Question

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

Answer 1

$\Rightarrow$ $\alpha$ and $\beta$ are roots of the equation $x^2-5x-3=0$

$\Rightarrow$ Here, $a=1,b=-5,c=-3$

$\Rightarrow$ $\alpha +\beta =\frac{-b}{a}=\frac{-(-5)}{1}=5$ ---( 1 )

$\Rightarrow$ $\alpha \beta =\frac{c}{a}=\frac{-3}{1}=-3$ ----- ( 2 )

Now,

$\Rightarrow$ $\frac{1}{2\alpha -3}+\frac{1}{2\beta -3}=\frac{2\beta -3+2\alpha -3}{(2\alpha -3)(2\beta -3)}$

$=\frac{2(\alpha +\beta )-6}{4\alpha \beta -6\alpha -6\beta +9}$

$=\frac{2\left(5\right)-6}{4(-3)-6(\alpha +\beta )+9}$ [ Using ( 1 ) and ( 2 ) ]

$=\frac{10-6}{-12-6\left(5\right)+9}$

$=\frac{4}{-33}$

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

$\Rightarrow$ $\frac{1}{2\alpha -3}.\frac{1}{2\beta -3}=\frac{1}{4\alpha \beta -6\alpha -6\beta +9}$

$=\frac{1}{4\alpha \beta -6(\alpha +\beta )+9}$

$=\frac{1}{4(-3)-6\left(5\right)+9}$

$=\frac{1}{-33}$

$\therefore$ $\frac{1}{2\alpha -3}.\frac{1}{2\beta -3}=\frac{-1}{33}$ --( 4 )

$\Rightarrow$ $x^2-(\frac{1}{2\alpha -3}+\frac{1}{2\beta -3})x+(\frac{1}{2\alpha -3}.\frac{1}{2\beta -3})=0$

By using ( 3 ) and ( 4 ),

$\Rightarrow$ $x^2-\frac{4}{33}-\frac{1}{33}=0$

$\Rightarrow$ $33x^2-4x-1=0$