x exceeds 3, y is imaginary. This means that there is no point on the locus whose ordinate exceeds 2 in absolute value and no point whose abscissa exceeds 3 in absolute value. If we plot the points whose coordinates are the pairs of values of x and y in the foregoing table and connect these points by a smooth curve, we shall have the graph of the equation. This curve is called an ellipse. SYSTEMS OF EQUATIONS IN TWO UNKNOWNS 113 84. Equation of a hyperbola. — We consider next the equa- tion 9 4 Solving for y, we get ?/ = ± fVic^ — 9. Hence there is no point on the locus whose abscissa is less than 3 in absolute value. Solving for x, we get x =±l Vy- + 4. Hence there is no limitation on the ordi- nates of the points of the locus, A table of corresponding values of x and y is as follows X 3 -3 4 4 -4 -4 5 5-5-5 6 6 - 6 -6 V §V7 -SVT §V7 -§^ § -i I -s -iV^ - Vb" 2Vu -2V3 This curve is called a hyperbola. Curves whose equations are of the form xy = k are also hyperbolas, but they are situated differently with re- spect to the coordinate axes. The fig- ure gives the graph of the equation xy=l. When k is positive, the two branches of the curve are in the first and third quadrants ; when k is negative, the two branches are in the second and fourth quadrants. 85. Equation of a parabola. — The graph of the equation y' = Sx is given in the figure. The curve is a parab- ola, but situated differently with respect to the coordinate axes from the parabola men- tioned in § 72. C^-. y^ UZ / 86. Equation of two straight lines. — If the equation is such that when the right member is zero the left member can be factored into two factors of the first degree, the locus is composed of two straight lines. 114 COLLEGE ALGEBRA Consider, for example, the equation Transposing and factoring, we get (x+2?/ + 3)(3x-y + 2)=0. Now any pair of values of x and y that make either of these factors zero must satisfy the equation and are therefore the coordinates of a point on the locus. And conversely the coordinates of any point on the locus when substituted for x and y respectively must make the product of these factors zero (10, Chap. I). Hence the locus of the given equation is the same as the loci of the two equations a; + 2 ?/ + 3 = 0, Sx-y + 2 = 0. 87. • These are the simpler forms of the equations of the second degree in two variables. But there are many others. The equations of circles, ellipses, hyperbolas, parabolas, and of two straight lines are of the second degree. EXERCISES Solve each of the following pairs of simultaneous equations. (Check your work by means of graphs?) ,v 1. S-f-'. h 5. y = 4:X-\-l, y^-Sx = 0. 2/ = 3aj + l. k- 6. xy=l, 2. xy = -2, V 4:X-2y + S = 0. 2xH-2/ + l = 0. 7. x'-^6xy-h9f = - 3. 9~4~ ' ^ + ^ = 1. 3^4 2/ + 3=K^ + l)- 8. 9+16"^' 4. f = 6x, / 4 5 SYSTEMS OF EQUATIONS IN TWO UNKNOWNS 115 16 ' 9. x" 3a; + 2?/ = 0. 10. / + a; = 0, 11. 3x2/4-4 = 0, M^^ 0. 12. 4 2/^ — 5 a; = 0, i/-2 = 3(x + l). 13. 4ic^ + 5a'2/ + a: = 0, 2aj-2/ + 4 = 0. 14. ^ + ^' 3 2 1, a; + 2/+l = 0. 15 ^_l!-i 2a; + 52/-fl = 0. IS. f + |=l, ^--^ = 1 9 16 17. ar^-2/2 = 0, 3 a; - 2 2/ = 4. 18. xy — 5 = 0, y = 2x-l. 19. 7a^ — 3a;2/ + 2a; = 0, 5a; + 2/ + 2 = 0. 20. — - ^_2^ 9 4 0, 0. 21. x^ — 5y = 0, x-y = l. ^\^^ 88. It is evident that the intersections of the line 2/ = 3 a; + & and the circle r* + y^ — 10 depend upon the value of 6, since this value determines the position of the locus of the equation y = Zx-\-h. If we eliminate y between the two equations, we get a;2+(3x + 6)2=10, 10 a:2 + 6 6a: + V^ - 10 = 0. If now h has a value that makes 36 62_40(52_io)>o, 116 COLLEGE ALGEBRA the roots of this equation are real and distinct and therefore the line cuts the circle in two points. But if h has a value that makes 36 62 _ 40(&2 _ 10) = 0, this equation has only one root and the line is tangent to the circle. If finally h has a value that makes 36 62 _ 40(62 _ io)< 0, this equation has imaginary roots and the line has no points in common with the circle. Now 36 62 - 40(62 _ 10) = - 4 6^ + 400. n then - 4 62 + 400 > 0, or 62 < 100, (See § 182.) the line cuts the circle. But in order that 62 be less than 100, 6 must have a value between - 10 and 10 ; that is, — 10 < 6 < 10. If - 4 62 + 400 = 0, 6 =±10, and if - 4 62 + 400 < 0, then 62 ^ loo, and either 6 < — 10 or 6 > 10. The foregoing results are summarized in the following table : - 10 < 6 < 10 Line cuts circle in two points. 6 = - 10, or 6 = 10 Line is tangent to circle. 6 <- 10, or 6 > 10 Line has no point in common with circle. Definition. — A literal constant that occurs in the equation of a curve is called a parameter. Thus in the equation ?/ = 3 ic + 6, 6 is a parameter. If the equation of a straight line contains a parameter, we can frequently determine, as in the example just given, the values of the parameter for which the line cuts the locus of a given equation of the second degree in two points ; the values of the parameter for which the line has only one point in com- mon with this locus ; and the values of the parameter for which the line has no points in common with this locus. SYSTEMS OF EQUATIONS IN TWO UNKNOWNS 117 EXERCISES Solve each, of the following pairs of equations. Find the values of the parameter for which the line has but one point in common with the curve. Find also at least one value of the parameter for which the line cuts the curve in two points, and at least one value for which the line and the curve have no common point. 1. a;2 + 2/2 = 16, A !-.-. 11. -x2 + 2/^ = 16, y = mx-\- 5. o» y = 2x-\-h. y = 3x-{-b. ^J' ^ + / = 16, 7. if • 9 16~ ' 12. xy = l, x-^y = k. 3. ^ + f = l, x-\-y = a. 13. y'=:3x, | + -^ = 1. 8. a,'2-|-/ = 49, y = mx -f 1. 2 m y = mx -\- 7. 4. ^ + f = % 9+16"^^ 14. xy + 4. = 0, y—4:=m(x — 5). 9. y = mx. 5. f'+.^=i, 10. y = h. xy = l, 15. ^ + t = o, 4 9 ' y = 3x-{-b. y = 2x + h. y = mx-{-3. 89. Case 11. Each equation contains each unknown only to the second power; that iSj there are no terms of the first degree and no terms containing xy. In this case we first solve the equations for a? and / and then from these values get x and y. The general procedure is illustrated in the following example : Solve the equations x^ + y'^ = 16, (1) ^+1^=1. (2) 25 9 Eliminating y"^, we get 16 x^ = 175, 118 COLLEGE ALGEBRA Eliminating x and solving for y, we get By substituting these values of x and y for x and y respectively in equations (1) and (2), we see that either value of x can go with either value of y. Thus, _ 5\/7 5 V7 -5V7 -5\/7 . '^ ~A ' A 1 ~A ' ~A ' 4 4 4 4 y = Hence there are four solutions to these equations and the loci intersect in four points. EXERCISES Solve each of the following pairs of equations. Check your work by means of graphs. Draw the graphs of the equations in Exs. 1-3 with reference to the same set of axes ; also those in Exs. 4-7 ; and those in Exs. 8, 9. 1. 0^^ + 2/^ = 16, 6. a;2 + 1/^ = 4, .. ' HO. iE! + -^' = l,_. 9 25 \^-^^ + t = l. ' ' 25 ^ 9 16^9* ' -'^5 9 ik ' >i 2. ^ + / = 9, 9 25 .7. a^'^y = 36, 11. x^ = 4.-f. 3. 9 25 ""^i+f-'. 12. 13. a^ + 2/' = 25. 2.-^2 + 32/2 = 60 4. ■^ + ^' = 1, 25 9 25 9 5 4 4 5 ar' + 2^ = 25. «• ^ + 1 = 1' 14. ^^--1=1, 5. ^ + f = % 16 9 9 4 ' ^ + 1 = 1. ^-^^ = 1. ^ + 2/^ = 1. 25 9 25 9 4 ^ SYSTEMS OF EQUATIONS IN TWO UNKNOWNS 119 15. ^-/=o, 17. 3)2 + ^2^25, 19. 4a:2==92/2, 16. a:2 + /=25, 5 4 5 4 20. 18. 3a:''^4-5/=32, .t2 = 2/^. 40:2 + 92/^ = 1. 0^^ + 2/^ = 1, 90. We have discussed in some detail the two simplest cases that can arise in the solution of sets of equations in two un- knowns solvable by quadratics. In general, the solution of a set of two equations in two unknowns is beyond the scope of this book. There are, however, certain ingenious special de- vices that can sometimes be used. We shall explain two of these. The others are not sufficiently important to the student to justify a discussion of them here. (a) Solve the equations x^ -\-y^ = 21^ (1) X + y = 3. (2) Equation (1) can be written (x + y)(x2-a-y + ?/2) = 27. (3) If, in the left member of (3) we substitute iov x + y its value as given by (2), we get or x^ + xy + y'^ = 9. (4) Now (2) and (4) can be solved in the manner explained under Case I. The graph of equation (1) is too complicated to be given here. (h) Solve the equations , xy^U. (2) If we multiply each member of (2) by 2 and add the members of the resulting equation to the corresponding members of (1), we get x2 -\-2xy + y'^ = M. (3) Hence x + y = 8, (4) or x-\- y =—^. (5) We shall get all the solutions of (1) and (2) by solving (2) and (4) and also (2) and (5). 120 COLLEGE ALGEBRA EXERCISES Solve each of the following pairs of equations. When neither of the equations is of third degree, check your results by means of graphs. When one of the equations is of the third degree, check your results by direct substitution in both the original equations. 1. ic2 + 2/2 = io, . 6. a^+16/ = 44, 11. y?-f=2, 0^2/ = 3. 0^2/4-1 = 0. x-y^\. 2. x^-f = % 1. «-^ + 2/^ = 64, 12. 16a?2 + 25/=9, x — y = 2. X -\-y = ^. xy = l. 3. So? + 21f=^l, 8. x'-f = l, 13. 64ar^ + ?/3 = 4, 2x + 3y = l. x-y=.l. 4,x-\-y = 2. 4. a;2_^ 1/2 = 20, 9. x' + y^=16, 14. x'-^y^^S, icy = S. ^y= 10. xy = 2. 5. 4a^ + 9?/2 = l, 10. ^ + f = % 15. a'?/ + 4 = 0, xy = 2. x + y = l. a;^ + 9 2/^ = 25. PROBLEMS rv Q « 1- The sum of the squares of two numbers is 193 and the iK product of the numbers is 84. What are the numbers ? 2. The sum of the squares of two numbers is 250 and the sum of the numbers is 22. What are the numbers ? 3. The difference of the cubes of two numbers is 819, and the difference of the numbers is 3. What are the numbers ? 4. It is known that the sum of the lengths of two cubical bins is 29 feet and that their combined capacity is 6119 cubic feet. What are the dimensions of each bin? 5. The area of a right triangle is 30 square inches and its hypotenuse is 13 inches. Find the lengths of the legs. SYSTEMS OF EQUATIONS IN TWO UNKNOWNS 121 6. The diagonal of a rectangular field containing 11^ acres is 80 rods. What are the dimensions of the field ? 7. Two circles with their centers on the same diameter of a third circle are tangent to each other externally and to the third circle internally, as in the figure. What must be the radii of the two inner circles in order that the sum of their areas be one half the area of the outer circle, the radius of the latter being 10 inches ? 8. What must be the radii of the inner circles of Problem 7 in order that the sum of their areas be three fourths that of the outer circle ? How will the sum of the circumferences of the inner circles compare with the circum- ference of the outer circle ? 9. What is the least possible combined area of two inner circles related to a circle of radius 10 inches in the way de- scribed in Problem 7 ? Hint. — Let this combined area be wa, and then find the least value of a for real values of the radii of the inner circles. 10. Two spheres with their centers on the same diameter of a third sphere are tangent externally to each other and in- ternally to the third sphere. What must be the radii of the two inner spheres in order that the sum of their volumes be three fourths the volume of the outer sphere, the radius of the latter being 10 inches ? 11. What is the least possible total volume of two inner spheres related to a sphere of radius 10 inches in the way de- scribed in Problem 10 ? 12. What is the least possible total surface of the inner spheres of Problem 11 ? 122 COLLEGE ALGEBRA 13. Being given that the sides AB, BC, and CA of the tri- angle ABC are 8 inches, 12 inches, and 15 inches respectively, find the altitude BD. Let BD=x and AD = y. Then DC=15-y, and we have the two equations x^ + y^ = 64, x'^ + (15 _ y)2 ^ 144. These equations do not come under any of the cases we have discussed. The method of solution is, however, obvious. When one altitude is known, the others can easily be found from the fact that the product of any side of a triangle and the altitude upon that side equals twice the area of the triangle. 14. Find the three altitudes of the triangle whose sides are 20, 24, and 30 respectively. 15. Find one altitude and the area of the triangle whose sides are 7 centimeters, 10 centimeters, and 15 centimeters respectively. CHAPTER IX PROGRESSIONS 91. Arithmetic progression. — A succession of numbers so related that each one is obtained by adding a fixed amount to the preceding one is called an arithmetic progression. The numbers forming the progression are called its terms. The fixed amount which must be added to any term to get the following one is called the common difference. 92. If «! is the first term of the progression and d the com- mon difference, the progression is Oi, ai + dj % + 2 (Z, ai 4- 3 c?, •• • The succession of dots after a-\-Sd means " and so forth." For example, 3, 7, 11, 15, 19, 23, 27; and 5, 3, 1, _ 1, - 3, - 5, - 7, - 9, - 11 are arithmetic progressions. In the former the common difference is 4 and in the latter it is — 2. If a^ denotes the wth term of an arithmetic progression, it is clear that , / *x , /t\ a„ = ai + (n-l)A (I) This formula involves the four numbers aj, d, n, and a„ ; and by means of it we can find any one of them when the other three are given. 93. The sum of n consecutive terms of an arithmetic pro- gression. — If we denote by s„ the sum of the first n terms of an arithmetic progression, then Sn = ai+[ai + d] + [ai +2c?]+ ... +[ai +(n - 3)d] + [ai +(n- 2)d^ + [ai +(n- 1) 2^^ ^ - ^\\ 11. In how many different relative positions can a party of six people seat themselves at a round table ? / '2- S 12. Given ^P^ = 12 • ^P, ; solve for n. 7 13. A railway signal has three arms and each arm can be put into two positions besides its position of rest. How many signals can be given by it? Every position of the arms, except that in which they are all at rest, forms a signal. 1^ 14. How many kinds of single trip local passenger tickets, good either way, will a railway company need for use on a division that has ten stations ? t/ C' . C <2. 15. In how many different relative positions can a party of five ladies and five gentlemen be seated at a round table, the ladies and gentlemen being seated alternately ? '2-g's' '-' 105. Distinguishable permutations of n things not all differ- ent. — The number of arrangements of the four letters of the word city is 4!, or 24; but the number of distinguishable arrangements of the six letters of the word Canada is not 6 ! . This is due to the fact that these six letters are not all dis- tinct. Any rearrangement of the three a's does not make a 138 COLLEGE ALGEBRA distinguishable difference in any of the arrangements. Hence for every arrangement of these letters there would be 3 !, or 6, arrangements if the letters were all distinct. The number of A t distinguishable arrangements is then — ^, or 120. o ! In general, for every arrangement of n things p of which are alike, there would be p ! arrangements if these n things were all distinct. Hence, Theorem. — If of n things p are alike, the number of per- n ' mutations of these things taken all together is —^ . p\ Corollary. — If of n things p are of one kind, q of another, r of another, and so on, the number of permutations of these things taken all together is '- . p\ ' q\ ' r\ '•' The proof is left to the student. EXERCISES 1. How many permutations are there of the letters of the word permutation ? f 'f / ^ ^) ^ 0'^-' 2. How many signals can be made by displaying twelve flags one above the other, of which four are white, three blue, one red, and the rest black ? / '3 1'^ .- 3. How many signals can be made by displaying the flags of Ex. 2 if the top flag must be red and the bottom one blue ? ^ ' ^ 4. In how many ways can a row of five white balls, four red ones, and two black ones be arranged ? Q^ ^3 5. Four dimes, three quarters, and three half-dollars are to be distributed among ten boys in such a way that each boy ^ ^ shall receive one coin. In how many ways can this be done ? 106. Combinations. — If in any set of things we are interested only in the things themselves and do not consider the order in which they g/my be arranged, we speak of the set as a combination. PERMUTATIONS AND COMBINATIONS 139 There is obviously only one combination of n things taken n at a time, but there is more than one combination of n things taken r at a time, if r is less than n. In certain problems it is neces- sary to determine this number. We shall accordingly derive a general formula for it, and shall represent it by the symbol „C,. 107. Number of combinations. — Consider first the combina- tions of 6 things taken 4 at a time. We can determine in the following way just how many of these there are. The 4 things of any one of these combinations can be ar- ranged in 4 ! different orders and these different arrangements give rise to 4 ! distinct permutations of the 6 things taken 4 at a time. Moreover every permutation of the 6 things 4 at a time is related to some one of these combinations in this way. 6^4 • ^ i — 6^4) and therefore e^* = 4! In general, since there are r! permutations of r things taken r at a time, it follows that for every combination of n things taken r at a time there are r ! distinct permutations of these n things taken r at a time. Moreover every permutation of these n things r at a time is related to some one of these combina- tions in this way. Hence, and therefore ^c, = '>{n-\){n-Z)...{n - r+\) ^ ^ If we multiply both numerator and denominator of this ex- pression for ^C^ by (ii — r) ! we get C = ^ (:n — 1) (n — 2) • . . (n — r 4- 1) {n — r) ! ^ n\ ,^. " ' r! '{n-r)l rl '(n-r)l' ^ ^ 108. Whenever we make a combination of r things from n things, there is left a combination of (n — r) th: , and two 140 COLLEGE ALGEBRA different combinations of r things from the same n things leave different combinations of {n — r) things. Hence, The number of combinations of n things taken r at a time is equal to the iiumber of combi7iations of n things taken n — r at a time. That is, ^O^ = „(7„_^. Prove this last statement directly from Formula (4). PROBLEMS 1. How many straight lines can be drawn through pairs of points selected from nine points no three of which are in a straight line? T,i>Q,i -^Ij<^i} ¥ic%(kf^ 2. If the lines of Ex. 1 are produced indefinitely and no 7t^ ' ■' two of them are parallel, how many triangles will be formed ?-p^ 3. If no four of a set of fourteen points lie in one plane, ^Q \ how many tetrahedra are there whose vertices are in this set ? t 4. How many planes are determined by the points of Ex. 3 ? A plane is determined by three points not in a straight line. 5. At a meeting of ten men each shakes hands with all the others. How many handshakes are there ? G. How many committees each consisting of five Republi- 1 7^8'/53,ns and four Democrats can be chosen from twenty Republi- cans and sixteen Democrats ? d^ 7. How many parallelograms are formed when a set of nine ' parallel lines is met by another set of eight parallel lines ? 8. In how many ways can seven ladies and four gentlemen I*" •» arrange a game of tennis, each side to consist of one lady and one gentleman ? 9. In how many ways can a baseball nine be formed from sixteen players, of whom nine can play only in the infield and jK seven only in the outfield? Each of the infield players can iVr^ play any one of the six infield positions, and each of the out- field players can play any one of the three outfield positions. //- ffc ui PERMUTATIONS AND COMBINATIONS 141 10. In how many ways can a committee of five men be se- / ; lected from a group of twelve men ? 11. How many sub-committees, each containing five mem- bers, of whom at least three shall be Democrats, can be formed from a committee of eight Democrats and five Republicans ? . 12. Given „P, = 20, „0, = 10; find n and r. ^t ^ s' r ~ 13. How many groups of three letters each can be formed from the letters of the alphabet ? 14. In how many ways can a picket of five men be formed from a company of 100 soldiers ? i ^^l^% '/ S 1^ O 15. How many combinations each consisting of four red balls, two white ones, and five black ones can be formed from seven red balls, six white ones, and nine black ones ? 16. In how many ways can a committee of four lawyers, three merchants, and one physician be formed from the stock- holders of a corporation consisting of ten merchants, eight lawyers, and five physicians.? /^. l^- 2^ o . 17. How manylairnngementsjof four consonants and three vowels can be made from eight consonants and five vowels ? 109. The binomial theorem for positive integral powers. — / Consider the product , , ,., , ,., , ,v of 3 equal binomial factors. The product of the first terms of all the factors, namely o?, is a term of the product. So also is the product of the first terms of any two of the factors and the second term of the remaining factor. This is a^h. Moreover there are three such terms in the product since the factor whose second term is used may be any one of the three factors. Hence when like terms are combined the product will contain the term 3 a^h. It is evident from the way this term was ob- tained that in it the coefficient of a}h is equal to the number of ways in which the term, h can be selected from the three given factors, and this number is fix- The product of the first term 142 COLLEGE ALGEBRA of any factor and the second terms of the other factors, namely aJy^j is also in the product ; and there are as many such terms as there are ways of selecting the term b from two of tlie three factors. Hence, when like terms are combined, the coefficient of a¥ is 3C2. Finally the product contains the product of the second terms of all the factors. Hence, (a + hf = a? + sC^a^b + Aa¥ + b\ Consider now the product (a + b){a + b) ...