# 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 $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$. 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 1

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 $\boxed{\textbf{(B) }-\frac19}$.

## Solution 2

Similar to Solution 1, we have $=81z^2+9z-6=0.$ The answer is the sum of the roots, which by Vieta's Formulas is $-\frac{b}{a}=-\frac{9}{81}=\boxed{\textbf{(B) }-\frac19}$.

~dolphin7

## Solution 3

Set $f\left(\frac{x}{3} \right) = x^2+x+1=7$ to get $x^2+x-6=0.$ From either finding the roots (-3 and 2), 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 $-\frac{1}{9}$ or $\boxed{\textbf{(B) }-\frac19}$.

## Solution 4

Since we have $f(x/3)$, $f(3z)$ occurs at $x=9z.$ Thus, $f(9z/3) = f(3z) = (9z)^2 + 9z + 1$. We set this equal to 7: $81z^2 + 9z +1 = 7 \Longrightarrow 81z^2 + 9z - 6 = 0$. For any quadratic $ax^2 + bx +c = 0$, the sum of the roots is $-\frac{b}{a}$. Thus, the sum of the roots of this equation is $-\frac{9}{81} = \boxed{\textbf{(B) }-\frac19}$.

## Video Solution 2

~ pi_is_3.14

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