Simplifying $\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}$ How do I simplify $\sqrt{(4+2\sqrt{3})}+\sqrt{(4-2\sqrt{3})}$?
I've tried to make it $x$ and square both sides but I got something extremely complicated and it didn't look right.  
I got $2\sqrt{3}$ on wolframalpha, but I'm not sure how is it possible?
Help would be appreciated!  Thanks!
 A: Since $(4+2\sqrt3)(4-2\sqrt3)=16-12=4$, try squaring:
$$
\begin{align}
\left(\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}\right)^2
&=(4+2\sqrt3)+(4-2\sqrt3)+2\sqrt{(4+2\sqrt3)(4-2\sqrt3)}\\
&=8+2\sqrt{16-12}\\[6pt]
&=12
\end{align}
$$
Therefore, $\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}=2\sqrt3$
A: Hint: Find the square of $1+\sqrt{3}$. ${}{}{}{}{}{}{}{}{}$
A: Write as $\sqrt{4+2\sqrt{3}} = a+b\sqrt{3}$. Now square both sides, equate real and radical part. This gives two equations in $a$ and $b$. Now eliminate $a$, solve for $b$. Goes perfect. Same for the other term.
A: \begin{align}
&\ \ \ \sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}
\\
\\
&=\sqrt{(\sqrt{3}+1)^2}+\sqrt{(\sqrt{3}-1)^2}\
\\
\\
&=\sqrt{3}+1+\sqrt{3}-1
\\
\\
&=\boxed{2\sqrt{3}}
\end{align}
A: You have the right idea in squaring and then taking the square root. 
Note that $k = (\sqrt{4 + 2\sqrt{3}} + \sqrt{4 - 2\sqrt{3}})^{2} = 4 + 2\sqrt{3} + 2\sqrt{(4+2\sqrt{3})(4-2\sqrt{3})} + 4 - 2\sqrt{3}$
But this is just: 
$k = 8 + 2\sqrt{16 - 12} = 12$
So 
$\sqrt{k} = 2\sqrt{3}$
A: You want to calculate $a+b$ where $a^2=4+2\sqrt3$ and $b^2=4-2\sqrt3$.
You could notice that $a^2$ and $b^2$ are roots of the quadratic equation
$$x^2-8x+4=0.$$
(To find this equation you just need to find appropriate coefficients in quadratic formula.)
Now from Viete's formula you get
\begin{align*}
a^2+b^2&=8\\
a^2b^2&=4
\end{align*}
which (together with the fact that $a,b>0$) leaves you with $ab=2$ and $$(a+b)^2=a^2+b^2+2ab=8+2\cdot3=12.$$
You get that $$a+b=\sqrt{12}=2\sqrt3.$$
A: Using the following formula:
$$\sqrt{a\pm \sqrt{b}}=\sqrt{\frac{a+c}{2}}\pm\sqrt{\frac{a-c}{2}}$$
where
$c=\sqrt{a^2-b}$