(a + 6) of n equal binomial factors. The product of the first terms of all the factors, namely, a% is a term of the product. So also is the product of the first terms of any n — 1 of the factors and the second term of the remaining factor. This is a""^6. More- over there are n such terms in the product since the factor whose second term is used may be any one of the n factors. Hence when like terms are combined the product will contain the term na'^'^b. It is evident from the way this term was ob- tained that in it the coefficient of a'*~^6 is equal to the number of ways in which the term b can be selected from the n given factors, and this number is „(7i. The product of the first terms of any n — 2oi the factors and the second terms of the remain- ing factors, namely, a^~^b^, is also in the product ; and there are as many such terms as there are ways of selecting the term b from two of the n factors. Hence when like terms are com- bined the coefficient of a'*"^^^ is „(72. In a similar way it can be seen that in the product the coefficient of a"~''6'", where r is any integer from 1 to w inclusive, is „(7^. Hence for any positive integral value of n, The statement of this relation is called the binomial theorem for positive integral powers. The right member of this iden- tity is called the expansion of (a + by. PERMUTATIONS AND COMBINATIONS 143 If we take into account the values of ^(7, for different values of r, as given in § 107, we can write the binomial theorem as follows : (a + &)^ = fl^ + na^-^b + "(^ ~ ^) 0^-2^2 4. ... 2 ! 110. The term whose coefficient is „(7^ is the (r-f l)th term from the left and the (n + 1 — r)th term from the right. It is called the general term since any term except the first one can be obtained from it by substituting in it the proper value of r. The term whose coefficient is „(7^ will be as far from the right end of the expansion as „(7^ is from the left end if r4-l=7i4-l — s; that is, if s = n — r. Hence the terms whose coefficients are ^C^ and „(7„._^ respect- ively are equidistant from the ends. But ^C^ = n^n-r (§ 108). This proves that the coefficients of any two terms equidistant from the ends of the expansion of (a-\-by are equal. 111. Greatest coefficient. — Since ^ _ 7i(n — l)(n — 2) '•' (n — r-^1) r I and ^ _ n(n-l) ( n-2) .-» (n-r-^2) ^^-^- F=^iy^ ' it follows that , ^ ^ __ n-r-\-l ^ r Hence ^C^ is greater than the preceding coefficient when n -r-\-l ^^ that is, when r< '!i±l. (See § 182.) r 2 If ?i is odd so that we can have r — ^ , the two coefficients 144 COLLEGE ALGEBRA „(7m+i and „(7n-i are equal and greater than any of the other 2 2 coefficients. If n is even, so that we cannot have r = ^ "^ , the greatest coefficient is ,,(7^, where r is the greatest integer less than Vl±1., This integer is -. Hence if n is even, JJn is the greatest coefficient in the ex- pansion of (a + hy. 112. The expansion of (a + ^j*" as given by (I) can be described in words as follows : It contains (n + 1) terms. Tlie first term is a" (or a^'b^) and the last is &" {or a°6"). The second term is na'^'^b. Tlie exponent of a in any term after the first is one less than it is in the preceding term. TJie exponent of h in any term after the first is one more than it is in the preceding term. TJie coefficient of any term after the first is obtained from the preceding term by dividing the product of the coefficient of this term and the exponent of a by one more than the exponent of b. It should be observed that the sum of the exponents of a and b in any term is n. The expansion of (a — by can be obtained from that of (a + by by putting — 6 in place of b, since a — b = a+(—b). Example 1.— Expand /?^ + ^V- Here a = — , 6 = -, and n = 6. 5-4.S (2rY (sy 5.4-3-2 /2r\/sy 5 • 4 • 3 . 2 . 1 /j?\5 ■^ 3! Uj Uj 4! V 3 JVj) "^ 5! UJ ' ^ 32r5 Wr^s 16r»8^ 8 r V , 2 rs^ . s^ 243 81 136 225 375 3125* u PERMUTATIONS AND COMBINATIONS 145 Example 2. —Expand (2 ic — 3 r/)^. Here a = 2ic, &= — 3y, and n =Q. Hence (2 a: - 3 yY = (2a;)6 + 6(2a;)5(-3y) + ^ (2 a;)4(_ 3 y)2 = 64 a;6 _ 576 r^?/ + 2160 cc^^a _ 4320 x^^/S + 4860 icV _ 2916 xy^ + 729 2/6. EXERCISES Expand the following expressions by the binomial theorem. 1. {x-^y)\ 11. {l+xy\ 2. (a-2y. 12. (x^+y^f, 3. (3-0^)^ 13. (a + 6)* -(a -6/. 4. (6 + iy. 14. (5a6-«-2a;y/. 5. (2a + 4cy. \ 15. /'^ + ^' 19. (m-^-n-y. 10. (.'c+ir. In each of the following exercises write the term asked for without finding any other term. 21. The 6th term of (a + by. 22. The 6th term of (a - by. 23. The 5th term of (3 a - 2 by. 24L. The 4th term of (4 x^y - 5 xy^^ iL. ir ^'^ 146 COLLEGE ALGEBRA 25. The 3d term of (1 - x)\ 26. The 8th term of (2 x + — V^* 27. Write the third term from each end of the expansion of (4a-7&)l ' 28. Write the middle term of (7 a^ - 4 Wf. 29. What is the coefficient of x^ in the expansion of {1—xy^ ? 30. Write the first five terms of the expansion of ( 1 -j- - V ^, 31. Are the coefficients in the expansion given in the result of the first illustrative example consistent with § 109 (I) ? Compute the values of the following correct to two decimal places : 32. (1.1)». Hint.— (1.1)8 =(1 + . 1)8 = 1 + 8(.l) + 28(.1)2 + 56(.1)3 + 70(.l)* + 56(.l)s + 28(.1)6 + 8(.1)-+(.1)8 = 1 + .8 + .28 + .056 + .007 = 3.14. The last four terms have been neglected since they can have no influ- ence on the first two decimal places. 3^. (3.1)9. ^34 , (^99)6, 35, (1.01)^ 36. (4.9)^ CHAPTER XI MATHEMATICAL INDUCTION 113. We consider in this chapter a method of proof, com- mon in mathematics, of a certain class of theorems in which an integer is in some way involved. We shall first apply the method in question to a simple theorem and shall then examine its essential features. 114. Theorem. — The sum of the first n even integers begin- ning with 2 is equal to the product of n and the next larger integer. I. It is easy to verify that the theorem is true for small values of n. For example, for w = 1, we have 2 = 1(1 -f- 1). II. Suppose that the theorem is true for 7i = r ; that is, ^^''^ 2 + 4 + 6+ ... +2r = r(r + l). (1) Adding 2 (r + 1) to each member of this supposed equation, we get 2 + 4 + 6+ ... +2r + 2(r + l)=r(r + l)+2(r + l) = (r + l)(r + 2). (2) Now (2) is true if (1) is true. But (2) is merely another form of the statement that the theorem is true for ?i = r + 1. If then the theorem is true for n = r, it is true for n = r-\-l. But we know from I that it is true for 7i = 1, and, therefore, it is true for 7i = 1 + 1 = 2. And since it is true for n = 2, it is true for n = 2 + 1 = 3. We can proceed in this way by suc- cessive steps until we get the proof that the theorem is true for any given integral value of n. 147 148 COLLEGE ALGEBRA 115. This method of proof is called mathematical induction. It is applicable to certain theorems in which a positive integer is in some way involved, and consists of two parts ; namely, the direct verification of the theorem for the smallest admis- sible value of the integer and the proof that if the theorem is true for one value of the integer, it is true for the next greater value. 116. Necessity of both parts of the proof. — The formula Z is true for n — \. It is also true for 71 = 2 and n = 3, but it is not true for all values of n, as may be seen by putting ti = 4. This gives 12 I 22 -L 32 4. 42 = 80 — 28 + 4 ^ 2g. 2 But this is obviously not true. On the other hand, the formula 124.22+ ... +n2 = |n(7i + l)(2n + l)+5 is true for n = r-\-l, provided it is true for n = r. For, if the formula is true for n = r, we have 1^ + 22+. ..+r^ = ^r(r + l)(2r + l)+5, and if we add (r + 1)^ to each member of the equation, we get 12 + 22+. ..+r2+(r + l)2 = ^r(r + l)(2r + l)+5+(r + l)2 = -J(r + l)(2r2 + r + 6r + 6) + 5 = i(^ + l)(2r2+7r + 6) + 5 = K^ + l)(^ + 2)(2r + 3) + 5 But this is what the original formula becomesjwlien we p uj n = r + 1. Hence if the formula were true for n = r, it would -^"be true for n = r + 1. J ih.^o ^-rr MATHEMATICAL INDUCTION 149 That this argument, however, does not establish the validity of the formula may readily be seen by testing the formula for n = l. This gives 12 = ^. 1(1 + 1)(2 + 1) + 5 = 1 + 5 = 6, which is obviously untrue. These two examples show that both parts of the proof as outlined are necessary in order to establish the truth of a theorem. r^^ EXERCISES .'W ^ -V|A>i. Establish the :following formulae by mathematical induction : /l. 1 + 3 + 5+ ...+(2M-l)=7i2. ^SL. ^ \^ 2. 12 + 22 + 32+ ... +7i2=i-7i(n + l)(2n + l). 3. 13_|_23 + 33H +n^ = l7i''(7i-\-iy. 4. a + a?' + ar^ + • . . + ar""^ = ^ ~ — ^ • 1 — r 5. 2 + 22 + 23+ ... +2" = 2(2'' -1). ^6. JL + ^+...+ 1 1.22.3 n(n + l) n-^1 7. 2.4+4.6 + 6.8+ ... +2n(2n + 2) = ^(2n+2)(2n+4). o 8. 1.2.3 + 2.3.4 + 3.4.5+ ...+n(n + l)(n + 2) = ^(n + l)(n + 2)(n + 3). 9. Prove that if n is a positive integer, a"" — b"" is divisible by a — 6. Hint, cf+i — &'-+i = a(a^ - b^) + b'-(a — b). The right member of this identity is divisible by a — b, if a*" — ft*" is divisible by a — b. 10. Prove that if n is a positive integer, a^" — 52™ ig divisible by a + 6. 150 COLLEGE ALGEBRA 117. The binomial theorem for positive integral powers. — The student is familiar with the fact that (a -\-by = a''-r2ab + h\ and that {a + l>f = a^ +^ o?h -\-^ ah'' -[-b\ Now these relations are special cases of a general theorem which tells us how to write out, or expand, any positive integral power of a binomial. This theorem is known as the binomial theorem. It says that if n is any positive integer (fl + hy=a-+ na^-^b-\- ^("~^) 0^-252^ « ! a(n-l)--(n-r+2) +1 1 (r-1)! _^_n(n-l)-(n-r+l)^„.,^ + ... + ft., r! The right member of this formula can be described as follows : Ji contains {n -\-l) terms. The first term is a"" (or a" 6°) and the last is 6** (or a^6"). The second term is na^^'^h. Tlie exponent of a in any term after the first is 07ie less than it is in the preceding term. The exponent of b in any term after the first is one more than it is in the preceding term. The coefficient of a,ny term after the first is obtained from the preceding term by dividing the product of the coefficient of this term and the exponent of a by one more than the exponent of b. It should be observed that the sum of the exponents of a and b in any term is n. MATHEMATICAL INDUCTION 151 118. Proof of the binomial theorem. — We shall prove the theorem by mathematical induction. In order to do this we observe in the first place that the theorem is obviously true for n = 1. (The student can also verify directly that it is true also for n = 2 and n = 3, although this is not necessary for the argument.) In the second place we assume that it is true for n = m* ; that is, we assume the truth of the statement that (a + by = a- + ma^-^h -h .•• + m(m- 1) ■•♦ (?ft -r + 1) ^„.-.^. + ••• + ma^'^-i + ?>"*. Then we multiply each member of this identity by a + &. This gives us = a'"+^+(m4-l)a"'6H — -f- ^ — ^^-^ — '^^ ^- — ^ ^J^-^a*" '^+^6'^ t\ ^ \-{m-\-l)ah'^-{-h'^^\ Now this expansion for (a + h)'^^'^ is true provided that the expansion for (a + 6)'" with which we started is true. More- over this expansion for (a + 6)"*+^ is in accordance with the binomial theorem. Hence, if the theorem is true for n ='m,\t is true for n = m -\-l. But we have already observed that it is true for n = 1. It is therefore true for all positive integral values of n. In deriving the expansion of (a + &)'"+^ we made use of the fact that * We do not say n= r because we are here using r for another purpose. 152 COLLEGE ALGEBRA m(m — 1 )...(m -r + 1) , m(m- . 1). ••• (m- -r + 2) _m(m - r! -1).. • (m — ?• + !) + ?'m(m -1)! -1).. • (m — r + 2) 7n(m - -1)... • (m — r! 7- + 2)[(m — r + 1) + rl rl _ (m + l)m(m — 1) « • • (m — r + 2) 119. The general term. — The term n(n-l)'"(n-r-{-l) ^„_,^, rl is the (r + l)t7i term in the expansion of (a + ft)**. It is called the general term of this expansion since it can be made equal to any term, except the first one, by assigning to r a suitable value. The expansion of (a — by can be obtained from that of (a -h by by putting — 6 in place of b, since a—b=a-{-(—b). For exercises on the binomial theorem, see those in § 112. CHAPTER XII COMPLEX NUMBERS - 120. In considering negative numbers for the first time it was found helpful to associate with every point of a given line a number which represented its distance from a given point (or origin) on the line. In order to distinguish between two points equally distant from the origin but on opposite sides of it we agreed to represent the distances of points on one side of the origin (say the right side) by positive numbers and the distances of points on the opposite side of the origin by negative numbers. The numbers that represent these points were called real numbers. Now a similar but more extensive association of num- bers with points will serve to make clear in a simple way the essential nature of a more complicated kind of numbers that are of great importance in mathematics; namely, the so-called imaginary, or com- plex, numbers. Consider two perpendic- ular lines X'X and Y' Y that intersect at 0, and let the real numbers represent the points of X'X, the ori- gin being taken at 0. We want to represent the points of the plane that are not on X'X by symbols which we can use in a way that resembles as closely as possible the use of the real numbers (the symbols that represent the points of X'X) in the operations of addi- tion, subtraction, multiplication, and division. We shall call these symbols numbers. 153 Y 5i+ Z-i- ■51- 154 COLLEGE ALGEBRA 121. In Chapter V we represented the points of T'Y, as well as those of X'X, by real numbers, but for our present purpose we must represent the points of the plane that are not on X'X by numbers different from the real numbers, since every real number represents a point on X'X, and we do not want any number to represent two points. The number by which we shall represent the point on Y'Y the ^unit's distance above will be called i. Any point on F'F whose distance from is a will be represented by the number ia. If a is positive, the point represented by ia is above 0, and if a is negative, the point is below 0. Thus, the point represented by 5^ (or, it 5) is on y F five units above 0, and the point represented by —5i is on Y'Y five units below 0. If we should mark on X'X the point M represented by the number a and should then proceed to lay off the point ib in the way just described, with the exception that we use the point M instead of for origin, we should get the point P which C is 6 units from M and above or below it according as b is posi- tive or negative. We shall say that the number representing F is the sum of a and ib and shall indicate it by the symbol a -\- ib. This gives us a definition of the sum of a real number and a number of the form ib that is closely analogous to the definition of the sum of two real numbers given in § 4. 122. Let now P be any point of the plane. Draw the per- pendicular PM from P to X'X. If 0M= a and MP = b the number that represents P is a 4- ib. We have at hand there- fore a system of numbers that will represent all the points of the plane, and conversely, every number of this form ; namely, a -f ib, represents a point of the plane. IT COMPLEX NUMBERS 155 123. Equal numbers. — We shall agree to say that two numbers are equal when they represent the same point. Hence if a-\-ib = c-\- id, we must have a = G, and h = d. Con- versely, if these last two inequalities hold, the numbers a + ib and c + id are equal. If the point represented by a + ih is on X'X, 6 := 0. If the point is the origin, a = 0, & = ; that is, if a + ib = 0, a = 0, and 6 = 0. 1. Write the that represent the points designated in the accom- panying figure. 2. Plot the points repre- sented by the following numbers : 3 + 1, 3 — 2*, —3 + 1, 5i, — ? , 1 + I, — 1 — 3 I, 4-22, 6, -3 + 4?, EXERCISES numbers Y 'D •B 'C f 'E 'G 'F Y' X 124. It was stated in § 120 that we wanted to be able to use these new numbers in a way resembling as closely as pos- sible the use of real numbers in the operations of addition, subtraction, multiplication, and division. Now the essential point in the use of real numbers is that it is governed by the assumptions of Chapter I. In the following paragraphs we shall accordingly describe combinations of these new num- bers which we shall call addition, subtraction, multiplication, and division respectively, and which will also be governed by these assumptions. It will be seen that these new operations are perfectly natural extensions of the corresponding opera- tions with real numbers. We have already described the addition of a real number and a number of the form ib. \fo 156 COLLEGE ALGEBRA 125. Addition and subtraction. — By the sum of the num- bers Xi + iyi and X2 + iy2 representing the points A and B re- spectively, we mean the number x + iy representing the point S which is obtained by laying oif B from A, instead of from 0, as origin. Thus, we draw ^iV equal, and parallel, to OM, and then NS equal and parallel to 3IB. The student can readily verify- that S is the fourth vertex of the parallelogram two of whose sides are OA and OB, except when 0, A, and B lie on a straight line. He can also see that r ^ ^!. r' /^' -^ M r' Hence, x = Xi + X2 and y=yi-\- {xi + iy{) + {X2 + iy2) = (xi + x^) + i{yx + yo). By the difference of x\ + iy\ and x<2, + i?/2, or (a;i + iy{) — {xi + 2*2/2) » we mean the number x -f- i?/ such that y^\ + «yi = (a; + iV) + {^1 + *y2) = (x + Xi) + i(y + ^2) . It follows from § 123 that xx^x^-xi and yi = y + y2' Hence, x = Xi — x^ and y = yi — yii and therefore {xi + i>i) - (^2 + 12/2) = (a*! - ajg) + i(2/i - 2/2). This difference can be obtained graphically by forming the sum of xi + iyi and — X2 + i(- 2/2). EXERCISES In each of the following exercises represent graphically the numbers in the parentheses, and their sum or difference, as the case may be : 1. (i + 2?:)H-(5-5i). 4. 3-(6-2r). 2. 2i-4i. 5. 3 2^-2. 3. (i_2i)-(2 + 50. 6. (6 + 20 + (5-4i). COMPLEX NUMBERS 157 7. (7-i)-(7-f-30. 8. (5-40 + (5-40. 9. (_3_3t)+4i. 10. 3 -(6 -6/). 11. (8+4i7-(8 + 4*). 12. (_3i+2i)-(-3i+i). 13. (V2-hV3 + (V2-V3 0. 14. (4^ + 3) + (4 + 3^). 15. (2 + 6 i)- (2 + 60- 16. (3-2*) + (2i-3). 17. (7-2i2 + (7 + 20. _ 18. (l+V5i) + (2 + 2V5i). 19. What is the relation between the points represented by 3 + 4 I and — 3 — 4 1' respectively ? Plot these points and also the points 3 — 4 1 and — 3 + 4 1. 20. Under what circnmstances is the sum of two complex numbers a real number ? 21. Apply the definitions of addition and subtraction in this article to two real numbers and compare the results with those obtained by applying the definitions of §§ 4 and 6. 126. Multiplication. — By the product of I and the number representing any point P we mean the number representing the point to which P is broug ht by_reyobdiig-jft P coun tgrclock- wise around through an angle of 90°. A'^-+- ^B2 Y' Thus, if the angle POQ is a right angle and OP = OQ, the number representing Q is equal to the num- ber representing P multiplied by i. In the figure let OBx = OB2 = OBs. If b represents the point J5i on X'X, the product of i and b repre- sents B2 on Y' Y. But we have al- ready represented this point by ib.^ Hence we say that ib is the product ' of i and b. 158 COLLEGE ALGEBRA The product of i and ih represents the point B^ on X'X. But - h also represents this point. Hence, i • ib =— b, or i^b=-b. This is equivalent to saying th at^, 1^""'^^ 1. 127. AVhen we consider that the product of two positive, or two negative, numbers is positive, it seems strange that we can have a number like i whose square is a negative number ; or, in other words, that we can have a square root of a negative number. The explanation is that this number is neither posi- tive nor negative, since it represents a point not on the axis X'X It is a new kind of number and it ought not to be surprising that it should have properties not possessed by the old numbers. If we had never heard of negative numbers, it would seem just as strange to talk about a number that is equal to 3 minus 5 as it does to talk about a number whose square is — 1. 128. Definitions. — Numbers of the form ib, where & is a real number, different from zero, are called pure imaginaries. Numbers of the form a -\- ib, where a and b are real numbers and b is not equal to zero, are called imaginary numbers. This term, imaginary, was applied at a time when it was supposed that these numbers had no existence and were not real. But with the conception of numbers that has been set forth here we see that they are just as real as the points of a plane, which they represent, and therefore just as real as the so-called real numbers. Since imaginary numbers are formed by a combination of two essentially different kinds of units ; namely, 1 and i, they are more properly called complex numbers. In the complex number a + ib, a is called the real part and ib the imaginary part. COMPLEX NUMBERS 159 In a real number the imaginary part is zero and in a pure imaginary the real part is zero. Complex numbers that differ only in the sign of their imagi- nary parts are said to be conjugate to each other. Thus 4 + 3 I and 4 — 3 i are conjugate, as are also a 4- ih and a — ib. A real number is its own conjugate. 129. Product of two complex numbers. — We are now in a position to define what we mean by the product of two com- plex numbers like a + ib and c + id. We agree to say that (a + ib) (c + id) = ac + ibc + iad + i'^bd = (ac — bd) + i(bc + ad). It will be observed that this product is just what it would be if the numbers a, b, c, d, and i were all real, with the excep- tion that it makes use of the fact that v^ = — 1. 130. Theorem. — The sum and the. product of two conjugate complex numbers are real numbers. For {a + ib) + (a — ib) = 2 a, and {a + ib) (a - ib) = a'^ -^ VK 131. Division of complex numoers. — -We define the division of a + ib by c -h id to be the operation of finding the number x + iy such that a-\-ib = {x-\- iy) (c + id). The indicated quotient . can be changed to the form X -\- iy by multiplying both the numerator and the denominator of the fraction by the conjugate of the denominator. Thus a 4- ib _ (a -f ib)(c — id) _ (ac + bd) + i (be — ad) c + id~ (c + id) (c — id) ~ c^ + d^ _ ac -f- b d . . be — ad This process justifies the statement of Assumption XII of Chapter I as applied to complex numbers. 160 COLLEGE ALGEBRA 132. We have now defined the four fundamental operations of addition, subtraction, multiplication, and division, as applied to complex numbers, in a way which can be shown to be con- sistent with the assumptions of Chapter I. Moreover when applied to numbers whose imaginary parts are zero, these operations are the same as the operations of addition, subtrac- tion, multiplication, and division respectively, as applied to real numbers. 133. Principal square root of negative numbers. — Since i^ and (— if are both equal to — 1, i and — i are square roots of — 1. We shall call i the principal square root of — 1 and rep- resent it by V— 1. Thus i = V— 1. By the principal square root of any negative number such as — a, where a is a positive number, we mean the product of i and the principal square root of a (see § 55). We shall use the symbol V — ct to indicate this principal square root. Thus V— a = y/a • i = Va V— 1. If b is also a positive number, then V— a • V— b = ■\/a • i • Vb • i = Va6 .