$k=-\sqrt{3}(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}})$ 
Let $k$ be equal to:
  $$k=-\sqrt{3} \left(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}} \right)$$

I am trying to simplify the expression and express $5+2\sqrt{6}$ and $5-2\sqrt{6}$ as squares. I don't think it's smart to multiply the brackets by $-\sqrt{3}$ at first. Is there any algorithm that I should know to express the expressions as squares?
 A: $\sqrt{5 \pm 2 \sqrt{6}} = \sqrt{3} \pm \sqrt{2}$
A: Squaring gives
\begin{eqnarray*}
k&=& -\sqrt{3} \left(\sqrt{5+ 2\sqrt{6}} + \sqrt{5- 2\sqrt{6}}   \right)   \\
k^2&=&3 \left(5+ 2\sqrt{6}+5- 2\sqrt{6} +2\sqrt{ (5+ 2\sqrt{6})( 5- 2\sqrt{6}) }     \right)   \\
k^2&=&3 \left(10 +2\sqrt{ 25- 4 \times{6} }     \right)   \\
k^2&=& 36   \\
\end{eqnarray*}
Now square root this and we recall we expect $k$ to be negative ... so $\color{red}{k=-6}$.
A: We have indeed that
$$5+2\sqrt{6} =(\sqrt 3 + \sqrt 2)^2$$
$$5-2\sqrt{6} =(\sqrt 3 - \sqrt 2)^2$$
and therefore
$$k=-\sqrt{3}(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}})=-\sqrt{3}(\sqrt 3 + \sqrt 2+\sqrt 3 - \sqrt 2)$$
A: You have $\sqrt{A \pm \sqrt{B}}=\sqrt{\frac{A+C}{2}}\pm \sqrt{\frac{A-C}{2}}$ where $C=\sqrt{A^2-B}$
So we can write $\sqrt{5+2\sqrt{6}}=\sqrt{5+\sqrt{24}}=\sqrt{\frac{5+1}{2}}+\sqrt{\frac{5-1}{2}}=\sqrt{3}+\sqrt{2}$
and $\sqrt{5-2\sqrt{6}}=\sqrt{5-\sqrt{24}}=\sqrt{\frac{5+1}{2}}-\sqrt{\frac{5-1}{2}}=\sqrt{3}-\sqrt{2}$
So substituting we get $k=-\sqrt{3}(\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2})=-6$
A: If we want $5+2\sqrt6$ to be a nice square $(a+b)^2=a^2+2ab+b^2$, most likely it is the case that $\sqrt6$ corresponds to $ab$. And, again if this is to be nice, that means either one of $a$ and $b$ is $1$ and the other is $\sqrt6$, or one is $\sqrt2$ and the other is $\sqrt3$. And it turns out that one of these two options works.
