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1984 AHSME Problems/Problem 13

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Problem

$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$ equals

$\mathrm{(A) \ }\sqrt{2}+\sqrt{3}-\sqrt{5} \qquad \mathrm{(B) \ }4-\sqrt{2}-\sqrt{3} \qquad \mathrm{(C) \ } \sqrt{2}+\sqrt{3}+\sqrt{6}-5 \qquad$

$\mathrm{(D) \ }\frac{1}{2}(\sqrt{2}+\sqrt{5}-\sqrt{3}) \qquad \mathrm{(E) \ } \frac{1}{3}(\sqrt{3}+\sqrt{5}-\sqrt{2})$

Solution

Multiply the numerator and the denominator by $\sqrt{2}-(\sqrt{3}+\sqrt{5})$ to get $\frac{2\sqrt{6}(\sqrt{2}-\sqrt{3}-\sqrt{5})}{(\sqrt{2}+(\sqrt{3}+\sqrt{5}))(\sqrt{2}-(\sqrt{3}+\sqrt{5})}$

$=\frac{2\sqrt{12}-2\sqrt{18}-2\sqrt{30}}{2-(\sqrt{3}+\sqrt{5})^2}$

$=\frac{4\sqrt{3}-6\sqrt{2}-2\sqrt{30}}{-6-2\sqrt{15}}$

$=\frac{2\sqrt{3}-3\sqrt{2}-\sqrt{30}}{-3-\sqrt{15}}$.

Now to get rid of the $\sqrt{15}$ in the denominator. Multiply the numerator and denominator by $-3+\sqrt{15}$ to get

$\frac{(2\sqrt{3}-3\sqrt{2}-\sqrt{30})(-3+\sqrt{15})}{(-3-\sqrt{15})(-3+\sqrt{15})}$


$=\frac{-6\sqrt{3}+9\sqrt{2}+3\sqrt{30}+2\sqrt{45}-3\sqrt{30}-\sqrt{450}}{9-15}$


$=\frac{-6\sqrt{3}+9\sqrt{2}+3\sqrt{30}+6\sqrt{5}-3\sqrt{30}-15\sqrt{2}}{-6}$


$=\sqrt{2}+\sqrt{3}-\sqrt{5}, \boxed{\text{A}}$.

Solution 2

Multiply the numerator and denominator by $\sqrt{2}+\sqrt{3}-\sqrt{5}$. We get $\frac{(2\sqrt{6})(\sqrt{2}+\sqrt{3}-\sqrt{5})}{(\sqrt{2}+\sqrt{3}+\sqrt{5})(\sqrt{2}+\sqrt{3}-\sqrt{5})}$

\[\frac{(2\sqrt{6})(\sqrt{2}+\sqrt{3}-\sqrt{5})}{2+3-5+2\sqrt{6}+\sqrt{10}-\sqrt{10}+\sqrt{15}-\sqrt{15}}=\frac{(2\sqrt{6})(\sqrt{2}+\sqrt{3}-\sqrt{5})}{2\sqrt{6}}=\sqrt{2}+\sqrt{3}-\sqrt{5}\] $\boxed{A}$

See Also

1984 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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