Simplify a square root in another square root

Question

How to simply $$\sqrt{2+\sqrt3}$$?

I tried to do it as $\sqrt{4+\sqrt3-2}$ but it didn't seem to work.
Is there any way to simplify it?

Answer 1

$$\sqrt{2+\sqrt3}=\frac{1}{\sqrt2}\sqrt{4+2\sqrt3}=\frac{1}{\sqrt{2}}\sqrt{1+2\sqrt3+3}=\frac{1+\sqrt3}{\sqrt2}$$

Answer 2

You can notice that $2 + \sqrt 3 =( x+y \sqrt3)^2 \implies x^2 + 3y^2 = 2 , 2xy = 1$.

This would give four solutions:

1 : $x = \pm\sqrt{\frac 32}, y = \pm\sqrt{\frac 16}$

2 : $x = y = \pm \sqrt{\frac 12}$

And hence $\sqrt{2 + \sqrt 3}$ can be any of the values $x + y \sqrt 3$ , where $x,y$ are as above. As per the definition of square root, the value that it would take is $x =y= \frac 1{\sqrt{2}}$

Answer 3

For yet another approach, let $a = \sqrt{2+\sqrt{3}}$ and $b = \sqrt{2-\sqrt{3}}$ and note that $a \gt b \gt 0\,$. Then:

$$\require{cancel} a^2+b^2 = (2+\bcancel{\sqrt{3}})+(2-\bcancel{\sqrt{3}}) = 4 \\ ab = \sqrt{(2+\sqrt{3})(2-\sqrt{3})}=\sqrt{2^2 - 3} = 1 $$

Then $(a+b)^2=a^2+b^2+2ab=6\,$, so $a+b=\sqrt{6}$ given that both are positive.

It then follows by Vieta's relations, that $a,b$ are the roots of $x^2 - \sqrt{6} + 1 = 0\,$ i.e. $\frac{1}{2}(\sqrt{6} \pm \sqrt{2})\,$, and $a$ being the largest root, it is $a = \sqrt{2+\sqrt{3}} = \frac{1}{2}(\sqrt{6} + \sqrt{2})\,$.