Question

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

• A. $14,8\sqrt{3}$
• B. $-14,-8\sqrt{3}$
• C. $14,-8\sqrt{3}$
• D. $-14,8\sqrt{3}$

Answer 1

The correct option is C $14,-8\sqrt{3}$

If $x=2-\sqrt{3}$

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

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

$=(2-\sqrt{3}+2+\sqrt{3}{)}^2-2=16-2=14$

And $x^2-\frac{1}{x^2}$

$=(2-\sqrt{3}{)}^2-(2+\sqrt{3}{)}^2=4+3-4\sqrt{3}-4-3-4\sqrt{3}=-8\sqrt{3}$