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If one real root of the quadratic equation $81 x^{2}+k x+256=0$ is cube of the other root, then a value of $\mathrm{k}$ is :
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Verified Answer
The correct answer is:
-300
Let $\alpha$ and $\beta$ be the roots of the equation, $81 x^{2}+k x+256=0$
Given $(\alpha)^{\frac{1}{3}}=\beta \Rightarrow \alpha=\beta^{3}$
$\because \quad$ Product of the roots $=\frac{256}{81}$
$\therefore \quad(\alpha)(\beta)=\frac{256}{81}$
$\Rightarrow \quad \beta^{4}=\left(\frac{4}{3}\right)^{4} \Rightarrow \beta=\frac{4}{3} \Rightarrow \alpha=\frac{64}{27}$
$\because$ Sum of the roots $=-\frac{k}{81}$
$\therefore \quad \alpha+\beta=-\frac{k}{81} \Rightarrow \frac{4}{3}+\frac{64}{27}=-\frac{k}{81}$
$\Rightarrow \quad k=-300$
Given $(\alpha)^{\frac{1}{3}}=\beta \Rightarrow \alpha=\beta^{3}$
$\because \quad$ Product of the roots $=\frac{256}{81}$
$\therefore \quad(\alpha)(\beta)=\frac{256}{81}$
$\Rightarrow \quad \beta^{4}=\left(\frac{4}{3}\right)^{4} \Rightarrow \beta=\frac{4}{3} \Rightarrow \alpha=\frac{64}{27}$
$\because$ Sum of the roots $=-\frac{k}{81}$
$\therefore \quad \alpha+\beta=-\frac{k}{81} \Rightarrow \frac{4}{3}+\frac{64}{27}=-\frac{k}{81}$
$\Rightarrow \quad k=-300$
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