Question

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

• A. $x^2-x-1=0$
• B. $x^2+x-1=0$
• C. $x^2+x+2=0$
• D. none of these

Answer 1

The correct option is A $x^2-x-1=0$

$\alpha ,\beta$ are the roots of the equation $x^2-3x+1=0$

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

Let $\frac{1}{\alpha -2}=y$

$\Rightarrow \alpha =2+\frac{1}{y}$

From (1), we get

${(2+\frac{1}{y})}^2-3(2+\frac{1}{y})+1=0$

$\Rightarrow \frac{(2y+1{)}^2}{y^2}-\frac{3(2y+1)}{y}+1=0$

$\Rightarrow y^2-y-1=0$

$\therefore$ The equation with roots $\frac{1}{\alpha -2},\frac{1}{\beta -2}$ is $x^2-x-1=0$

Hence, option A.