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

The correct option is D $34$

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, ${\alpha }^2{\beta }^{-2}+{\alpha }^{-2}{\beta }^2=\frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}=\frac{{\alpha }^4+{\beta }^4}{(\alpha \beta {)}^2}$

$\Rightarrow ({\alpha }^2+{\beta }^2{)}^2=6^2$

$\Rightarrow {\alpha }^4+{\beta }^4+2{\alpha }^2{\beta }^2=36$

$\Rightarrow {\alpha }^4+{\beta }^4+2=36$

$\Rightarrow {\alpha }^4+{\beta }^4=34.....\left(i\right)$

$\Rightarrow \frac{{\alpha }^4+{\beta }^4}{(\alpha \beta {)}^2}=\frac{34}{(-1{)}^2}=34$

[Putting value of ${\alpha }^4+{\beta }^4=34$ from Equation (i)].