Difference between revisions of "1952 AHSME Problems/Problem 35"

Revision as of 07:53, 1 May 2016

Problem

With a rational denominator, the expression $\frac {\sqrt {2}}{\sqrt {2} + \sqrt {3} - \sqrt {5}}$ is equivalent to:

$\textbf{(A)}\ \frac {3 + \sqrt {6} + \sqrt {15}}{6} \qquad \textbf{(B)}\ \frac {\sqrt {6} - 2 + \sqrt {10}}{6} \qquad \textbf{(C)}\ \frac{2+\sqrt{6}+\sqrt{10}}{10} \qquad\\ \textbf{(D)}\ \frac {2 + \sqrt {6} - \sqrt {10}}{6} \qquad \textbf{(E)}\ \text{none of these}$

Solution

Let $k=\sqrt{2}+\sqrt{3}$ Then $\frac{\sqrt{2}}{k-\sqrt{5}}\implies \frac{\sqrt{2}(k+\sqrt{5})}{k^2-\sqrt{5}^2}\implies\frac{\sqrt{2}k+\sqrt{10}}{k^2-5}\implies \frac{\sqrt{2}(\sqrt{2}+\sqrt{3})+\sqrt{10}}{(\sqrt{2}+\sqrt{3})^2-5}\implies \frac{2+\sqrt{6}+\sqrt{10}}{2\sqrt{6}}\implies\frac{2\sqrt{6}+6+\sqrt{60}}{2}\implies \frac{\sqrt{6}+3+\sqrt{15}}{6}\fbox{A}$

1952 AHSC (Problems • Answer Key • Resources)
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@@ Line 12: / Line 12: @@
 Let <math>k=\sqrt{2}+\sqrt{3}</math>
-Then <math></math>
+Then <math>\frac{\sqrt{2}}{k-\sqrt{5}}\implies \frac{\sqrt{2}(k+\sqrt{5})}{k^2-\sqrt{5}^2}\implies\frac{\sqrt{2}k+\sqrt{10}}{k^2-5}\implies \frac{\sqrt{2}(\sqrt{2}+\sqrt{3})+\sqrt{10}}{(\sqrt{2}+\sqrt{3})^2-5}\implies \frac{2+\sqrt{6}+\sqrt{10}}{2\sqrt{6}}\implies\frac{2\sqrt{6}+6+\sqrt{60}}{2}\implies \frac{\sqrt{6}+3+\sqrt{15}}{6}\fbox{A}</math>
-<math>\fbox{}</math>
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Difference between revisions of "1952 AHSME Problems/Problem 35"

Revision as of 07:53, 1 May 2016

Problem

Solution

See also