Difference between revisions of "2000 AMC 12 Problems/Problem 15"
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== Solution == | == Solution == | ||
| − | 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>\frac{-1}{9}\ \mathrm{(B)}</math>. | + | 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{\frac{-1}{9}\ \mathrm{(B)}}</math>. |
| − | Alternative solution: When we have <math>0=81z^2+9z-6</math>, we just use Vieta's and get the sum is <math>\frac{-9}{81}=\frac{-1}{9}\ \mathrm{(B)}</math> | + | Alternative solution: When we have <math>0=81z^2+9z-6</math>, we just use Vieta's and get the sum is <math>\frac{-9}{81}=\boxed{\frac{-1}{9}\ \mathrm{(B)}}</math> |
== See also == | == See also == | ||
Revision as of 18:50, 4 July 2012
- 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
Let 
; then 
. Thus 
, and 
. These sum up to 
.
Alternative solution: When we have 
, we just use Vieta's and get the sum 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 | ||
