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.
 A: $\surd$ 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 $\sqrt{\vphantom{\big)}4-2\sqrt{3}}=\sqrt{2+\frac 12\sqrt{4}}-\sqrt{2-\frac 12\sqrt{4}}=\sqrt{3}-\sqrt{1}=\sqrt{3}-1$
A: $\sqrt{4-2\sqrt{3}}$ is a solution.  So too is $-\sqrt{4-2\sqrt{3}}$
These particular expressions can be simplified (others not): 


*

*the $2\sqrt{3}$ might suggest something of the form  $ (y-\sqrt{3})^2=4-2\sqrt{3}$ where $y=x+\sqrt{3}$

*so solve $y^2-2y\sqrt{3}-1+2\sqrt{3} = 0$ 

*giving $y=1$ or $y=2\sqrt{3}-1$, 

*leading to $x=1-\sqrt{3}$ or $x=\sqrt{3}-1$
A: Notice that $$4=3+1=\sqrt3^2+1^2$$
and you have a remarkable identity.
A: 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 $a = \frac{1}{2}\big((a+b)-(b-a)\big)=\frac{1}{2}\left(2 \sqrt{3} - 2\right)=\sqrt{3}-1\,$.
A: Note that $4-2\sqrt{3}$ is simply $\left(\sqrt{3}-1\right)^2$. Thus, $\sqrt{4-2\sqrt{3}}$ can be simplify as: $$\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1$$
