Difference between revisions of "2000 AMC 12 Problems/Problem 15"
Sumanpreet (talk | contribs) (→Solution 4) |
Sumanpreet (talk | contribs) (→Solution 4) |
||
| Line 18: | Line 18: | ||
==Solution 4== | ==Solution 4== | ||
| − | 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 | + | 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: |
<math>81z^2 + 9z +1 = 7 \Longrightarrow 81z^2 + 9z - 6 = 0</math>. For any quadratic <math>ax^2 + bx +c = 0</math>, the sum of the roots is <math>-\frac{b}{a}</math>. Thus, the sum of the roots of this equation is <math>-\frac{9}{81} = \boxed{\textbf{(B) }-\frac19}</math>. | <math>81z^2 + 9z +1 = 7 \Longrightarrow 81z^2 + 9z - 6 = 0</math>. For any quadratic <math>ax^2 + bx +c = 0</math>, the sum of the roots is <math>-\frac{b}{a}</math>. Thus, the sum of the roots of this equation is <math>-\frac{9}{81} = \boxed{\textbf{(B) }-\frac19}</math>. | ||
Revision as of 15:44, 12 July 2020
- 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 
.
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 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 
.
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. 
