Simplify $\sqrt{6-\sqrt{20}}$ My first try was to set the whole expression equal to $a$ and square both sides. $$\sqrt{6-\sqrt{20}}=a \Longleftrightarrow a^2=6-\sqrt{20}=6-\sqrt{4\cdot5}=6-2\sqrt{5}.$$
Multiplying by conjugate I get $$a^2=\frac{(6-2\sqrt{5})(6+2\sqrt{5})}{6+2\sqrt{5}}=\frac{16}{2+\sqrt{5}}.$$
But I still end up with an ugly radical expression.

In this particular problem, you can pretty much guess the answer.
$$\sqrt{6-\sqrt{20}}=\sqrt{6-2\sqrt{5}}$$
Now, suppose that the $-2\sqrt{5}$ was the middle term of a perfect square trinomial, where $x = \sqrt{5}$. In other words, that middle term is $-2x$.
What would the first and last term look like? Obviously it would be $x^2$ and $1$ respectively.
$x^2 - 2x +1 = (x-1)^2$
Substituting $\sqrt{5}$ for $x$ we have...
$$x^2 - 2x +1 = (x-1)^2$$
$$\sqrt{5}^2 - 2\sqrt{5} +1 = (\sqrt{5}-1)^2$$
$$5 - 2\sqrt{5} +1 = (\sqrt{5}-1)^2$$
$$6 - 2\sqrt{5} = (\sqrt{5}-1)^2$$
and taking the square root of both sides we have $$\sqrt{6-2\sqrt{5}} = \sqrt{5}-1$$
It's almost like somebody just make up that problem to work out cleanly like that. ;)
