Difference between revisions of "2000 AIME II Problems/Problem 13"

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(Added solution)
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== Solution ==
 
== Solution ==
{{solution}}
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We may factor the equation as:
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<math>
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\begin{align*}
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2000x^6+100x^5+10x^3+x-2&=0\\
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2(1000x^6-1) + x(100x^4+10x^2+1)&=0\\
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2[(10x^2)^3-1]+x[(10x^2)^2+(10x^2)+1]&=0\\
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2(10x^2-1)[(10x^2)^2+(10x^2)+1]+x[(10x^2)^2+(10x^2)+1]&=0\\
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(20x^2+x-2)(100x^4+10x^2+1)&=0\\
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\end{align*}
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</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:
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<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>
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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>
  
 
== See also ==
 
== 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}}

Revision as of 01:37, 27 November 2007

Problem

The equation $2000x^6+100x^5+10x^3+x-2=0$ has exactly two real roots, one of which is $\frac{m+\sqrt{n}}r$, where $m$, $n$ and $r$ are integers, $m$ and $r$ are relatively prime, and $r>0$. Find $m+n+r$.

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 $100x^4+10x^2+1\ge 1>0$ for real $x$. Thus the real roots must be the roots of the equation $20x^2+x-2=0$. By the quadratic formula the roots of this are: $x=\frac{-1\pm\sqrt{1^2-4(-2)(20)}}{40} = \frac{-1\pm\sqrt{1+160}}{40} = \frac{-1\pm\sqrt{161}}{40}$

Thus $r=\frac{-1+\sqrt{161}}{40}$, and so the final answer is $-1+161+40 = \boxed{200}$

See also

2000 AIME II (ProblemsAnswer KeyResources)
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
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