Question

Let $\alpha ,\beta$ be roots of the equation $a{x}^2+bx+c=0$ and $\Delta =b^2-4ac$. If $\alpha +\beta ,{\alpha }^2+{\beta }^2,{\alpha }^3+{\beta }^3$ are in $GP$, then

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

Answer 1

Answer: (C)

$c\cdot \Delta =0$

Since, $\alpha ,\beta$ be roots of the equation $a{x}^2+bx+c=0$
$\Rightarrow \alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}$
Now, ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$
$\Rightarrow {\alpha }^2+{\beta }^2=\frac{b^2-2ac}{a^2}$ .....(1)
Also, ${\alpha }^3+{\beta }^3=(\alpha +\beta )({\alpha }^2+{\beta }^2-\alpha \beta )$
$\Rightarrow {\alpha }^3+{\beta }^3=(-\frac{b}{a})\left(\frac{b^2-3ac}{a^2}\right)$ ....(2)
Given, $\alpha +\beta ,{\alpha }^2+{\beta }^2,{\alpha }^3+{\beta }^3$ are in GP
$\Rightarrow ({\alpha }^2+{\beta }^2{)}^2=(\alpha +\beta \left)\right({\alpha }^3+{\beta }^3)$
Put the values from (1) and (2),we get
$\Rightarrow (b^2-2ac{)}^2=b^2(b^2-3ac)$
$\Rightarrow a{b}^{2}c-4a^2{c}^2=0$
$\Rightarrow ac(b^2-4ac)=0$
Since, $a\ne 0$
$\Rightarrow c.\Delta =0$ (Given, $\Delta =b^2-4ac$).