Difference between revisions of "2000 AMC 12 Problems/Problem 15"
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==Solution 2== | ==Solution 2== | ||
| − | 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> or <math>\boxed{(B)}</math> |
==Solution 3== | ==Solution 3== | ||
Revision as of 20:50, 6 May 2018
- The following problem is from both the 2000 AMC 12 #15 and 2000 AMC 10 #24, so both problems redirect to this page.
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. 
