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:
 +
 
<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 $2000x^6+100x^5+10x^3+x-2=0$ 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:

$2000x6+100x5+10x3+x2=02(1000x61)+x(100x4+10x2+1)=02[(10x2)31]+x[(10x2)2+(10x2)+1]=02(10x21)[(10x2)2+(10x2)+1]+x[(10x2)2+(10x2)+1]=0(20x2+x2)(100x4+10x2+1)=0$ (Error compiling LaTeX. Unknown error_msg)

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