2000 AMC 12 Problem 15 — Solution | LIVE by Po-Shen Loh

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2000 AMC 12 Problem 15

Below is the professionally curated solution for Problem 15 of the 2000 AMC 12, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AMC 12 solutions, or check the answer key.

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Concepts:function substitution Vieta’s Formulas

Difficulty rating: 1650

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 $f(3z) = 7.$

$-\dfrac{1}{3}$

$-\dfrac{1}{9}$

$0$

$\dfrac{5}{9}$

$\dfrac{5}{3}$

Solution:

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

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.

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