<s>[INST] 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.
 [/INST] Let $\sqrt{6 - \sqrt{20}} = \sqrt{a} - \sqrt{b}$, where $a$ and $b$ are rational numbers.  Squaring both sides of the equation 
$$\sqrt{a} - \sqrt{b} = \sqrt{6 - \sqrt{20}}$$
yields
\begin{align*}
a - 2\sqrt{ab} + b & = 6 - \sqrt{20}\\
a - 2\sqrt{ab} + b & = 6 - 2\sqrt{5}
\end{align*}
Matching rational and irrational parts yields the system of equations 
\begin{align*}
a + b & = 6 \tag{1}\\
-2\sqrt{ab} & = -2\sqrt{5} \tag{2}
\end{align*}
Solving equation 2 for $b$ yields 
\begin{align*}
-2\sqrt{ab} & = -2\sqrt{5}\\
\sqrt{ab} & = \sqrt{5}\\
ab & = 5\\
b & = \frac{5}{a}
\end{align*}
Substituting $5/a$ for $b$ in equation 1 yields 
\begin{align*}
a + \frac{5}{a} & = 6\\
a^2 + 5 & = 6a\\
a^2 - 6a + 5 & = 0\\
(a - 1)(a - 5) & = 0
\end{align*}
Hence, $a = 1$ or $a = 5$.  
If $a = 1$, then $b = 6 - a = 5$, in which case 
$$\sqrt{6 - \sqrt{20}} = \sqrt{1} - \sqrt{5} < 0$$
which is impossible since the principal square root of a positive number must be positive.  
Thus, $a = 5$ and $b = 6 - a = 1$, so
$$\sqrt{6 - \sqrt{20}} = \sqrt{5} - \sqrt{1} = \sqrt{5} - 1$$
Check:  Observe that $\sqrt{5} - 1 > 0$.  Moreover, 
$$(\sqrt{5} - 1)^2 = 5 - 2\sqrt{5} + 1 = 6 - 2\sqrt{5} = 6 - \sqrt{20}$$
Hence, 
$$\sqrt{6 - \sqrt{20}} = \sqrt{5} - 1$$
as claimed.
</s>