Question

If $\alpha ,\beta$ are the roots of $a{x}^2+bx+c=0$, then $-\frac{1}{\alpha },\frac{1}{\beta }$ are the roots of:

• A. $a{x}^2-\sqrt{b^2-4ac}.x+c=0$
• B. $c{x}^2-\sqrt{b^2-4ac}.x-a=0$
• C. $c{x}^2+bx+a=0$
• D. $a{x}^2-bxc=0$

Answer 1

Answer: (B)

$c{x}^2-\sqrt{b^2-4ac}.x-a=0$

Since $\alpha ,\beta$ are the roots of $a{x}^2+bx+c=0$,
So, $\alpha +\beta =\frac{-b}{a}$ and $\alpha \beta =\frac{c}{a}$

Here,

$\frac{-1}{\alpha }+\frac{1}{\beta }=\frac{\alpha -\beta }{\alpha \beta }=\frac{\sqrt{\frac{b^2}{a^2}-4\frac{c}{a}}}{\frac{c}{a}}=\frac{\sqrt{b^2-4ac}}{c}$

$\frac{-1}{\alpha \beta }=\frac{-a}{c}$

We thus get the equation as

$c{x}^2-\left(\sqrt{b^2-4ac}\right)x-a=0$