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2000 AMC 12 Problem 15

Problem 15 of 25IntermediateAlgebra

Let ff be a function for which f ⁣(x3)=x2+x+1.f\!\left(\dfrac{x}{3}\right) = x^2 + x + 1. Find the sum of all values of zz for which f(3z)=7.f(3z) = 7.

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Solution

Setting x3=3z\dfrac{x}{3} = 3z gives x=9z,x = 9z, so f(3z)=(9z)2+9z+1=81z2+9z+1=7. \begin{aligned} f(3z) &= (9z)^2 + 9z + 1 \\ &= 81z^2 + 9z + 1 \\ &= 7. \end{aligned} This rearranges to 81z2+9z6=0.81z^2 + 9z - 6 = 0. By the sum-of-roots formula, the sum of the values of zz is 981=19.-\dfrac{9}{81} = -\dfrac{1}{9}. Thus, the correct answer is B.

More practice

Concepts: function · substitution · Vieta’s Formulas

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.