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

Complex Numbers and Quadratic Equations

Solution:

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