Difference between revisions of "2000 AIME II Problems/Problem 13"
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== Problem == | == Problem == | ||
− | The equation <math>2000x^6+100x^5+10x^3+x-2=0</math> has exactly two real roots, one of which is <math>\frac{m+\sqrt{n}}r</math>, where <math>m</math>, <math>n</math> and <math>r</math> are integers, <math>m</math> and <math>r</math> are relatively prime, and <math>r>0</math>. Find <math>m+n+r</math>. | + | The [[equation]] <math>2000x^6+100x^5+10x^3+x-2=0</math> has exactly two real roots, one of which is <math>\frac{m+\sqrt{n}}r</math>, where <math>m</math>, <math>n</math> and <math>r</math> are integers, <math>m</math> and <math>r</math> are relatively prime, and <math>r>0</math>. Find <math>m+n+r</math>. |
== Solution == | == Solution == | ||
− | We may factor the equation as: | + | We may factor the equation as:{{ref|1}} |
− | < | + | <cmath> |
\begin{align*} | \begin{align*} | ||
2000x^6+100x^5+10x^3+x-2&=0\\ | 2000x^6+100x^5+10x^3+x-2&=0\\ | ||
Line 13: | Line 13: | ||
(20x^2+x-2)(100x^4+10x^2+1)&=0\\ | (20x^2+x-2)(100x^4+10x^2+1)&=0\\ | ||
\end{align*} | \end{align*} | ||
− | </ | + | </cmath> |
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: | ||
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− | + | <cmath>x=\frac{-1\pm\sqrt{1^2-4(-2)(20)}}{40} = \frac{-1\pm\sqrt{1+160}}{40} = \frac{-1\pm\sqrt{161}}{40}.</cmath> | |
+ | 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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+ | <br /> | ||
+ | {{note|1}} A well-known technique for dealing with symmetric (or in this case, nearly symmetric) polynomials is to divide through by a power of <math>x</math> with half of the polynomial's degree (in this case, divide through by <math>x^3</math>), and then to use one of the substitutions <math>t = x \pm \frac{1}{x}</math>. In this case, the substitution <math>t = x\sqrt{10} - \frac{1}{x\sqrt{10}}</math> gives <math>t^2 + 2 = 10x^2 + \frac 1{10x^2}</math> and <math>2\sqrt{10}(t^3 + 3t) = 200x^3 - \frac{2}{10x^3}</math>, which reduces the polynomial to just <math>(t^2 + 3)\left(2\sqrt{10}t + 1\right) = 0</math>. Then one can backwards solve for <math>x</math>. | ||
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+ | == See also == | ||
{{AIME box|year=2000|n=II|num-b=12|num-a=14}} | {{AIME box|year=2000|n=II|num-b=12|num-a=14}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Revision as of 11:54, 30 August 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:^{[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 .
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 |