Now, if we substitute for x in these functions, the number given below, the signs of the results will be as follows: Here the three roots of this equation are seen to be 1, 2, 3, and no change of sign occurs by the substitution for x of any number less than 1; but when p exceeds 1, there is a change of sign in the original equation from to, by which one variation is lost. When p=2, two of the functions disappear simultaneously, showing that 2 is a root of the second derived function as well as of the original equation, and a second variation of sign is lost. Also, when p becomes equal to 3, a third variation is lost; and there are no farther changes of sign arising from the substitution of any numbers between 3 and + ∞. There are three changes of sign of the primitive function, two of the first auxiliary function, and one of the second auxiliary function; but no variation is lost by the change of sign of any of the auxiliary functions; while every change of sign of the primitive function occasions a loss of one variation. Ex. 2. How many real roots has the equation Hence this equation has but one real root, and consequently must have two imaginary roots. Ex. 3. How many real roots has the equation 2x3 -7x2+10x+10=0? 2x3 -7x2+10x + 10. - 6x2- 14x+10; or 2x3-3x2- 7x+5. Hence the four roots of this equation are real. If we try different values for x, we shall find that When x=-3, the signs are + − + − +, giving 4 variations; Hence this equation has two positive roots between 2 and 3; one negative root between 0 and -1; and one negative root between 2 and 3. Ex. 4. How many real roots has the equation x3 7x+7=0? Ans. Three: viz., two between 1 and 2, and one between — 3 and -4. Ex. 5. How many real roots has the equation 2x1 13x2+10x 19 0? Ans. Two: one positive and one negative. Ex. 6. How many real roots has the equation x536x372x2· - 37x+72=0? Ans. Three. SECTION XX. SOLUTION OF NUMERICAL EQUATIONS. (311.) WE will first consider the method of finding the integral roots of an equation, and will begin with forming the equation whose roots are 2, 3, 4, and 5. This equation must consist of the factors (x − 2) (x — 3) (x — 4) (x — 5) = 0. If we perform the multiplication (which is most expeditiously done by the method of detached coefficients shown in Art. 64), we obtain the equation We know that this equation is divisible by x-5. Let us perform the division by the method of detached coefficients shown in Art. 80. Supplying the powers of x, we obtain for a quotient This operation may be still farther abridged, as follows: Represent the root 5 by a, and the coefficients of the given equation by A, B, C, D, .... V. We first multiply a by A, and subtract the product from B; the remainder, 9, we multiply by a, and subtract the product from C; the remainder, + 26, we multiply again by a, and subtract from D; the remainder, 24, we multiply by a, and, subtracting from V, nothing remains. If we take care to change the sign of a, we may substitute addition for subtraction in the above statement; and if we set down only the successive remainders, the work will be as follows: Multiply A by a, and add the product to B; set down the sum, multiply it by a, and add the product to C; set down the sum, multiply it by a, and add the product to D, and so on. If the equation is exactly divisible, the final product will be equal to the last term V, taken with a contrary sign. The coefficients above obtained are the coefficients of a cubic equation whose roots are 2, 3, 4. The equation may therefore be divided by x-4, and the operation will be as follows: 1-926-24 | 4 These, again, are the coefficients of a quadratic equation whose roots are 2 and 3. Dividing again by x-3, we have 1-5+613 which are the coefficients of the binomial, factor x- -2. These three operations of division may be exhibited together as follows: 1 — 14+71 — 154120 | 5, first divisor. (312.) The method here explained will enable us to find all the integral roots of an equation. For this purpose, we make trial of different numbers in succession, all of which must be divisors of the last term of the equation. If any division leaves a remainder, we reject this divisor; if the division leaves no remainder, the divisor employed is a root of the equation. Thus, by a few trials, all the integral roots may be easily found. Ex. 2. Find the seven roots of the equation x2+x6- 14x5-14x+49x3 +49x2. 36x-36 0. We take the coefficients separately, as in the last example, and try in succession all the divisors of 36, both positive and negative, rejecting such as leave a remainder. The operation is as follows: 11-14-14+49 +49-36-36 12-12-26 +23+ 72+36 14-4-34-45-18 1+7+17 +17+6 1+6+11+ 6 1+5+ 6 1+3 Hence the seven roots are, 1, first divisor. - 1, fifth divisor. Ex. 3. Find the six roots of the equation x65x5-81x1 — 85x3 +964x2 + 780x-1584 = 0. 1+ 5—81— 85+964 + 730-1584 1+10-35-300-396 1+16+61 + 66 1+14 +33 1+11 1. 4. 6. 2. 3. 11. The six roots, therefore, are, 1, 4, 6, — 2, — 3, 11. Ex. 4. Find the five roots of the equation x+6x-10x3- 112x2-207x-110-0. |