Question

If $\frac{1}{\sqrt{\alpha}}$ and $\frac{1}{\sqrt{\beta}}$ are the roots of the equation, $ax^{2} + bx +1 = 0 \left(a ^{ }\ne 0, a, b \in R\right)$, then the equation, $x\left(x + b^{3}\right) + \left(a^{3} — 3abx\right)$ = 0 has roots :

• A. $\alpha^{3/2}$ and $\beta^{3/2}$
• B. $\alpha \,\beta^{1/2}$ and $\alpha^{1/2}\,\beta$
• C. $\sqrt{\alpha \,\beta }$ and $\alpha\,\beta$
• D. $\alpha^{-\frac{3}{2}}$ and $\beta^{-\frac{3}{2}}$

Answer 1

Answer: (A)

$\frac{1}{\sqrt{\alpha}}+\frac{1}{\sqrt{\beta}}=-\frac{b}{a} also \frac{1}{\sqrt{\alpha\beta }}=\frac{1}{a}$
$ \Rightarrow \sqrt{\alpha}+\sqrt{\beta}=-b$
now $x \left(x + b^{3}\right) + a^{3} - 3abx$
$=x^{2} + \left(b^{3} - 3ab\right) x + a^{3} $
$= x^{2}+ b \left(b^{2} - 3a\right) x +a^{3}$
$=x^{2}-\left(\sqrt{\alpha }+\sqrt{\beta }\right)\left\{\alpha+\beta+2\sqrt{\alpha\beta }-3\sqrt{\alpha \beta }\right\}x+\alpha\beta \sqrt{\alpha\beta }$
$=x^{2}-\left(\alpha\sqrt{\alpha }+\beta\sqrt{ \beta }\right)+\alpha\beta\sqrt{\alpha \beta }$
$\Rightarrow $ roots are $\alpha \sqrt{\alpha}$ and $\beta \sqrt{ \beta}$