Question

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

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

Answer 1

Answer: (C)

$14,-8\sqrt{3}$

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

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

$(x^2-\frac{1}{x^2})=(x-\frac{1}{x})(x+\frac{1}{x})=(4\times -2\sqrt{3})=-8\sqrt{3}$