Can the nested radical expression $\sqrt{8 + \sqrt{48}}$ be simplified? Solving a pythagorean problem I came up with the following radical: 
$$C = \sqrt{1^2+(\sqrt{3} + 2)^2}$$
I can simplify this to: 
$$\sqrt{8+4\times\sqrt{3}} = \sqrt{8+\sqrt{48}}$$
Is there a way to simplify this further? I'm not satisfied with the double radical, but the sum is an obstacle to further simplification, am I correct? 
 A: $$8+\sqrt{48}=8+2\sqrt{12}=2+2\sqrt{2}\sqrt{6}+6=(\sqrt{2}+\sqrt{6})^2$$
A: A nested radical $\sqrt{u + \sqrt{v}}$ can be simplified if $u^2 - v$ is a perfect square.  Since $$8^2 - 48 = 64 - 48 = 16 = 4^2$$ this nested radical can be simplified.  
Suppose that 
$$\sqrt{8 + \sqrt{48}} = \sqrt{a} + \sqrt{b}$$
where $a$ and $b$ are rational numbers.  Squaring both sides of the equation yields
$$8 + \sqrt{48} = a + b + 2\sqrt{ab}$$
Matching rational and irrational parts yields the system of equations
\begin{align*}
a + b & = 8 \tag{1}\\
2\sqrt{ab} & = \sqrt{48} \tag{2}
\end{align*}
Since $\sqrt{48} = 4\sqrt{3}$, we obtain
\begin{align*}
2\sqrt{ab} & = 4\sqrt{3}\\
\sqrt{ab} & = 2\sqrt{3}\\
ab & = 12\\
b & = \frac{12}{a}
\end{align*} 
Substituting $12/a$ for $b$ in equation 1 yields 
\begin{align*}
a + \frac{12}{a} & = 8\\
a^2 + 12 & = 8a\\
a^2 - 8a + 12 & = 0\\
(a - 2)(a - 6) & = 0
\end{align*}
Hence, $a = 2$ or $a = 6$.  If $a = 2$, then $b = 6$, so we obtain
$$\sqrt{8 + \sqrt{48}} = \sqrt{2} + \sqrt{6}$$
If $a = 6$, then $b = 2$, which yields the same solution.
