2000 AMC 10 Problems/Problem 24

Problem

Let $f$ be a function for which $f\left(\frac{x}{3}\right)=x^2+x+1$ . Find the sum of all values of $z$ for which $f(3z)=7$ .

$\mathrm{(A)}\ -\frac{1}{3} \qquad\mathrm{(B)}\ -\frac{1}{9} \qquad\mathrm{(C)}\ 0 \qquad\mathrm{(D)}\ \frac{5}{9} \qquad\mathrm{(E)}\ \frac{5}{3}$

Solution

Solution 1

In the definition of $f$ , let $x=9z$ . We get: $f(3z)=(9z)^2+(9z)+1$ . As we have $f(3z)=7$ , we must have $f(3z)-7=0$ , in other words $81z^2 + 9z - 6 = 0$ .

One can now either explicitly compute the roots, or use Vieta's formulas. According to them, the sum of the roots of $ax^2+bx+c=0$ is $-\frac ba$ . In our case this is $-\frac{9}{81}=\boxed{-\frac 19}$ .

(Note that for the above approach to be completely correct, we should additionally verify that there actually are two distinct real roots. This is, for example, obvious from the facts that $81>0$ and $f(0)-7 < 0$ .)

Solution 2

$f(\frac{x}{3}) = x^2 + x + 1 = [3(\frac{x}{3})]^2 + 3(\frac{x}{3}) + 1$ . Multiply both sides by 9 to get rid of the denominators.

$f(x) = 9x^2 + 3x + 1$

so $f(3z) = 9(3z)^2 + 3(3z) + 1 = 81z^2 + 9z + 1$

Now, if $f(3z) = 7$ , then $81z^2 + 9z + 1 = 7$ The subtract 7 from both sides we get:

$81z^2 + 9z - 6 = 0$ . Divide by 3 and we get:

$27z^2 + 3z - 2 = 0$

The solutions for $z$ are $\frac{2}{9}$ and $- \frac{1}{3}$

so the sum is $\frac{2}{9} - \frac{1}{3} = \boxed{- \frac{1}{9}}$ .

Solution 3

Put $x=9z$ and we get:

$f(3z) = 81z^{2}+9z+1 =7$ .

Now, by Vieta's formula we get that:

$\sum z = (-1)^{1}\cdot \frac{ 9}{81}= \boxed{-\frac{1}{9}\implies B}$ .

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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