2000 AMC 12 Problems/Problem 15

The following problem is from both the 2000 AMC 12 #15 and 2000 AMC 10 #24, so both problems redirect to this page.

Problem

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

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

Solution 1

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

Solution 2

Similar to Solution 1, we have $=81z^2+9z-6=0.$ The answer is the sum of the roots, which by Vieta's Formulas is $-\frac{b}{a}=-\frac{9}{81}=\boxed{\textbf{(B) }-\frac19}$ .

~dolphin7

Solution 3

Set $f(\frac{x}{3}) = x^2+x+1=7$ to get $x^2+x-6=0.$ From either finding the roots or using Vieta's formulas, we find the sum of these roots to be $-1.$ Each root of this equation is $9$ times greater than a corresponding root of $f(3z) = 7$ (because $\frac{x}{3} = 3z$ gives $x = 9z$ ), thus the sum of the roots in the equation $f(3z)=7$ is $-\frac{1}{9}$ or $\boxed{\textbf{(B) }-\frac19}$ .

Solution 4

Since we have $f(x/3)$ , $f(3z)$ occurs at $x=9z.$ Thus, $f(9z/3) = f(3z) = (9z)^2 + 9z + 1$ . We set this equal to 7:

$81z^2 + 9z +1 = 7 \Longrightarrow 81z^2 + 9z - 6 = 0$ . For any quadratic $ax^2 + bx +c = 0$ , the sum of the roots is $-\frac{b}{a}$ . Thus, the sum of the roots of this equation is $-\frac{9}{81} = \boxed{\textbf{(B) }-\frac19}$ .

Video Solution

https://m.youtube.com/watch?v=NyoLydoc3j8&feature=youtu.be

Video Solution 2

https://youtu.be/3dfbWzOfJAI?t=1300

~ pi_is_3.14

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 12 Problems and Solutions

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
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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