Equality with roots

We have two positive numbers with the same squares so they are equal.

The key idea is to recognize the expression under the radical as a perfect square trinomial.
Observe that:

$\sqrt{5 + 2\sqrt{6}} = \sqrt{(\sqrt{2})^{2} + 2\cdot\sqrt{2}\cdot\sqrt{3} + (\sqrt{3})^{2}}$

This matches the identity $a^2 + 2ab + b^2 = (a+b)^2$ with $a = \sqrt{2}$ and $b = \sqrt{3}$, so:

$\sqrt{5 + 2\sqrt{6}} = \sqrt{\left(\sqrt{2} + \sqrt{3}\right)^2} = \left|\sqrt{2} + \sqrt{3}\right| = \sqrt{2} + \sqrt{3}$

The last step uses $\sqrt{X^2} = |X|$ for all real $X$, and then the absolute value resolves to $\sqrt{2} + \sqrt{3}$ itself since $\sqrt{2} + \sqrt{3} > 0$.