square in a square

$\sqrt {2+ \sqrt{3}} \\ = \sqrt{ \cfrac{3+2 \cdot \sqrt{3}+1}{2}} \\ = \cfrac{ \sqrt{ (\sqrt{3}+1)^2}}{\sqrt{2}} \\ = \cfrac{ \sqrt{3}+1}{\sqrt{2}}$

Similarly, we have

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

Hence,

$x=\cfrac{ \sqrt{3}+1}{\sqrt{2}}+ \cfrac{\sqrt{3}-1}{\sqrt{2}} \\ x= \cfrac{2 \cdot \sqrt{3}}{\sqrt{2}} \\ a=x^2= \boxed{6}$

$x=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}$ $x^2=(\sqrt{2+\sqrt{3}})^2+2(\sqrt{2+\sqrt{3}})(\sqrt{2-\sqrt{3}})+(\sqrt{2-\sqrt{3}})^2$ $x^2=2+\sqrt{3}+2(\sqrt{4-3})+2-\sqrt{3}$ $x^2=2+2+2=\boxed{6}$

hey dude, you might wanted to ask $x^{2}$ but you have put up the picture asking the value of x... so better change it!

now coming to the question, in this we have got to find $x^{2}$ rather than x..

da