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?

$\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y}$
$a+\sqrt{b}=x+y+2\sqrt{xy}$
$a+\sqrt{b}=(x+y)+\sqrt{4xy}$
Then:
$a=x+y$ and $b=4xy$
$a^2=x^2+y^2+2xy$
$a^2-b=(x-y)^2$
If $x>y$ then:
$\sqrt{a^2-b}=x-y$
Let's take $c=\sqrt{a^2-b}$ then $c=x-y$
$c=x-y$ and $a=x+y$ then:
$x=\frac{a+c}{2}$ and $y=\frac{a-c}{2}$
Then:
$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+c}{2}}+\sqrt{\frac{a-c}{2}}$
Now applly it for $\sqrt{7+4\sqrt{3}}$.
$\sqrt{7+4\sqrt{3}}=\sqrt{7+\sqrt{48}}$
$c=\sqrt{49-48}$
$c=1$
$\sqrt{7+\sqrt{48}}=\sqrt{\frac{8}{2}}+\sqrt{\frac{6}{2}}$
Then:
$\sqrt{7+4\sqrt{3}}=2+\sqrt{3}$
solved!
