How do you simplify this square root of sum: $\sqrt{7+4\sqrt3}$?

If you want $\sqrt{7+4\sqrt{3}}=a+b\sqrt d$ (where $a$ and $b$ are rational numbers, and $d$ is a square-free integer) then $7+4\sqrt{3} = a^2+db^2+2ab\sqrt d$, which yields the natural choice $d=3$.

Then $a^2+3b^2=7$ and $2ab=4$ can be solved in the rational numbers, and you find $a=2,b=1$ (because $a^2+3 \cdot (2/a)^2=7$ has solutions $±2$ and $±\sqrt 3$).

${\sqrt {7 + 4{\sqrt3}}} = {\sqrt {7 + \sqrt{48}}}$

The latter can be solved by the formula:

$\sqrt {a + \sqrt{b} } = \sqrt{ {a + \sqrt{a^2 -b}}\over2} +\sqrt{ {a - \sqrt{a^2 -b}}\over2} $

$$\sqrt { 7+4\sqrt { 3 } } =\sqrt { 7+2\cdot 2\sqrt { 3 } } =\sqrt { { \left( \sqrt { 3 } \right) }^{ 2 }+{ 2 }^{ 2 }+4\sqrt { 3 } } =\sqrt { { \left( \sqrt { 3 } +2 \right) }^{ 2 } } =\sqrt { 3 } +2\\ \\ $$