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

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== Solution ==
 
== Solution ==
 
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>\frac{-1}{9}\ \mathrm{(B)}</math>.
 
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>\frac{-1}{9}\ \mathrm{(B)}</math>.
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Alternative solution: When we have <math>y=81z^2+9z-6</math>, we just use Vieta's and get the sum is <math>\frac{-9}{81}=\frac{-1}{9}</math>
  
 
== See also ==
 
== See also ==

Revision as of 11:36, 26 May 2009

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

Alternative solution: When we have $y=81z^2+9z-6$, we just use Vieta's and get the sum is $\frac{-9}{81}=\frac{-1}{9}$

See also

2000 AMC 12 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AMC 12 Problems and Solutions