Finding the square root of $a \pm b\sqrt c$.

Question

I have an exercise that says :

Simplify $\sqrt{16+2\sqrt{55}}$.

Please, I need a vivid explanation. Can anyone help?

Answer 1

$$16+2\sqrt{55} = 11 +5 +2\sqrt{5} \sqrt{11} = ( \sqrt{11}+ \sqrt{5})^2$$

Thus $$\sqrt{16+2\sqrt{55}}=\sqrt{11}+ \sqrt{5}$$

Answer 2

We have $a^2+2ab+b^2 =(a+b)^2$. If we're lucky, we can use that. Assuming that the $2\sqrt{55}$ term corresponds in some nice way to the $2ab$ term, then most likely we have $a=\sqrt5,b=\sqrt{11}$. And, oh, look: $16=\sqrt5^2+\sqrt{11}^2$.