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
$\Rightarrow 2\left(\frac{1}{2}\left(\frac{1}{y}+3\right)\right)^{2}+4\left(\frac{1}{2}\left(\frac{1}{y}+3\right)\right)-5=0$
$\Rightarrow 2 \frac{(1+3 y)^{2}}{4 y^{2}}+\frac{4(1+3 y)}{2 y}-5=0$
$\Rightarrow (1+3 y)^{2}+4 y(1+3 y)-5 \times 2 y^{2}=0$
$\Rightarrow 1+9 y^{2}+6 y+4 y+12 y^{2}-10 y^{2}=0$
$\Rightarrow 11 y^{2}+10 y+1=0$
Hence, the required equation is $11 x^{2}+10 x+1=0$