Difference between revisions of "2000 AMC 12 Problems/Problem 11"

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==Problem==
 
==Problem==
Two non-zero real numbers, <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>?
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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==
[[WLOG]], let a=1. Solving for b, we have b=1/2, and <math>\frac {a}{b} + \frac {b}{a} - ab=2\Rightarrow \text{(E)}</math>
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<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 Also==
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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>.
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==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 16:59, 3 January 2008

Problem

Two non-zero real numbers, $a$ and $b,$ satisfy $ab = a - b$. Which of the following is a possible value of $\frac {a}{b} + \frac {b}{a} - ab$?

$\text{(A)} \ - 2 \qquad \text{(B)} \ \frac { - 1}{2} \qquad \text{(C)} \ \frac {1}{3} \qquad \text{(D)} \ \frac {1}{2} \qquad \text{(E)} \ 2$

Solution

$\frac {a}{b} + \frac {b}{a} - ab = \frac{a^2 + b^2}{ab} - (a - b) = \frac{a^2 + b^2}{a-b} - \frac{(a-b)^2}{(a-b)} = \frac{2ab}{a-b} = 2 \Rightarrow \text{(E)}$.

Alternatively, we could test simple values, like $(a,b)=\left(1, \frac{1}{2}\right)$, which would yield $\frac {a}{b} + \frac {b}{a} - ab=2$.

See also

2000 AMC 12 (ProblemsAnswer KeyResources)
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