^— — y/ab. The student is warned against saying V— a . V—b = V— a • -- b = Vab. EXERCISES Reduce each of the following expressions to the standard form a -\- ih. 1. (3-2^)(4-5^). ^ 1 2. (2 + 1)2. ' i' 3. i'. Jidi VI i p4M ^y^S. (2 + V^(7 - V~i). 4- **• djJvi^ ivY***^ P Hint. 5. i\ k .;; iJi S^ . A'-*'^ . 2 + V^l = 2 + V3 I, 6. ^^^ |i' >^x |#^ '^ h^ ^^Ljr^^/~^ = 'J-2L COMPLEX NUMBERS 161 18. 5 * . — 8 i. 3+i 2+i 2 ^ ^ 3_^,. 2 + 1 12. ^ __ . ^, 1-V3i -1 + V3"i 1 + V3i / ^^- 2 2 13. (2-0(2 + i)(3 + 2i). 22^ V-7+V-8 14 1 + ^' V-9+V-2 1-^- 23. ^ 15. (V2 4-V5^)(lV2^-V5). 16. (V^ + V^i)(^V^ + V^). V3 + V2i 25. What is the relation between the points represented by two conjugate numbers ? Solve the following equations, plot the points represented by the roots, and show graphically that the sum of the roots of each equation is zero. 26. ar5-l = 0. 28. ic*-l = 0. 27. 0)3+1 = 0. 29. x«-l = 0. 30. Show that 1 -f i is a root of the equation 31. Show that ^ — ^i is a root of the equation 36 a^. + 36 a;2 - 47 a; + 50 = 0. 162 COLLEGE ALGEBRA 134. The polar representation of complex numbers. — There is another method of representation of complex numbers that is sometimes useful. Consider the complex number x-\-iy that represents the point P of the figure. If OP=r and Z.X0P=6, then X = r cos 6, (1) y = r sin 6, (2) ^ + f = A (3) and therefore x-{-iy = r cos 6 + ir sin d = r (cos $ -\-i sin 6). This representation of a complex number in terms of r and 6 is called the polar representation of the number. The angle 9 is called the amplitude, or argument, and the distance OP, or r, is called the modulus, or absolute value, of the complex number. It follows from formula (3) that r = -yjx^ + 2/^ is the modulus of the number x + iy. It is always considered positive. It follows from (1) and (2) that the argument 6 oi x-\- iy is given by the formulae iC x = cos~^-= cos~^ = sin~^^ = sm~^ ^ ■y/x^ -j- 2/2 For example, the modulus of 2 — 3 lis r = VlS and the argument is 5^' 2 3 d = cos-i = sin-i — - Vis COMPLEX NUMBERS 163 135. Theorem. — Tlie^ m odulus of the prod uct of two numbers is equal to the product of their modul i and the argument of the pr oduct is equ al to the sum of their arguments. For r(cos 6-{-i sin 0) • 7*'(cos 6' + i sin 0') = {r cos 6 + ir sin ^)(/ cos 0' + ii-' sin 6') — rr' cos 6 cos 6' — rr' sin 6 sin 6' + i(r/ sin 6 cos 6' + r/ cos 6 sin 6') = r/[(cos 6 cos ^' — sin 6 sin ^') + i(sin cos ^'H- cos ^ sin $')'] = r/[cos((9 + ^') + i sin ((9 + 6')\ 136. Since r(cos ^ + i sin ^) _ r(cos 6 -\-i sin ^)(cos ^^ — ^^ sin 6^) /(cos ^' + i sin 6') ~ /(cos ^ + i sin ^')(cos 6'— i sin 6*') _ r[cos(^-^-)+.sin(^-^0 ] ^ .^ ^ ^ _ , . ^ _ , /(cos^^'+sin^^') /*- ^ ^^ ^ ^-'' we have the Theorem. — The modulus of the quotient of two numbers is equal to the quotient of their moduli and the argument of the quotient is equal to the argument of the dividend minus the argu- ment of the divisor. 137. De Moivre»s Theorem. — If the two factors in § 135 are equal, we have [r (cos e + i sin 0)^ = r\cos 2 + i sin 2 0). This is a special case of the following more general formula [r(cos 6 -\- i sin 0)y = r"(cos nd -f i sin nO). (I) The statement of this relation is known as De Moivre's Theo- rem. In the case that ti is a positive integer this theorem can be proved by means of mathematical induction in connection 164 COLLEGE ALGEBRA with the theorem of § 135. The details of the proof are left to the student. He should go through them with great care. The proof of (I) when n is the reciprocal of a positive in- teger, say — , is as follows : m Let e = TYKJ). Then ill (cos e+i sin d)m = (cos m

^ J . • fe-{-S' 360° , oaf^o\ • fO-{-S' 360°\ and sm —^ h q • 360 = sm — — — )• \ n J \ f^ J If mi and mg are any two values of m and if cos ^±^360: ^ ^^^ g + m,360°^^^ ^.^ 6±m,S^ n n n . i9 + m2 360° ,, ^ + mi 360° -..^ . ^ + m2 360° = sm — ■ — , then — ' — can diner from — ' — ■ n n n only by a multiple of 360°, since two angles which have the same sines and cosines can differ only by a multiple of 360°. But this difference is ^^i ~ '^v ^j^^j ^g j^q^. ^^ multiple of n 360° when mi and mg are unequal integers less than n and not less than 0. Therefore the following values of m give all the distinct num- bers of the form (1) : 0, 1, 2, • • -, n — 1. This proves the following Theorem. — Jf n is a positive integer, any number different from zero has exactly n distinct nth roots. Since all the nth roots have the same modulus, and since the \ 360° • \ argument of any one of them increased by is the argu- n ment of another one, we see that the points represented by the nth roots of a; + iy all lie on the circumference of a circle whose center is at the origin and divide this circumference into n equal parts. 166 COLLEGE ALGEBRA EXERCISES Find the modulus and the argument of each of the following numbers and plot the number. 1. 2 + 5/. ' 3. 6-V^~2. 5. 2 + L 2. -3 1. 4. 8. 6. l-^. '■'■ _ "l^:- 8. (7 + V-10)(3+V"^=^). 11- 5 (cos 30° + 1 sin 30°). Simplify the following products : 13. 2 (cos 10° + I sin 10°) • 5 (cos 25° + 1 sin 25°). 14. - 5 (cos 30° + i sin 30°) • (cos 15° + ^ sin 15°). 15. (1 + if. 19. [i(cos 180° + i sin 180°)]^. 16. 7if^-\-^\ 20. [3(cos60° + isin60°)]0 ,, /I _'iV3VV2^ .V2X ''• (-«30° + .-sin30°)3. \2 2 j\ 2 2 J 22. [2 (cos 40° + ^ sin 40°)]'. 18. [4(cos20°+isin20°)]^ 23. [4(cosl20° + isinl20°)]3. 24. Find all the cube roots of 1 + i. Here r = Vl + 1 = V2, sin ^ = cos ^ = V2 Hence, 6 = 45°, and 1 + I = V2 (cos 45° + * sin 45°) . We shall get all the cube roots of 1 + i by giving to m in this expres- sion the successive values 0, 1, 2. For m = 0, we get v^2(cos 15° + isin 15°). For m = 1, we get v^2(cos 135° + isin 135°). For w = 2, we get v/2(cos255° + isin 255°). COMPLEX NUMBERS 167 25. Find and plot all the cube roots of t^ + ^-tt— 26. Find and plot all the fourth roots of \^' 1 — ^ 27. Find and plot all the cube roots of — i. 28. Solve the equation ar' — 32 = 0, and plot the roots. Here x^ = 32. Now 32 = 32 (cos 0° + i sin 0^). TT 6/sH a/ 0° + m • 360° , . . 0° + m • 360°\ Hence v32 = 2 cos ^ h \ sin — ^^—^ ), V 5 5 / where m = 0, 1, 2, 3, 4. Hence the roots are 2, 2 (cos 72° + i sin 72°) , 2(cos 144° + i sin 144°), 2(cos216°-f isin216°), 2(cos288° + isin288°). Solve the following equations and plot the roots: 29. a^ - 1 = 0. 30. a;*' + 1 = 0. Compare the method of solution used in Exs. 29 and 30 with that used in Exs. 26, 27, § 133. 31. a;*-16 = 0. 32. x'-X^^. 33. aj3 + 27 = 0. 34. What is the relation between the roots of Equation 30 and those of Equation 33 ? Simplify each of the following quotients : \-i 3_2V-4 ^^ V2J-^V2^ ^^ 6(cos60°-f isin60°) V3H-t ' 3(cos20° + *sin20°)' 1 41. cos 40° - i sin 40° CHAPTER XIII THEORY OF EQUATIONS 139. The student is familiar with the solution of equations of the first and second degrees in one unknown; that is, he knows how to identify numbers described by such equations. It happens, however, that many of the descriptions met with in the applications of mathematics are in the form of equations of higher degrees, and it is far more difficult to identify num- bers from such descriptions than from linear or quadratic equa- tions. We shall make no attempt to find all the roots of such equations, but shall confine our attention to the problem 'of finding their real roots. 140. This latter problem naturally divides itself into two parts : 1. The problem of finding the rational roots. 2. The problem of finding the approximate values of irra- tional real roots. This is the most important problem considered in this book and the purpose of this chapter is to show how to solve it. Everything given in this chapter has a direct bearing on this solution. 141. We shall use such symbols as f(x), F(x), and Q(x) to represent polynomials in x. These symbols are read "/of x," " large F of a?," and " Q oi x " respectively. If /(a?) stands for afpc^'-^a^x*'-^ -{-•.• +«„_!« + a„; that is, if f(x) = aoaJ** + aia;""^ 4- ••• + a«-ia; + «„, we shall use the symbol /(c) to stand for the number obtained by putting c for x in this polynomial. 168 THEORY OF EQUATIONS 169 Thus if f(x)=^x^ + x^-5x + 2, then /(2) =4 . 23 4- 22 - 5 . 2 + 2 = 28, and /(O) =4.0 + 0-5.0 +2 = 2. EXERCISES Determine 7i and «„, ctj, ••• , a„in each of the following cases : 1. f(x)=-2x*-\-7a^-Sx^-5x-6. 2. f(x) = x^-{-x'-l. 3. f(x) = 5x^ + 3a^-\-10x. 4. f(x) = x^-x*-\-2a^ + 5. 5. f(x)=x'-l. 6. f(x) = (x-i-l)(x + 2)(x + S), 7. In each of the preceding exercises find /(I), /(— 2), and /(O). 142. Remainder Theorem. — When f (x) is divided by x — r the remainder isf{7-). This remainder must be a constant since it is of lower de- gree in X than the divisor, which is of the first degree. The quotient must be a polynomial in x. We shall represent it by Q(x) and the remainder by E. Then f(x) = Q(x){x -r)-\- R. If in this identity we put x = r, we get /(r) = Q{r)(r - r) + R. But Q(r) (r — r) = 0, since r—r = 0. Hence R = f(r). The student should observe that the expression for B does not contain X and that therefore B is the same number, no matter what value is given to X. Corollary. — If r is a root of the equation fix) = 0, then x — r is a factor of fix), and conversely. For, by the theorem, the rehiainder obtained in dividing /(ic) by a; — r is f{r), and by hypothesis /(r) = 0. 170 COLLEGE ALGEBRA Conversely, if f(x) = Q (x)(x — r), then f(r) = Q (r) (r — r) = 0, and r is a root of the equation f(x) = 0. (See § 65.) This corollary is sometimes referred to as the Factor Theorem. i EXERCISES Perform the following divisions and check your work by means of the Remainder Tlieoi-em. 1. Q^-^^.x'^ + ^x-lOhj x-2, 2. 2 ic^ — 7 a^ — a; + 4 by a; -t 1. 3. a?* + a^ + a;2 + a; + 1 by cc. 4. 7a;^ + 2x^ + 8a;3 — 6ar^ — 5a; + 4.by a^— 1. 5. a;^ + 1 by a; + 3- 6. a;^ — ar^ + 1 by a; — 1. 7. a^-6a?2-3a; + 2by a;+.5. 8. 3 a^ + 5 a^- 6a; -2 by a; + 2. 143. Synthetic division. — In looking for the roots of the equations considered in this chapter we shall have frequent occasion to find the values of polynomials in x for given values of X. The Remainder Theorem tells us that these values are equal to the remainders obtained by dividing the polynomials by expressions of the form x — r. It is therefore important to be able to find these remainders as quickly as possible. Consider the process of dividing 2a:^— 5a?^4-3a; — 4bya; — 3. 2a^-5a;2+3a;-4|a;-3 2:x?-Qx' 2ar2+a;4-6 ar^-f 3 a; X?- -3a; 6a;- - 4 6a;- -18 14 If we wish to shorten the work of this division as much as possible, we observe that it would be sufficient to write merely the coefficients of the various powers of x in the dividend and THEORY OF EQUATIONS 171 the quotient and that we could omit the first term of the divisor without causing confusion, since all the divisors we are considering have the same first term. Moreover only the first term of each partial remainder need be brought down. Then our work could be arranged as follows : 2-5 + 3-41-3 2-6 2+1+6 1 1-3 6 6-18 14 We note further that it is unnecessary to write the coef- ficients of the quotient, since these are equal respectively to the first (or left hand) coefficients of the dividend and the successive remainders. The coefficients of the first terms of the successive partial products may also safely be omitted. The arrangement of the work will then be as follows : 2-5 + 3-41-3 -6 1 -3 6 -18 14 We could put 3 in place of — 3 if we remember to add the partial products instead of subtracting them. It is desirable to do this because what we are really trying to find is the value of the polynomial when 3 is put in place of x. More- over all the partial products can be written on one line. The work can therefore be arranged as follows : 2-5 + 3- 4|3^ + 6 + 3 + 18 2 + 1+6 + 14 172 COLLEGE ALGEBRA The successive terms in the last line reading from left to right, except the last one, are the coefficients of the descending powers of x in the quotient. The last term is the remainder. Thus if we divide 2x3— 5aj2 + 3aj — 4 by x — 3 we get the quotient 2x2 + X + 6 and the remainder 14. Hence this polynomial equals 14 when 3 is put for x. This shortened form of division is known as Synthetic Division. 144. Rule for synthetic division. — In order to divide the polynomial f(x) by x — r, arrange f{x) according to descending powers of x. If f(x) = a^x'^ + a^x^''^ + ••• + a„_ia; -f a„, supply zeros in the places of the coefficients of the missing terms and write the successive coefficients in a row beginning with aQ. Bring down aQ, multiply it by r and add the j^roduct to a^ ; mul- tiply the resulting sum by r and add the product to a^. Continue this process of multiplying each sum by r and adding the product to the next coefficient until a product has been added to the last coefficient. The last resulting sum is the remainder. The numbers in the roiv of sums, except the last one, are the coefficients of the successive terms of the quotient beginning with the term containing x""'^. 145. Application of synthetic di- vision. — This method of finding the value of f{x) for a given value of x can be used to good advantage in finding points on the loci of equa- tions of the form y = f{x). If f(a) — b, the point {a, b) is on the locus of the equation y=f(x). Thus the locus of y=x'^—6x-{-b passes through the points (0, 5), (1, 0), (2, — 3), etc. 146. Graphical solution of equations. — The abscissae of the points that are common to the locus of the equation y=f(x) and the ii?-axis are the real roots of the equation f(x) = 0. This THEORY OF EQUATIONS 173 suggests at once that we can find the real roots oif(x) = graphi- cally by drawing the locus of the equation y=f(x) and measur- ing the abscissae of the points common to the locus and the a;-axis. If the degree of f(x) is greater than 2, the computation necessary in order to draw the curve enables us to make a fair approximation to the roots without the figure. Nevertheless, a consideration of the figure often gives us important informa- tion concerning the roots. This is illustrated in the proof of the following Theorem. — If a and h are real numbers and f(a) and f(b) have opposite signs, there is at least mie real root of the equation f(x) = between a and b. If /(a) and/(6) have opposite signs, the locus of the equation y =f(x) is on opposite sides of the ic-axis at the points whose abscissae are a and b respec- tively. Now this locus is an unbroken curve and does not turn back on itself (see figure), since for each value of x there is only one value for y, and therefore it must cross the a>axis at least once between the point whose abscissa is a and the point whose abscissa is b. This proves the theorem. This argument rests upon the assumption that the locus of the equa- tion y =f(^x) is an unbroken curve. The assumption can be proved, but the proof is too difficult to be given here. EXERCISES Find the real roots of the following equations graphically ; 1. a^-\-3x + 2 = 0. 5. x^-^4:X^ + 2x-3 = 0. 2. a;2-4a;+4 = 0. 6. a^ -\- Aa^ -\-5 x-^2= 0. 3. a^ + 2«2_|_2a; + l = 0. 7. 2 a.-^ -3 a;^- 12 «-{- 1 = 0. 4. a^-6a^-7x-6 = 0. 8. a^ ^2x^-2x-S = 0. '] 174 COLLEGE ALGEBRA 147. The fundamental theorem of algebra. — Every equa- tion of the form f(x) = has at least one root. This theorem, which is usually referred to as the fundamen- tal theorem of algebra, was first proved by Karl Friedrich Gauss, a German mathematician, in 1799. The proof is too difficult to be given here.* "^N r 148. Theorem. — Any polyrtomial f(x) of degree n has linear factors of the form x — r, where r is a real or a complex number. By the fundamental theorem the equation f(x) = has at least one root, say r-^. Then by the factor theorem f(x) is divisible by x—r^, and if /(x) = »(,«'' -h a^a;'*-^ -{- ... -f a„, we have, a^- + a,x--^-{- ■" +a, = (x-r,)(a,x--'+b,x--'+ ... +&„_i). Again by the fundamental theorem the equation a.x^-'' + b^x--' -f- - + &n-i = has at least one root, say rg. Then using the factor theorem again we see that a^x^-^-j-b,x^-'-\- ... -{-b„_,=(x- r,)(a,x^-'-\- c,x--'-\- ... 4-c„_2), and that therefore f(x) = {x - r^)(x - r,)(a,x--'--^ c,x--'-{- ... + c,.^). By continuing this argument we see that we shall finally get f(x) expressed as the product of n linear factors. Thus f(x) = ao(a; - r;)(x - r^) .•• (x - r^). (1) This shows that rj, rg, .••, r„ are roots of the equation ♦ The student who is interested will find a proof in Fine's College Algebra, p. 588. THEORY OF EQUATIONS 175 These numbers may not all be distinct. If f{x) = a^{x - ViY^ix - ra) "2 ... (x - r^)«*, we agree to say that ?\ is a root of multiplicity n^ and to count it Ui times. Thus rj is a root of multiplicity rii and we count it ?ii times ; rg is a root of multiplicity Wg and we count it n.^ times ; and so on. We have then the following Corollary. — Every equation of degree n has at least n roots. 149. Theorem. — No equation of degree n has more than n distinct roots. For suppose that the equation is aoic''+ aiOf'^ 4- agic^'^H- ... -\-a^ = 0, where ao ^ ^^ The preceding theorem shows that the left member can be written in the form «o(aJ - r{)(x - rg) ..• (« - O. If now r were a root of this equation and distinct from r^ r^i '"} r„, we should have «oO' - ri)(r - r^) "• {r - r„) = 0. But the first factor, ao, of the left member of this relation is by hypothesis not zero. Moreover none of the other factors is zero, since, by hypothesis r is different from rj, rg, •-, r„. But no product can be zero unless at least one of the factors is zero (10, Chapter I). Hence the supposition that r can be a root of /(a?) = is wrong, and the theorem is proved. Corollary. — Every equation of degree n has exactly n roots in the sense explained in § 148. Corollary. — If two polynomials in one variable of degrees not greater than n are equal to each other for more than n values of the variable the coefficients of like powers of the variable in the two polynomials are equal, and conversely. If aoa;"+ aj^c^-^ + - + a„=M"+ ^la:""' + .-. + b^ 176 COLLEGE ALGEBRA for more than n values of x, then the relation is satisfied by more than n values of x. If some of the coeffi- cients besides the last one in the left member of this relation were not zero, we should have an equation of degree not greater than n with more than n roots. But the theorem shows that this is impossible. Hence all these coefficients are zero, and therefore the last coefficient is also zero. That is, a^-h^ = 0, or a^^b^; ai — bi = 0, or a^ = &i ; a^ — &„ = 0, or a^^ = b^. The converse is obvious. 150. Relations between roots and coefficients. — If r^, 7\, •••, r„ are the roots of the equation x'>' + aix'^-'^ + ■•' +an-iX + an = 0, (ao = 1) then (§ 148) ic" + aix"-i + ••• + ttn-ix + an= (ix — ri)(x - r^) ••• (x— Vn). If we perform the indicated multiplications in the right member of this identity, we shall get a polynomial which is equal to the left member of the identity for all values of x. Hence the coefficients of corresponding powers of x in the two polynomials must be equal (§ 149), and we have ai=-ri- r2- ••• -r„. (1) «2 = riVi + nn H + nvn + r2r3 H h r„_ir„. (2) as = — rir2r3 - nr^n — • • • - r„_2r„_irn. (3) a„ = (- l)«rir2 •••»•„. (n) The second member of (1) is the sum of all the roots with their signs changed. The second member of (2) is the sum of the products of the roots with their signs changed taken THEORY OF EQUATIONS 177 two at a time. The second member of (3) is the sum of the products of the roots with their signs changed taken three at a time. And so on. It should be noted that these relations hold only when the coefficient of the highest power of x is 1. If it is not 1, we must divide each member of the equation by it before apply- ing these relations. ^ f (.^ 151. Imaginary roots. — Although we are interested here only in the real roots of an equation, the following theorem concerning the appearance of imaginary roots will be useful. Theorem. — If c + id is a root of an equation with real coeffi- cients, c — id is also a root. Let the equation be a^x^ 4- aia;"-^ H h a„= 0, where a^, a^, •••, o„ are real numbers, and let f(x) be a symbol for the left member of this equation. The product of the two factors x—{g + id) and x — (c — id) is x^ — 2cx + c^ -\- d^. If now we divide f(x) by this product, we shall get a quotient which we can represent by the symbol Q{x), and a remainder which must be of degree less than 2. ^^^^ f{^) = Q(^) («' - 2ca? + c2 4- d"") 4- ra + r ', where r and / are certain real constants."* Now this relation holds for all values of x, and therefore for x — c-\- id. But by hypothesis /(c + id) = 0. Therefore 0=Q{c-hid){G'+2icd-d^-2 = 0. 22.20.^ + 40. = !. ^ 21. 2x^-1 ;^-Wx^+x-S = 0. 28. Za^+x'^-2y^-4.x'-Qx-2 23. a.'3+2V = 0. ^ ^^ 24. 9a.*-4x^-6a.24-9a. + 18 29. 3a;*4-a^ + a; + 5 = 0. = 0. 30. -2a.3_|_4^^^6a. + 8 = 0. 31. What are the roots of ar^ + 2ar — 2 a. + 3 = 0, it being given that 3 is a root of ar' — 2 a?- — 2 a; — 3 = ? 32. What are the roots of ar^ + 4 a.^ — 7 a; — 10 = 0, it being given that — 2 is a root of a;^ — 4 a;- — 7 a; -|- 10= ? Form equations whose roots are the roots of the following equations, each diminished by the number placed opposite the respective equation. Then plot the loci of the equations obtained by putting the left members of the original equa- tions equal to y, and show the effect on the ?/-axis of each transformation. 33. a.2_53._^g^Q^ I ZS. x^-.^x + Q = 0, -1. 34. ar^_3a.-2-a. + 3 = 0, 2. 39. x2 + 3a; + 2 = 0, 2. 35. a.3-l = 0, 1. 40. a.2-2a; + l = 0, 1. 36. a,-3 + 6a.2+9a; + 20=0, 3. 41. x^- 3a;- + 3a.-l = 0, 1. 37. a.^ + 6a. = 8 + ar^ 2. 42. ar^- Gaj^-lla; - 6 = 0, 2. 43. What relations do the loci of the equations y = —f(x) and y=f(—x) bear to the locus of y=f(x)? Apply your answer to the cases in which f(x) is each of the polynomials in Exs. 33-37 in turn. ^w - :^'l and that jp and q have no common divisor except 1. Then we have the relation qn qu 1 gn . Multiplying both members by ^""^ and transposing, we get Q Now the right member of this relation is an integer, and therefore if ^ were a root of the given equation, p" would be divisible by q. But this is impossible, since p and q have no common divisor except 1 and q>l. Hence every rational root of the equation must be an integer. If r is an integral root of the equation, we have r" + air"-i-|- agr^-^H 1- a„ = 0, or r""-^ + air"-2 + cfgr^-^ H \- a„_i = - ^ . r Now the left member of this relation is an integer and there- fore — is an integer. This is equivalent to saying that a" is divisible by r. This completes the proof of the theorem. 