• # question_answer For the equation $3{{x}^{2}}+px+3=0,\,p>0$ if one of the root is square of the other, then p is equal to  [IIT Screening 2000] A) $\frac{1}{3}$ B) 1 C) 3 D) $\frac{2}{3}$

$\alpha +{{\alpha }^{2}}=-\frac{p}{3}$ and $\alpha .{{\alpha }^{2}}=1.$ So$\alpha =1,$$\omega ,\,{{\omega }^{2}}$. If $\alpha =1,p<0$. If $\alpha =\omega \,\,\text{or}\,\,{{\omega }^{2}},$ we have $\omega +{{\omega }^{2}}=-\frac{p}{3}\,$ Þ  $-1=\frac{-p}{3}\,\,\,\,\,\Rightarrow \,\,p=3$.