2000 AMC 12 Problems/Problem 15
- The following problem is from both the 2000 AMC 12 #15 and 2000 AMC 10 #24, so both problems redirect to this page.
Contents
Problem
Let be a function for which . Find the sum of all values of for which .
Solution 1
Let ; then . Thus , and . These sum up to .
Solution 2 (Similar)
This is quite trivially solved, as , so . has solutions and . Adding these yields a solution of .
~ icecreamrolls8
Solution 3
Similar to Solution 1, we have The answer is the sum of the roots, which by Vieta's Formulas is .
~dolphin7
Solution 4
Set to get From either finding the roots (-3 and 2), or using Vieta's formulas, we find the sum of these roots to be Each root of this equation is times greater than a corresponding root of (because gives ), thus the sum of the roots in the equation is or .
Solution 5
Since we have , occurs at Thus, . We set this equal to 7:
. For any quadratic , the sum of the roots is . Thus, the sum of the roots of this equation is .
Note
All solutions that apply Vieta must check if the discriminant is zero, which in this case it isn't.
Video Solutions
https://m.youtube.com/watch?v=NyoLydoc3j8&feature=youtu.be
https://www.youtube.com/watch?v=ZbcJ0ja5TJ8 ~David
Video Solution 2
https://youtu.be/3dfbWzOfJAI?t=1300
~ pi_is_3.14
See also
2000 AMC 12 (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 12 Problems and Solutions |
2000 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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.