Q. If one root is square of the root of the equation $x^{2} + px + q = 0$, then the relation between $p$ and $q$ is
A) $p^{3} - (3p - 1)q + q^{2} = 0$
B) $p^{3} - q(3p + 1) + q^{2} = 0$
C) $p^{3} + q(3p - 1) + q^{2} = 0$
D) $p^{3} + q(3p + 1) + q^{2} = 0$
Explanation
(a)
Let roots of the equation $x^{2} + px + q = 0$ be $\alpha\$and$\ \alpha^{2}$
$\therefore\ \alpha + \alpha^{2} = - p$ and $\alpha^{3} = q$
$\Rightarrow \ \alpha(\alpha + 1) = - p$
$\Rightarrow \ \alpha^{3}\left\lbrack \alpha^{3} + 1 + 3\alpha(\alpha + 1) \right\rbrack = - p$
$\Rightarrow q(q + 1 - 3p) = - p^{3}$
$\Rightarrow p^{3} - (3p - 1)q + q^{2} = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.