Question

For the equation $3x^2 + px + 3 = 0, p > 0$, if one of the root is square of the other, then $p$ is equal to

Answer 1

Let $\alpha, \alpha^2$ be the roots of $3x^2 + px + 3$.
$\therefore \alpha + \alpha^2 = - p/3$ and $\alpha^3 = 1$
$\Rightarrow ( \alpha - 1)(\alpha^2 + \alpha + 1) = 0$
$\Rightarrow \alpha = 1$ or $\alpha^2 + \alpha = 1$
If $\alpha = 1 ,p =- 6$
which is not possible as $p > 0$
If $\alpha^2 + \alpha = -1 $
$\Rightarrow -p/3=-1$
$\Rightarrow p = 3$.