Difference between revisions of "2000 AMC 10 Problems/Problem 15"
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==Solution== | ==Solution== | ||
| − | <math>ab=a-b</math> | + | <math>\begin{align*} |
| + | \frac{a}{b}+\frac{b}{a}-ab &= \frac{a^2+b^2}{ab}-ab \\ | ||
| + | &= \frac{-a^2b^2+a^2+b^2}{ab}\\ | ||
| + | \end(align*}</math> | ||
| − | <math> | + | Substituting <math>ab=a-b</math>, we get |
| − | <math>\frac{-a^2+2ab-b^2+a^2+b^2}{ab}=2</math> | + | <math>\begin{align*} |
| + | \frac{-(a+b)^2+a^2+b^2}{ab} &= \frac{-a^2+2ab-b^2+a^2+b^2}{ab} \\ | ||
| + | &= \frac{2ab}{ab} \\ | ||
| + | &= 2 \\ | ||
| + | \end{align*}</math> | ||
| − | E | + | <math>\boxed{\text{E}}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2000|num-b=14|num-a=16}} | {{AMC10 box|year=2000|num-b=14|num-a=16}} | ||
Revision as of 00:23, 11 January 2009
Problem
Two non-zero real numbers, 
and 
, satisfy 
. Find a possible value of 
.
Solution
$\begin{align*} \frac{a}{b}+\frac{b}{a}-ab &= \frac{a^2+b^2}{ab}-ab \\ &= \frac{-a^2b^2+a^2+b^2}{ab}\\ \end(align*}$ (Error compiling LaTeX. Unknown error_msg)
Substituting , we get
$\begin{align*} \frac{-(a+b)^2+a^2+b^2}{ab} &= \frac{-a^2+2ab-b^2+a^2+b^2}{ab} \\ &= \frac{2ab}{ab} \\ &= 2 \\ \end{align*}$ (Error compiling LaTeX. Unknown error_msg)
See Also
| 2000 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
