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

The correct option is C $-300$

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