Difference between revisions of "2000 AIME II Problems/Problem 13"
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-Grizzy | -Grizzy | ||
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+ | ==Solution 3== | ||
+ | Notice the original expression can be written as <math>2000x^6+100x^5-200x^4+200x^4+10x^3-20x^2+20x^2+x-2</math>. | ||
+ | |||
+ | Which equals to <math>(20x^2+x-2)(100x^4+10x^2+1)=0</math> | ||
+ | |||
+ | So our solution is to find what is the root for <math>20x^2+x-2=0</math> since the determinant of <math>100t^2+10t+1<0</math>(Let <math>x^2=t</math>) | ||
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+ | By solving the equation, we can get that <math>x = \frac{-1 \pm \sqrt{161}}{40}</math> for a final answer of <math>-1 + 161 + 40 = \boxed{200}</math> | ||
+ | |||
+ | ~bluesoul | ||
==Video solution== | ==Video solution== |
Revision as of 03:51, 23 December 2021
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]
Now for real . Thus the real roots must be the roots of the equation . By the quadratic formula the roots of this are:
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 .
Solution 2 (Complex Bash)
It would be really nice if the coefficients were symmetrical. What if we make the substitution, . The the polynomial becomes
It's symmetric! Dividing by and rearranging, we get
Now, if we let , we can get the equations
and
(These come from squaring and subtracting , then multiplying that result by and subtracting ) Plugging this into our polynomial, expanding, and rearranging, we get
Now, we see that the two terms must cancel in order to get this polynomial equal to , so what squared equals 3? Plugging in into the polynomial, we see that it works! Is there something else that equals 3 when squared? Trying , we see that it also works! Great, we use long division on the polynomial by
and we get
.
We know that the other two solutions for z wouldn't result in real solutions for since we have to solve a quadratic with a negative discriminant, then multiply by . We get that . Solving for (using ) we get that , and multiplying this by (because ) we get that for a final answer of
-Grizzy
Solution 3
Notice the original expression can be written as .
Which equals to
So our solution is to find what is the root for since the determinant of (Let )
By solving the equation, we can get that for a final answer of
~bluesoul
Video solution
https://www.youtube.com/watch?v=mAXDdKX52TM
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.