1973 AHSME Problems/Problem 13

Problem

The fraction $\frac{2(\sqrt2+\sqrt6)}{3\sqrt{2+\sqrt3}}$ is equal to

$\textbf{(A)}\ \frac{2\sqrt2}{3} \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ \frac{2\sqrt3}3 \qquad \textbf{(D)}\ \frac43 \qquad \textbf{(E)}\ \frac{16}{9}$

Solution

Squaring the expression and taking the positive square root (since numerator and denominator are positive) of the result yields $\sqrt{\left(\frac{2(\sqrt2+\sqrt6)}{3\sqrt{2+\sqrt3}}\right)^2}$ $\sqrt{\frac{4(2+4\sqrt{3}+6)}{9(2+\sqrt{3})}}$ $\sqrt{\frac{4(8+4\sqrt{3})}{9(2+\sqrt{3})}}$ $\sqrt{\frac{16}{9}}$ $\frac43$

The answer is $\boxed{\textbf{(D)}}$ .

1973 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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1973 AHSME Problems/Problem 13

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