How to show that $\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$ 
$\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$

How to change $\sqrt{2+\sqrt{3}}$ into  $\dfrac{\sqrt{6}+\sqrt{2}}{2}$
 A: 
$$\sqrt{2+\sqrt3}=\sqrt{\frac{4+2\sqrt3}{2}}=\sqrt{\frac{(1+\sqrt3)^2}{2}}=\frac{1+\sqrt3}{\sqrt2}$$
Can you end it now?
A: You can use the identity
\begin{equation}
\sqrt{a+\sqrt{b}} = \sqrt{\frac{a+\sqrt{a^2-b}}{2}} + \sqrt{\frac{a-\sqrt{a^2-b}}{2}}.
\end{equation}
A: Hint: $$\left( \frac{\sqrt{6} + \sqrt{2}}{2}\right)^2 = \frac{6 + 2 \sqrt{6} \sqrt{2} + 2}{4} = ?$$
A: Assume $$\sqrt{a+\sqrt b}=\sqrt c+\sqrt d$$
where $a,b,c,d$ are all rational.
Then
$$a+\sqrt b=c+2\sqrt{cd}+d.$$
By identification
$$\begin{cases}a=c+d,\\b=4cd.\end{cases}$$
From this,
$$4c^2+4cd-4ac=4c^2-4ac+b=0.$$
This is a quadratic equation in $c$, with roots
$$c=\frac{a\pm\sqrt{a^2-b}}2.$$
So for a rational solution to exist, $a^2-b$ must be a perfect rational square.

With $a=2,b=3$, 
$$\sqrt{2+\sqrt3}=\sqrt{\frac12}+\sqrt{\frac32}=\frac{\sqrt2+\sqrt6}2.$$
A: If you want to change $\sqrt{2+ \sqrt{3}}$ into $\dfrac{\sqrt{6}+\sqrt{2}}{2}$, then go through the following calculation steps from bottom to top. If you want to change $\dfrac{\sqrt{6}+\sqrt{2}}{2}$ into $\sqrt{2+ \sqrt{3}}$, then go through the following calculation steps from top to bottom. Either way we get equality $\sqrt{2+ \sqrt{3}}=\dfrac{\sqrt{6}+\sqrt{2}}{2}$.
\begin{align*} 
\dfrac{\sqrt{6}+\sqrt{2}}{2} &= \sqrt{\left(\dfrac{\sqrt{6}+\sqrt{2}}{2}\right)^2} 
&& a = \sqrt{a^2} \text{ for } a \geq 0
\\&= \sqrt{\dfrac{(\sqrt{6}+\sqrt{2})^2}{2^2}} 
&& \left(\dfrac{a}{b}\right)^2 = \dfrac{a^2}{b^2}
\\&= \sqrt{\dfrac{(\sqrt{6})^2+2\cdot\sqrt{2}\cdot\sqrt{6}+(\sqrt{2})^2}{4}}
&& (a+b)^2 = a^2+2\cdot a\cdot b+b^2
\\&= \sqrt{\dfrac{6+2\cdot\sqrt{2}\cdot\sqrt{6}+2}{4}} 
&& a = (\sqrt{a}) \text{ for } a \geq 0
\\&= \sqrt{\dfrac{6+\sqrt{4}\cdot\sqrt{2}\cdot\sqrt{6}+2}{4}} 
&& 2 = \sqrt{4}
\\&= \sqrt{\dfrac{6+\sqrt{4\cdot 2\cdot 6}+2}{4}} 
&&   \sqrt{a}\cdot \sqrt{b}\cdot \sqrt{c} = \sqrt{a\cdot b\cdot c}
\\&= \sqrt{\dfrac{6+\sqrt{48}+2}{4}} 
&&  4\cdot 2\cdot 6=8\cdot6=48
\\&= \sqrt{\dfrac{6+\sqrt{3\cdot 16}+2}{4}} 
&&  48 = 3 \cdot 16
\\&= \sqrt{\dfrac{6+\sqrt{3}\cdot \sqrt{16}+2}{4}} 
&&    \sqrt{a\cdot b} = \sqrt{a}\cdot \sqrt{b} 
\\&= \sqrt{\dfrac{6+\sqrt{3}\cdot 4+2}{4}} 
&&    \sqrt{16} = \sqrt{4^2} = 4
\\&= \sqrt{\dfrac{6+4 \cdot \sqrt{3}+2}{4}} 
&&   a \cdot b = b\cdot a
\\&= \sqrt{\dfrac{8+4 \cdot \sqrt{3}}{4}} 
&&   6+2=8
\\&= \sqrt{\dfrac{8}{4}+\dfrac{4 \cdot \sqrt{3}}{4}} 
&&   \dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}
\\&= \sqrt{2+ \sqrt{3}} 
\end{align*}
A: $ (\sqrt u + \sqrt v)^2 = u+v + 2\sqrt{u\,v }$ 
So at first bring in $2$ in front of the radical sign.
$$\sqrt{2+\sqrt3}=\sqrt{\frac{4+2\sqrt3}{2}}$$
Next find two factors for $3$ whose sum is $4$
They are easy to guess. They are $3$ and $1$; 
(... or else you need to solve another quadratic equation)
$$ \sqrt{\frac{(\sqrt3 +1)^2}{2}}=\sqrt{  \frac{2(\sqrt3 +1)^2}{4}} $$
where the denominator is made to be $4$ so that while square rooting, $2$ would stay there.
Next manipulate numerator to make them  squares of two terms and make denominator also a square
$$\sqrt{  \frac{(\sqrt{2})^2(\sqrt3 +1)^2}{2^2}} $$
that  is now ( cancelling out squares and square roots) and multiplying two terms in the numerator:
$$ \dfrac{\sqrt 6 +\sqrt 2}{2}.$$
