Difference between revisions of "2000 AMC 12 Problems/Problem 15"

Revision as of 16:54, 30 July 2019

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 $f(x/3) = x^2 + x + 1$ . 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 $\boxed{\textbf{(B) }-\frac19}$ . (We can also use Vieta's formulas to find the sum more quickly.)

Solution 2

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 $\boxed{\text{(B)}}$

Solution 3

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 $-\frac{9}{81} = \boxed{-\frac{1}{9}} \Longrightarrow \boxed{\text{(B)}}$

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

@@ Line 7: / Line 7: @@
 ==Solution 1==
-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>\boxed{-\frac{1}{9}\ \mathrm{(B)}}</math>. (We can also use [[Vieta's formulas]] to find the sum more quickly.)
+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>\boxed{\textbf{(B) }-\frac19}</math>. (We can also use [[Vieta's formulas]] to find the sum more quickly.)
 ==Solution 2==

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