Question
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
A.
$$\frac{1}{3}$$
B.
1
C.
3
D.
$$\frac{2}{3}$$
Answer :
3
Solution :
Let $$\alpha ,{\alpha ^2}$$ be the root of $$3{x^2} + px + 3.$$
$$\eqalign{
& \therefore \,\,\alpha + {\alpha ^2} = - \frac{p}{3}\,\,{\text{and }}{\alpha ^3} = 1 \cr
& \Rightarrow \,\,\left( {\alpha - 1} \right)\left( {{\alpha ^2} + \alpha + 1} \right) = 0 \cr
& \Rightarrow \,\,\alpha = 1\,\,{\text{or }}{\alpha ^2} + \alpha = - 1 \cr
& {\text{If }}\alpha = 1,p = - 6\,\,{\text{which is not possible as }}\,p > 0 \cr
& {\text{If }}{\alpha ^2} + \alpha = - 1 \cr
& \Rightarrow \,\, - \frac{p}{3} = - 1 \cr
& \Rightarrow \,\,p = 3. \cr} $$