Question

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

• A. -81
• B. 100
• C. -300
• D. 144

Answer 1

Answer: (C)

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