Question

If $x=\sqrt{7-4\sqrt{3}}$, then $x+\frac{1}{x}$ is equal to:

• A. 2
• B. $3\sqrt{7}$
• C. 4
• D. $4\sqrt{7}$

Answer 1

Answer: (C)

$7-4\sqrt{3}=2^2+(\sqrt{3}{)}^2-4\sqrt{3}$
$=(2-\sqrt{3}{)}^2$
$\therefore x=\sqrt{7-4\sqrt{3}}=2-\sqrt{3}$
and $\frac{1}{x}=\frac{1}{2-\sqrt{3}}=2+\sqrt{3}$
Thus, $x+\frac{1}{x}=4$