Question

If $x=\sqrt{6}-\sqrt{5}$, then $x^2+\frac{1}{x^2}=$?

• A. $11$
• B. $1$
• C. $22$
• D. $\sqrt{30}$

Answer 1

The correct option is C $22$

$x=\sqrt{6}-\sqrt{5}$.
$\therefore \frac{1}{x}=\frac{1}{\sqrt{6}-\sqrt{5}}\times \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}=\frac{\sqrt{6}+\sqrt{5}}{6-5}=\sqrt{6}+\sqrt{5}$.
$\therefore {(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2x\times \frac{1}{x}$
$⟹$ ${(\sqrt{6}-\sqrt{5}+\sqrt{6}+\sqrt{5})}^2=x^2+\frac{1}{x^2}+2$
$⟹$ ${\left(2\sqrt{6}\right)}^2=x^2+\frac{1}{x^2}+2$
$⟹$ $4\times 6=x^2+\frac{1}{x^2}+2⟹x^2+\frac{1}{x^2}=24-2=22$.

Therefore, option $C$ is correct.