2000 AMC 10 Problem 24

1. Understand the rule. The rule tells us $f$ of $\frac{x}{3}$, not $f$ of $x$. For example, with $x = 3$: $f(1) = 9 + 3 + 1 = 13$. With $x = 6$: $f(2) = 36 + 6 + 1 = 43$.
2. Find $f$ of a plain number. To get $f(t)$, we need $\frac{x}{3} = t$, which means $x = 3t$. Put that into the rule:
$$f(t) = (3t)^2 + 3t + 1 = 9t^2 + 3t + 1.$$
Check: $f(1) = 9 + 3 + 1 = 13$. It matches step 1.
3. Now use the input $3z$. Replace $t$ by $3z$:
$$f(3z) = 9(3z)^2 + 3(3z) + 1 = 81z^2 + 9z + 1.$$
4. Set it equal to $7$.
$$81z^2 + 9z + 1 = 7 \quad\Longrightarrow\quad 81z^2 + 9z - 6 = 0.$$
5. Solve. Divide by $3$: $27z^2 + 3z - 2 = 0$. This factors as $(9z - 2)(3z + 1) = 0$, so $z = \frac29$ or $z = -\frac13$.
6. Add them. $\frac29 - \frac13 = \frac29 - \frac39 = -\frac19$.

The answer is $\boxed{\textbf{(B)}\ -\frac{1}{9}}$.

A rule written as "$f$ of some expression equals ..." only tells you $f$ at that expression. Rewrite it first as a formula for $f(t)$ with a plain input $t$; after that, any input is just a substitution. When a question asks only for the sum of the solutions of $az^2 + bz + c = 0$, you can skip solving: the sum is always $-\frac{b}{a}$, here $-\frac{9}{81} = -\frac19$.

Work with the original rule directly. $f(3z)$ means $\frac{x}{3} = 3z$, so $x = 9z$. The equation $f(\frac x3) = 7$ is $x^2 + x + 1 = 7$, or $x^2 + x - 6 = 0$, which gives $x = 2$ or $x = -3$ (sum $-1$). Each $z$ is $\frac{x}{9}$, so the $z$ values add to $-\frac19$.

We are told what $f$ does to $x/3$, not to $x$. Put $t = x/3$, so $x = 3t$ and
$$f(t) = (3t)^2 + 3t + 1 = 9t^2 + 3t + 1.$$
Now set $t = 3z$:
$$f(3z) = 9(3z)^2 + 3(3z) + 1 = 81z^2 + 9z + 1.$$
The condition $f(3z) = 7$ becomes $81z^2 + 9z - 6 = 0$. This quadratic has two real roots (its discriminant $81 + 4\cdot81\cdot6$ is positive), and by Vieta their sum is $-\frac{9}{81} = -\frac{1}{9}$.

The answer is $\boxed{\textbf{(B)}\ -\frac{1}{9}}$.