1973 AHSME Problems/Problem 13

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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 $\boxed{\textbf{(D)}}$.

See Also

1973 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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