Question

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

• A. $14$
• B. $-14$
• C. $2$
• D. $-2$

Answer 1

Answer: (A)

$14$

$\frac{1}{x}=\frac{1}{2-\sqrt{3}}=\frac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})}=2+\sqrt{3}$

Now,

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

$\Rightarrow x^2+\frac{1}{x^2}=(2-\sqrt{3}+2+\sqrt{3}{)}^2-2=16-2=14$