Question

In the quadratic equation $a{x}^2+bx+c=0,\Delta =b^2-4acand\alpha +\beta ,{\alpha }^2+{\beta }^2,{\alpha }^3+{\beta }^3,$ are in G.P. 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

The correct option is C $c\Delta =0$

In the quadratic equation $a{x}^2+bx+c=0$
$\Delta =b^2-4acand\alpha +\beta =-\frac{b}{a}.\alpha \beta =\frac{c}{a}{\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =\frac{b^2}{a^2}-\frac{2c}{a}=\frac{b^2-2ac}{a^2}and{\alpha }^3+{\beta }^3=-\frac{b^3}{a^3}-\frac{3c}{a}(-\frac{b}{a})=-\left(\frac{b^3-3abc}{a^3}\right)Given\alpha +\beta ,{\alpha }^2+{\beta }^2,{\alpha }^3+{\beta }^{3}areinG.P.\Rightarrow -\frac{b}{a},\frac{b^2-2ac}{a^2},-\frac{(b^3-3abc)}{a^3}areinG.P.\Rightarrow {\left(\frac{b^2-2ac}{a^2}\right)}^2=\frac{b}{a}\left(\frac{b^3-3abc}{a^3}\right)\Rightarrow b^4+4a^2{c}^2-4a{b}^{2}c=b^4-3a{b}^{2}c\Rightarrow 4a^2{c}^2-a{b}^{2}c=0\Rightarrow ac\Delta =0\Rightarrow c\Delta =0(\because Inquadratic$ a $\ne 0)$