# How to simplify $\sqrt{2+\sqrt{3}}$ $?$

Simplify $$\dfrac{2\left(\sqrt2 + \sqrt6\right)}{3\sqrt{2+\sqrt3}}$$

The answer to this question is $$\frac{4}{3}$$ in a workbook.

How would I simplify $$\sqrt{2+\sqrt3}$$ $$?$$ If it was something like $$\sqrt{3 + 2\sqrt2}$$ , I would have simplified it as follows:
$$\sqrt{3 + 2\sqrt2}$$
$$=$$ $$\sqrt{(\sqrt2)^2 + 2(\sqrt2)(1) + (1)^2}$$
$$=$$ $$\sqrt{(\sqrt2 + 1)^2}$$
$$=$$ $$\sqrt2 + 1$$

But I can't simplify $$\sqrt{2+\sqrt3}$$  like that as $$2+\sqrt3$$ is can't be written as squares of two numbers. Is there any other method?

• See the latter half of this answer. Mar 17 '19 at 7:45

Note that$$\left(\frac{2\left(\sqrt2+\sqrt6\right)}{3\sqrt{2+\sqrt3}}\right)^2=\frac{4\left(8+4\sqrt3\right)}{9\left(2+\sqrt3\right)}=\frac{16}9=\left(\frac43\right)^2.$$
$$\sqrt 2+\sqrt 6=\sqrt 2(1+\sqrt 3)=\sqrt{2(1+\sqrt 3)^2}=\sqrt{2(4+2\sqrt 3)}=2\sqrt{2+\sqrt 3}$$
$$\sqrt{2+\sqrt 3}=\frac 1{\sqrt 2}\sqrt{4+2\sqrt 3}=\frac 1{\sqrt 2}\sqrt{(\sqrt 3+1)^2}=\frac{\sqrt 3+1}{\sqrt 2}=\frac{\sqrt 6+\sqrt 2}2$$
$$\left(\frac{2\left(\sqrt2+\sqrt6\right)}{3\sqrt{2+\sqrt3}}\right)= \frac{(2√2+2√6)\sqrt{2+\sqrt{3}}}{3(2+√3)}=\frac{8+4√3}{2+√3}=\frac{4}{3}.$$