Integer value of the given radical: $\sqrt{2+\sqrt{5}-\sqrt{6-3\sqrt{5}+\sqrt{14-6\sqrt{5}}}}$

Question

What is the value of $$\sqrt{2+\sqrt{5}-\sqrt{6-3\sqrt{5}+\sqrt{14-6\sqrt{5}}}}$$ I don't know how to simplify it?

Answer 1

HINT:

$$14-6\sqrt5=-2\cdot\sqrt5\cdot3+3^2+(\sqrt5)^2=(3-\sqrt5)^2$$

Similarly, $$6-3\sqrt5+\sqrt{14-6\sqrt5}=(\sqrt5-2)^2$$

Answer 2

You convert $$a+b\sqrt5$$ to a square by

$$(a'+b'\sqrt5)^2=a'^2+5b'^2+2a'b'\sqrt5.$$

The trick is to factor $b$ and try all combinations of $b=2a'b'$, checking if $a=a'^2+5b'^2$.

For the innermost radical, $6=2\cdot1\cdot3$ or $2\cdot3\cdot1$. The second decomposition works.

Then the middle radical becomes $\sqrt{9-4\sqrt5}$, yielding $2\cdot2\cdot1$ or $2\cdot1\cdot2$.

The outer radical is then $\sqrt4$.