Difference between revisions of "2000 AMC 12 Problems/Problem 15"
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==Solution 1== | ==Solution 1== | ||
− | Let <math>y = \frac{x}{3}</math>; then <math>f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y+1</math>. Thus <math>f(3z)-7=81z^2+9z-6=3(9z-2)(3z+1)=0</math>, and <math>z = -\frac{1}{3}, \frac{2}{9}</math>. These sum up to <math>\boxed{ | + | Let <math>y = \frac{x}{3}</math>; then <math>f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y+1</math>. Thus <math>f(3z)-7=81z^2+9z-6=3(9z-2)(3z+1)=0</math>, and <math>z = -\frac{1}{3}, \frac{2}{9}</math>. These sum up to <math>\boxed{\textbf{(B) }-\frac19}</math>. (We can also use [[Vieta's formulas]] to find the sum more quickly.) |
==Solution 2== | ==Solution 2== |
Revision as of 16:54, 30 July 2019
- The following problem is from both the 2000 AMC 12 #15 and 2000 AMC 10 #24, so both problems redirect to this page.
Contents
[hide]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 . (We can also use Vieta's formulas to find the sum more quickly.)
Solution 2
Set to get From either finding the roots 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 3
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
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.