190 COLLEGE ALGEBRA 160. We are now in a position to solve the main problem of this chapter; namely, to find the real roots of an equation with rational coefficients. We consider first the problem of finding the rational roots of such an equation. 161. Rational roots. — If the equation is of the form ttox- + Oia;"-' + a^--' + . . . 4- a„ = 0, (1) where aQ ^ 0, we first form a new equation whose roots are 711 times the roots of this equation, and find the least positive value of m for which this new equation has integral coefficients when the coefficient of its highest power of x is 1. Suppose that the new equation for this value of m is a;" + b^x^-'^ + h^""-^ H h &„ = (2) in which the coefficients are integers. If equation (1) has a rational root, equation (2) has an integral root which is a divisor of 6„. We therefore determine by synthetic division what positive and negative divisors of 6„ are roots of equation (2). If we divide every such root by m, we shall get all the rational roots of equation (1). When one root r of (2) has been found, divide the left mem- ber of (2), which we will call F(x), by x — r. F{x) = (x-r)Q(x). Every other root of (2) is a root of Q(x)=0, and these other roots can best be obtained from this last equation. We shall refer to it as the depressed equation since it is of lower degree than (1). Sometimes the labor of finding these roots can be shortened by determining the maximum number of positive and negative roots by means of Descartes's rule of signs. THEORY OF EQUATIONS 191 Example. — Find the rational roots of 9x4 + 12x3+ 10x2 + 3;- 2 = 0. (1) The equation whose roots are m times the roots of this can he written in the form , ^ „ The least positive value of m that will make these coefficients integers is 3. For this value of m the equation is x* + 4 x3 + 10 x2 + 3 X - 18 = 0. (2) "We know from § 158 that this equation has just one positive root and from § 159 that this root is a divisor of 18, if it is rational. Now the positive divisors of 18 are 1, 2, 3, 6, 9, 18 ; and we find by synthetic division that 1 is a root. 1+4 + 10+ 3- 18 U_ -f- 1 4- 5 + 15 + 18 1 + 5 + 15 + 18+ All the other roots are roots of the depressed equation a;3 + 5ic2 4.i5a; + i8 = o. This equation has no positive root, and any negative root it may have is a negative divisor of 18. Try - 1. 1 + 5 4. 15 4. 18 | -1 _1- 4-11 1+4 + 11+ 7 Hence — 1 is not a root. Try - 2. 1 + 5+15 + 18 [ -2 -2- 6-18 1 + 3+9+0 Hence — 2 is a root, and the other roots are roots of the depressed and therefore imaginary. Since 1 and — 2 are all the rational roots of (2) , i and — | are all the rational roots of (1), since the roots of (2) are the roots of (1) each mul- tiplied by 3. It is not necessary to perform any extra work to get the successive de- pressed equations since the left member of each one is obtained inciden- tally in finding a root of the preceding equation. 192 COLLEGE ALGEBRA EXERCISES Find all the rational roots of each of the following equa- tions : \ 1. a^4.a;2 4-aj4-l = 0. 1 11. 6 a^-2x^ -\-Ax-l = 0. 2. 2a;3 4.a;2_^a;-l = 0. ' 12. 4.a^-l()x'^ -9x -}-S6=0. 3. a:^ + 5a^-2x-\-2 = 0. 13. 4:X^ + S x" - x-2 = 0. 4. x^-5a^-\-10x-Tx-2 = 0. 14. 6x' + 2a^ + 5 = 0. 5. a^-\.x'-16x + 20=:0. 15. 4.x^-5x-6 = 0. 6. 2 a.-3 + 9 3.-2 + 110; + 3 = 0. 16. 3ar^+16 a-^+lS a;-20=0. 7. x^-^Sx^ = a^. 17. 2a;3 + 3i»2 + 5a; + 2 = 0. 8. a;* + ic3_|_3j2_j_^_^;L^0^ ;L8. a;^ + 3 ic2 + 2 = 0. 9. a;3_3a^ + 5^. + 4 = 0. 19. x^-3 x^-{-S x^-3 xi-2=0. 10. 8a;3_4a.2_2aj + i = o. 20. a^-32 = 0. 162. Irrational roots. — The first step in finding an irra- tional real root of f(x) = is to find two consecutive integers between which this root lies. This can be done by making use of the theorem of § 146. The following method for computing approximately the ir- rational real roots of an equation is based upon this theorem. It is known as Horner's Method, from the name of its in- ventor. 163. Positive irrational roots. — The essential features of this method can best be explained by means of an example. Example. — Find the irrational real roots of the equation 0^ + 3 a;2 - 2 a? - 5 = 0. (1) 1. The left member, which we represent by f(x), has one variation of sign and therefore the equation has just one posi- tive root. 2. In order to determine the location of this positive root, we substitute for x in f{x) successive integral values of x be- THEORY OF EQUATIONS 193 ginning with 0. We see directly that /(O) = — 5 and indirectly by means of synthetic division that /(I) = — 3 and/(2) = 11. l + 3_2-5[l_ 1+3- 2- b\2_ + 1+4 + 2 +2 + 10+16 1 + 4+2-3 1 + 5+ 8 + 11 Hence by the theorem of § 146, the equation has a root be- tween 1 and 2. 3. Form a new equation whose roots are equal to the roots off(x)=0 each diminished by 1. 1 + 3 -2 -5[1_ + 1 +4 +2 1+4 +2 + 1 +6 1 + 5 + 1 1 + 6 + 7 This new equation is 0,^ + 6 0-^4- 7 a;- 3 = 0, (2) and it has a root between and 1. 4. For a; = 0, .1, .2, ,3, and .4 the left member of this equa- tion equals respectively —3, —2.239, —1.352, —.333, and .824, as may be seen from the following computations : 1 + 6+7-3 y. 1+6+7-3 [^ + .1 + .61 + .761 + .2 + 1.24,+ 1.648 1 + 6.1 + 7.61 - 2.239 1 + 6.2 + 8.24 - 1.352 1+6 +7 -3 Lii 1 + ^ +7 -3 [A^ + .3 + 1.89 + 2.667 + .4+2.56 + 3.824 1 + 6.3 + 8.89- .333 1 + 6.4+9.56+ .824 Hence equation (2) has a root between .3 and .4. 5. Form a new equation whose roots are the roots of (2) each diminished by .3. 194 COLLEGE ALGEBRA 1 + 6+7 -3 |_^ .3 + 1.89 + 2.667 1 + 6.3 + 8.89 + .3 + 1.98 — .333 1 + 6.6 .3 + 10.87 1 + 6.9 The new equation is a? +- 6.9 x" + 10.87 X - .333 = (3) and it lias a root between and .1. 6. We find by trial that this root lies between .03 and .04. .03 .04 1+ 6.94 + 11.1476 + .112904 7. Form a new equation whose roots are the roots of (3) each diminished by .03. 1+6.9 + 10.87 - .333 1_^ .03 + .2079 + ..332337 1 + 6.9 + 10.87 - .03 + .2079 + .333 •332337 1 + 6.93 + 11.0779 - 1+6.9 + 10.87 - + .04+ .2776 + .000663 .333 .445904 1 + 6.93+11.0779 + .03+ .2088 - .000663 1 + 6.96 + .03 + 11.2867 1 + 6.99 The new equation is 0? + 6.99 x" + 11.2867 x - .000663 = (4) and it has a root between and .01. 8. We find by trial that this root lies between and .001. 1 + 6.99 + 11.2867 - .000663 | .001 + .001 + .006991 + .011293691 1 +6.991 + 11.293691 + .010630691 Obviously the left member of (4) is negative for ic = 0. THEORY OF EQUATIONS 195 9. Since each root of (4) is .03 less than a root of (3), the latter has a root between .03 and .031. Since each root of (3) is .3 less than a root of (2), the latter has a root between .33 and .331. Since each root of (2) is 1 less than a root of (1), the latter has a root between 1.33 and 1.331. Hence we have found a root of the original equation correct to two decimal places. If the digit in the third decimal place had been 5 or larger, the root to two decimal places would have been 1.34. In general, in order to get a root to r decimal places, it is necessary to determine whether the digit in the (r + l)th place is less than 5, or not. In practice the body of this work can be compactly arranged as follows : 1 + 3 _ 2 -5 \1 +1+4 +2 L§ 1+4 + 1 + 2 + 6 -3 1 + 5 + 1 + 7 1 + 6 + .3 + 7 + 1.89 -3 + 2.667 1 + 6.3 + .3 + 8.89 + 1.98 - .333 1+6.6 + .3 + 10.87 1+6.9 + .03 + 10.87 - + .2079 + .333 .332337 |.03 1 + 6.93 + 11.0779 + .03 .2088 1+6.961 + 11.2867 .03 1 1 + 6.99 + 11.2867 - .000663 .000663 164. By an obvious continuation of this method we can com- pute this root to any number of decimal places. The higher powers of a number between and 1 are less than the number itself. Hence if an equation is known to have a root between and 1, the terms containing the higher \^ 196 COLLEGE ALGEBRA powers of the unknown will be relatively unimportant, and the root can be determined approximately by neglecting these higher powers and solving the resulting equation of the first degree. The nearer the root is to zero, the closer this approximation will be. Applying this principle to (4) we see that .00006 is an approximate value of one of its roots, and this suggests that this root lies between and .001, as was stated. But it must be kept clearly in mind that this is only a suggestion, and that the suggestion must be tested. In a similar way we get the suggestions that .03 and .4 are approximate roots of (3) and (2) respectively. These considerations enable us to do away with many syn- thetic divisions that otherwise would be necessary. If in our efforts to locate a root between two consecutive in- tegers we divide /(a;) hj x — r, where r is positive, and find that the remainder and all the coefficients of the quotient are posi- tive, we may conclude that r is greater than any real root of the equation f(x) = 0. For if we divide f(x) by a? — rj, where Vi > r, all these coefficients after the first one, as well as the remainder, will be greater than they were and therefore the remainder will not be zero. The value of this observation may be seen in step 3 of the example worked above. The sugges- tion just described tells us that the root we are looking for is approximately .4, and we accordingly divide the left member of equation (2) by a; — .4. 1 + 6 +7 -3 |_^ + .4 4-2.56 + 3.824 1+6.4 + 9.56+ .824 Since all these sums are positive, we know that the root cannot lie between .4 and .5, and we accordingly divide by X — .3 and note the sign of the remainder. There are other considerations by means of which this process can be shortened, but we shall not take them up here. THEORY OF EQUATIONS 197 When one irrational root has been found to the required degree of approximation, an entirely new start must be made in finding the next one. On the other hand, when a rational root has been found, the equation can be depressed (§ 161) and the finding of the remaining real roots, both rational and irrational, is thereby made correspondingly easier. 165. Negative irrational roots. — In order to find the nega- tive irrational roots of f{x) — 0, find the positive irrational roots of /(— a;)=0. These with their signs changed are the roots sought. 166. Summary. — In order to find all the real roots of an equation f(x) = 0, in which f(x) is a polynomial with rational numerical coefficients, proceed as follows : 1. Find all the rational roots by the method described in § 161. When each rational root has been found depress the cor- responding equation. 2. See if the last depressed equation after all the rational roots have been found has any positive rootSy and determine by synthetic division two consecutive integers between which such a root lies. 3. Form a new equation tchose roots are the roots of this equa- tion each diminished by the smaller of these tico integers. 4. Tlie residting equation has a root between and 1. Find by synthetic division the two consecutive tenths between which the root lies. 5. Form a new equation whose roots are the roots of this equa- tion each diminished by the smaller of these tenths. 6. Tlie residting equation has a root between and .1. Find by synthetic division two consecutive hundredths between which this root lies. 7. Form a neiv equation whose roots are the roots of this equa- tion each diminished by the smaller of these hundredths. 8. If the root is required to r decimal places ^ continue this pro- cess until r -f 1 decimal places have been determined. 198 COLLEGE ALGEBRA 9. Add up the amounts by ichich the roots of the successive equations have been diminished. This sum, with the figure in the rth decimal place increased by 1 in case the figure in the (r + l)th place is 5 or more, is the root sought to the required degree of ap- proximation. 10. If there are other positive irrational roots, find each of them in the same way. 11. In order to find the negative irrational roots, fiyid the posi- tive irrational roots off{—x) = a7id change the sign of each one. 12. Make use of all the information that is obtainable from Descartes' s rule of signs as to the number of positive and of neg- ative roots. If in finding the irrational roots of f{x) = by Horner's Method we have any reason to suspect the existence of two roots between a and a + 1, we should examine the sign of f{x) for x — a, x = a-\-^, and x = a-[-l; and it may be necessary to examine the sign of f{x) for values of x still closer together. EXERCISES AND PROBLEMS Find the values of the real roots of the following equations correct to two decimal places : 1. 2a^ + 4aj2-10a; + 3 = 0. 2. 5a^-3a;2- 6x4-3 = 0. 3. x4_^a^-7a;2-8a;+20=0. 4. 2a^-aj3-a;2-a;-3 = 0. 5. 4 0^3 ^3 a;- 3. 11. How deep will a cork sphere 4 inches in diameter sink in water, the specific gravity of the cork being .2 ? Hint. — The volume of a spherical segment of one base is given by the formula . o , , q where x is the altitude of the segment and ri is the radius of its base. 6. 2a;3_3^2^^^0. 7. 8a;4 + 8ic2_ 1^ = 0. 8. 8a;4 = 8a;2_3. 9. x^-2x-2=^0. 10. a^-4a;-2 = 0. THEORY OF EQUATIONS 199 12. How deep will a sphere of pine of specific gravity i sink in water? 13. If the specific gravity of ice is .9, how much of a sphere of ice 2 feet in diameter would protrude above the water in which the ice is floating ? 14. The volume of a box 10 x 12 x 15 inches is to be in- creased 50 cubic inches by adding the same amount to each di- mension. What should this amount be ? 15. How thick should a hollow spherical shell be whose in- ner radius is 3 inches in order to contain 40 cubic inches ? 16. If a is the cosine of an angle and x is the cosine of one third of this angle, then 4 a^ = 3 a; + a. What is the cosine of an angle of 20° ? Hint. —The cosine of 60° is |. 17. A house may be bought for $3400 cash, or in annual in- stallments of $1000 each, payable 1, 2, 3, and 4 years from date. What is the annual interest rate implied in this offer ? Hint. — The amount of i$3400 for 4 years should equal the sum of the amount of the first payment for 3 years, the amount of the second pay- ment for 2 years, the amount of the third one for 1 year, and the fourth payment. Hence, if x is the rate of interest, 3400(1 4- xy - 1000(1 + xy - 1000(1 + x)-^ - 1000(1 +x)- 1000 = 0, or, 17(l + x)4-5(l + x)3-6(l4-a:)2-5(H-a;) -5 = 0. 18. A house may be bought for $2800 cash, or in annual installments of $1000 each payable 1, 2, and 3 years from date. What is the annual rate of interest implied in this offer ? 19. An open box is to be made from a rectangular piece of tin 18 inches long and 10 inches wide, by cutting out equal squares from the corners and turning up the sides. How large should these squares be in order that the box contain 168 cubic inches ? 200 COLLEGE ALGEBRA 20. Find the cube root of 3. Hint. — Find the approximate value of the real root of the equation 21. Find the cube root of — 17. 22. Find the fifth root of 10. 23. Find the cube root of 115. Find the points of intersection of the following curves : 24. a;2 4-2/2 = 10, 25. x^-^y^ = 9, 26. 4:x'^ + 7f = 16, y = x'^ + x-\-l. y — x'^ — x. x + 5y = y'^ -{-2. CHAPTER XIV DETERMINANTS 167. In Chapter V we used determinants of the second and third orders to advantage in the solution of systems of linear equations in two and three unknowns respectively. Analogous symbols, which are called determinants of order n can be used in the solution of systems of n linear equations in n unknowns, where n is any positive integer. For values of n greater than 3 there is a greater advantage in this use of determinants than there is in the simple cases in which n= 1, 2, or 3. The symbol order and by definition was called a determinant of the second = tti^a — a^i ; likewise, by definition of a determinant of the third order, We wish to define a determinant of order w, where n is any positive integer, in a way that will be consistent with these definitions of determinants of order two and three respectively. 168. Inversions of numbers. — But we must first introduce the notion of an inversion. We use this term to describe the presence in an arrangement of positive integers of a greater integer before a smaller one. Thus, in the arrangement 1, 2, 4, 5, 6, 3, the integers 4, 5, and 6 each appear before the smaller one 3, and there are in this arrangement there- fore three inversions. There are no inversions in the arrangement 1, 2, 3, 4, 5, 6 ; and there are two in the arrangement 2, 1, 3, 5, 4, 6. 201 202 COLLEGE ALGEBRA Theorem. — If in an arrangement of positive integers a7iy two of the integers he iyiterchanged, the number of inversions is in- creased or diminished by an odd number. Consider first the effect of interchanging two adjacent inte- gers a and b. Every integer that preceded both a and h in the original arrangement will precede them in the new arrangement, and every integer that followed them originally will follow them in the new arrangement. Moreover, the only relative positions that are changed are those of a and 6, and the effect of this change of relative positions is to increase or decrease the number of inversions by 1 according as a is less than, or greater than, b. That is, the effect of the interchange of two adjacent integers is to increase or decrease the number of in- versions by 1. Suppose now that there are Tc integers between a and b. Then b can be brought to the original position of a by A; + 1 interchanges of adjacent integers, and after this has been done a can be brought to the original position of 6 by A: interchanges of adjacent integers. These interchanges do not affect the relative positions of the integers other than a and b. The desired interchange of a and b has then been brought about by 2 A; -f- 1 interchanges of adjacent integers. In general, some of these interchanges, say x of them, have each caused an increase of 1 in the number of inversions and the remaining 2k+l — x have caused a decrease of 1 each. The net result has been an increase or a decrease of the number of inver- sions equal to the difference of these two numbers. But this difference, which is either 2k-{-l — 2x or 2x — 27c—l, is an odd number for all possible values of x. The theorem is there- fore proved. 169o Inversions of letters. — In an arrangement of letters of the alphabet the presence of a letter before one that pre- cedes it in the alphabetical order is also called an inversion. Consider now the symbols aj, Oa? ^^3, • • • ; &i, &2> ^3) • •• I ^i, Cj, Cg, • • • ; and so onj and select a set of these in which neither any DETERMINANTS 203 letter nor any subscript occurs twice. We have then the following Theorem. — The number of inversions in the letters when there are no inversions in the subscripts and the number of inversions in the subscripts when there are no inversions in the letters are both even or both odd. Consider first the special case aibzCidc^e^. Here there are no inversions in the letters. If we interchange dg and e^ in order to bring the letter with the greatest subscript to the last place, we get the arrangement a^^^^ie^d^. In doing this we have changed the number of inversions in the letters and the number of inversions in the subscripts each by an odd number. If now we interchange a^ and eg in order to bring the letter with the next to the greatest subscript to the next to the last place, we shall again change the number of inversions in the letters and the number of inversions in the subscripts each by an odd number. Finally, by the interchange of e^ and q we get the arrange- ment Cib^e^fi^dr^^ and this last step changed the number of inver- sions in the letters and the number of inversions in the sub- scripts each by an odd number. In the final arrangement there are no inversions in the subscripts. The number of inversions in the subscripts has been changed three times, by an odd number each time, and the number of inversions in the letters has been changed the same number of times, by an odd number each time. Hence the number of inversions in the subscripts in the original arrangement and the number of inversions in the letters in the final arrangement are both odd. In general we can change the arrangement in which there are no inversions in the letters to the arrangement in which there are no inversions in the subscripts by a certain number of interchanges of letters. Each of these interchanges increases or diminishes the number of inversions in the subscripts by an odd number (§ 168), and also increases or diminishes the num- ber of inversions in the letters by an odd number. Hence the number of inversions in the letters when there are no inver- 204 COLLEGE ALGEBRA sions in the subscripts and the number of inversions in the subscripts when there are no inversions in the letters are both even or both odd according as the number of these successive interchanges is even or odd. 170. Definition of a determinant. — In an array of n^ num- bers in n rows and n columns form all possible products ofn fac- tors each, taking as factors one number, and but one, from each row and column. Arrange the factors of each product in the order of the columns in which they occur and change the signs of those products in which the arrangement of the integers representing the respective rows presents an odd number of inversions. The algebraic sum of these products with their signs thus modi- fied is called a determinant of the nth order. It is represented by the array inclosed between two vertical lines. Thus, hi ^1 1 = ai62 - «2&i, I a-z &2 I = ai&2C3 + dibzCx + azbiCz — azb^Ci— a^biCz ■— ciibzC-2, ai bi ci| a^ hz C2 as bs cs 2 4 - -1 3 6 2 4 7 6 = 2.6.5+3-7-(-l)+4.4.2-4.6.(-l)-2.7.2-3.4.5 = 7. Here, for example, the sign of the product 3 • 4 • 5 is changed because when its factors are arranged in the order of the columns in which they occur there is an inversion in the order of the rows in which they occur, this order being 2, 1, 3. 