Question

Let $f (x) = ax^2 + bx + c, a \ne 0$ and $\Delta =b^2 -4ac.$ If $\alpha + \beta, \alpha^2+ \beta^2$ and $\alpha ^3+ \beta^3$ are in GP, then

• A. $\Delta \ne 0$
• B. $b\Delta = 0$
• C. $c\Delta = 0$
• D. $ bc \ne 0$

Answer 1

Answer: (C)

Since, $ (a+\beta) , (a^2+\beta^2), (a^3+\beta^3)$ are in GP
$\Rightarrow (a^2+\beta^2)^2 =(a+\beta) (a^3+\beta^3) $
$\Rightarrow a^4+\beta^4+2a^2 \beta^2 = a^4+ \beta^4 + a\beta^3+\beta a^3 $
$\Rightarrow a\beta (a^2 +\beta^2 - 2a\beta) = 0 $
$\Rightarrow a\beta (a-\beta)^2 = 0 $
$\Rightarrow a\beta = 0$ or $ a = \beta$
$\Rightarrow \frac{c}{a} = 0 $ or $ \Delta = 0$
$\Rightarrow c \Delta = 0 $