Difference between revisions of "2000 AIME II Problems/Problem 13"
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Thus <math>r=\frac{-1+\sqrt{161}}{40}</math>, and so the final answer is <math>-1+161+40 = \boxed{200}</math> | Thus <math>r=\frac{-1+\sqrt{161}}{40}</math>, and so the final answer is <math>-1+161+40 = \boxed{200}</math> | ||
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{{AIME box|year=2000|n=II|num-b=12|num-a=14}} | {{AIME box|year=2000|n=II|num-b=12|num-a=14}} |
Revision as of 19:36, 18 March 2008
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:
$
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
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 |