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Difference between revisions of "2000 AMC 12 Problems/Problem 15"
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{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #15]] and [[2000 AMC 10 Problems|2000 AMC 10 #24]]}}
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== Problem ==
 
== Problem ==
 
Let <math>f</math> be a [[function]] for which <math>f(x/3) = x^2 + x + 1</math>. Find the sum of all values of <math>z</math> for which <math>f(3z) = 7</math>.
 
Let <math>f</math> be a [[function]] for which <math>f(x/3) = x^2 + x + 1</math>. Find the sum of all values of <math>z</math> for which <math>f(3z) = 7</math>.
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2000|num-b=14|num-a=16}}
 
{{AMC12 box|year=2000|num-b=14|num-a=16}}
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{{AMC10 box|year=2000|num-b=23|num-a=25}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 23:07, 26 November 2011

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 $f$ be a function for which $f(x/3) = x^2 + x + 1$. Find the sum of all values of $z$ for which $f(3z) = 7$.

$\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3$

Solution

Let $y = \frac{x}{3}$; then $f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y+1$. Thus $f(3z)-7=81z^2+9z-6=3(9z-2)(3z+1)=0$, and $z = \frac{-1}{3}, \frac{2}{9}$. These sum up to $\frac{-1}{9}\ \mathrm{(B)}$.

Alternative solution: When we have $0=81z^2+9z-6$, we just use Vieta's and get the sum is $\frac{-9}{81}=\frac{-1}{9}\ \mathrm{(B)}$

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
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