Question

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

• A. $∆\ne 0$
• B. $\frac{b}{∆}=0$
• C. $c∆=0$
• D. $bc\ne 0$

Answer 1

The correct option is C

$c∆=0$

Explanation for the correct option:

The given quadratic equation is

$a{x}^2+bx+c=0$

And $\alpha$ and $\beta$ be the roots

Therefore,

Sum of roots $\alpha +\beta =-\frac{b}{a}$,

Product of roots $\alpha \beta =\frac{c}{a}$

And also, given that $\alpha +\beta ,{\alpha }^2+{\beta }^2$ and ${\alpha }^3+{\beta }^3$ are in GP.

$$\begin{gathered} \because {({\alpha }^2+{\beta }^2)}^2=({\alpha }^3+{\beta }^3)(\alpha +\beta ) \\ \Rightarrow {\alpha }^4+{\beta }^4+2{\alpha }^2{\beta }^2={\alpha }^4+{\beta }^4+\alpha \beta ({\alpha }^2+{\beta }^2) \\ \Rightarrow 2{\alpha }^2{\beta }^2=\alpha \beta ({\alpha }^2+{\beta }^2) \\ \Rightarrow 2\alpha \beta +2\alpha \beta =({\alpha }^2+{\beta }^2)+2\alpha \beta \\ \Rightarrow {(\alpha +\beta )}^2=4\alpha \beta \\ \Rightarrow \left(\frac{b^2}{a^2}\right)=4\left(\frac{c}{a}\right)\left[\because \alpha +\beta =-\frac{b}{a}and\alpha \beta =\frac{c}{a}\right] \\ \Rightarrow b^2=4ac \\ \Rightarrow b^2–4ac=0 \\ \Rightarrow ∆=0\left[\because ∆=b^2-4ac\right] \\ \Rightarrow c∆=0 \end{gathered}$$

Hence, the correct option is (C)