Question

If $\alpha \ne \beta$ 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,
$x^2-\left(\frac{\alpha }{\beta }+\frac{\beta }{\alpha }\right)x+\frac{\alpha }{\beta }\times \frac{\beta }{\alpha }=0$
$\Rightarrow x^2-\frac{\left\{(\alpha +\beta {)}^{-}-2\alpha \beta \right\}}{\alpha \beta }x+1=0$
$\Rightarrow$ $x^2-\frac{(16-4)}{2}x+1=0$
$\Rightarrow$ $x^2-6x+1=0$