Question

If $\alpha$ and $\beta$ are the roots of the equation $2x^2+4x-5=0,$ then the equation whose roots are $\frac{1}{2\alpha -3}$ and $\frac{1}{2\beta -3}$ is

• A. $x^2+10x-11=0$
• B. $11x^2+10x+1=0$
• C. $x^2+10x+11=0$
• D. $11x^2-10x+1=0$

Answer 1

Answer: (B)

Substituting $y=\frac{1}{2\alpha -3},$ we get,
$2\alpha =\frac{1}{y}+3\Rightarrow \alpha =\frac{1}{2}\left(\frac{1}{y}+3\right)$
Since, $\alpha$ is a root of the given equation
$\Rightarrow 2{\left(\frac{1}{2}\left(\frac{1}{y}+3\right)\right)}^2+4\left(\frac{1}{2}\left(\frac{1}{y}+3\right)\right)-5=0$
$\Rightarrow 2\frac{(1+3y{)}^2}{4y^2}+\frac{4(1+3y)}{2y}-5=0$
$\Rightarrow (1+3y{)}^2+4y(1+3y)-5\times 2y^2=0$
$\Rightarrow 1+9y^2+6y+4y+12y^2-10y^2=0$
$\Rightarrow 11y^2+10y+1=0$
Hence, the required equation is $11x^2+10x+1=0$