Difference between revisions of "2000 AIME II Problems/Problem 13"
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== Solution == | == Solution == | ||
We may factor the equation as: | We may factor the equation as: | ||
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<math> | <math> | ||
\begin{align*} | \begin{align*} | ||
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\end{align*} | \end{align*} | ||
</math> | </math> | ||
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Now <math>100x^4+10x^2+1\ge 1>0</math> for real <math>x</math>. Thus the real roots must be the roots of the equation <math>20x^2+x-2=0</math>. By the [[quadratic formula]] the roots of this are: | Now <math>100x^4+10x^2+1\ge 1>0</math> for real <math>x</math>. Thus the real roots must be the roots of the equation <math>20x^2+x-2=0</math>. By the [[quadratic formula]] the roots of this are: | ||
<math>x=\frac{-1\pm\sqrt{1^2-4(-2)(20)}}{40} = \frac{-1\pm\sqrt{1+160}}{40} = \frac{-1\pm\sqrt{161}}{40}</math> | <math>x=\frac{-1\pm\sqrt{1^2-4(-2)(20)}}{40} = \frac{-1\pm\sqrt{1+160}}{40} = \frac{-1\pm\sqrt{161}}{40}</math> |
Revision as of 11:37, 27 November 2007
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:
$\begin{align*} 2000x^6+100x^5+10x^3+x-2&=0\\ 2(1000x^6-1) + x(100x^4+10x^2+1)&=0\\ 2[(10x^2)^3-1]+x[(10x^2)^2+(10x^2)+1]&=0\\ 2(10x^2-1)[(10x^2)^2+(10x^2)+1]+x[(10x^2)^2+(10x^2)+1]&=0\\ (20x^2+x-2)(100x^4+10x^2+1)&=0\\ \end{align*}$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)
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
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 |