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== | ||
| − | {{AHSME box|year=1950|num-b=9|num-a=11}} | + | {{AHSME 50p box|year=1950|num-b=9|num-a=11}} |
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
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 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
