2000 AMC 10 Problem 24

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

A rule given as $f(\text{expression}) = \ldots$ defines $f$ on that expression, so first solve for a formula $f(t)$ in terms of a plain variable. After that, any input is just a substitution. Asking for the sum of the solutions is a signal to use Vieta rather than the quadratic formula.

• Plugging 3z in for x on the right side (as if f(x) = x^2 + x + 1), instead of first finding f(t) for a general input t; this yields $9z^2 + 3z - 6 = 0$ and sum $-\tfrac{1}{3}$, choice (A).
• Solving the quadratic fully and making an arithmetic error; Vieta avoids that work.