Question

If one root of the equation $x^{2} + p x + q = 0$ is the square of the other, then

• A. $p^{3} + q^{2} - q \left(\right. 3 p + 1 \left.\right) = 0$
• B. $p^{3} + q^{2} + q \left(\right. 1 + 3 p \left.\right) = 0$
• C. $p^{3} + q^{2} + q \left(\right. 3 p - 1 \left.\right) = 0$
• D. $p^{3} + q^{2} + q \left(\right. 1 - 3 p \left.\right) = 0$

Answer 1

Answer: (D)

$x^{2}+px+q=0\,$
Let, $\alpha , \alpha ^{2}$ are the roots of the given equation.
$\Rightarrow \alpha +\alpha ^{2}=-p$
and $\alpha \cdot \alpha ^{2} = q$
$\Rightarrow \alpha ^{3} = q$
$\because \alpha + \alpha ^{2} = - p$
$\Rightarrow\left(\alpha+\alpha^{2}\right)^{3}=-p^{3}$
$\Rightarrow \alpha^{6}+\alpha^{3}+3 \alpha \cdot \alpha^{2}\left(\alpha+\alpha^{2}\right)=-p^{3}$
$\Rightarrow q^{2}+q+3 \alpha^{3}\left(\alpha+\alpha^{2}\right)=-p^{3}$
$\Rightarrow q^{2}+q+3 q(-p)=-p^{3}$
$\Rightarrow p^{3}+q^{2}+q-3 p q=0\,$
$\Rightarrow p^{3}+q^{2}+q(1-3 p)=0\,$