171. Definitions. — The products described in the definition with their proper signs are called the terms of the determinant. When a determinant is written out in full as the algebraic sum of its terms, it is said to be expanded. The numbers in the array from which the terms are formed are called the elements of the determinant. DETERMINANTS 205 172. Properties of determinants. — I. There are n\ terms in the expansion of a determinant of order n. This is an immediate consequence of the fact that there are n ! permutations of the n rows taken n at a time (see § 104). When n is greater than 3, n ! is so large that it is not prac- ticable to find the value of the determinant by the direct pro- cess of expansion. And besides the number of the terms there is the difficulty of determining which of the products described in the definition should have their signs changed. These dif- ficulties can be avoided by the application of some of the fol- lowing properties of determinants. II. A determinant is not changed if the corresponding columns and rows are interchanged. Since each term of a determinant of order n is the product, with a possible change of sign, of n factors, no two of which occur in the same row or the same column, it is evident that the interchange of corresponding rows and columns will have no effect on the terms except perhaps to change the signs of some of them. If the original form of the determinant is «! 6i Ci • • 0,2 &2 C2 • • dn bn Cn •• any of its terms is a product of n factors so arranged that there are no inversions in the letters and the sign of this prod- uct is changed or not according as the number of inversions in the subscripts is odd or even. The corresponding term of the new determinant is the product of the same factors so ar- ranged that there are no inversions in the subscripts and the sign of the product is changed or not according as the number of inversions in the letters is odd or even. But by the theo- rem of § 169, the number of inversions in the subscripts is 206 COLLEGE ALGEBRA ai &i Cl 052 62 C2 = as 63 C3 here odd or even according as the number of inversions in the letters is odd or even. Hence the determinant is not changed by the interchange of corresponding rows and columns. The student should verify directly that 3 C3 DETERMINANTS 207 V. The value of a determinant is zero if the corresponding ele- ments of two of its rows (or columns) are the same. Let D represent the determinant. Then by IV D is changed into — D by the interchange of any two rows (or columns). On the other hand, if these two rows (or columns) are the two identical ones, D is not affected by their interchange. Hence D = -D, 2D = 0, and D = 0. For example, «i t>i ci I ai bi ci I = aibics + ai&sCi -f a^hiCi — asbiCi — aibsCi — aibiCs = 0. as 63 C3 1 VI. Hie value of the determinant is zero if the elements of any row {or column) are m times the corresponding elements of any other roiv (or column). We know from III that the value of such a determinant is m times the value of a determinant the corresponding elements of two of whose rows (or columns) are the same, and we know from V that the value of this latter determinant is zero. For example, ai &i Ci a2 &2 C2 ma2 W162 niC2 — maibiCi + ma2&2Ci + maibiCi — maib^Ci — maibiCi — ma2&iC2 = 0. VII. If two determinants are the same except possibly for the elements of a certain row (or column)^ their sum is equal to a de- terminant in which the elements of this row (or column) are the sum.s of the corresponding elements of the two determinants and the elements of the other roivs are the same as in the two determinants. For exaa iple, ai 61 ci a'l &i ci «i + a'l bi ci 0,2 &2 C2 + a'2 &2 C2 = a2 + a'2 62 C2 as bs Cs a's bs Cs as + a's bs Cs 208 COLLEGE ALGEBRA That this is so is obvious when we consider that every term of the third determinant is the sum of two parts which are respec- tively the corresponding terms of the first two determinants. VIII. The value of a determinant is not changed if each ele- ment of any row (or column) multiplied by any number m be added to the corresponding elements of any other row (or column). By VII the resulting determinant is equal to the sum of two determinants of which one is the original determinant and the other has the value zero by VI. For example, ai + mci 61 ci Gz + WIC2 62 C2 as + mcs 63 C3 173. Minors of a determinant. — If in a determinant of order n we omit the elements of any row and any column, the remaining elements with their relative positions unchanged form a determinant of order n — 1 which is called the minor of the element in this row and this column. Thus, if in the determinant ai 61 Ci = tti &2 Co 0,3 &3 C3 + mci 61 Ci ai 61 ci TOC2 62 C2 = a2 &2 C2 WC3 &3 C3 «3 h C3 «i &i Cl di a2 62 C2 d2 as 63 C3 ds a4 64 C4 d4 we omit the elements of the second row and the third column the remain- ing elements form the determinant Oi 61 di as 63 ds a^ &4 d^ which is the minor of C2. If an element of a determinant is represented by a small letter with a subscript, we represent its minor by the corre- sponding capital letter with the same subscript. DETERMINANTS 209 ai &i Cl a2 &2 C2 as bs C3 62 C2 — a^ 61 Cl + «3 &1 Cl 63 C3 &3 C3 &2 C2 174. The expansion of a determinant of the third order can be arranged as follows : = ai62C3 + ciibiCi + a3&iC2 — 03&2C1 — O2&1C3 — ai&3C2 = ai(&2C8 — &3C2) — ttiibiCs — 63C1) + a3(&iC2 — 62C1) = ai = ai-4i — a2^2 + «3^3- That is, a determinant of the third order can be expressed in terms of determinants of the second order. In a similar way a determinant of order n can be expressed in terms of deter- minants of order n — 1. The following theorem states in de- tail how this can be done. Theorem. — Multiply the elements of any row (or column) of a determinant by their respective minors, and if the sum of the number of the roiv and the number of the column in which the ele- ment occurs is odd, change the sign of the resulting product. The algebraic sum of these products with their signs thus changed is equal to the determinant. Represent the determinant by D, the element in the upper left hand corner by Oj, and any other element by s^. 1. If the elements in the terms of D and Ai are arranged as described in § 170, the coefficient of a^ in any term of Z> is a term of Ai, since the row and column in which ai occurs pre- cede all the other rows and columns. This conclusion is based upon the fact that in any arrangement of a series of integers beginning with 1 the number of inversions is the same as it is in this arrangement with the 1 omitted. Thus 135462 and 35462 contain the same number of inversions. Conversely, any term of Ai is the coefficient of aj in some term of D. 210 COLLEGE ALGEBRA 62C3 — 53C2, which is ai &i Cl det erm inant az &2 C2 «3 63 C3 &2 C2 , or Ai. &3 C3 the coefficient of ai is 2. Consider the element s^ in the ith row and Jth column of D. It can be brought to the first row by i — 1 interchanges of adjacent rows, and then to the first column by J — 1 inter- changes of adjacent columns, Avithout affecting the relative positions of the elements of D not contained in the original ith row or the original Jth column. The effect of these inter- changes is to change the sign of the determinant (^ — 1 4-J— 1) times; that is, (i-\-j — 2) times. This number is odd or even according as i-\-j is odd or even. If then D' denotes the determinant resulting from these changes, iy=(-iy+JD. If we multiply both members of this equation by (— !)'+■', we get ( _ iy+^- 2/ = (_ iyi+2j d = D, since (-If +2^=1. Since now the relative positions of the elements not in the ith row or the Jth column of D are not affected by these changes, the minor of s^ in U is the same as its minor in D. And from (1) we see that the coefficient of s^ in jy is Si- Hence the coefficient of s^ in Z) is (— Vf'^^ Si. For example, in the determinant 63. Here i = 3 and j = 2. Now «i hi a^ 62 as ^3 0^2 as a4 C3 Cl C2 C4 consider the element 63 as Cz dz 61 ai Cl dx 62 a2 C2 d2 64 a4 C4 d. DETERMINANTS 211 From (1) we know that the coefficient of 63 in this last determinant is and therefore the coeflBcient of 63 in the original deter- ai ci di ai C2 d-i , and theref( ai C4 di ai ci di minant is — a2 C2 di as C3 di We shall get all the terms in the expansion of D by taking the sum of all the terms containing each of the elements of a given row or column, and no term will appear more than once, since no term in the expansion of D can contain two elements from the same row or column. This proves the theorem. In order to see clearly the meaning of this theorem consider its appli- cation to a determinant of the fourth order : ai &i Cl di a2 62 C2 do as &3 C3 di ai hi C4 di 61 Cl = 03 &2 C2 bi Ci + C3 ds 175. By means of successive applications of the theorem of the preceding article we can make the expansion of a determi- nant depend finally upon the expansion of a series of determi- nants of the third order. This process, simplified by means of the properties given in § 172 in a way we shall explain by means of examples, is the one most frequently used for the expansion of a determinant. Example 1. — Expand the determinant 2 12 1 3 1 4 7 7 3 2 G G 5 -1 212 COLLEGE ALGEBRA If we subtract the elements of the second column from the correspond- ing elements of the first column, the resulting determinant is, by § 172, VIII, where w = - 1, equal to D. Hence 1 1 2 1 D = 3 1 7 3 4 2 • , by § 174, 6 5 -1 3 1 4 1 D = 7 3 2 = -2 3 -10 = _ 42 + 90 = 48 6 5 -1 -9 5 -21 The last determinant here is obtained from the preceding one by an application of § 172, VIII. The advantage of the preliminary transformation is due to the fact that it puts D into a form in which all but one of the elements of a cer- tain column are zero. In the expansion of a determinant in such a form by § 174 all the terms except the first one are equal to zero. Example 2 .— Expand the determinant 9 13 17 -4 D = 18 27 35 -8 30 44 21 10 12 15 2 • 9 13 17 - 4 9 13 4-4 D = 30 12 1 1 44 21 10 15 2 = 1 30 44 -23 10 12 15-15 2 9 4 -4 3 4-4 3 4 = 30 -23 10 = 3 10 - 23 10 = 3 10 -23 -13 12 -15 2 4-15 2 4 -15 -13 = 3(897 - 208 - 585 + 520) = 1872. Here we multiplied the elements of the first row of the original deter- minant by 2 and subtracted the products from the corresponding elements DETERMINANTS 213 of the second row. Then we subtracted the elements of the second column of the resulting determinant from the corresponding elements of the third column. If the determinant has an element equal to ±1, we can easily, by applying § 172, VIII, make all the other elements in the same row, or the same column, zero. Then by the theorem of § 174 the determinant is equal to a determinant of lower order. If the determinant has no element equal to ± 1, a suitable application of § 172, VIII, will transform it into one with at least one element equal to ± 1. When we get to determinants of the third order, it is best to expand them in full unless there is an obvious way to shorten the process. The student should always be on the lookout to see if the elements of any row (or column) are equal to w times the corresponding elements of any other row (or column). In this case the determinant equals zero (§ 172, VI). If the elements of any row (or column) are integers with a common factor, § 172, III, can be applied to advantage. 176. Theorem. — If the elements of a determinant D are polynomials in x, and if D = when x = a, then x— a is a factor of the expansion of D. This is an immediate consequence of the Factor Theorem (§ 142, Cor.), since the expansion of D is a polynomial in x. For example, if D = x^ X 1 4 2 1 1 1 1 it is evident that Z) = when a; = 1 and when a; = 2 (§ 172, V). Hence D is divisible by x — 1 and by a; — 2. Moreover it is clear from § 174 that D is of the second degree in x, and that the coefficient of x^ is 2 1 1 1 Hence X2 X 1 4 2 1 1 1 1 = (^-l)(^-2). 214 COLLEGE ALGEBRA EXERCISES 1. How many inversions are there in the arrangement 1, 7, 5, 8, 6, 2, 4, 3 ? In the arrangement 7, 5, 8, 6, 2, 4, 3 ? 2. How many inversions are there in the subscripts of ag ^5 Cq di 67 /4 ^3? How many inversions in the letters of ^1 «2 93 fi ^5 Co 67 ? Compare § 169. 3. Determine by inspection whether the following two determinants are equal or not: 3 -4 1 3 6 1 -6 6 2 7 9 -4 2 3 1 3 8 4 > 7 8 7 6 7 3 1 9 4 3 4. Show without expanding the determinants that 3 14 2 7 5 4 9 8-10 1 6 5 8 3 12 -f- 3 2 14 7 5 9 4 8-10 6 1 5 8 12 3 0. 5. Express the following sum as a single determinant of the fourth order : 15 2 6 1 5 4 8 + 2 rr i 3 10 8 8 2 2 2 6 1 5 4 8 7 7 3 10 2 8 2 2 Find the values of the following determinants : 6. 3 15 16 29 6 7 33 14 6 30 32 58 4 2 6 6 7. 3 4-2 4-3 8 2 8-3 DETERMINANTS 215 3 4 10 -6 8 5 10. 1 3 4 5 2 3 2 4 1 1 1 2 1 4 7 6 5 2 13 7 2 5 3 4 14 4 6 5 11. f Factor the following determinants : 12. a^ a^ X 1 1111 8 4 2 1 8 4 2. 1 2T ST 3 ^ 13. 3 + a; 4: — x x-\-l 5 2 3 6 1 14. Show that 111 X y z X^ y^ 2;2 is divisible hy x — y,y — z, and z — x. 15. Show that X 2/ 1 2 3 1 1 4 1 = when x=2, 2/ = 3 ; and also when x = — l, 2/ = 4. 16. Show that y 1 2 1 5 1 = is the equation of the straight line through the points (—1, — 2) and (3, 5). 216 COLLEGE ALGEBRA 17. Find three sets of values of x, y, and z for which X y z \ 5 3-21 4 111 6 1 = 0. 177. Solutions of systems of linear equations by means of determinants. — It was stated at the beginning of this chapter that determinants of order n could be used in the solution of systems of n linear equations in n unknowns. We are now in a position to see in detail how this is done. We shall consider the solution of a system of four linear equations in four unknowns. The method used is applicable to the general case. Let the given equations be axx + hxy + c^z + d\V3 = A^i, (1) a^x + 622/ + Ciz -}- d%w = ki, (2) aax -\- bsy + Csz + dsw = ks, (3) ttiX + biy + c^z + diW = k^. (4) The determinant formed by the coefficients of the unknowns, that is, «1 bi Cl di aa &2 C2 d2 as 63 cs dz a4 h C4 d. is called the determinant of the system of equations. We shall represent it by D, and shall assume that the equations are such that D=^0. If there is a set of values of x, y, z, and w that satisfy these equations, these values can be found in the following way : Multiply each member of equations (1), (2), (3), and (4) by Aij —A2, As, and —A^ respectively. This gives QiAix + biAiy + c\Aiz + diA\w = — QiAiX — b^A^y — CiA4. (10) (11) (12) (13) Now the right members of these four equations are equal to kx 61 Cl di k2 62 C2 d2 ks 63 Cs dz > ki 64 Ci di «i A-i Cl dx rt2 A:2 C2 d2 as A:3 Cs ds 1 04 ki Ci di ax bx kx dx a2 62 k2 d2 as 63 ks ds 1 Qi hi ki di ax bx cx kx a2 62 C2 ki as 63 Cs ks ai 64 Ci ki respectively. Therefore, if there is a set of values of x, y, z, and w that satisfy equations (1), (2), (3), and (4), these values must be given by the formulae 218 COLLEGE ALGEBRA ^•l hi Cl ^1 k2 62 C2 d2 ks &3 C3 dz k. &4 C4 d. ai ki Cl c?i a2 k2 C2 d2 as ks C3 2) + ksiaiAs - biBs + c^Cs - diDs) + kA(— OiAa + biBA — CiCa-\- 4) _ In the numerator of this fraction the coefficient of ki is D and the coefficients of A^s, k^, and kA -are «1 61 Cl di ai &i Cl d. as &3 C3 ds ttA ?>4 C4 dA ai 61 Cl d. a2 &2 C2 d2 a\ &i Cl di ' aA 64 C4 dA ai bi Cl d. a2 62 C2 d2 as 63 C3 ks ai &i Cl di respectively. Each of these last three determinants is equal to zero. Hence the fraction is equal to ki. That is, these values of x, y, z, and w satisfy equation (1). In a similar way DETERMINANTS 219 it can be shown that these values also satisfy equations (2), (3), and (4). The student should observe that these formulae for the solu- tion of equations (1), (2), (3), and (4) are closely analogous to the formulae for the solution of systems of two or three linear equations in two or three unknowns respectively. They apply with slight and obvious modifications to the solutions of sys- tems of n linear equations in n unknowns. 178. Inconsistent equations. — The argument by means of which these solutions were obtained breaks down in case i)=0. In this case the left members of equations (10), (11), (12), and (13), § 177, are all zero. Hence, unless the numerators of formula (14) are all zero, we have a contradiction, and the original equations, (1), (2), (3), and (4), are inconsistent. This argument can be extended to a system of n linear equations in n unknowns. We conclude therefore that if the determinant D of a system of n linear equations in n unknowns is zero, the equations are inconsistent unless all the determinants in the formulce analogous to (14) are zero. Example. — Show that the equations -2x + 4y + 5z-{-2w = 3, Sx-6y — 6z + iw = 6, 8x + 6y-\-z + 5w = 4:j are inconsistent. The equations may be inconsistent when D and all the de- terminants in the formulae analogous to (14) are zero. This question is, however, too complicated to be discussed here. 179. Homogeneous equations. —Equations (1), (2), (3), and (4) are said to be homogeneous when their right members are all zero. In this case, ii D^O, the only values of the unknowns that satisfy the equations are ic = 0, 2/=0, 2 = 0, and to = 0. If Z) = 0, there are other solutions. An analogous thing holds for a system of n linear homogeneous equations in n unknowns. 220 COLLEGE ALGEBRA EXERCISES Solve the following systems of equations by means of formulae (14) or similar formulae : 7. Sx-{-y-{-w=:20, 8z-^6x-\-w = ^0, Sz-\-x-\-4:y = 30, 5z + 8y + Sw = 50, 8. .4 + 2J5 + 3(7+42) = l, 4^+54-2 (7+3 J9+3 = 0, 3 A+A B-\-C-j-2 D+6 = 0, 2^+35+4 C+D-f 2 = 0. 9. A+2 B-hS C-D-2 E=5, 2^+J5+3 a-2Z>+^=4, B-h2C-2D-{-E=2, -B-C-\-2D + E = 0, 2B + SC-\-D-E = e. 4. 2x-{-Sy-z-^S = 0j 2x-3y + 3z = 2, — x-{-2y-i-5z=:5. 5. 3^ + 2^-0+D=2, A-{-3B+G-D = 5, A-{-B = 4:, 11. 3x + 2y-5z=:5, 0+D = 6, 4.x-6y-\-2z=:-10, 2x-^10y-12z = S. 6. —4:X-\-5y-^4:Z — 3w = 0, jx-{-y-\-z + w==34:, 12. A-5B-6C=2, 4:X-10y-\-4.z-w = 6, -3^-45 + 100=7, x-6y-Sz-{-5w=-ld. 10J. + 25-36O = -l. 1. 2x-y-^z = 5, y — 2z — 5w = — S, x + y + z-{-iv = 2, x — 2y-{-5iv = 6. 2. 5x — 2y + z + 3w — 7 = 0, 2a; + 4?/ + 32 + 2w = 8, x — y-~z-{-2v = 4:f 3x + y + 5z-\-4:W = 9. 3. x-\-y = a, y + z^h, z-\-w= c, 2w-{-x=^d. 1. x + y-\-z x — y-\-z 2x->f-3y- — w = -2W: = a, = 5, = Z, y-\-z = m. CHAPTER XV INEQUALITIES . 180. Definition of inequality. — We have considered in some detail how to identify numbers that are described by means of equations. In certain important problems, however, the infor- mation we are given concerning the numbers involved is more indefinite than that contained in equations. It merely tells us that one thing i^ greater or less than another. Such a state- ment is called an inequality. Inequalities are used only in connection with real numbers, and the real number a is greater than the real number h if a— 6 is positive, and a is less than 6 if a— 6 is negative. (§ 3.) The expression a ^ 6 means that a is greater than, or equal to, 6, and a^h means that a is less than, or equal to, h. Two inequalities like a>h and c > d, in each of which the greater number appears on the same side, are said to be alike in sense. On the other hand, a>6 and c is an unconditional inequality. Such a description gives us no information whatever about the numbers involved. Many inequalities on the other hand are satisfied only by certain numbers. These are called conditional inequalities. Thus a; + 1 > is a conditional inequality, since not every real number when added to 1 gives a sum greater than 0. 221 222 COLLEGE ALGEBRA 182. Properties of inequalities. — In order to identify as closely as possible the numbers described by means of an in- equality we proceed very much as in the solution of an equa- tion. The justification of such a procedure is found in the following theorems : I. Tlie sense of an inequality is not changed if both sides are increased or decreased by the same number. For if a>b, then a — b = 7c, where Zc is a positive number, and a-]-n — b — n = k, w^here n is any real number, positive or negative ; or (a 4- n) — (6 -f n) = k. Hence a-\-n>b -\-n. This demonstration applies, of course, to inequalities of the opposite sense since the statement that a>^ is equivalent to the statement that b that was referred to in § 181 is equivalent to the inequahty x >— 1. II. The sense of an inequality is not changed if both sides are multiplied or divided by the same positive nurnber. For if a > b, and m is any positive number, then a — b = 1c, where A; is a positive number, and am — bm = km. But since k and m are both positive, km is positive, and therefore am > bm. III. The sense of an inequality is reversed if both sides are multiplied or divided by the same negative number. The proof of this is left to the student. INEQUALITIES > 223 It follows from Theorem III that the sense of an inequality- is reversed if the signs of all its terms are changed. Thus, if a > ft, then —a< — b. 183. Conditional inequalities. — For our purpose the most important conditional inequalities are those involving one un- known to the first or second degree. These are called linear and quadratic inequalities respectively. By Theorem I of the preceding article, every inequality in- volving one unknown can be put into the form f(x) > 0. If f(x) is linear in x, the inequality is of the form ax + b>0. Hence a; > , a or x< , a according as a is positive or negative. If f(x) is a quadratic in x, we have ax^ -f- &a; -}- c > 0. Now by § 148, ax^ -{- bx -\- c = a (x — r^) (x — rg), where Vi and rg are the roots of the equation aoi^ -\-bx-\-c = 0. If Vi and rg are imaginary, ax^ -i-bx-^c has the same sign for all values of x, since if it had opposite signs for x = l and x = m, the equation would have a real root between I and m (§ 146). This sign is the same as the sign of c inasmuch as ax^ -{-bx-{-c = c when x = 0. Since the condition for imaginary roots is (§ 71) 6^— 4 ac< 0, we see that a and c must have the same sign when the roots are imaginary. Hence in this case ax^ -\-bx-{-c has the same sign as a for all values of x. ^ 224 COLLEGE ALGEBRA If Vi = r2, (6^ — 4 ac = 0), we have aa^ + 6a; + c = a (a; — 7\y. Hence aoi^ -{-bx-hc has the same sign as a for all values of X, except x = o-^. For this value ax^ -\- bx -\- c = 0. If Vi and 7*2 are real (6^ — 4 ac > 0) and Vi < r2 we have ax^ -{- bx -^ c = a(x — ri) (x — r^. Then when a is positive aaj^ + 6a; + c is negative for all values of x between i\ and rg and positive for all other values except 1\ and rg. When a is negative, aa^ + 6a; -f c is positive for all values of x between rj and r^, and negative for all other vahies except rj and rg. The following table exhibits these results in convenient form : a &2 - 4 ac ax"^ + bx-^ c + + + + + + + , except for x = n. — for values of x between n and r^ ; + for all other values, except n and Vi. -, except for a; = n. + for values of x between n and r^ ; — for all other values, except ri and Vi. Example. — For what values of a: is a;^ — 10 > 5 + a: — a;^ ? By Theorem I this inequality is equivalent to 2 x2 - X - 16 > 0. Here n =— f, r ± V-- All we can conclude is that either and these two inequalities lead to the results already obtained. 