Question

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

Answer 1

Given:

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

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

Lets do rationalize the denominator,

Since, the $2-\sqrt{3}$ is Rationalizing factor of $2+\sqrt{3}$, multiply both numerator and the denominatoe with $2-\sqrt{3}$.

$=\frac{1\times (2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}$

$=\frac{2-\sqrt{3}}{(2{)}^2-({\sqrt{3}}^2)}$

$=\frac{2-\sqrt{3}}{4-3}$

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

$\therefore \frac{1}{x}=2-\sqrt{3}$ ---(2)

Now,

$x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4$