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.
![\[\frac{\sqrt3 - \sqrt2}{\sqrt3}\cdot \frac{\sqrt3 + \sqrt2}{\sqrt3 + \sqrt2} = \frac{1}{3+\sqrt6}.\]](http://latex.artofproblemsolving.com/e/5/d/e5d99a6de0d2cb312b7b00969908d5192b313c5e.png)
The denominator is 
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
