<s>[INST] 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.
 [/INST] Let $\,a=\sqrt{4-2\sqrt{3}}\,$, $\,b=\sqrt{4+2\sqrt{3}}\,$ and note that $b \gt a \gt 0\,$, then:
$$\require{cancel}
\begin{cases}
a^2+b^2 = 4-\cancel{2\sqrt{3}} + 4+\cancel{2\sqrt{3}} = 8 \\
ab = \sqrt{(4-2\sqrt{3})(4+2\sqrt{3})} = \sqrt{16 - 2 \cdot 3} = 2
\end{cases}
$$
It follows that:
$$
\begin{cases}
\begin{align}
a+b &= \sqrt{a^2+b^2+2ab}=\sqrt{12}=2 \sqrt{3} \\
b-a &= \sqrt{a^2+b^2-2ab}=\sqrt{4} = 2
\end{align}
\end{cases}
$$
Therefore $a = \frac{1}{2}\big((a+b)-(b-a)\big)=\frac{1}{2}\left(2 \sqrt{3} - 2\right)=\sqrt{3}-1\,$.
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