Finding the square root $\sqrt{4-2\sqrt{3}}$ How do you solve this: 
$$X^2=4-2\sqrt{3}$$
I don't really know how to solve for $X$. I initially got $X=\sqrt{4-2\sqrt{3}}$ but I don't know how to simplify it.

$\surd$ simplifies if what's inside is a square.
Since $(x+y)^2=x^2+2xy+y^2$ we will search for $\begin{cases} x^2+y^2=4 \\ 2xy=-2\sqrt{3}\end{cases}\iff\begin{cases} x^2y^2+y^4=4y^2 \\ xy=-\sqrt{3}\end{cases}$
The quadratic equation $\big[y^4-4y^2+3=0\big]$ gives $y^2=1$ or $3$ and this leads to $x+y=\pm(1-\sqrt{3})$

If you want to explore further some methodology, I invite you to read this interesting paper :
Jeffrey-Rich : Simplifying square roots of square roots by denesting
Applying theorem $\mathbf{(4.12)}$ : $\sqrt{\vphantom{\big(}X\pm Y}=\sqrt{\frac 12X+\frac 12\sqrt{X^2-Y^2}}\pm\sqrt{\frac 12X-\frac 12\sqrt{X^2-Y^2}}$

With $X=4$ and $Y=2\sqrt{3}$
Gives $\sqrt{\vphantom{\big)}4-2\sqrt{3}}=\sqrt{2+\frac 12\sqrt{4}}-\sqrt{2-\frac 12\sqrt{4}}=\sqrt{3}-\sqrt{1}=\sqrt{3}-1$