184. Graphical interpretation of linear and quadratic in- equalities. — The values of x for which aa; + 6 > are the abscissae of the points on the locus of the equation y = ax-{-h that are above the a^axis, and the values of x for which aa; + 6 < are the abscissae of the points on this locus that are below the rc-axis. The value of x for which y = ax-{-b is the abscissa of the point of inter- section of this locus and the cc-axis. Thus, a glance at the locus of the equation y = Sx-\-6 tells us for what values of x the expression 3 X + 5 > 0, = 0, or < 0. In a similar way, an inspection of the locus of the equation y = ax"^ -h bx + c tells us for what values of x the expression ax^ + 6a; + c > 0, = 0, or < 0. Suppose we wished to solve graphically the inequality considered in the latter part of the preceding article. We draw the locus of the equation y = 2x^ — x— 16 and note the ahscissse of the points on the locus that are above the a;-axis. 226 COLLEGE ALGEBRA A+ -V EXERCISES 1. Does the proof given in Theorem II of § 182 cover the case in which both members of the inequality are divided by the same positive number ? 2. Prove that a? -\-y^>2xy, whenever x and y are unequal real numbers. Hint. — This inequality is by Theorem I equivalent to 3. Show that x^->2 ^x - O"^ > ^y + y^ ■\- ^^• What if X and y are equal ? 5. Which is the greater, the arithmetic mean, or the posi- tive geometric mean, of two unequal positive numbers ? Find for what values of x the following inequalities hold and interpret your results graphically : 6. 2 a; — 5 < a; + 2. 7. 8. Ta;-f 2a;>3ar^ + 5a; + l-f4«2_33. 9. 5a? + 6<0. 12. ar^>4. 13. (2a;-3)2<9. <0. 14. a;(a;-2) =0. 11. _4a;2_p20x-25<0. i>0. X 2 -1 3a; X 3a; 4 < 2 5 5 2 1 f>0. 10. a^ + 2 X + 5 ] = 0. l<0. INEQUALITIES 227 15. 3 2-1 X 2 X x-\-l — ^xx — 2 3x > r>o. 16. (;2x-\-5)(x-2)\ =0. [<0. f >1. 17. (x-2f X 2x 4 7 1 -2 -6 21. 10-x>2a?. x + ^ 22. 1. 1<1. 23. 18. 25a^ + 18a;4-90<0. 19. {x-r)(x-2){x-3)>Q. 24. -A 20. x{x-{-lf<0. x + 2 2a;-l 3 a; - 4 >0. <0. x — b >0. = 0. -l)3 - (2 a; + 6)' 18. i^±3 g 3ar'-5o0 jg ^ + 3 (3 a; -5)^(2 a; + 1) (a^-9)(a!-3) 10. ,^±1 . 20. 6 (a^-4)2 (a;^ + 3a;H-2)' 234 COLLEGE ALGEBRA 190. If we apply the preceding methods to the decomposi- tion of ; , we get (a^ + l)(a; + d) 20a;-f-40 _l + 7i l-7i 2 (a^ + l)(a; + 3) x-\-i x — i a; +3 The appearance of imaginary coefficients in this decomposi- tion can be avoided by not carrying the decomposition so far. Thus if we retrace our steps in part by expressing the sum of the first two fractions of this decomposition as a single fraction, ^^^^^ 200^ + 40 ^ 20^ + 14 2 (a^ + l)(a; + 3) a;2 + l a; + 3* Here there are no imaginary coefficients in the final form and the denominators are of the first or second degree. If the coefficients of the original fraction are real, the decomposition can always be stopped at a point at which the coefficients are real and the denominators are of the first or the second degree or powers of such expressions. By § 148, we know that any polynomial, as g {x), of degree n is the product of n factors. Moreover if the coefficients of g {x) are real, the imaginary factors (that is, those with imaginary coefficients) occur in conjugate pairs (§ 151), if at all. The part of the decomposition that corresponds to such a pair of imaginary factors of the first degree will be of the form A B x — a — ih x — a-\-ib But this is equal to (A + B)x-^(-Aa-Ba-\- iAb - iBb) 3f-2ax-\-a^-]-b^ Then in the decomposition we can put the single fraction AiX 4- Bi x^-2ax + a^-\- W' where A^=A-\-B and B^= — Aa — Ba-{- iAb — iBb. PARTIAL FRACTIONS 235 It can be shown in general that if n^ is the highest power of x — a — ib and of x — a + ib by which the denominator is divis- ible, then the sum of the partial fractions whose denominators are the different powers oi x — a — ib and of x — a -\-ib is equal to the sum of fractions of the form ^ A^x-hB^^ A„^_,x 4- ^n,-i .. (a;2 -2ax-\- a^+ b^y^' (x" - 2 aa; + a^ + 62)n,-i' '"> A,x -f B, x^-2ax + a'^-\-b^' where the ^4's and ^'s do not contain x. Then in the original decomposition in place of a pair of frac- tions whose denominators are conjugate imaginaries we can place a single fraction whose numerator is of the first degree with undetermined coefficients and whose denominator is a real factor of g (x) of the second degree or a power of such a factor. The resulting equations for determining the undetermined con- stants in the numerators of the partial fractions will have only real coefficients and will be of the first degree. . Hence these undetermined constants must have real values. This partial decomposition of a rational fraction has there- fore the advantage of involving only real coefficients, and it is sufficient for the uses to which decomposition into partial fractions is put. Example 1. — Decompose ^-i into real partial fractions. (x2 +!)(:*: + 3) Put 20 a; + 40 _ Ax -\- B C (x2 4-l)(a; + 3) x^ -^ 1 x + s" Clearing of fractions we get 20 x + 40 = {Ax + 5)(x + 3) + 0(^2 + 1). Equating coeflBcients of like powers of x, = A+C, 20 = SA-\-B, 40 = 35 + 0. 236 COLLEGE ALGEBRA Whence A = 2, B = li, C = — 2, and the decomposition is 20 X + 40 ^ 2 a; + 14 2 (a;2+l)(x + 3) x^+1 x + s' Q y K Example 2. — Decompose into real partial fractions. (x2 + a: + l)2(x + 1) 3x-5 Ax-\-B , Cx+ D , E {X^ + X + lf{X +1) (iC2 + X + 1)'-^ x^ + X+1 X + 1 Clearing of fractions, 3x-5 = (^x + -B)(x+ l) + (Cx + i>)(x2+a;+ l)(x+l)+^(x2+x+l)2. Equating the coefficients of like powers of x, G + E=0, 2C+D + 2E = 0, A-\-2C+2D-\-SE = 0, A-\-B-^C+2D + 2E = S, B + D-\-E = -5. Whence J. = 8, 5 = 3, 0=8, Z)=0, E=-S; and the decomposition is 3x-5 ^ 8x + 3 8x 8_ (x2 + x + l)2(x+ 1) (x2 + x + l)2 a:2 + x+l x + l' EXERCISES Decompose the following fractions into real partial frac- ^tions : 6. '^+^ ' 1. 2x^-5 ^ 2. Zx-2 *K a^_5aj2 + 2a; + l a^ + 5a;2 -1-4 d 4. a^ + a;2 + a? (a;2 + 2aj + 5)(a;-l)2 4 a;2 -{- a; — 6 6 8. (05^+ 1)^ PARTIAL FRA.CTIONS 237 11 a^ 4- 11 15 x' + x + 2 (2a; + 3)(»2 + x-f-3)' * ^^x'^^x + l ,^ a;2 + 2a; + 3 ,. 4a^4-4a;2 4- 8 cc + 6 lU. • Id. — • a?"-! x<+2a;2 11 ^±^- "• T^- 3a;-l 18. -2x^-9a;-10 /^ -f 1)2* cc^ + 6 a;2 + 5 a; 13. 4a;H-5 .o 2a^4-14 2/4-81 a^-2ic3 + l CHAPTER XVII LOGARITHMS 191. Definition of irrational exponents. — In § 58 it was stated that irrational exponents could be used in a way con- sistent with the five fundamental laws of exponents. This statement is based upon the meaning we attach to these exponents. We cannot give a complete justification of it here, but shall indicate briefly what this meaning is. In the first place the student should recall that a decimal fraction can always be written as a common fraction, and that accordingly a number with a decimal exponent can be ex- pressed in a way with which he is familiar. For example, Now, as he can readily verify, the numbers 1, 1.4, 1.41, 1.414, 1.4142, ..., which are obtained in the process of finding the approximate square root of 2, are such that their successive squares are closer and closer to 2. As a matter of fact, it can be shown that when this sequence is continued indefinitely the successive numbers approach a certain definite limit, which is the positive square root of 2. (See § 202 for the definition of a limit.) It can be shown further that the numbers of the corre- sponding sequence a\ a}\ a'*\ a}'^\ a^-^^V •••, also approach a limit. It is this limit that we denote by the symbol a^~^. 238 LOGARITHMS 231 In general, every irrational number n is the limit of a sequence of rational numbers Wu ^2, %, .••, and the numbers of the sequence a**!, a**!, a**3, ... approach a limit, which we indicate by the symbol a^ 192. Definition. If a>0 and a' = h, x is said to be the logarithm of b, and b the antilogarithm of x, to the base a. The symbol for the logarithm of b to the base a is log^ b. Thus, 25 = 32, and therefore 5 is the logarithm of 32, and 32 the anti- logarithm of 5, to the base 2, or 5 = log2 32. EXERCISES 1. log39= ? log93= ? log6l= ? log,a= ? \og,a^= ? 2. Complete the following table : ^ Number Base Logarithm 243 3 2 8 10 3 5 4 2 16 343 3 3 i 3 A1 ^. ^ '.^ ^^^.'> 193. We can take for the base a any positive number except 1. Theorem. — For any positive number b and any positive numr ber a, except 1, there is a real number x such that a^^b. 240 COLLEGE ALGEBRA In other words, every positive number has a real logarithm with respect to any positive base except. 1. The proof of this theorem is too difficult for this book, and tjLfi/efore is omitted. Negative numbers have no real logarithms. It can be shown that the logarithm of a negative number to a positive base is imaginary. We shall confine our attention here to real jogarithms. 194. The logarithms of all positive numbers with respect to any given positive base different from 1 form a system of real numbers that possess the following properties : I. Tlie logarithm of the product of two or more numbers is equal to the sum of the logarithms of these numbers. For, if log« b = x, log« c = y, or a' — b and a^ = c, then by multiplication a'^+y = be. But this is only another way of saying . log^ 6c = a; + 2/ = ^^^a & + log„ c. This proves the theorem as far as concerns the product of two numbers. The completion of the proof is left to the student. II. Tlie logarithm of the quotient of two numbers is equal to the logarithm of the dividend minus the logarithm of the divisor. For, if log„ b = x, log« c = y, or a' =b, a^ = c, then by division, a*"" = - • c But this is only another way of saying log« - = x-y = log^ b - loga c. c LOGARITHMS 241 III. If m is any real number, the logarithm of b"^ is equal to m times the logarithm of b. For, if loga b = x, or a^ = by then we have, by equating the mth powers of the two members of this equation, ^«.^^„^ Hence log^ b'^ = mx = m » log^ b. If in III m is equal to -, where n is an integer, we have, n the logarithm of the positive nth root of a positive number is equal to the logarithm of the number divided by n. EXERCISES Given log^ 2 = .3010, logio 3 = .4771, logio 7 = .8451 logg 7 = 1.7712, logg 11 = 2.1827; find: 1. logio 14. Hint. — 14 = 2 • 7. Hence logio 14 = logio 2 + logio 7. 2. logaV^. 7. logio V2«.7^ 3. logio^v^. 8. logio 2058. 4. logsJ^^ 9. log3V7. 5. logio V2i. 10. loga^^TT. 6. log3^5^5929. 195. The systems of logarithms most frequently used. — For reasons that are explained in the Differential Calculus it is convenient for certain purposes to use as base of the system of logarithms an irrational number whose approximate value is 2.71828. Logarithms to this base are called natural logarithms. On the other hand, for purposes of numerical computation, there is an advantage in using the base 10. The reason for this will be explained in § 198, I. Logarithms to the base 10 are called common logarithms. 242 COLLEGE ALGEBRA 196. Change of base. — Since different systems of logarithms are in use, it is important to know how to change from one system to another. The following theorem explains how this can be done. Theorem. — The logarithm of a number to the base b is equal to the product of its logarithm to the base a and the logarithm of a to the base b. If log„ X = u and logj x = v', that is, if a''=x and b" = x, then a" = 5". Hence a = b'^, or - = logj a. u Hence v = U' logj a, or logj a; = log« aj • logj a. This proves the theorem. The following theorem enables us to express this relation in another form, which is sometimes useful. Theorem. — The logarithm of a to the base b is the reciprocal of the logarithm of b to the base a. For, if logj, a = x, or a = 6* ; I 1 then a"" — b. or log„ 6 = - . X Hence log. a = :; • log«6 The relation proved in the preceding theorem can now be written thus: 1 197. Common logarithms. — We shall confine our atten- tion in the rest of this chapter to common logarithms and it will therefore be unnecessary to indicate the base in the LOGARITHMS 243 symbol for a logarithm. Thus we shall write log 2 instead of logio 2. Now it is easy to see what the logarithm of 100, or of 1000, is ; but it is not easy to determine the logarithms of most numbers, say 20, for example. The approximate value of the logarithm of such a number (that is, of a number that is not an integral power of 10) can be determined only by a series of computa- tions that are too long and complicated to be explained here. The integral part of a logarithm is called the characteristic of the logarithm and can be determined by inspection in a way that we shall explain (§ 198). The fractional part of the log- arithm is called the mantissa and is given in a table. Thus, the approximate value of the logarithm of 20 being 1.3010, the characteristic is 1 and the mantissa .3010. By virtue of their properties given in § 194, logarithms are extremely useful in shortening numerical computation, and it is for this reason we are considering them here. Suppose, for example, we want the fifth root of 20. We have already seen that j^ggo = 1.3010. But by § 194, III, log v/20 = ^ log 20. Hence log v^20 = .2602. Now the number whose logarithm is .2602 can be determined approxi- mately from the table in a way to be explained in § 198, II. And when this is determined we shall have the approximate value of \/20. The student would appreciate the advantage in this use of loga- rithms if he were to attempt to find the fifth root of 20 directly. The operations of multiplication, division, and raising to powers can also often be shortened by the use of logarithms. But for this numerical work it is necessary to use a table of logarithms. 198. Use of a logarithmic table. — We proceed therefore to explain the two ways in which the table can be used. The procedure is based upon the fact that if a and b are greater than 1 and b is greater than a then log b > log a. 244 COLLEGE ALGEBRA- I. To find the logarithm of a number. — The table in this book is so arranged that the mantissa of the logarithm of a num- ber of three digits can be taken out directly. To do this we look in the first column for the first two digits (counting from the left) and then over on this row to the column headed by the right-hand digit. The number in this row and column is the mantissa we are looking for. It is understood that all the numbers in the body of the table are to be preceded by the decimal point. Thus, the mantissa of the logarithm of 483 is .6839. The characteristic is determined without the use of the table from the following considerations: 10« = 1, orlogl = 0. 10^ = 10, or log 10 = 1. 102=100, or log 100 = 2. 103 = 1000, or log 1000 =3. Hence the logarithm of any number between 1 and 10 lies between and 1, and has therefore the characteristic 0. Moreover every such number (when written as an integer or a decimal) has one digit to the left of the decimal point. The logarithm of any number between 10 and 100 is between 1 and 2 and has the characteristic 1. Such a number has two digits to the left of the decimal point. In general, since 10** is the smallest integer with n-\-l digits the characteristic of the logarithm of a number greater than 1 is one less than the number of digits to the left of the decimal point in the number. We agree to say that the numbers 2015, 201,500, 2.015, and .0002015 have the same sequence of digits. In general, two numbers that have the same digits in the same order and dif- fer only in the position of the decimal point and in the ciphers that may be necessary to indicate the position of the .decimal point are said to have the same sequence of digits. LOGARITHMS 245 The position of the decimal point in a number can be changed any number of places to the right or left by multi- plying or dividing the number by some integral power of 10. Moreover the logarithm of any integral power of 10 is an integer. For example, log 432 = 2.6355 ; that is, 432 = 1026355. Dividing each member of this equation by 10, we get 43.2 = 101-6355. Hence, log 43.2 = 1.6355. In a similar way we see that log 4.32 = 0.6355, log .432 = .6355-1, log .0432 = .6355 -2, log .00432 = .6355 -3, From this we see that, for example, log .432 = .6355 - 1 = - .3645. But a logarithm is usually not written in this way, but thus, 9.6355 - 10. Similarly, we say log .0432 = 8.6355 -10, log .00432 = 7.6355 - 10. This agreement always to write the logarithm in such a way as to have its mantissa positive is based on grounds of con- venience in the use of the tables. If there are n ciphers immediately following the decimal point in a number less than 1, the characteristic will be taken as — n — 1. It is convenient to write this 9 — 71 — 10, 246 COLLEGE ALGEBRA If we agree to write the logarithms of numbers less than 1 in the way indicated here, we can make the following statements : The mantissa of the logarithm of a number is always positive. Tlie logarithms of two numbers that have the same sequence of digits have the same mantissa and differ only in their charac- teristics. The characteristic of the logarithm of a number less than 1 is equal to 9 — w — 10, where n is the number of ciphers immedi- ately following the decimal point in the number. This character- istic is written in two parts. Tlie first part, 9 — 7i, is written at the left of the mantissa and the — 10 at the right. Thus log .00432 = 7.6355 - 10. Here w = 2 and 9 — w = 7. It is clear from what has been said that in looking in the table for the mantissa of the logarithm of a number we do not need to pay any attention to the position of the decimal point in the number, — the characteristic is the only thing about the logarithm that is affected by the position of the decimal point. The finding of the logarithm of a number of more than three digits from this table is not so simple. The method of pro- cedure is best illustrated by an example. Find log 5421. From the tahle log 5420 = 3. 7340, log 5430 = 3.7348. These two logarithms differ by .0008 and correspond to numbers that differ by 10. Now the numbers 5420 and 5421 differ by 1 and we assume that the difference in their logarithms is .1 of the difference in the loga- rithms of 5420 and 6430 ; that is, .1 of .0008. This is .00008. Hence log 5421 = 3.7340 + .00008 = 3.7341. If we were looking for the logarithm of 5427, we should add .7 of .0008 to the logarithm of 5420 since the difference be- LOGARITHMS 247 tween 5427 and 5420 is .7 of the difference between 5430 and 5420. In doing this work it is customary, in the interests of brevity, to omit the decimal point in giving the difference between the logarithms of two numbers. Thus, in the ex- ample just given, we say that the difference between the logarithms of 5420 and 5430 is 8, instead of .0008. We have assumed that the difference between the loga- rithms of two numbers is proportional to the difference between the numbers. This is not strictly true. How- ever, when the difference between the numbers is small, as it is in these cases, the error due to this assumption is very slight and can safely be neglected, since most of the logarithms given in the table are themselves only approximations. In using foar-place tables the student should keep the anti- logarithms in all his calculations down to four significant fig- ures. The point of this remark is illustrated by the fact that 43.526 and 43.53 have the same four-place logarithm. From the table log 43. 600 = 1.6395 log43.500 = 1.6385 Difference in logs. = 10 The difference between 43.600 and 43.526 is .26 of the difference be- tween 43.500 and 43.600, and .26 of the difference between the logarithms of these last two numbers is 3. Hence log 43.526 = 1.6388. On the other hand the difference between 43.500 and 43.53 is .3 of the difference between 43.500 and 43.600, and .3 of the difference between the logarithms of these numbers is 3, Hence log 43.53 = 1.6388. When the digits to the right of the fourth place (counting from the left) are dropped off, increase the digit in the fourth place by 1 if the digit in the fifth place is greater than 5, or if the fourth digit is odd and the fifth one is equal to 5 ; in other cases leave the fourth one unchanged. 248 COLLEGE ALGEBRA EXERCISES Find the logarithms of the following numbers : 1. 461. 8. 24.86. 15. 3. 2. .0024. 9. 2562. 16. 12. 3. 500000. 10. 14380. 17. 58.64. 4. .000005. 11. .06473. 18. 5437. 5. 1934. 12. 374.6. 19. .6892. 6. 7.832. 13. 8001. 20. 40.15. 7. .6294. 14. 19.03. II. To find the antilogarithm of a logarithm. — If the man- tissa of the given logarithm is found in the table, it is an easy matter to find the antilogarithm. Suppose, for example, we wish to find the antilogarithm of 7.9258 — 10. Now .9258 appears in the table as the mantissa of the log- arithms of those numbers whose sequence of digits is 843. Since the characteristic is 7 — 10, the antilogarithm must be less than 1 and there must be two ciphers immediately following the decimal point. Hence antilog (7.9258 - 10) = .00843, or log .00843 = 7.9258 - 10. When the given mantissa is not found in the table the matter is not so simple. The method of procedure in such cases is illustrated by the following example : Find antilog 2.6959. In this part of the tables when a four-place number is in- creased by 10, the mantissa of its logarithm is increased by 9. Hence we* assume that if a mantissa is increased by 1, the number it corresponds to is increased by ^'. Since the man- tissa 6959 is obtained by increasing the mantissa 6955 by 4, the antilogarithm of 6959 should be that of 6955 increased by 4x^ = 4. Hence the given mantissa must correspond to numbers whose sequence of digits is 4964. Since the charac- LOGARITHMS 249 teristic is 2, the antilogarithm must be greater than 1, and it must have three digits to the left of the decimal point. Hence antilog 2.6959 = 496.4, or log 496.4 = 2.6959. The difference between the mantissae of the logarithms of two consecutive numbers is called a tabular difference. The preceding example suggests the following rule for find- ing the antilogarithm of a logarithm whose mantissa does not appear in the table. Rule. — Find two consecutive mantissce in the table between which the given mantissa lies, and get the difference between them. Multiply the difference between the smaller of these mantissce and the given mantissa by 10 and divide the product by the tabular difference just found. Annex the quotient, expressed in the nearest integer, to the sequence of three digits corresponding to the smaller mantissa of the table. In the resulting sequence of digits place the decimal point as indicated by the given characteristic. EXERCISES Find the antilogarithms of the following : 1. 