The ponder of a square root within a square root(with integers inside)

$\sqrt{31+12\sqrt3}$

=$\sqrt{2^2+12\sqrt3+(3\sqrt3)^2}$

=$\sqrt{(2+3\sqrt3)^2}$

=$2+3\sqrt3$

So, I came across a interesting question, that is, to simplify:sqrt(31+12*sqrt(3)) We are told to show the answer in a+b*sqrt(3) format, so i wonder how is it possible to simplify to this extent.

$simplify\;\;\;\sqrt{31+12\sqrt3}\\ \mbox{We need to express }31+12\sqrt3 \mbox{ as a perfect square.}\\ (a+b)^2=a^2+2ab+b^2\\ 2ab=12\sqrt3\\ ab=6\sqrt3\\ Let\;\;b=c\sqrt3\\ (a+c\sqrt3)^2=a^2+2ac\sqrt3+(c\sqrt3)^2=31+12\sqrt3\\ (a+c\sqrt3)^2=a^2+2ac\sqrt3+3c^2\qquad=31+12\sqrt3\\ ac=6\\ a^2+3c^2=31\\ Try\;\;a=2\;\; and\;\; c=3\\ ac=6\;\; tick\\ a^2+3c^2=2^2+3*3^2=4+27=31\;\;tick\\ so\\ (2+3\sqrt3)^2=31+12\sqrt3\\ \sqrt{31+12\sqrt3}=\sqrt{(2+3\sqrt3)^2}=2+3\sqrt3$