Maths Olympiad Prep

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Algebra Difficulty 2.0 Junior Find the answer

Let ff be a function for which f(x3)=x2+x+1f\left(\dfrac{x}{3}\right) = x^2 + x + 1. Find the sum of all values of zz for which f(3z)=7f(3z) = 7.
(A) −1/3(B) −1/9(C) 0(D) 5/9(E) 5/3\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3

Multiple choice: answer with the letter of the option you want.

Solution

Let y=x3y = \frac{x}{3}; then f(y)=(3y)2+3y+1=9y2+3y+1f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y+1. Thus f(3z)−7=81z2+9z−6=3(9z−2)(3z+1)=0f(3z)-7=81z^2+9z-6=3(9z-2)(3z+1)=0, and z=−13,29z = -\frac{1}{3}, \frac{2}{9}. These sum up to (B) −19\boxed{\textbf{(B) }-\frac19}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.