Question

Let $\alpha$ and $\beta$ be the zeros of $f\left(x\right)=a{x}^2+bx+c,a\ne 0$ and $\Delta =b^2-4ac$. If $\alpha +\beta ,{\alpha }^2+{\beta }^2$ and ${\alpha }^3+{\beta }^3$ are in G.P., then :

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

Answer 1

Answer: (B), (C)

The correct options are
B $b.\Delta =0$
C $c.\Delta =0$

$f\left(x\right)=a{x}^2+bx+c,\alpha ,\beta arerootsofequation\therefore \alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}{\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =\frac{b^2}{a^2}-\frac{2c}{a}=\frac{b^2-2ac}{a^2}{\alpha }^3+{\beta }^3=(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )=\frac{-b^3}{a^3}-\frac{3c}{a}\left(\frac{-b}{a}\right)=\frac{3abc-b^3}{a^3}given,\alpha +\beta ,{\alpha }^2+{\beta }^2,{\alpha }^3+{\beta }^{3}areinG.P\therefore ({\alpha }^2+{\beta }^2{)}^2=(\alpha +\beta \left)\right({\alpha }^3+{\beta }^3\left)\right(\because theyareinG.P)\therefore \frac{(b^2-2ac{)}^2}{a^4}=(\frac{-b}{a}\left)\right(\frac{3abc-b^3}{a^3})\therefore b^4+4a^2{c}^2-4a{b}^{2}c=-3a{b}^{2}c+b^4\Rightarrow b^2-4ac=0\Rightarrow \Delta =0\therefore b\Delta =0\&c\Delta =0$

Answer B&C