Difference between revisions of "1980 AHSME Problems/Problem 27"
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== Solution == | == Solution == | ||
| − | Lets set our original expression equal to <math>x</math>. So <math>\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}} = x</math>. Cubing this gives us <math>x^3 = (\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}})^3 = 5 + 2\sqrt{13} + 5 - 2\sqrt{13} + 3(\sqrt[3] {5+2\sqrt{13}}*\sqrt[3]{5-2\sqrt{13}})(\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}) = 10 - 9x</math> | + | Lets set our original expression equal to <math>x</math>. So <math>\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}} = x</math>. Cubing this gives us <math>x^3 = \left(\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\right)^3 = 5 + 2\sqrt{13} + 5 - 2\sqrt{13} + 3\left(\sqrt[3] {5+2\sqrt{13}}*\sqrt[3]{5-2\sqrt{13}}\right)\left(\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\right) = 10 - 9x</math> |
| − | So we have <math>x^3 + 9x - 10 = 0</math>. We can easily see that 1 is a root of this polynomial. By synthetic division, the new polynomial is <math>x^2 + x + 10</math>, which has no real roots. Thus <math>\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}} = 1</math>. Since 1 is not A-D, our answer is <math>\fbox{E}</math> | + | So we have <math>x^3 + 9x - 10 = 0</math>. We can easily see that 1 is a root of this polynomial. By synthetic division, the new polynomial is <math>x^2 + x + 10</math>, which has no real roots. Thus <math>\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}} = 1</math>. Since 1 is not A-D, our answer is <math>\fbox{E}</math>. |
== See also == | == See also == | ||
Latest revision as of 17:25, 5 June 2015
Problem
The sum ![$\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$](http://latex.artofproblemsolving.com/1/7/b/17bdbabd42c80acc2d8e8e6dadc7b7afd91968a2.png)
equals
![$\text{(A)} \ \frac 32 \qquad \text{(B)} \ \frac{\sqrt[3]{65}}{4} \qquad \text{(C)} \ \frac{1+\sqrt[6]{13}}{2} \qquad \text{(D)}\ \sqrt[3]{2}\qquad \text{(E)}\ \text{none of these}$](http://latex.artofproblemsolving.com/1/b/7/1b7ee865b061b3278bcf81aca65d35e47d85f783.png)
Solution
Lets set our original expression equal to 
. So ![$\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}} = x$](http://latex.artofproblemsolving.com/9/c/e/9cecd7675a7254a9278c19b23b20fa795ca6550c.png)
. Cubing this gives us ![$x^3 = \left(\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\right)^3 = 5 + 2\sqrt{13} + 5 - 2\sqrt{13} + 3\left(\sqrt[3] {5+2\sqrt{13}}*\sqrt[3]{5-2\sqrt{13}}\right)\left(\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\right) = 10 - 9x$](http://latex.artofproblemsolving.com/e/0/7/e078f81e11883bf6cfe7763f06fe3a1bb1b17fdd.png)
So we have 
. We can easily see that 1 is a root of this polynomial. By synthetic division, the new polynomial is 
, which has no real roots. Thus ![$\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}} = 1$](http://latex.artofproblemsolving.com/d/9/7/d9705b5ec9fe8c15bc936ab279062156348c7e2c.png)
. Since 1 is not A-D, our answer is 
.
See also
| 1980 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 26 |
Followed by Problem 28 | |
| 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. 
