Question

If $x=\frac{3-\sqrt{13}}{2}$, find the value of $x^2+\frac{1}{x^2}$.

Answer 1

Given that, $x=\frac{3-\sqrt{13}}{2}$

$\therefore \frac{1}{x}=\frac{2}{3-\sqrt{13}}$ [Taking reciprocal on both sides]

We know that, $(a+b{)}^2=a^2+b^2+2ab$

$\therefore {(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2$

$\Rightarrow x^2+\frac{1}{x^2}={(x+\frac{1}{x})}^2-2$

Now, $x+\frac{1}{x}=\frac{3-\sqrt{13}}{2}+\frac{2}{3-\sqrt{13}}=\frac{(3-\sqrt{13}{)}^2+4}{2(3-\sqrt{13})}$

$\Rightarrow \frac{9+13-6\sqrt{13}+4}{2(3-\sqrt{13})}=\frac{26-6\sqrt{13}}{2(3-\sqrt{13})}$

$\Rightarrow \frac{2\sqrt{13}(\sqrt{13}-3)}{2(3-\sqrt{13})}=\frac{-2\sqrt{13}(3-\sqrt{13})}{2(3-\sqrt{13})}=-\sqrt{13}$

$\Rightarrow x+\frac{1}{x}=-\sqrt{13}$

and hence, ${(x+\frac{1}{x})}^2=(-\sqrt{13}{)}^2=13$

$\therefore x^2+\frac{1}{x^2}={(x+\frac{1}{x})}^2-2$

$\Rightarrow 13-2=11$

Hence, if $x=\frac{3-\sqrt{13}}{2}$, then the value of $x^2+\frac{1}{x^2}$ is $11$.