Question

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

• A. $\frac{2}{3}$
• B. $1$
• C. $3$
• D. $\frac{1}{3}$

Answer 1

The correct option is C

$3$

$\alpha +{\alpha }^2=-\frac{p}{3}$ and $\alpha .{\alpha }^2=1$

So, $\alpha =1,\omega ,{\omega }^2.$ If $\alpha =1,p<0$

If $\alpha =\omega$ or ${\omega }^2,$ we have $\omega +{\omega }^2=-\frac{p}{3}\Rightarrow -1=\frac{-p}{3}\Rightarrow p=3$