2000 AMC 12 Problems/Problem 15

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 $\boxed{-\frac{1}{9}\ \mathrm{(B)}}$ .

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

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 12 Problems and Solutions

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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All AMC 10 Problems and Solutions

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2000 AMC 12 Problems/Problem 15

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