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>.
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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:
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We may factor the equation as:{{ref|1}}
  
<math>
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<cmath>
 
\begin{align*}
 
\begin{align*}
 
2000x^6+100x^5+10x^3+x-2&=0\\
 
2000x^6+100x^5+10x^3+x-2&=0\\
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(20x^2+x-2)(100x^4+10x^2+1)&=0\\
 
(20x^2+x-2)(100x^4+10x^2+1)&=0\\
 
\end{align*}
 
\end{align*}
</math>
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</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:
<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>
 
  
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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<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>
  
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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>.
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 +
 +
<br />
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{{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}}
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[[Category:Intermediate Algebra Problems]]

Revision as of 11:54, 30 August 2008

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:[1]

\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*}

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



^ A well-known technique for dealing with symmetric (or in this case, nearly symmetric) polynomials is to divide through by a power of $x$ with half of the polynomial's degree (in this case, divide through by $x^3$), and then to use one of the substitutions $t = x \pm \frac{1}{x}$. In this case, the substitution $t = x\sqrt{10} - \frac{1}{x\sqrt{10}}$ gives $t^2 + 2 = 10x^2 + \frac 1{10x^2}$ and $2\sqrt{10}(t^3 + 3t) = 200x^3 - \frac{2}{10x^3}$, which reduces the polynomial to just $(t^2 + 3)\left(2\sqrt{10}t + 1\right) = 0$. Then one can backwards solve for $x$.

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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