Let $f(x) = ax^{2} + bx + c,\ a
eq 0$ and $\mathrm{\Delta} = b^{2} - 4ac$. If $\alpha + \beta,\ \alpha^{2} + \beta^{2}$ and $\alpha^{3} + \beta^{3}$ are in GP, then

Question

Let $f(x) = ax^{2} + bx + c,\ a 
eq 0$ and $\mathrm{\Delta} = b^{2} - 4ac$. If $\alpha + \beta,\ \alpha^{2} + \beta^{2}$ and $\alpha^{3} + \beta^{3}$ are in GP, then

Dr. Abubacker · Accepted Answer

(c)
Since, $(\alpha + \beta),\ (\alpha^{2} + \beta^{2})$ and $(\alpha^{3} + \beta^{3})$ are in GP.
$\left( \alpha^{2} + \beta^{2} \right)^{2} = (\alpha + \beta)(\alpha^{3} + \beta^{3})$
$\Rightarrow \ \alpha^{4} + \beta^{4} + 2\alpha^{2}\beta^{2} = \alpha^{4} + \beta^{4} + \alpha\beta^{3} + \beta\alpha^{3}$
$\Rightarrow \ \ \alpha\beta\left( \alpha^{2} + \beta^{2} - 2\alpha\beta \right) = 0$
$\Rightarrow \ \ \alpha\beta(\alpha - \beta)^{2} = 0$
$\Rightarrow \ \alpha\beta = 0$ or $\alpha = \beta$
$ie,\ \frac{c}{a} = 0$ or $\mathrm{\Delta} = 0$
$\Rightarrow c\mathrm{\Delta} = 0$

Q. Let $f(x) = ax^{2} + bx + c,\ a \neq 0$ and $\mathrm{\Delta} = b^{2} - 4ac$. If $\alpha + \beta,\ \alpha^{2} + \beta^{2}$ and $\alpha^{3} + \beta^{3}$ are in GP, then

Explanation

Reviewed by Dr. Abubacker

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