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== Problem ==
Let <math>f</math> be a [[function]] for which <math>f(x/3) = x^2 + x + 1</math>. Find the sum of all values of <math>z</math> for which <math>f(3z) = 7</math>.

<math>\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3</math>

== Solution ==
Let <math>y = \frac{x}{3}</math>; then <math>f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y+1</math>. Thus <math>f(3z)-7=81z^2+9z-6=3(9z-2)(3z+1)=0</math>, and <math>z = \frac{-1}{3}, \frac{2}{9}</math>. These sum up to <math>\frac{-1}{9}\ \mathrm{(B)}</math>.

Alternative solution: When we have <math>0=81z^2+9z-6</math>, we just use Vieta's and get the sum is <math>\frac{-9}{81}=\frac{-1}{9}</math>

== See also ==
{{AMC12 box|year=2000|num-b=14|num-a=16}}

[[Category:Introductory Algebra Problems]]