Showing that $\sqrt{2+\sqrt3}=\frac{\sqrt2+\sqrt6}{2}$ So I was playing around on the calculator, and it turns out that $$\sqrt{2+\sqrt3}=\frac{\sqrt2+\sqrt6}{2}$$
Does anyone know a process to derive this?
 A: $$\sqrt{2+\sqrt3}=\sqrt{\frac{4+2\sqrt3}{2}}=\sqrt{\frac{3+2\sqrt3.1+1}{2}}=\sqrt{\frac{(\sqrt3+1)^2}{2}}=\frac{\sqrt3+1}{\sqrt{2}} =\frac{\sqrt6+\sqrt{2}}{2}$$
A: Both sides are clearly positive, and squaring the left hand side shows that
$$\begin{align}  \left(\frac{\sqrt{2}+\sqrt{6}}{2}\right)^2&=\frac{\sqrt{2}^2+2\sqrt{2}\sqrt{6}+\sqrt{6}^2}{2^2}\\&=\frac{2+4\sqrt{3}+6}{4}\\&=2+\sqrt{3}.
\end{align}$$
A: To simplify a square root
$$\sqrt{x+\sqrt y}$$
to the form $\sqrt a+\sqrt b$, you are looking for $a$ and $b$ with
$$a+b+2\sqrt{ab}=x+\sqrt y.$$
If $x$ and $y$ are rational, and $a$ and $b$ should be rational (and $y$ isn't the square of a rational), you want
$$a+b=x,\qquad 4ab=y.$$
You can solve for $a$ and $b$ as the roots of the quadratic
$$t^2-tx+\frac y4.$$
In this particular case, the quadratic $t^2-2t+\frac 32$ has two rational roots, and so you can find the square root in this form.
A: The simpliest way is to square both sides and then compare them.
Formally, consider $\left(\frac{\sqrt{2}+\sqrt{6}}{2}\right)^{2}=\left(\frac{8+2\sqrt{12}}{4}\right)=2+\sqrt{3}$.
Taking square-root on both sides yields $\frac{\sqrt{2}+\sqrt{6}}{2}=\sqrt{2+\sqrt{3}}$.
