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 properties of quadratic equations,the product o

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 properties of quadratic equations,the product of the roots is given by $\alpha \cdot \alpha^3 = \frac{256}{81}$.This simplifies to $\alpha^4 = \frac{256}{81}$,which gives $\alpha = \pm \frac{4}{3}$.The sum of the roots is given by $\alpha + \alpha^3 = -\frac{k}{81}$.Case $1$: If $\alpha = \frac{4}{3}$,then $\frac{4}{3} + (\frac{4}{3})^3 = -\frac{k}{81} \implies \frac{4}{3} + \frac{64}{27} = -\frac{k}{81} \implies \frac{36 + 64}{27} = -\frac{k}{81} \implies \frac{100}{27} = -\frac{k}{81} \implies k = -300$.Case $2$: If $\alpha = -\frac{4}{3}$,then $-\frac{4}{3} - \frac{64}{27} = -\frac{k}{81} \implies -\frac{100}{27} = -\frac{k}{81} \implies k = 300$.Comparing with the given options,the correct 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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