Question

If $\alpha ,\beta$ are the roots of equation $a(x^2-1)+2bx=0$, then the equation whose roots are $2\alpha -\frac{1}{\beta }\&2\beta -\frac{1}{\alpha }$ is

• A. $a{x}^2+2bx-a=0$
• B. $a{x}^2+6bx+9a=0$
• C. $b{x}^2+6ax-9b=0$
• D. $a{x}^2+6bx-9a=0$

Answer 1

The correct option is D

$a{x}^2+6bx-9a=0$

Given that $\alpha \&\beta$ are the roots of the equation $a(x^2-1)+2bx=0$.
$\Rightarrow \alpha +\beta =\frac{-2b}{a}\&\alpha \beta =-1$

We need to find the equation whose roots are $2\alpha -\frac{1}{\beta }\&2\beta -\frac{1}{\alpha }$.

Sum of roots:
$S=2\alpha -\frac{1}{\beta }+2\beta -\frac{1}{\alpha }$
$S=2(\alpha +\beta )-(\frac{1}{\alpha }+\frac{1}{\beta })$
$S=2(\alpha +\beta )-\frac{\alpha +\beta }{\alpha \beta }$
$S=\frac{-4b}{a}+\frac{-2b}{a}=\frac{-6b}{a}$
$\Rightarrow S=\frac{-6b}{a}$

Now, Product of Roots:
$P=2(\alpha -\frac{1}{\beta })(2\beta -\frac{1}{\alpha })$
$P=4\alpha \beta -2-2+\frac{1}{\alpha \beta }$
$P=-9$

The required quadratic equation would be of the form:
$x^2-Sx+P=0$
$x^2-(\frac{-6b}{a})x-9=0$
$a{x}^2+6bx-9a=0$