Q. Let $\alpha,\beta$ are the roots of $f(x) = ax^{2} + bx + c,\ a \neq 0$ and $\mathrm{\Delta} = b^{2} - 4ac.\$If $\alpha + \beta,\ \alpha^{2} + \beta^{2}\ and\ \alpha^{3} + \beta^{3}$ are in GP, then

A) $\mathrm{\Delta}\ \neq 0$
B) $b\mathrm{\Delta} = 0$
C) $c\mathrm{\Delta} = 0$
D) $bc\ \neq 0$
Explanation

(c)
$\because(\alpha + \beta),\ \left( \alpha^{2} + \beta^{2} \right),\ (\alpha^{3} + \beta^{3})$ are in GP.
$\Rightarrow \$ $\left( \alpha^{2} + \beta^{2} \right)^{2} = (\alpha + \beta)(\alpha^{3} + \beta^{3})$
$\Rightarrow \ \alpha^{4} + \beta^{4} + 2\alpha^{2}\beta^{2} = \alpha^{4} + \beta^{4} + \alpha\beta^{3} + \beta\alpha^{3}$
$\Rightarrow \ \alpha\beta\left( \alpha^{2} + \beta^{2} - 2\alpha\beta \right) = 0$
$\Rightarrow \ \alpha\beta(\alpha - \beta)^{2} = 0$
$\Rightarrow \ \alpha\beta = 0$ or $\alpha = \beta$
$\Rightarrow \ \frac{c}{a} = 0$ or $\mathrm{\Delta} = 0$
$\Rightarrow c\mathrm{\Delta} = 0$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.