6. Resolve 15x+19x+6 into simple factors. Ans. 15(x+3)(x+3). 7. Resolve cx'—2ax+cox-2 ac into simple factors. 2a Ans. c (x+c). 306. The same principle also enables us to construct an equation whose roots shall be any given quantities. This is done by multiplying together the two binomial factors, which, according to the principle in question, the required equation must contain. 1. Find the equation whose roots shall be and - Ž Factors, x + 1, 1 Product, x+7-6=0, or, 63*+2—1= 0, Ans. EXAMPLES. 1. Find the equation whose roots shall be 6 and -15. Ans. x+93-90 = 0. 2. Find the equation whose roots shall be 3 and -15. Ans. x+123-45 = 0. 3. Find the equation whose roots shall be 16 and 9. Ans. x_25x+144 = 0. 4. Find the equation whose roots shall be 84 and -1. Ans. x_83_-84 = 0. 5. Find the equation whose roots shall be į and — . 1 Ans. x' = 0. 2 9 6. Find the equation whose roots shall be ; and -4. 1 Ans. x 0. 2 7. Find the equation whose roots shall be į and 4. Ans. 8x*—6x+1= 0. 8. Find the equation whose roots shall be 2a and Ans. m-(20—c).+2ac = 0. 17x 56 DISCUSSION OF THE FOUR FORMS. 307. In the general equation x' +2ax = 6, the coefficient of x, as well as the absolute term, may be either positive or negative. Hence, to represent all the varieties, with respect to signs, we must employ the four forms, as follows: 2+2ax = +6, (1) X-2ax = +6, (2) a+2ax = -6, (3) 20'2ax = -6. From these equations we obtain x = -a+Va+6, (1) a = +a+Va+b, (2) x = -a+Va–b, (3) 3= ta+Va*_6. (4) We may now consider what conditions will render these roots real or imaginary, positive or negative, equal or unequal. 308. Real and imaginary roots. In the first and second forms, the quantity a'+b, under the rad ical, is positive, and the radical quantity is therefore real. But in the third and fourth forms, the quantity a-6, under the radical, will be negative when b is numerically greater than aʼ; in which case the radical quantity is imaginary. Hence, 1.-In each of the first and second forms, both roots are always real. 2.-In each of the third and fourth forms, both roots are imaginary when the absolute term is numerically greater than the square of one half the coefficient of x; otherwise they are real. 309. Positive and negative roots. Since a+b>aand a'-b<a', we have Vã+6> a and Va_6 <a It follows, therefore, that the signs of the roots in the first and second forms will correspond to the signs of the radical; but the signs of the roots in the third and fourth forms will correspond to the signs of the rational parts. Hence, 1.- In each of the first and second forms, one root is positive and the other negative. 2.- In the third form both roots are negative, and in the fourth form both roots are positive. 310. Equal and unequal roots. It is obvious that in the first and second forms the two roots are always unequal; for in each of these forms, one root is numerically the sum of a rational and a radical part, and the other the difference of the same parts. The same may be said of the third and fourth forms, if we except the case where a b; in which case the roots are eqnal, and we have, for the third form, x = -a+0=—a or -a, and for the fourth form, x = ta+0 = ta or ta. Hence, 1.-In each of the first and second forms, the two roots are always unequal. 2.-In each of the third and fourth forms, the roots will be equal when the absolute term is numerically equal to the square of one half the corfficient of x; otherwise they will be unequal. In the first and third forms, the negative root consists of the sum of the rational and radical parts; while in the second and fourth forms, the positive root consists of the sum of the two parts. Hence, if we exclude the case of equal roots, 3.- In the first and third forms the negative root is numerically greater than the positive. 4 - In the second and fourth forms, the positive root is numerically greater than the negative. The principles which we have now established, respecting the roots of quadratic equations, are all that are of importance, either theoretically or practically. DISCUSSION OF PROBLEMS. or 100 X 24; Or, 311. In the solution of particular problems involving quadratics, we shall find that in certain cases both roots of the equation will answer the conditions of the problem, while in other cases only one of the roots is admissible. The reason is, that the algebraic expression is more general in its meaning than ordinary language; and thus the equation which represents the conditions of the given problem, will sometimes be found to represent the conditions of other analogous problems. 1. A man bought a horse for a certain price. Now if he sells him for $24, he will lose as much per cent. as the horse cost; required the price of the horse. X® Let x denote the price. Then x x will be the loss, if 100 he sells him for $24. Hence, we have x2 100 ac -100. = -2400; 20_100x+2500 = 100; whence, +10; or, 60 or 40. Both values of x fulfill the conditions. For, 60 X.60 = 36; and 60—36 = 24. 40 X.40 = 16; and 40-16 = 24. 2. A person bought a number of sheep for $240; if he haa bought 8 more for the same sum, each sheep would have cost $1 less. How many sheep did he purchase ? 240 Let <= the number of sheep purchased; then = cost of Had he purchased 8 sheep more, the cost of one would have 240 240 240 been Hence, ; 3+8 reducing, * +8x = 1920.; a* +80+16 = 1936; or, 3+4= +44; whence, * = 40 or 48. -50 = one. 1 = 2+8 In this case only the first value of x is admissible. The negative result, –48, is numerically the answer to the problem which would be formed by substituting in the above, the word more for the word less, and the word less for the word more. INTERPRETATION OF IMAGINARY RESULTS. 312. We have seen that when the absolute term of a quadratic is negative, and numerically greater than the square of one half the coefficient of the second power of the unknown quantity, the roots of the equation will be imaginary. Now the imaginary roots will always satisfy the equation, and it is necessary to ascertain what they indicate respecting the conditions of the problem which the equation represents. 1. Let it be required to divide 20 into two such parts, that their product shall be 140. Let x = one part; then 20-2 the other. Hence, x(20—~) = 140; or, 29_202 = -140, x'—20x+100 = -40, 2—10 = +V-40, * = 10+V-40. The result is imaginary; how shall it be interpreted ? Recurring to the problem, we find that the greatest possible product that can be formed by multiplying together two parts of 20, is 10x10= 100, the product of the halves of 20. Thus we find that the problem is impossible. 2. A farmer would inclose 50 square rods in rectangular form, by a fence whose entire length shall be 24 rods. Required the length and breadth of the inclosure. Let x = the length, and y = the breadth; then x+y = 12, and xy = 50; x−y= +2V-14; whence, x = 6+1 -14, y = 65V 14. |