Here is how I solved this:

1) Follow what Bunuel did up to a point

Bunuel wrote:

LM wrote:

\(\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}\) lies between

A) 1 & 2

B) 2 & 3

C) 3 & 4

D) 4 & 5

E) 5 & 6

Multiply both nominator and denominator of \(\frac{2}{5+2*\sqrt{6}}\) by \({5-2*\sqrt{6}}\) and apply \((a+b)(a-b)=a^2-b^2\): \(\frac{2*(5-2*\sqrt{6})}{(5+2*\sqrt{6})(5-2*\sqrt{6})}=\frac{2*(5-2*\sqrt{6})}{25-24}=2*(5-2*\sqrt{6})\).

2) multiply out \(2(5-2\sqrt{6}) = 10-4\sqrt{6}\)

3) \(\sqrt{96} = 4\sqrt{6}\)

4) \(\sqrt{4\sqrt{6}+10-4\sqrt{6}} = \sqrt{10}\)

5) 3² = 9 and 4² = 16, so \(\sqrt{10}\) is between 3 and 4

Answer is

C