2000 AMC 12 Problems/Problem 15: Difference between revisions

Revision as of 19:04, 4 January 2008

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

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 $\frac{-1}{9}\ \mathrm{(B)}$ .

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

@@ Line 8: / Line 8: @@
 == See also ==
-{{AMC12 box|year=2000|num-b=8|num-a=10}}
+{{AMC12 box|year=2000|num-b=14|num-a=16}}
 [[Category:Introductory Algebra Problems]]

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2000 AMC 12 Problems/Problem 15: Difference between revisions

Revision as of 19:04, 4 January 2008

Problem

Solution

See also