If one real root of the quadratic equation 81 | vedclass

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(D) Let the roots of the quadratic equation $81x^2 + kx + 256 = 0$ be $\alpha$ and $\alpha^3$.From the relation between roots and coefficients:Sum of

Vedclass · Accepted Answer

$-300$

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(D) Let the roots of the quadratic equation $81x^2 + kx + 256 = 0$ be $\alpha$ and $\alpha^3$.From the relation between roots and coefficients:Sum of roots: $\alpha + \alpha^3 = -\frac{k}{81}$  $(1)$Product of roots: $\alpha \cdot \alpha^3 = \alpha^4 = \frac{256}{81}$  $(2)$From $(2)$,$\alpha^4 = (\frac{4}{3})^4$,so $\alpha = \pm \frac{4}{3}$.Case $1$: If $\alpha = \frac{4}{3}$,then $\alpha + \alpha^3 = \frac{4}{3} + (\frac{4}{3})^3 = \frac{4}{3} + \frac{64}{27} = \frac{36+64}{27} = \frac{100}{27}$.Substituting into $(1)$: $\frac{100}{27} = -\frac{k}{81} \implies k = -\frac{100 	imes 81}{27} = -300$.Case $2$: If $\alpha = -\frac{4}{3}$,then $\alpha + \alpha^3 = -\frac{4}{3} - \frac{64}{27} = -\frac{100}{27}$.Substituting into $(1)$: $-\frac{100}{27} = -\frac{k}{81} \implies k = 300$.Since $300$ is not in the options,the value is $-300$.

If one real root of the quadratic equation $81x^2 + kx + 256 = 0$ is the cube of the other root,then a value of $k$ is

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