Difference between revisions of "2000 AIME II Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | {{ | + | We may factor the equation as: |
| + | <math> | ||
| + | \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*} | ||
| + | </math> | ||
| + | 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> | ||
== 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 
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:
$\begin{align*}
2000x^6+100x^5+10x^3+x-2&=0\\
2(1000x^6-1) + x(100x^4+10x^2+1)&=0\\
2[(10x^2)^3-1]+x[(10x^2)^2+(10x^2)+1]&=0\\
2(10x^2-1)[(10x^2)^2+(10x^2)+1]+x[(10x^2)^2+(10x^2)+1]&=0\\
(20x^2+x-2)(100x^4+10x^2+1)&=0\\
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
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 
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 | ||