2.65^2. 11. 2.0312. 2. 9.9805-10. 12. 3. 3. 1.8457. 13. 8. 4. 8.1644-10. 14. .8354. 5. 7.8162-10. 15. .1870. 6. 1.6245. 16. 1.8125. 7. .1287. 17. 9.8449-10. 8. 9.1287-10. 18. 7.3950-10. 9. 1.1287. 19. 2.6045. 10. 8.9970-10. 20. 8.4857-10. 250 COLLEGE ALGEBRA No. 1 2 3 4 5 6 7 8 9 lO 0000 0043 cx)86 0128 0170 0212 0253 0294 0334 0374 II 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 13 "39 1173 1206 1239 1271 1303 1335 13^7 1399 1430 14 1461 1492 1523 ^553 1584 1614 1644 1673 ^703 1732 15 1 761 1790 1818 1847 1875 1903 1931 1959 1987 2014 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 18 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 23 3617 3636 3655 3674 3692 371^ 3729 3747 3766 3784 24 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 26 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 27 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 28 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 32 5051 5065 5079 5092 5105 5"9 5132 5145 5159 5172 33 5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 34 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 35 5441 5453 5465 5478 5490 5502 55H 5527 5539 5551 36 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 37 5682 5694 5705 5717 5729 5740 5752 5763 5775 5786 38 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 39 59" 5922 5933 5944 5955 5966 5977 5988 5999 6010 40 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 44 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 45 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 47 ^.r' 6730 6739 6749 6758 6767 6776 6785 6794 6803 48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 49 6902 691 1 6920 6928 6937 6946 6955 6964 6972 6981 50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 51 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 53 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 54 7324 7332 7340 7348 7356 7364 7372 7380 7388 7396 "So, 1 2 3 4 5 6 7 8 9 LOGARITHMS 251 No. 1 2 3 4 6 6 7 8 9 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 56 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 59 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 66 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 72 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 89 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 91 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 94 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 9 No. 1 2 3 4 5 6 7 8 252 COLLEGE ALGEBRA 199. Numerical computation by means of logarithms. — It is of great importance for the student to be able to perform fairly complicated numerical computations with accuracy. He must not think that a numerical error is of small conse- quence so long as his work has been done on the correct prin- ciple. In all but the simplest computations he will find it worth while, in the interests of accuracy, to give attention to the proper arrangement of his work. The following general arrangement is suggested : Find the value of 5.25x89.46, 17.35 X 38.61 log 5.25 = 0.7202 log 17.35 = 1.2393 log 89.46= 1.9516 log 38.61 = 1.5867 log 5.25 X 89.46 = 12.6718 - 10 log 17.35 x 38.61 = 2.8260 log 17.35 X 38.61 = 2.8260 log 5.25x89.46^ 9.8458-10 ^ 17.35 X 38.61 5.25 X 89.46 17.35 X 38.61 .7012. In this work we changed the form of log 5.25 x 89.46 by adding and subtracting 10 in order to avoid a negative form for the logarithm of the required quotient. It is best to lay out all the work before looking up any logarithms. Thus, in the preceding problem the work was arranged in the follow- ing way before any logarithms were filled in : log 5.25 = 0. log 17.35 = 1. log 89.46 = 1. log 38.61 = 1. log 17.35 X 38.61 = log 5.25 log 17.35 X 89.46 = X 38.61 = lo£r ^-25 X 89.46 ^°^ 17.35 6.26 X 38.61 X 89.46 17.35 X 38.61 It is suggested that the student find the value of this frac- tion without the use of logarithms and compare his result with that obtained aboVe. LOGARITHMS 253 The following example illustrates the use of an important artifice in logarithmic computation : Find v^.OST log .05 = 8.6990 - 10. .-. log VM = 2.S997 - i^. Now it is inconvenient to have the fraction -^ occurring here, and we can avoid it by writing log .05 = 28.6990 - 30. Then log v^.'OS = 9.5663-10. The general procedure in such cases is to arrange the loga- rithms in such a way that the final logarithm will appear in the standard form. The principal advantage in the use of logarithms for per- forming the operations of multiplication, division, and raising to an integral power depends upon one's facility and accuracy in the use of logarithmic tables. But in the extraction of a root there is a further important and obvious advantage. See for example the illustration on page 243. EXERCISES AND PROBLEMS Perform the following indicated operations by the use of logarithms : 1. .372x4.26x37.1. 9. (6.21)'(9.432)5-- (46.74)1 2. 98.41x39.72. 3/ 3.96x4.87 3. (9.23)2(.761)3. * \13.9x (5.67)2* 4. 1246 X 923.4 X. 001672. ii. (.8762)2(9.436)2 x (.7582)5. 5- ^2. 12^ V6943. 6. 9762 -^ 48.75. 13. ^^^2 3762 X 7931 7963 8. ^100. * ^4.932 254 COLLEGE ALGEBRA (12)3(496)'^ X 976.4 17. ViO. * 5621 X 498.6 x 71.34' ^g ,^/j4 3/ 19.3 x(47.21)^(5.931)^ ^^- ^2^- \ 4500 20. i/iU 16. 4500 20. V(468/. 21. What is the weight in tons of a marble sphere 2i feet in diameter if a cubic foot of water weighs 62.36 pounds and the speciJBc gravity of marble is 2.7 ? 22. When a weight of m grams is attached to the free end of a suspended brass wire, the wire stretches S centimeters. If I is the length and r the radius of the wire in centimeters, g = 980, and k = 1051 . 10^ then irr'k Compute S when m = 750, I = 164, and r = .4. 23. What will $850 amount to in 10 years at 4 %, interest being compounded annually ? This example affords a good illustration of the limitations of a four- place logarithmic table. We have A - 850 (1.04)io, and therefore log A = log 850+ 10 log 1.04. Now all that we can be sure of in a four-place table is that no logarithm can be more than .00005 too much or too little. Accordingly in this ex- pression log 850 is subject to an error not greater than .00005 and 10 log 1.04 to an error not greater than .0005, and therefore log A as computed in this way cannot be more than .00005 -\- .0005(= .0006) too much or too little. The computed value of log A is .3.0994 and hence its true value lies between 3.1000 and 3.0988. The true value of A therefore lies between $ 1260 and 1 1255, while the computed value is 1 1257. A six-place table shows that the true value of A lies between 1 1258.21 and $ 1258.18. A seven-place table would give A correct to the nearest cent. In practice the nature of the problem determines the number of places that should be given in the table used. The greater the number of places given in the table the greater the accuracy of the computations. 24. What sum will amount to $1500 in 5 years at 3%, in- terest being compounded annually ? LOGARITHMS 255 25. If a string weighing .0098 grams per centimeter is stretched between two bridges 60 centimeters apart by a weight of 10 grams, how many vibrations will it make per second ? ' (See Ex. 1, p. 105.) 26. The weight w in grams of a cubic meter of aqueous vapor saturated at 15° is given by the formula ^^ ^ 1293 X 12.7 X 5 ^ , "" (1 + J^%)760x8- Compute w. 27. The volume v in litres of three kilograms of mercury at 85° is given by the formula 13.6 V 5550y Compute V. 28. The time t of oscillation of a pendulum of length I centi- meters is given by the formula Find the time of oscillation of a pendulum 74.36 centimeters long. 29. What must be the length of a pendulum in order that its time of oscillation be one second ? 30. The velocity -y of a body that has fallen s feet is given by the formula / ,.. .. What is the velocity acquired by a body falling 29 feet 7 inches ? 200. Exponential equations. — An exponential equation is an equation that involves the unknown in an exponent. The simplest exponential equations can be solved by inspection. For example, if 4== = 16, it is immediately obvious that a = 2. 256 COLLEGE ALGEBRA Many exponential equations that cannot be solved by inspec- tion can readily be solved by the use of logarithms in the way illustrated in the following example : Solve 2'' = 5. Since the logarithms of equal numbers must be equal, we can equate the logarithms of the two members of this equation. In this way we get log 2=« = log 5, or ic log 2 = log 5. Hence ^ = f^g^ ^ ..6990 ^ 2.32. log 2 .3010 Note. — The student should observe the difference between -^^ and ,5 log 2 log-. EXERCISES AND PROBLEMS Solve the following equations : 1. 3^ = 4. 3. (1.03)^ = 2. 6. (1.02)^ = 5. 2. 5^ = 126. 4. 6'=' = 21. 6. 72''+i = 12. 7. In how many years will a sum of money double itself at 3 %, interest being compounded annually ? In one year $ 1 will amount to $ 1.03 ; at the end of two years the amount will be 1.03 x 1.03 =(1.03)2; and at the end of x years the amount of % 1 will be {\my. Hence x must be such that (1.03)* = 2. (See Ex. 3.) 8. In how many years will a sum of money double itself at 3 %, interest being compounded semiannually ? 9. In how many years will $ 1000 amount to % 1338 SitQfoj interest being compounded annually ? If p dollars amount to a dollars in n years at r%, compounded an- nually, then « = K' + i5o)- 10. Solve this equation for^^, r, and n in turn. When solving for n, and for r when 7i > 2, it is best to use logarithms. 11. What must be the rate of interest in order that $2500 may amount to $2814.00 in four years, interest being com- pounded annually ? CHAPTER XVIII VARIATION 201. Constants and variables. — Many of the letters that we use in mathematics represent only one number throughout a given discussion. They are called constants. A letter, on the other hand, that represents different numbers in one discussion is called a variable. Thus, if t represents the time that has elapsed since a given moment, it is a variable, since this time is changing. Also in the equation y = x^-\-Qx-\-^ the letters x and y are variables, since they represent any two numbers that are related to each other in the way described by the equation. In some problems two or more variables are involved, and in such cases there is usually a connection between the variables. The particular form of this connection depends upon the nature of the problem, but there are certain forms that occur much more frequently than others. These are described in the following paragraphs. If two variables are so related that their ratio is constant, either one is said to vary directly as the other. Sometimes the word " directly " is omitted, and we say that one of the variables varies as the other. The two expressions have the same meaning. If a body is moving at a uniform rate, the distance passed over varies as the time, since - = a constant ; namely, the rate. The symbol oc when placed between two numbers indicates that one varies as the other. Thus, doct means that d varies as t^ or that- = A;, where k is a constant. 257 258 COLLEGE ALGEBRA If two variables are so related that their product is constant, either one is said to vary inversely as the other. If x varies inversely as y, it varies directly as the reciprocal of y, since if xy — k, then^ = fc, or x = -. We can accordingly express that X varies inversely as y by the symbol icoc-. If a body moving at a uniform rate of r feet per minute goes a distance of d feet in t minutes, then d = rt. In order to get over the distance of d feet in a different time the rate must be changed in such a way that the product of the new rate and the new time still equals d. Hence, under these cir- cumstances, the rate varies inversely as the time. If three variables are so related that the ratio of the first one to the product of the other two is constant, the first one is said to vary jointly as the other two. Thus, if x varies jointly as y and z, then — = k, ov x = kyz, where A; is a constant. yz The first variable is said to vary directly as the second one and inversely as the third if it is equal to the product of a constant and the ratio of the other two. Thus, x varies directly as y and inversely as z if x = -^, where Z: is a constant. If the first variable is equal to the product of a constant and the ratio of the second to the square of the third, it is said to vary directly as the second and inversely as the square of the third. In this case x = r^. Thus, the attractive force that draws z^ ' two particles of matter together varies directly as the product of their masses and inversely as the square of the distance between them. It is common in physics to meet with two variables so re- lated that one varies inversely as the square of the other. In k this case x = -. VARIATION 259 Example. — 11 xccy, what is the value of x when y = 20? We are not given enough information about the relation between x andy to enable us to answer this question, for all we are told is that x = ky, where k is some constant whose value is not given. But if we are given the additional information that x = 10 when y = 8, we can find the value of k, and then answer the question. Since x = ky, and ic = 10 when ?/ = 8, we have 10 = 8 k, and k = ^. Hence x = ky, and therefore a; = 25 when y = 20. PROBLEMS 1. If X varies inversely as y and x = 4 when y = 9, what is the value of x when y = 15? 2. If X varies jointly as y and z and x=7 when y = 5 and z = 12, what is the value of x when ?/ = 14 and z = 22? 3. If ic Qc — , and x = 15 when y = 4 and 2; = 14, what is the z^ value of x when y = 20 and 2; = 18 ? 4. If x varies inversely as the square of y, what will be the effect on y of doubling the value of a; ? 5. The distance passed over by a body falling from rest varies directly as the square of the number of seconds during which it has fallen, and in 3 seconds it falls approximately 144.9 feet. How far will it fall in 5 seconds ? 6. The velocity acquired by a body falling from rest varies directly as the number of seconds during which it has fallen, and its velocity at the end of 2 seconds is 64.4 feet per sec- ond. What is its velocity at the end of 6 seconds ? 7. The weight of an object above the surface of the earth varies inversely as the square of its distance from the center of the earth. If an object weighs 50 pounds at the sea level, what would be its weight on top of a mountain a mile high ? Assume that the radius of the earth is 4000 miles. I '5 ± 260 COLLEGE ALGEBRA -j 8. When an object is taken below the surface of the earth its weight varies directly as its distance from the center of the earth. If an object weighs 200 pounds at the surface, how much would it weigh midway between the center and the surface ? How much would it weigh at the center ? ^ 9. Would an object that weighs 100 pounds at the surface of the earth weigh more 5 miles above the surface than it does 5 miles below the surface, or less ? J 10. The time of vibration of a simple pendulum varies di- rectly as the square root of its length. If the time of vibra- tion of a pendulum 39.14 inches long is one second, how long must a pendulum be in order that its time of vibration shall be two seconds ? ^^ 11. A pendulum two meters long makes 61650 vibrations in a day. Find the length of a pendulum whose time of vibra- tion is one second ? 12. Represent graphically the relation between two vari- ables when one varies directly as the other. 13. Represent graphically the relation between two vari- ables when one varies inversely as the other. 14. Represent graphically the relation between two vari- ables when one varies directly as the square of the other. 15. The amount of light received on a page from a given source varies directly as the size of the page and inversely as the square of its distance from the source of the light. One page is twice as large as another one and twice as far from the source of the light. Which page receives the more light ? 16. A metallic sphere whose radius is 3 inches weighs 32 pounds. How much will a sphere of the same material whose radius is 5 inches weigh, if the volume of a sphere varies di- rectly as the cube of its radius ? CHAPTER XIX INFINITE SERIES 202. Limit of a variable. — Consider the sum of the first n U terms of the geometric progression whose first term is 1 and y * whose common ratio is ^. s = Lziffi! = 2 — m»-\ " 1_JL ^ \2) ' / — ^ The value of s„ depends upon the number of terms repre- sented by it ; that is, upon the value of n. Hence if we give different values to n, s„ becomes a variable. If we fix in mind any small positive number, say tot^ttto"? an easy computation will show that s„ differs from 2 by less than this number if n = 21, and the difference will also be less than To oTo ou" ^^^ ^^^ greater values of n. It is immaterial how small the positive number is that we have in mind. As soon as we fix upon it, we can take enough terms of the series so that the sum of these terms or of any greater number differs from 2 by less than this number. We express these facts by saying that 2 is the limit of s^ as n increases without limit. The relation between s„ and 2 can be exhibited graphically by considering the successive values of s„ as abscissae of points. Of course we cannot mark all of these points, since there is an unlimited number of them. But a few of the first ones are sufficient to indicate that they are getting successively nearer and nearer to 2. 1 li 2 H h- 1 1— HH If l2_generalj we say that a variable x approaches a constant fe as a limit if its law of variation is such that, when we fix in 261 262 COLLEGE ALGEBRA mind any positive number whatever, the difference between x and k will become and remain less in ahsolute value than this lmml5er. The student should observe the importance of the word remain in this definition. In the preceding illustration, for example, the absolute value of the difference between s„ and Ifi becomes less than any positive number we can think of. As a matter of fact, it actually becomes when n = 6. But when n increases from G the absolute value of this difference increases and cannot for any following value of x be less than -^^. We therefore do not say that s„ approaches If^ as a limit. It follows from what has just been said that a variable can have only one limit. The student should note that the limit of a variable is al- ways a constant. In the illustration the variable is increasing and is therefore always less than its limit. But some variables are decreasing and therefore always greater than their limits. If, for example, S represents the area of a regular polygon circum- scribed about a circle, and if we increase without limit the number of sides of the polygon, /S' is a variable which decreases toward the area of the circle as its limit. Still other variables are increasing part of the time and de- creasing part of the time. For example, the sum s„ of the first n terms of the geometric progres- sion whose first term is 1 and whose common ratio is — ^. Here the limit is |, and s„ is alternately greater than and less than the limit. Some variables do not approach any limit, as, for example, the variable t described in § 201. INFINITE SERIES 263 203. Convergent and divergent series. — A series with an unlimited number of terms is called an infinite series. If the sum of the first n terms of an infinite series approaches a limit as n increases without limit, the series is said to be convergent. The geometric progression referred to in § 202 is a conver- gent series, and in general any geometric progression with an unlimited number of terms and a common ratio greater than —1 and less than 1 is a convergent series (see § 100). But not e very- convergent series is a geometric progression, as we shall see in § 206. If as n increases without limit, s„ does not approach a limit, the series is said to be divergent. Example 1. — Consider the series 1 + 2 + 3 + --+W + ..-. If we think of any positive number M, no matter how large, s„ > M for all values of n that are greater than M. Hence s„ does not approach a limit and the series is divergent. Example 2. — Consider the series 1_1 + 1_1+ ... ^(_l)n+l+ .... Here Sn is alternately 1 and for successive values of n and therefore does not approach a limit. Hence the series is divergent. There is an important difference between these two series. In the first one Sn can be made as great as we please by taking n sufficiently great, while in the second one Sn never exceeds 1. 204. The general term of a series. — An infinite series is not fully described until we are told how to form any given term. This information is usually supplied by means of the nth, or general, term. It is important therefore that the student should be able to write down any term of the series when the general term is given. Example. — Write down the first five terms of the series whose nth term " 2»(2Ll) - These terms are: 1 ^ 1 1 and 1 2-l'4.3«.58.7 10-9 264 COLLEGE ALGEBRA EXERCISES Write down the first three terms and the (ri + l)th term of the series whose nth. term is : 6. 1. 1 2n 2. 1 3. 1 n{7i + l) 4. 1 nl 5. n 3«' 2 71-1 — 7. — 71 8 1 (^-2)- , n 2» 9. (-1)» 10. (-1) /v.2n-l +1 ^ (2 n - 1) ! n+1 _'fi__ (2n)!* SERIES ALL OF WHOSE TERMS ARE POSITIVE 205. The only problem in connection with infinite series that will be considered in this book is that of determining whether a given series is con-vei-gent or divergent. This is an extremely difficult problem, and we shall discuss only the simpler phases of it. We shall suppose at first that the series we are dealing with have all their terms positive, and we shall make use of the following Fundamental principle. — If a variable always i7icreases (o?