How do you simplify this square root of sum: $\sqrt{7+4\sqrt3}$? I came around this expression when solving a problem.
$$\sqrt{7+4\sqrt{3}}$$
WolframAlpha says it equals $2+\sqrt{3}$. We can confirm it like this
$$\left(2+\sqrt{3}\right)^2  \;=\;  4+4\sqrt{3} + 3  \;=\;  7 + 4\sqrt{3}.$$
However, the only way I can think of how to simplify that expression in hand is guessing. Is there a better way of calculating square root of a sum like that one?

A very funny way to solve this which can be generalized goes as follow:
We define $X=\sqrt{7+4\sqrt{3}}$ and $Y= \sqrt{7-4\sqrt{3}}$. Note that $X\pm Y>0$. Now we form $X+Y$ and $X-Y$

$(X+Y)^2=X^2+Y^2+2XY=7+4\sqrt{3}+7-4\sqrt{3}+2\sqrt{49-48}=16\Longrightarrow X+Y=4$
$(X-Y)^2=X^2+Y^2-2XY=7+4\sqrt{3}+7-4\sqrt{3}-2\sqrt{49-48}=12\Longrightarrow X-Y=2\sqrt{3}$ 
  The last step is to solve the system:$$\begin{cases} 
X+Y=4\\
X-Y=2\sqrt{3}\\ 
\end{cases}$$
$X=2+\sqrt{3}$ and $Y=2-\sqrt{3}$

