Question

If $\alpha ,\beta$ are roots of $x^2-5x-3=0$, then the equation with roots $\frac{1}{2\alpha -3}$ and $\frac{1}{2\beta -3}$ is

• A. $33x^2+4x-1=0$
• B. $33x^2-4x+1=0$
• C. $33x^2-4x-1=0$
• D. $33x^2+4x+1=0$

Answer 1

The correct option is A $33x^2+4x-1=0$

For $x^2-5x+3=0$

$\alpha +\beta =5;\alpha \beta =3$

Let $r=\frac{1}{2\alpha -3},s=\frac{1}{2\beta -3}$

$r+s=\frac{1}{2\alpha -3}+\frac{1}{2\beta -3}=\frac{2\beta -3+2\alpha -3}{(2\alpha -3)(2\beta -3)}=\frac{-4}{33}$

$rs=\frac{1}{2\alpha -3}\times \frac{1}{2\beta -3}=\frac{-1}{33}$

Then the equation is $33x^2+4x-1=0$.