Difference between revisions of "2000 AMC 12 Problems/Problem 11"
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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>? | 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>\textbf{(A)} \ - 2 \qquad \textbf{(B)} \ \frac - | + | <math>\textbf{(A)} \ - 2 \qquad \textbf{(B)} \ \frac { -1 }{2} \qquad \textbf{(C)} \ \frac {1}{3} \qquad \textbf{(D)} \ \frac {1}{2} \qquad \textbf{(E)} \ 2</math> |
==Solution 1== | ==Solution 1== |
Revision as of 11:23, 1 April 2022
- The following problem is from both the 2000 AMC 12 #11 and 2000 AMC 10 #15, so both problems redirect to this page.
Contents
[hide]Problem
Two non-zero real numbers, and satisfy . Which of the following is a possible value of ?
Solution 1
.
Another way is to solve the equation for giving then substituting this into the expression and simplifying gives the answer of
Solution 2
This simplifies to . The two integer solutions to this are and . The problem states than and are non-zero, so we consider the case of . So, we end up with
Solution 3
Just realize that two such numbers are and . With this, you can solve and get
Video Solution
https://www.youtube.com/watch?v=7-RloNHTnXM
Video Solution
https://youtu.be/ZWqHxc0i7ro?t=6
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=8nxvuv5oZ7A&t=3s
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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.