Difference between revisions of "1986 AHSME Problems/Problem 5"
(Created page with "==Problem== Simplify <math>\left(\sqrt[6]{27} - \sqrt{6 \frac{3}{4} }\right)^2</math> <math>\textbf{(A)}\ \frac{3}{4} \qquad \textbf{(B)}\ \frac{\sqrt 3}{2} \qquad \textbf{(C)}...") |
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==Solution== | ==Solution== | ||
| − | + | We have <math>\sqrt[6]{27} = (3^{3})^{\frac{1}{6}} = 3^{\frac{1}{2}} = \sqrt{3}</math> and <math>\sqrt{6 \frac{3}{4}} = \sqrt{\frac{27}{4}} = \frac{3 \sqrt{3}}{2}</math>, so the answer is <math>(\sqrt{3} - \frac{3 \sqrt{3}}{2})^{2} = (-\frac{\sqrt{3}}{2})^{2} = \frac{3}{4}</math>, which is <math>\boxed{A}</math>. | |
== See also == | == See also == | ||
Latest revision as of 17:09, 1 April 2018
Problem
Simplify ![$\left(\sqrt[6]{27} - \sqrt{6 \frac{3}{4} }\right)^2$](http://latex.artofproblemsolving.com/f/2/7/f27c3f23f04eb150d43b50f763804f7a63e2affa.png)
Solution
We have ![$\sqrt[6]{27} = (3^{3})^{\frac{1}{6}} = 3^{\frac{1}{2}} = \sqrt{3}$](http://latex.artofproblemsolving.com/e/e/c/eecc8058dd5a4bbb12c47888360d8004ae0946e0.png)
and 
, so the answer is 
, which is 
.
See also
| 1986 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
