Question

If $\alpha ,\beta$ are the roots of the equations $x^2-2x-1=0$, then what is the value of ${\alpha }^2{\beta }^{-2}+{\alpha }^{-2}{\beta }^2$

• A. -2
• B. 0
• C. 30
• D. 34

Answer 1

Answer: (D)

Since, $\alpha$ and $\beta$ are the roots of the equation $x^2-2x-1=0$, then
Sum of roots, $\alpha +\beta =2$ and
product of the roots $\alpha \beta =-1$
Since, $(\alpha +\beta )={\alpha }^2+{\beta }^2+2\alpha \beta$
$\Rightarrow 4={\alpha }^2+{\beta }^2-2$
$\Rightarrow {\alpha }^2+{\beta }^2=6$
Now, ${\alpha }^2{\beta }^{-2}+{\alpha }^{-2}{\beta }^2=\frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}=\frac{{\alpha }^4+{\beta }^4}{(\alpha \beta {)}^2}$
$\Rightarrow {\left({\alpha }^2+{\beta }^2\right)}^2=6^2$
$\Rightarrow {\alpha }^4+{\beta }^4+2{\alpha }^2{\beta }^2=36$
$\Rightarrow {\alpha }^4+{\beta }^4+2=36$
$\Rightarrow {\alpha }^4+{\beta }^4=34\dots$ (i)
$\Rightarrow \frac{{\alpha }^4+{\beta }^4}{(\alpha \beta {)}^2}=\frac{34}{(-1{)}^2}=34$
[Putting value of ${\alpha }^4+{\beta }^4=34$ from Equation (i)].