If the roots of the equation x^5-40x^4-Px^3-R | vedclass

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(C) Let the roots of the equation be $a/r^2, a/r, a, ar, ar^2$. From Vieta's formulas,the product of the roots is $a^5 = S$. The sum of the reciprocal

Vedclass · Accepted Answer

$32$

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(C) Let the roots of the equation be $a/r^2, a/r, a, ar, ar^2$. From Vieta's formulas,the product of the roots is $a^5 = S$. The sum of the reciprocals of the roots is $\frac{1}{a/r^2} + \frac{1}{a/r} + \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} = \frac{r^2+r+1+1/r+1/r^2}{a} = 10$. Also,the sum of the roots is $a/r^2 + a/r + a + ar + ar^2 = 40$. Factoring out $a$,we get $a(1/r^2 + 1/r + 1 + r + r^2) = 40$. Dividing the sum of roots by the sum of reciprocals: $\frac{a(1/r^2 + 1/r + 1 + r + r^2)}{(1/a)(1/r^2 + 1/r + 1 + r + r^2)} = \frac{40}{10} = 4$. This simplifies to $a^2 = 4$,so $a = 2$ or $a = -2$. Since $a^5 = S$,we have $S = 2^5 = 32$ or $S = (-2)^5 = -32$. Thus,$|S| = 32$.

If the roots of the equation $x^5-40x^4-Px^3-Rx-S=0$ are in geometric progression and the sum of the reciprocals of the roots is $10$,then $|S|=$

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