Question

$\text{If}x=6-\sqrt{35}\text{find}{x}^2+\frac{1}{x^2}$

Answer 1

$x=6-\sqrt{35}$

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

$\Rightarrow x=6-\sqrt{35}$

$\Rightarrow \frac{1}{x}=\frac{1}{6-\sqrt{35}}\times \frac{6+\sqrt{35}}{6+\sqrt{35}}$

$=\frac{6+\sqrt{35}}{{\left(6\right)}^2+{\left(\sqrt{35}\right)}^2}$

$=6+\sqrt{35}$

$\therefore x+\frac{1}{x}=6-\sqrt{35}+6+\sqrt{35}=12$

$\therefore x^2+\frac{1}{x^2}={(x+\frac{1}{x})}^2-2={12}^2-2=142.$

Hence, the answer is $142.$