2006 Cyprus MO/Lyceum/Problem 6

Problem

The value of the expression $K=\sqrt{19+8\sqrt{3}}-\sqrt{7+4\sqrt{3}}$ is

$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 4\sqrt{3}\qquad\mathrm{(C)}\ 12+4\sqrt{3}\qquad\mathrm{(D)}\ -2\qquad\mathrm{(E)}\ 2$

Suppose that $19 + 8\sqrt{3}$ can be written in the form of $(a+b\sqrt{3})^2$ , in order to eliminate the square root.

Then $19 = a^2 + 3b^2$ and $2ab\sqrt{3} = 8\sqrt{3} \Longrightarrow ab = 4$ , and we quickly find that $19 + 8\sqrt{3} = (4+\sqrt{3})^2$ .

Doing the same on the second radical gets us $(2 + \sqrt{3})^2$ .

Thus the expression evaluates to $\sqrt{(4+ \sqrt{3})^2} - \sqrt{(2 + \sqrt{3})^2} = 4 + \sqrt{3} - 2 - \sqrt{3} = 2\ \mathrm{(E)}$ .