<s>[INST] How to evaluate $\sqrt{5+2\sqrt{6}}$ + $\sqrt{8-2\sqrt{15}}$? My exams are approaching fast and I found this question in one of the unsolved sample papers.
I tried squaring the whole term but couldn't work out the answer.
I am a ninth grader so please try to explain in simple terms.
 [/INST] It's worth noting that $(\sqrt a \pm \sqrt b)^2 = (a+b) \pm 2\sqrt{ab}$.
So, if the answer to $\sqrt{M \pm 2\sqrt N}$ is going to be of the form $
\sqrt a \pm \sqrt b$ then we need $N=ab$ and $M = a+b$.
$\sqrt{5+2\sqrt{6}}$
Note that $6 = 2 \times 3$ and $5 = 2 + 3$.
Hence $$(\sqrt 3 + \sqrt 2)^2 = 5+2 \sqrt 6$$

$\sqrt{8-2\sqrt{15}}$
Note that $15 = 5 \times 3$ and $8 = 5 + 3$.
Hence $$(\sqrt 5 - \sqrt 3)^2 = 8 - 2 \sqrt{15}$$
\begin{align}
   \sqrt{5+2\sqrt{6}} + \sqrt{8-2\sqrt{15}}
   &= (\sqrt 3 + \sqrt 2) + (\sqrt 5 - \sqrt 3) \\
   &= \sqrt 5 + \sqrt 2
\end{align}

Added 1/18/2021
Another way of looking at $\sqrt{M + 2\sqrt N} = \sqrt a + \sqrt b$.
\begin{align}
   \sqrt{M + 2\sqrt N} &= \sqrt a + \sqrt b \\
   M + 2 \sqrt N &= (a+b) + 2\sqrt{ab} \\
\hline
   M &= a + b \\
   N &= ab \\
\hline
  x^2 - Mx + N &= (x-a)(x-b) \\
  \{a,b\} &= \dfrac{M \pm \sqrt{M^2-4N}}{2}
\end{align}
So if $\sqrt{5 + 2\sqrt 6} = \sqrt a + \sqrt b$
$$\{a,b\} = \left\{\dfrac{5 \pm \sqrt{5^2 - 4(6)}}{2}\right\} = \{3,2\}$$
$$x^2-5x+6 = (x-2)(x-3)$$
So
$$\sqrt{5 + 2\sqrt 6} = \sqrt 2 + \sqrt 3 $$
</s>