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 $\boxed{-\frac{1}{9}\ \mathrm{(B)}}$ . (We can also use Vieta's formulas to find the sum more quickly.)

Alternative solution: Set $f(\frac{x}{3}) = x^2+x+1=7$ to get $x^2+x-6=0.$ From either finding the roots or using Vieta's formulas, we find the sum of these roots to be $-1.$ Each root of this equation is $9$ times greater than a corresponding root of $f(3z) = 7$ (because $\frac{x}{3} = 3z$ gives $x = 9z$ ), thus the sum of the roots in the equation $f(3z)=7$ is $-\frac{1}{9}.$

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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