2006 Cyprus MO/Lyceum/Problem 6

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Problem

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

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

Solution

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 $\sqrt{(4+ \sqrt{3})^2} - \sqrt{(2 + \sqrt{3})^2} = 4 + \sqrt{3} - 2 - \sqrt{3} = 2\ \mathrm{(E)}$.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 5
Followed by
Problem 7
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