Difference between revisions of "2000 AMC 12 Problems/Problem 15"
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Set <math>f(\frac{x}{3}) = x^2+x+1=7</math> to get <math>x^2+x-6=0.</math> From either finding the roots or using Vieta's formulas, we find the sum of these roots to be <math>-1.</math> Each root of this equation is <math>9</math> times greater than a corresponding root of <math>f(3z) = 7</math> (because <math>\frac{x}{3} = 3z</math> gives <math>x = 9z</math>), thus the sum of the roots in the equation <math>f(3z)=7</math> is <math>-\frac{1}{9}.</math> | Set <math>f(\frac{x}{3}) = x^2+x+1=7</math> to get <math>x^2+x-6=0.</math> From either finding the roots or using Vieta's formulas, we find the sum of these roots to be <math>-1.</math> Each root of this equation is <math>9</math> times greater than a corresponding root of <math>f(3z) = 7</math> (because <math>\frac{x}{3} = 3z</math> gives <math>x = 9z</math>), thus the sum of the roots in the equation <math>f(3z)=7</math> is <math>-\frac{1}{9}.</math> | ||
− | =Solution 3== | + | ==Solution 3== |
Since we have <math>f(x/3)</math>, <math>f(3z)</math> occurs at <math>x=9z.</math> Thus, <math>f(9z/3) = f(3z) = (9z)^2 + 9z + 1</math>. We set this equal to 7: | Since we have <math>f(x/3)</math>, <math>f(3z)</math> occurs at <math>x=9z.</math> Thus, <math>f(9z/3) = f(3z) = (9z)^2 + 9z + 1</math>. We set this equal to 7: | ||
Revision as of 16:05, 20 January 2017
- 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
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.