Question

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

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

Answer 1

Answer: (C)

$x^2-x-1=0$

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

$\therefore \frac{1}{\alpha -2}+\frac{1}{\beta -2}=\frac{\beta -2+\alpha -2}{(\alpha -2)(\beta -2)}$

$=\frac{\alpha +\beta -4}{\alpha \beta +4-2\lambda -2\beta }$

$=\frac{3-4}{1+4-2}$

$=\frac{-1}{-1}=1$

$\therefore$ Sum of new roots $=1$

$\therefore \left(\frac{1}{\alpha -2}\right)\left(\frac{1}{\beta -2}\right)=\frac{1}{\alpha \beta +4-2\alpha -2\beta }$

$=\frac{1}{1+4-2}\left(3\right)=\frac{1}{5-6}=-1$

$\therefore$ Equation

$x^2-x-1=0$