2000 AIME II Problems/Problem 13

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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 $-1+161+40 = \boxed{200}$

2000 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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