Difference between revisions of "2000 AIME II Problems/Problem 13"
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Revision as of 19:31, 4 July 2013
Problem
The equation 
has exactly two real roots, one of which is 
, where 
, 
and 
are integers, and are relatively prime, and 
. Find 
.
Solution
We may factor the equation as:[1]
![\begin{align*} 2000x^6+100x^5+10x^3+x-2&=0\\ 2(1000x^6-1) + x(100x^4+10x^2+1)&=0\\ 2[(10x^2)^3-1]+x[(10x^2)^2+(10x^2)+1]&=0\\ 2(10x^2-1)[(10x^2)^2+(10x^2)+1]+x[(10x^2)^2+(10x^2)+1]&=0\\ (20x^2+x-2)(100x^4+10x^2+1)&=0\\ \end{align*}](http://latex.artofproblemsolving.com/f/6/8/f68c428309181a5078fbf763f5af924e1d567b7e.png)
Now 
for real 
. Thus the real roots must be the roots of the equation 
. By the quadratic formula the roots of this are:
![\[x=\frac{-1\pm\sqrt{1^2-4(-2)(20)}}{40} = \frac{-1\pm\sqrt{1+160}}{40} = \frac{-1\pm\sqrt{161}}{40}.\]](http://latex.artofproblemsolving.com/7/4/a/74ad52c87470980a9a2d6a16385a2625d99ca75b.png)
Thus 
, and so the final answer is 
.
^ A well-known technique for dealing with symmetric (or in this case, nearly symmetric) polynomials is to divide through by a power of with half of the polynomial's degree (in this case, divide through by 
), and then to use one of the substitutions 
. In this case, the substitution 
gives 
and 
, which reduces the polynomial to just 
. Then one can backwards solve for .
See also
| 2000 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
