# 1980 AHSME Problems/Problem 27

## Problem

The sum $\sqrt {5+2\sqrt{13}}+\sqrt{5-2\sqrt{13}}$ equals $\text{(A)} \ \frac 32 \qquad \text{(B)} \ \frac{\sqrt{65}}{4} \qquad \text{(C)} \ \frac{1+\sqrt{13}}{2} \qquad \text{(D)}\ \sqrt{2}\qquad \text{(E)}\ \text{none of these}$

## Solution

Lets set our original expression equal to $x$. So $\sqrt {5+2\sqrt{13}}+\sqrt{5-2\sqrt{13}} = x$. Cubing this gives us $x^3 = \left(\sqrt {5+2\sqrt{13}}+\sqrt{5-2\sqrt{13}}\right)^3 = 5 + 2\sqrt{13} + 5 - 2\sqrt{13} + 3\left(\sqrt {5+2\sqrt{13}}*\sqrt{5-2\sqrt{13}}\right)\left(\sqrt {5+2\sqrt{13}}+\sqrt{5-2\sqrt{13}}\right) = 10 - 9x$ So we have $x^3 + 9x - 10 = 0$. We can easily see that 1 is a root of this polynomial. By synthetic division, the new polynomial is $x^2 + x + 10$, which has no real roots. Thus $\sqrt {5+2\sqrt{13}}+\sqrt{5-2\sqrt{13}} = 1$. Since 1 is not A-D, our answer is $\fbox{E}$.

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