2000 AMC 12 Problems/Problem 5
(Redirected from 2000 AMC 10 Problems/Problem 9)
- The following problem is from both the 2000 AMC 12 #5 and 2000 AMC 10 #9, so both problems redirect to this page.
Contents
Problem
If , where , then
Solution
When is negative so and .
Thus .
Solution 2 (guess and check/desperation)
If you did not find that slick Solution 1, all hope is not lost. We could still guess and check our way to the right answer.
We first plug in , and get that too. Hence , eliminating choices and .
We then plug in , and get . Therefore, . The answer is negative, eliminating . Furthermore, , so choice is false. Hence, the answer must be , which upon checking indeed still holds true. -Monkey_King
Video Solution by Daily Dose of Math
https://youtu.be/albUhCOwv3Y?si=4XcusOEp70EA6XKr
~Thesmartgreekmathdude
See also
2000 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 |
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 4 |
Followed by Problem 6 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.