Difference between revisions of "1950 AHSME Problems/Problem 10"
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<cmath>\frac{\sqrt3 - \sqrt2}{\sqrt3}\cdot \frac{\sqrt3 + \sqrt2}{\sqrt3 + \sqrt2} = \frac{1}{3+\sqrt6}.</cmath> | <cmath>\frac{\sqrt3 - \sqrt2}{\sqrt3}\cdot \frac{\sqrt3 + \sqrt2}{\sqrt3 + \sqrt2} = \frac{1}{3+\sqrt6}.</cmath> | ||
− | The denominator is <math>\boxed{\mathrm{(D)}\text{ } 3+\sqrt6 . | + | The denominator is <math>\boxed{\mathrm{(D)}\text{ } 3+\sqrt6 }.</math> |
==See Also== | ==See Also== |
Latest revision as of 10:37, 20 December 2018
Problem
After rationalizing the numerator of , the denominator in simplest form is:
Solution
To rationalize the numerator, multiply by the conjugate.
The denominator is
See Also
1950 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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All AHSME Problems and Solutions |
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