Question

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

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

Answer 1

Answer: (D)

$p^{3} + q^{2} + q(1 - 3p) = 0$

Answer 2

Let α and $\alpha^{2}$ be the roots then $\alpha + \alpha^{2} = - p$, $\alpha.\alpha^{2} = q$

Now $(\alpha + \alpha^{2})^{3} = \alpha^{3} + \alpha^{6} + 3\alpha^{3}(\alpha + \alpha^{2})$ ⇒ $- p^{3} = q + q^{2} - 3pq$ ⇒ $p^{3} + q^{2} + q(1 - 3p) = 0$