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 to Solution 1, we have The answer is the sum of the roots, which by Vieta's Formulas is
.
~dolphin7
Solution 3
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 4
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
.
Video Solutions
https://youtu.be/qR85EBnpWV8 https://m.youtube.com/watch?v=NyoLydoc3j8&feature=youtu.be
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.