Question

If $x=\frac{1}{\sqrt{3}+1}$ then the value of $x+\frac{1}{x}$ is

Answer 1

Given:

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

so, $x+\frac{1}{x}=\left(\frac{1}{\sqrt{3}+1}\right)+(\sqrt{3}+1)$

On solving it, we get

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

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

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