Radical problem: $\sqrt{7+2\sqrt{12}}+\sqrt{7-2\sqrt{12}}$

Question

What is the value of $$\sqrt{7+2\sqrt{12}}+\sqrt{7-2\sqrt{12}}$$

Answer 1

With $x=\sqrt{7+2\sqrt{12}}$ and $y=\sqrt{7-2\sqrt{12}}$, we have $x^2+y^2=14$ and $xy=\sqrt{7^2-2^2\cdot12}=1$, so $(x+y)^2=14+2\cdot1=16$ and thus $x+y=4,$ since $x,y$ are positive.
In general, we'd have $$\sqrt{a+b\sqrt{c}}+\sqrt{a-b\sqrt{c}}=\sqrt{2a+2\sqrt{a^2-b^2c}}.$$

Answer 2

Since $(2\pm\sqrt3)^2=7\pm4\sqrt3=7\pm2\sqrt{12}$,$$\sqrt{7+2\sqrt{12}}+\sqrt{7-2\sqrt{12}}=2+\sqrt3+2-\sqrt3=4.$$