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

Answer: 3

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 ${\alpha }^2+\alpha =-1$
$\Rightarrow -p/3=-1$
$\Rightarrow p=3$.