- at least never decreases) a7id never gets greater than a given nu7nher M, then it approaches a limit which is either M or less tha7i M. 206. Tests for convergence and divergence. — The following four tests for series with positive terms are the simplest and most important ones. I. Comparison test for convergence. — Let «i -I- «2 H +«„+••• INFINITE SERIES 265 he an infinite series with positive terms. If a second infinite series with positive terms ,, , . , , %. , \ is convergent, and if every term of this latter series is greater than, or equal to, the corresponding term of the first series, then the first series is convergent. Denote by s„ and S^ the sums of the first n terms of these two series respectively. As n increases without- limit S^, by hypothesis, approaches a limit. Call this limit B. Since the series have positive terms, s^ increases as n in- creases, but for no value of n is it greater than S„, which in turn is always less than B. Hence s^ 2, since, 1 < 1 1.2.3. ..w 2.2. ..2 (w — 1) factors Hence series (1) is convergent. Example 2. Consider the series 1 + 1+J_^_...+ 1 4.... (3) After its first term this. series is the same as series (1). Hence if s„ and Sn denote the sums of the first n terms of series (1) and series (3) respec- tively, we have o 1 . « 266 COLLEGE ALGEBRA Now we saw in Ex. 1 that as n increases without limit, s„, and therefore also Sn-i, approaches a limit. Hence Sn approaches a limit and series (3) is convergent. In considering the convergence of a series it is frequently desirable, as in Ex. 2, to investigate the new series obtained by dropping off the first few terms of the given series. Sup- pose that we drop off the first r terms and that the sum of these terms is s. Then if S^ and s^ denote the sums of the first n terms of the given series and the new series respectively, we have S^=s-{-s^_„ for values oi n> r. Now s is a constant, and therefore either one of the variables Sn and s„_^ approaches a limit if the other one does. Hence the two series are either both convergent or both divergent. The stiident is reminded that s„ and s„_^, where r is a constant, represent different stages of the same variable. The conclusion just reached applies also to series some of whose terms are negative. In order to use Test I we must know some convergent series with positive terms in order to have a basis for comparison ; and the more of these we know the wider the range of usefulness of this test will be. EXERCISES Show that the following series are convergent ; 1 . 1 1- 1 1 1 1 (2!/ (3!/ 1 ' 2. hh- -^. + + (niy • 2"^l + 22 "^1-1-33 "^'*"*'l + n" \\o \^ INFINITE SERIES 267 U. 7. i_4.i_| 1 ± 1 . ' 3! 5! {2n + l)l V 3 4-2! 5-3! (n + 2)-nl 207. II. Comparison test for divergence. — Let ai + «2H !-«„+ ••• be an infinite series with positive terms. If a second infinite series loith positive terms i\ , , , , r A is divergent^ and if every term of this latter series is less than, or equal to, the corresponding term of the first series, then the first series is divergent. Denote by s„ and S^ the sums of the first n terms of these two series respectively. Since S^ increases as n increases and does not approach a limit, it must increase beyond all limit (see fundamental principle). But for any value of n it is equal to, or less than, s„. Hence s^ increases without limit and the first series is divergent. In order to have a useful basis for comparison in the appli- cation of this test we shall prove that the series is divergent. 1 + - + -+... +- + 268 COLLEGE ALGEBRA Consider the n successive terms of this series, 1+ 1 +...+ 1 n n + 1 2?i — 1 Every one of these terms is greater than — - and their sum 11 ^^ is therefore greater than n • -— = -<, Now the series contains 2n 2 an unlimited number of groups of terms like this with no terms in common. If then, we take n so great that we can form ten million groups of this kind, s„ will exceed five million ; and by- taking n sufficiently large we can make s„ exceed any number that we had in mind. Hence s„ does not approach any limit and the series is divergent. This series is known as the harmonic series (see § 101). We are now ready to illustrate the application of Test II. Example. — Consider the series l + -A_4._A. + ...+_2jLzJ_+.... 'h 1.22.3-Tw(«-1) i We observe that JjLnl_ = ^ n - 1\ \_ , 71 {n — 1) w — 1 n that is, that the ?ith term of this series is equal to times the nth. 71—1 term of the harmonic series for all values of n greater than 1. But when w>l, > 1. Hence, by the comparison test, this series is divergent. ** "~ III. If the nth term of a series does not approach zero as n in- creases without limit, the series is divergent. If a„ is the rith term of the series, then «n = Sn - S«-l ; and if the series is convergent, s„ and s„_i approach the same limit as n increases without limit. Hence a„ approaches the limit zero. If a„ approaches the limit zero as n increases without limit, it does not follow that the series is convergent. The harmonic series, for example, is divergent. ^^ INFINITE SERIES 269 EXERCISES Show that the following series are divergent : 2. l + 2! + 3!H-. .. + %! + •••. 3. l + -L + -i4--- + -i=4--. V2 V3 Vw 4. 2+1 + 1 + ... + '-^ + .... 2 4 2w fi ^.L^-l.. ..L^ + n 2 O 1 + 71^ 208. Since in the application of the next test we shall have occasion to look for the limit of certain fractions, we shall ex- plain here, by means of examples, how to proceed in such cases. Example 1. — Suppose that we want to know what happens to the frac- tion ^^ "*" when n increases without limit. 3w+ 1 We observe in the first place that both the numerator and the denomi- nator increase without limit. But this gives us no information about the value of the fraction, since a fraction may have any value except 0, and have its numerator and denominator as large as we please. If, how^ever, we first change the form of the fraction by dividing both the numerator and denominator by n, we shall be able to get the information we want. 2 + 5 2n + 5 n^ 3w + l~3 l' n Now as n increases without limit the numerator of this last fraction approaches the limit 2 and the denominator approaches the limit 3, since - and - both approach zero. Hence the limit of the fraction is |. n n ^ 270 COLLEGE ALGEBRA Example 2. — Consider the limit of ^— ^t — ^ j" as n increases without limit. "" n^ + Sn + S 2w-6 1+5+4 2 6 The numerator of this last fraction approaches the limit 1 and the de- nominator approaches the limit zero. Hence the fraction increases without limit. Q J, 4 Example 3. — Consider the limit of ^ — : — ^ r as w increases with- out limit. 3w-4 8 Ji2 _ 7 u _ 2 w2 _ 7 M - 2 §_1 n n^ ''2 7 2 The numerator of this last fraction approaches the limit zero and the de- nominator approaches the limit 8. Hence the fraction approaches the limit zero. EXERCISES Find the limit of each of the following fractions as n in- creases without limit. 1. 2. f n-f 1 n n + 2' \ 3"' 3(n+l)2' (n+1)! (2n-l)! . ' (2n + l)! 7. 8. 9. n • n 10. 271-1 2n + l* n^ + n-f-3 n^ — 71 — 3 7n-f-4 n^ + l INFINITE SERIES 271 209. IV. The ratio test. — If the terms of the Irifinite series «i4-«2H 1-«„4- ••• are all positive, and if as n increases without limit the ratio -^ approaches a limit I, the series is * (a) convergent when Z < 1, (6) divergent when Z > 1. Ifl — 1, the series may be either convergent or divergent. (a) 1<1. Select some number r that lies between I and 1 and is there- fore less than 1. Since -^ approaches the limit I as n increases without limit, we can select a positive interger m so large that -^, for all values of n greater than m, differs from I by an amount less in absolute value than r—l. (§pe the definition of limit, § 202.) _ ^ For these values of n ^^ < r. ' l i ' a„ Hence, ^ < r, or a^+, < ra^+,, a w+l Y-t\^ a ~^1. As in (a), we can select a positive integer m so large that -^, for all values of n greater than m, differs from I by an amount less in absolute value than l—\. For these values of w, >1. Hence, ^2^1^ or a^+2>««.+i, a«+i a. >1, or a„,+3>«m+2>am+l, ''m+2 Since these inequalities hold for all positive integral values of p, the series is divergent by Test III. The harmonic series is an example of a divergent series for which l — \\ and the series J_ + ^ + ... + _J_ + ... 1.22.3 n(7i + l) is an example of a convergent series for which l — \. That this series is convergent may be seen by writing s„ in the form : •-{'4)-(i-i)— e-.-i,)--^- INFINITE SERIES 273 Example. — Consider the series l + i + 2Q)2+...+(w-l)(i)«-i-f ... ««+i _ ng-r and the limit of this ratio as n in- «n (n-l)(i)-l 3(71-1) creases without limit is |. Hence the series is convergent. ^ What has been said here about series all of whose terms are positive applies with slight, obvious modifications to series all of whose terms are negative. In what way is it necessary to modify the fundjamental priticiple in order that it shall apply to a variable that is always decreasing ? EXERCISES Test the following series for convergence or divergence : Q2 Qn 2. 1 + 2 + 3+...+ » 2! 3! 4! ' (n + l)l 1 2' 3' n' 3 -t I I . . . I -I- . . . 4 2.3 3.4 4.5 (,, + l)(n + 2) 5. 14-1+1+. .. + ii + .... 2 22 23 2" 6 1 2 ' Tt? — IV 7. 1 + 2. | + 3-.(|)^ + ... +»(!)»-' + 9 2 1 2' 2' 1 2""' 1 ■ 1 . 3 ' 3 . 3' 5 . 3' ' ' (2»-l).3»-' ' . .,|,|,...,L-,.... 274 COLLEGE ALGEBRA SERIES WITH POSITIVE AND NEGATIVE TERMS 210. Alternating series. — A series whose terms are alter- nately positive and negative is called an alternating series. The following test applies to these series : V. An alternating series is convergent if the absolute value of every term is less than that of the preceding term and if the limit of the nth term is zero as n increases without limit. Thus the alternating series is convergent. We shall omit the proof of this test. 211. Absolutely convergent series. — The symbol \a\ repre- sents the absolute value of a. VI. Tlie series «! + ag + ag + • • • + a,, + . • • is convergent if the series l«il + l«2l + k3H — +| 1. Ifl = l the series may be either convergent or divergent. (a) I < 1. By Test IV the series is absolutely convergent and therefore, by Test VI, it is convergent. (b) I > 1. In this case a^ cannot approach as w increases without limit, and therefore, by Test III, the series is divergent. That the series may be either convergent or divergent when Z = 1 is shown by the examples cited under Test IV. The general ratio test includes Test IV as a special case. There are, of course, many other tests, which cannot be given here. EXERCISES Test the following series for convergence or divergence : ^- 3 4 2 + 5 3 +^ ^> ;7T2 »+ • Q3 05 Q2r»-1 3! 5! ^ ^ (271-1)!^ Q2 Q4 q2n 5. l-A, + _L-...+(-l)n+l_L+.... V2 V3 -^n 93 95 92»-l 3! 5! ^^ ^ (2n-l)!^ .,-.(>.!)■ 276 COLLEGE ALGEBRA 213. Power series. — A series of the form where the c's are constants, is called a power series. A power series may converge for all values of x, or it may diverge for all values of x except 0. These are the extreme cases. Usually such a series converges for some values of x and diverges for the other values. Test VII is the most useful one for determining the values of X for which a power series converges. Example. Consider the series ^__^+_^_ ... + (_ i)n+i — ^^aii — ^ ... 1 33.3 3S.5 ^ ' 32^-1 (2 w- 1) Here ttw+l 32»-i(2n-l) a;2(2n-l) 32(2w + l) 32«+i(2n + l) x2«-i The limit of this fraction as n increases without limit is — . 32 ^ Hence the series is convergent for those values of x for which — < 1, or for which — 3 < sc < 3 ; and the series is divergent for those values of x for which — > 1. This inequality is satisfied whenever a: < — 3 or a; >3. 32 The test gives no information as to the convergence or divergence of the series when x= — ^ or 3. These cases require a special investigation. It is easy to see directly that this particular series is convergent for either of these values of x. The values of x for which the series is convergent form what is called the interval of convergence of the series. Thus, the interval of convergence of the series just considered extends from — 3 to 3. We can indicate our conclusions graphically by making the interval of convergence heavier than the rest of the axis. -3 3 1 1 \ Divergent Convergent Divergent V •INFINITE SERIES 277 EXERCISES Determine the interval of convergence of each of the following series, and represent it graphically : rvA /v,5 /v.2n-l w 2! 4! ^ ' {2n)\ ^3 /V.5 ^.2n-l 5. l + x + x^^ (-«"'^H . 7. 0, + - + ^+...+-+.... 8. 9 + 10^ + 35^ 2«l+5» + 2^_ APPENDIX FORMULAE FROM SOLID GEOMETRY 1. Section of a pyramid. — If a plane be drawn parallel to the base of a pyramid V-ABC, cutting the pyramid in the section DEF, and if JSi and S2 represent the areas of ABC and DEF respectively, then „ S2 VK' where VH and F/rare the distances of the base and the cutting plane respectively from the vertex. (For figure, see p. 37.) 2. Surface and volume of a sphere. — If S represents the surface, and Fthe volume, of a sphere of radius r, then 8 = 4:177^, and V=^Trr^. 3. Volume of a spherical segment of one base. — If a sphere be divided into two parts by a plane, either part is called a spherical segment of one base. The section made by the plane is a circle and is called the base of the segment. If we draw the radius of the sphere through the center of the base, the length of that part of the radius which lies between the base and the surface of the sphere is called the altitude of the segment. If V represents the volume of the segment, x its altitude, and ri the radius of its base, then i; = 1. TTXri^ -\- i TTX^. 278 APPENDIX 279 FORMULiE FROM PHYSICS 1. Distance passed over by a falling body. — The distaDce s passed over by a falling body in t seconds is given in feet by the formula — 16 /^ 2. Velocity of a falling body. — The velocity v of a body that has fallen s feet is given in feet per second by the formula i; = V64.3 5. The formulae in 1 and 2 are both approximate, but the latter Is the more exact. 3. Velocity of a projectile. — The velocity v of a projectile at any moment in feet per second is given by the formula V = Vi^o^ — 64 y, where -^o is the initial velocity and y is the height of the pro- jectile at this moment in feet above the level of the starting point. 4. Number of vibrations made by a stretched wire. — The number n of vibrations made by a stretched wire is given by the formula __^^ 1 /980 M where I is the length in centimeters between the bridges, M the weight in grams of the stretching weight, and 7n the weight in grams of the wire per centimeter of length. 5. Oscillation of a pendulum. — The time t of oscillation of a pendulum of length I centimeters is given in seconds by the formula . 6. The lever. — The point of support of a lever is called the fulcrum. 280 APPENDIX There are different kinds of levers, but in all of them two applied forces are in equilibrium (the effect of the weight of the lever being neglected) when the product of the first force and the distance of its point of application from the fulcrum is equal to the product of the second force and the distance of its point of application from the fulcrum. fA=fA' The two forces must be applied in the same direction when their points of application are on opposite sides of the fulcrum, as in a teeter board. The forces must be applied in opposite directions when their points of application are on the same side of the fulcrum. ^ ' —d ^ — \ y^ When two forces /i and f^ act in the same direction on a lever at the distances d^ and cZg respectively from the fulcrum and on the same side of it, the combined effect is/iC?i +/2C?2. The effect of the weight of a uniform lever is the same as if the entire weight were concentrated at the middle point. 7. Center of gravity. — If the centers of gravity of two bodies of weights TFi and W^ respectively lie on the ic-axis and have the abscissae a and h respectively, then the abscissa c of the center of gravity of the two together is given by the formula 8. Specific gravity. — The ratio of the weight of a volume of a given substance to the weight of the same volume of water is called the specific gravity of the substance. Thus, a cubic inch of platinum weighs approximately 22 times as much as a cubic inch of water, and we accordingly say that the specific gravity of platinum is approximately 22. 9. Principle of Archimedes. — A body immersed in a liquid loses a part of its weight equal to the weight of the displaced liquid. INDEX (Numbers refer to pages.) Abscissa, 2, 48, Absolute value, 2, 162, 274. Addition, definition of, 3. definition of, for imaginary num- bers, 154, 156. Algebraic solution of linear equations in two or more unknowns, 51. Amplitude of a number, 162. Antecedent of a ratio, 33. Antilogarithm, 239. Argument of a number, 162. Arithmetic means, 124. Arithmetic progression, 123. Associative law of addition, 3. Associative law of multiplication, 7. Assumptions for addition, 3. Assumptions for multiplication, 7. Axes, coordinate, 48. Axis, 1. Base, change of, 242. of system of logarithms, 239. Binomial theorem, 141, 150. Cancellation, law of, for addition, 4. for multiplication, 7. Circle, equation of, 66. Coefficients, relations between roots and, 176. Combinations, 138. number of, 139. Complex numbers, 153, 158. Conjugate numbers, 159. Consequent of ratio, 33. Constants, definition of, 257. Continuation of signs, 186. Convergent series, definition of, 263. Coordinates of a point, 48. De Moivr6's Theorem, 163. Dependent equations, 53, 59. Descartes's Rule of Signs, 187. Determinants, definition of, 204. elements of, 204. minors of, 208. of order n, 201. of second order, 55. of third order, 62. properties of, 205-208. solution of equations by means of, 216. terms of, 204. Difference, definition of, 4. Discriminant of a quadratic, 91. Distributive law of multiplication, 7. Divergent series, definition of, 263. Dividend, definition of, 8, 16. Division, definition of, 8, 16. Divisor, definition of, 8, 16. Equations, definition of, 38. dependent, 53, 59. exponential, 255. homogeneous, 219. inconsistent, 53, 59, 219. independent, 60. in form of quadratics, 103. involving fractions, 96. involving radicals, 99. linear, 38. of condition, 38. of first degree, 38. quadratic, 84. rational integral, 38. systems of, 47, 110. transformations of, 178-183. with given roots, 85. Ellipse, 112. 281 282 INDEX Exponents, fractional, 73. fundamental laws of, 73. irrational, 238. negative, 73. zero, 75. Extremes of a proportion, 33. Factorial n, 136. Factor, rationalizing, 82. Factor Theorem, 170. Factors, 7, 19. Fourth proportional, 34. Fractions, definitions and principles, 27. partial, 228. Function, rational, 81. rational integral, 81. Fundamental theorem of algebra, 174. General term in binomial expansion, 143, 152. Geometric means, 128. Geometric progression, 127. Graphical interpretation of Trans- formation III, 183. Graphical representation, 47. Graphical solution of equations, 51, 93, 95, 110-120, 172. Graphical solution of linear and quad- ratic inequalities, 225. Greater than, meaning of, 2. Greatest coeflEicient in binomial ex- pansion, 143. Harmonic progression, 133. Highest common factor, 21. Euclid's Method, 24. ' HoTl^er's Method, 192. Hyperbola, 113. Identity, 17, 38. Imaginary numbers, 153, 158. Imaginary roots, 177. Inequalities, definition of, 221. conditional and unconditional, 221, 223. properties of, 222. Infinite geometric progression, 131. Inversions, of letters, 202. of numbers, 201. Irrational roots, Homer's Method for finding approximate values of, 192. Less than, meaning of, 2. Limit, definition of, 131, 261. Locus of an equation, 50. Logarithms, change of base, 242. characteristic of, 243. common, 241, 242. definition of, 239. mantissa of, 243. natural, 241. table of, 250, 251. use of table of, 243. Lowest common denominator, 30. Lowest common multiple, 22. Mean proportional, 34. Means of a proportion, 33. Minor of determinant, 208. Minuend, definition of, 4. Modulus of a number, 162, Multiplication, associative law of, 7. definition of, 6. definition of, for imaginary num- bers, 157, 159. Numbers, commensurable, 33. complex, 153, 158. equal, 155. imaginary, 153, 158. incommensurable, 33. negative, 2. positive, 2. pure imaging^ry, 158. rational, 2. real, 2. representation of points by, 1, 47, 153. Ordinate, 48. Origin, 1, 48. Parabola, 94, 113. Parameter, 116. Parentheses, 5. removal and insertion of, 10. Permutations, definition of, 134. of n things not all different, 137. of n things r at a time, 135. INDEX 283 Polar representation of complex numbers, 162. Polynomials, addition and subtrac- tion of, 12. division of, 16. homogeneous, 14. multiplication of, 14. prime, 21. Principal root, 74, 160. Products, special, 16. Proportion, by alternation, 34. by composition, 35. by composition and division, 35. by division, 35. by inversion, 34. definition of, 33. Quotient, definition of, 8, 16. Radicals, addition and subtraction of, 79. coefficient of, 78. index of, 78. multiplication of, 80. radicand of, 78. similar, 79. simplification of, 78, 79. Ratio, definition of, 33. of geometric progression, 127. Rationalizing factor, 82, Rational roots of an equation, 168, 190. Ratio test for infinite series, 271, 275. Real number, 2, 159. Reciprocal of a number, 7. Remainder, definition of, in division, 16, 17. definition of, in subtraction, 4. Remainder Theorem, 169. Repeating decimals, 132. Representation of points by numbers, 1, 48, 153, 154. . Roots, of an equation, 39. Roots, imaginary, 177. irrational, 192. negative irrational, 197. number of, 175. positive irrational, 192. rational, 190. relation between, and coefficients, 176. Roots of numbers, 164, 165. principal, 74, 160. Series, absolutely convergent, 274. alternating, 274. comparison test for convergence of, 264. comparison test for divergence of, 267. convergent, 263. divergent, 263. general ratio test for, 275. general term of, 263. limit of certain geometric, 131. power, 276. ratio test for, 271. with positive terms, 264. with positive and negative terms, 274. Sign, continuations and variations in, 186. Signs, Descartes's Rule of, 187. Straight lines, equation of two, 113, 114. Subtraction, definition of, 4. Subtrahend, definition of, 4. Synthetic division, 170-172. Systems of equations, 47, 110, 216. Tables of logarithms, 250, 251. Tabular difference, 249. Third proportional, 34. Variable, definition of, 257. limit of, 131, 261. Variations in sign, 186. OCT FEB 25 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. ^j Renewed books are subject to immediate recall. 2.1 m s- ^i^^^- K0V3Q1959 MAT 20^ug' 22N 14fan'6 l L0i JAN 1 1 -:g1 ^''. ' ^I ' ji^^SS i^J' REC'D LD ISJuifSO 0CT2 5'63-9AM LD 21-100W-12 '4 uii M LD 21A-50m-4,'59 (A17248l0)476B General Library University of California Berkeley ^rr^ ^IpTT" -/ , 9c "V YB 1 7268 M306040 ■ QAI5^ ■ F ^% THE UNIVERSITY OF CALIFORNIA LIBRARY Hi ^