<s>[INST] 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?
 [/INST] In general, there is a formula where given the expression $\sqrt{X\pm Y}$, one can rewrite that into $\sqrt{\frac {X+\sqrt{X^2-Y^2}}{2}}\pm\sqrt{\frac {X-\sqrt{X^2-Y^2}}{2}}$ where $X$ and $Y$ can be any number and $X>Y$.
In your case, we have $X=7$ and $Y=4\sqrt{3}$ because $7<4\sqrt{3}$. Plugging the values into the formula, we get: $$\sqrt{\frac {7+\sqrt{7^2-(\sqrt{48})^2}}{2}}+\sqrt{\frac {7-\sqrt{7^2-(\sqrt{48})^2}}{2}}\tag{1}$$
The radical $\sqrt{7^2-(\sqrt{48})^2}$ simplifies into $1$ so $(1)$ becomes $$\sqrt{\frac {7+1}{2}}+\sqrt{\frac {7-1}{2}}$$ which further simplifies into $$\sqrt{4}+\sqrt{3}\iff\boxed{\sqrt{3}+2}$$

Extra: This works on any nested radical under a square root. For example: Denest $\sqrt{5\sqrt{3}+6\sqrt{2}}$. Setting $X=5\sqrt{3}$ and $Y=6\sqrt{2}$, gives us $$\sqrt{\frac {5\sqrt{3}+\sqrt{3}}{2}}+\sqrt{\frac {5\sqrt{3}-\sqrt{3}}{2}}\tag{1}$$
Simplifying gives us our denesting. $$\sqrt[4]{27}+\sqrt[4]{12}$$
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