Question

If $(\alpha +\sqrt{\beta })$ and $(\alpha -\sqrt{\beta })$ are the roots of the equation $x^2+px+q=0$, where $\alpha ,\beta ,p$ and $q$ are real, then the roots of the equation $(p^2-4q)(p^2{x}^2+4px)-16q=0$ are

• A. $(\frac{1}{\alpha }+\frac{1}{\sqrt{\beta }})$ and $(\frac{1}{\alpha }-\frac{1}{\sqrt{\beta }})$
• B. $(\frac{1}{\sqrt{\alpha }}+\frac{1}{\beta })$ and $(\frac{1}{\sqrt{\alpha }}-\frac{1}{\beta })$
• C. $(\frac{1}{\sqrt{\alpha }}+\frac{1}{\sqrt{\beta }})$ and $(\frac{1}{\sqrt{\alpha }}-\frac{1}{\sqrt{\beta }})$
• D. $(\sqrt{\alpha }+\sqrt{\beta })$ and $(\sqrt{\alpha }-\sqrt{\beta })$

Answer 1

The correct option is A $(\frac{1}{\alpha }+\frac{1}{\sqrt{\beta }})$ and $(\frac{1}{\alpha }-\frac{1}{\sqrt{\beta }})$

Since, $(\alpha +\sqrt{\beta })$ and $(\alpha -\sqrt{\beta })$ are the roots of the equation $x^2+px+q=0$
$\therefore$ Sum of roots $=-p$
$\Rightarrow (\alpha +\sqrt{\beta })+(\alpha -\sqrt{\beta })=-p$
$\Rightarrow 2\alpha =-p\Rightarrow \alpha =-\frac{p}{2}$
and

Product of roots $=q$
$(\alpha +\sqrt{\beta })(\alpha -\sqrt{\beta })=q$
$\Rightarrow {\alpha }^2-\beta =q$
$\Rightarrow \beta ={\alpha }^2-q={(-\frac{p}{2})}^2-q=\frac{p^2}{4}-q$
$\Rightarrow p^2-4q=4\beta$ ..... $\left(i\right)$

The given equation is
$(p^2-4q)(p^2{x}^2+4px)-16q=0$ ....... $\left(ii\right)$

$\Rightarrow 4\beta (p^2{x}^2+4px)-16({\alpha }^2-\beta )=0[\because {\alpha }^2-\beta =q]$

$\Rightarrow \beta (p^2{x}^2+4px)-4({\alpha }^2-\beta )=0[\because {\alpha }^2-\beta =q]$

$\Rightarrow \beta (4{\alpha }^2{x}^2-8\alpha x)-4({\alpha }^2-\beta )=0$

$\Rightarrow {\alpha }^2\beta x^2-2\alpha \beta x+\beta ={\alpha }^2$

$\Rightarrow (\alpha x\sqrt{\beta }-\sqrt{\beta }{)}^2={\alpha }^2$

$\Rightarrow \alpha x\sqrt{\beta }-\sqrt{\beta }=\pm \alpha$

$\therefore x=\frac{1}{\alpha }\pm \frac{1}{\sqrt{\beta }}$

$\Rightarrow (\frac{1}{\alpha }+\frac{1}{\sqrt{\beta }})$ and $(\frac{1}{\alpha }-\frac{1}{\sqrt{\beta }})$ are the roots of the given equation $\left(ii\right)$