An algebra problem by shiv kumar

First, we take away the radical and we have $5+2\sqrt6$ . Now, for this to be represented in $\sqrt a+\sqrt b$ , we need to find $(\sqrt a+\sqrt b)^2=a+b+2\sqrt{ab}$ . Since a,b are integers, a must be equal to 1 or 2 while b is equal to either 4 or 3. Checking 2 and 3: $(\sqrt2+\sqrt3)^2=2+2\sqrt6+3\Rightarrow 5+2\sqrt6$

We can write this in the following way as well.

$\sqrt{5+2\sqrt{6}}$

$=\sqrt{2+3+2\sqrt{6}}$

$=\sqrt{(\sqrt{2})^{2}+(\sqrt{3})^{2}+2\times\sqrt{2}\times\sqrt{3}}$

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

$\sqrt{5+2\sqrt{6}}=\sqrt{2}+\sqrt{3}=\sqrt{a}+\sqrt{b}$

Therefore, the answer is: $a+b=2+3=\boxed{5}$

$x=5+2\sqrt{6}$

$\sqrt{x}=\sqrt{a}+\sqrt{b}$

$x=a+b+2\sqrt{ab}$

$5+2\sqrt{6}=(a+b)+2\sqrt{ab}$

Comparing both sides, a+b=5