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.

2000 AMC 12 Problem 15 — Solution

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15.

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 $f(3z) = 7.$

$-\dfrac{1}{3}$

$-\dfrac{1}{9}$

$0$

$\dfrac{5}{9}$

$\dfrac{5}{3}$

Setting $\dfrac{x}{3} = 3z$ gives $x = 9z,$ so $\begin{aligned} f(3z) &= (9z)^2 + 9z + 1 \\ &= 81z^2 + 9z + 1 \\ &= 7. \end{aligned}$

This rearranges to $81z^2 + 9z - 6 = 0.$

By the sum-of-roots formula, the sum of the values of $z$ is $-\dfrac{9}{81} = -\dfrac{1}{9}.$

Thus, the correct answer is B.