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  4. If one real root of the quadratic equation 81x2 + kx + 256 = 0 is cube of the other root, then a value of k is

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

JEE MainJEE Main 2019Complex Numbers and Quadratic Equations

Solution:

$81 x^2 + kx + 256 = 0 ; x = \alpha , \alpha^3$
$\Rightarrow \alpha^{4} = \frac{256}{81} \Rightarrow \alpha =\pm\frac{4}{3} $
Now $ - \frac{k}{81} = \alpha+\alpha^{3} = \pm \frac{100}{27} $
$ \Rightarrow k = \pm 300$