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 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.
