Find the value of $\sqrt{2+\sqrt{3}} +\sqrt{2-\sqrt{3}}$.

$\sqrt{6}$

Find the value of $\sqrt{2+\sqrt{3}} +\sqrt{2-\sqrt{3}}$

Multiply and divide the equation by 2

= $\sqrt{2+\sqrt{3}} +\sqrt{2-\sqrt{3}}$

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

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

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

= $\frac{2\sqrt {3}}{\sqrt {2}}$ = $\sqrt{6}$