square root of $\frac{2+\sqrt{3}}{4}$ 
Find the square root of $\frac{2+\sqrt{3}}{4}$.

Attempt:
I'm thinking of equating it to $a + bi$. And then finding the root. Is that right?
 A: For convenience we will work with $4(2+\sqrt3)$ to avoid fractions.
Let us search a solution of the form $a+b\sqrt3$, giving
$$(a+b\sqrt3)^2=a^2+3b^2+2ab\sqrt3=8+4\sqrt3.$$
By identifying $2ab$ and $4$, we have $b=\dfrac2a$ and
$$a^2+\frac{12}{a^2}+4\sqrt3=8+4\sqrt3.$$
By inspection (or resolution of a quadratic equation), $a^2=2$, so that
$$\left(\sqrt2+\frac2{\sqrt2}\sqrt3\right)^2=8+4\sqrt3,$$
$$\left(\frac{\sqrt2+\sqrt6}4\right)^2=\frac{2+\sqrt3}4.$$
A: $\frac{2+\sqrt{3}}{4}$ is a positive number so it has a real square root.
Notice first that:
$$\sqrt{\dfrac{2+\sqrt{3}}{4}}=\dfrac{\sqrt{2+\sqrt{3}}}{2}$$
 so what you have to find is $\sqrt{2+\sqrt{3}}$. A technique is to try to find what are $a,b\in\mathbb{R}$ where $2+\sqrt{3}=(a+b)^2$ so that $2+\sqrt{3}=|a+b|$.
You have $$(a+b)^2=a^2+b^2+2ab$$.
The technique in such cases is to try to get $a^2+b^2=2$ and $2ab=\sqrt{3}$.
Notice that $b=\frac{\sqrt{3}}{2a}$
If you put it in $a^2+b^2=2$ you'll have an equation with one variable $a$.
A: To add to the other answers, notice that:
$$\frac{\sqrt{2 + \sqrt{3}}}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4},$$
where I've used zezanjee's hint
$$2 + \sqrt{3} = \frac{(\sqrt{3} + 1)^2}{2}$$
in the comments.
