Difference between revisions of "2000 AMC 12 Problems/Problem 11"
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==Problem== | ==Problem== | ||
| − | Two non-zero real | + | Two non-zero [[real number]]s, <math>a</math> and <math>b,</math> satisfy <math>ab = a - b</math>. Which of the following is a possible value of <math>\frac {a}{b} + \frac {b}{a} - ab</math>? |
<math>\text{(A)} \ - 2 \qquad \text{(B)} \ \frac { - 1}{2} \qquad \text{(C)} \ \frac {1}{3} \qquad \text{(D)} \ \frac {1}{2} \qquad \text{(E)} \ 2</math> | <math>\text{(A)} \ - 2 \qquad \text{(B)} \ \frac { - 1}{2} \qquad \text{(C)} \ \frac {1}{3} \qquad \text{(D)} \ \frac {1}{2} \qquad \text{(E)} \ 2</math> | ||
==Solution== | ==Solution== | ||
| − | + | <math>\frac {a}{b} + \frac {b}{a} - ab = \frac{a^2 + b^2}{ab} - (a - b) = \frac{a^2 + b^2}{a-b} - \frac{(a-b)^2}{(a-b)} = \frac{2ab}{a-b} = 2 \Rightarrow \text{(E)}</math>. | |
| − | ==See | + | Alternatively, we could test simple values, like <math>(a,b)=\left(1, \frac{1}{2}\right)</math>, which would yield <math>\frac {a}{b} + \frac {b}{a} - ab=2</math>. |
| + | |||
| + | ==See also== | ||
{{AMC12 box|year=2000|num-b=10|num-a=12}} | {{AMC12 box|year=2000|num-b=10|num-a=12}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 17:59, 3 January 2008
Problem
Two non-zero real numbers, 
and 
satisfy 
. Which of the following is a possible value of 
?
Solution
.
Alternatively, we could test simple values, like 
, which would yield 
.
See also
| 2000 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 10 |
Followed by Problem 12 |
| 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 12 Problems and Solutions | |
