The roots of the equation x^5 - 40x^4 + px^3 | vedclass

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Question

(D) Let the roots of the equation be $\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$. From the sum of roots,$\frac{a}{r^2} + \frac{a}{r} + a + ar + ar^2 = 4

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(D) Let the roots of the equation be $\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$. From the sum of roots,$\frac{a}{r^2} + \frac{a}{r} + a + ar + ar^2 = 40$. From the sum of reciprocals,$\frac{r^2}{a} + \frac{r}{a} + \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} = 10$. This can be written as $\frac{1}{a} (r^2 + r + 1 + \frac{1}{r} + \frac{1}{r^2}) = 10$. Also,the sum of roots is $\frac{a}{r^2} (1 + r + r^2 + r^3 + r^4) = 40$. Dividing the sum of roots by the sum of reciprocals: $\frac{a(1+r+r^2+r^3+r^4)}{(1/a)(1+r+r^2+r^3+r^4)/r^2} = \frac{40}{10} = 4$. This simplifies to $a^2 r^2 = 4$,so $ar = \pm 2$. The product of the roots is $s = (\frac{a}{r^2})(\frac{a}{r})(a)(ar)(ar^2) = a^5$. Since $a$ is the middle term,from the sum of roots: $a(1 + (r + \frac{1}{r}) + (r^2 + \frac{1}{r^2})) = 40$. Using $ar = \pm 2$,we find $a = 2$ or $a = -2$. If $a = 2$,$s = a^5 = 2^5 = 32$. If $a = -2$,$s = (-2)^5 = -32$. Thus,$|s| = 32$.

The roots of the equation $x^5 - 40x^4 + px^3 + qx^2 + rx + s = 0$ are in $G.P.$ The sum of their reciprocals is $10$. Then the value of $|s|$ is

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