Question

Let $f(x)=a{x}^2+bx+c,a\ne 0$ 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 $\Delta =0$
$\Rightarrow c\Delta =0$