Question

In the quadratic equation $a{x}^2+bx+c=0,a\ne 0,\Delta =b^2-4acand\alpha +\beta ,{\alpha }^2+{\beta }^2,{\alpha }^3+{\beta }^3$ are in GP where $\alpha ,\beta$ are the root of $a{x}^2+bx+c=0$ then

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

Answer 1

Answer: (C)

$c\Delta =0$

$\alpha +\beta ,{\alpha }^2+{\beta }^2,{\alpha }^3+{\beta }^3$ are in G.P.

$\therefore ({\alpha }^2+{\beta }^2{)}^2=(\alpha +\beta \left)\right({\alpha }^3+{\beta }^3)$

$\therefore {\alpha }^4+{\beta }^4+2{\alpha }^2{\beta }^2={\alpha }^4+{\beta }^4+{\alpha }^3\beta +\alpha {\beta }^3$

$\therefore {\alpha }^2{\beta }^2-{\alpha }^3\beta -\alpha {\beta }^3+{\alpha }^2{\beta }^2=0$

$\therefore {\alpha }^2\beta (\beta -\alpha )-\alpha {\beta }^2(\beta -\alpha )=0$

$\therefore ({\alpha }^2\beta -\alpha {\beta }^2)(\beta -\alpha )=0$

$\therefore \alpha \beta (\alpha -\beta )(\beta -\alpha )=0$

$\therefore \alpha \beta (\alpha -\beta {)}^2=0$

$\therefore \alpha =0$ or $\beta =0$ or $\alpha -\beta =0$

Case (1) $\alpha =0$ or $\beta =0$,

$⟹x=0$ is a solution of given equation

$\therefore a(0{)}^2+b(0)+c=0$

$\therefore c=0$

Case (2)$\alpha -\beta =0$

$⟹\alpha =\beta$

So, the equation has equal roots.

$\therefore \Delta =b^2-4ac=0$

$\therefore c=0$ or $\Delta =0$

$\therefore c\Delta =0$