Question

If $\alpha \neq \beta $ but, $\alpha ^{2}=4\alpha -2$ and $\beta ^{2}=4\beta -2$ , then the quadratic equation with roots $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$ is

• A. $x^{2}-4x+2=0$
• B. $x^{2}-6x+1=0$
• C. $x^{2}+6x-1=0$
• D. $x^{2}+4x-2=0$

Answer 1

Answer: (B)

Since, $\alpha ^{2}=4\alpha -2,\beta ^{2}=4\beta -2$
$\Rightarrow $ $\alpha ,\beta $ are the roots of $x^{2}=4x-2$
$\Rightarrow $ $\alpha +\beta =4,\alpha \beta =2$
The quadratic equation with roots $\frac{\alpha }{\beta },\frac{\beta }{\alpha }$ is,
$x^{2}-\left(\frac{\alpha }{\beta } + \frac{\beta }{\alpha }\right)x+\frac{\alpha }{\beta }\times \frac{\beta }{\alpha }=0$
$\Rightarrow x^{2}-\frac{\left\{(\alpha+\beta)^{-}-2 \alpha \beta\right\}}{\alpha \beta} x+1=0$
$\Rightarrow $ $x^{2}-\frac{\left(\right. 16 - 4 \left.\right)}{2}x+1=0$
$\Rightarrow $ $x^{2}-6x+1=0$