Showing that $\sqrt{2+\sqrt3}=\frac{\sqrt2+\sqrt6}{2}$

Question

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?

Answer 1

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.

Answer 2

$$\sqrt{2+\sqrt3}=\sqrt{\frac{4+2\sqrt3}{2}}=\sqrt{\frac{3+2\sqrt3.1+1}{2}}=\sqrt{\frac{(\sqrt3+1)^2}{2}}=\frac{\sqrt3+1}{\sqrt{2}} =\frac{\sqrt6+\sqrt{2}}{2}$$

Answer 3

Both sides are clearly positive, and squaring the left hand side shows that $$\begin{align} \left(\frac{\sqrt{2}+\sqrt{6}}{2}\right)^2&=\frac{\sqrt{2}^2+2\sqrt{2}\sqrt{6}+\sqrt{6}^2}{2^2}\\&=\frac{2+4\sqrt{3}+6}{4}\\&=2+\sqrt{3}. \end{align}$$

Answer 4

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