Difference between revisions of "2000 AMC 12 Problems/Problem 15"
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== Problem == | == Problem == | ||
− | Let <math>f</math> be a | + | Let <math>f</math> be a function for which <math>f\left(\dfrac{x}{3}\right) = x^2 + x + 1</math>. Find the sum of all values of <math>z</math> for which <math>f(3z) = 7</math>. |
<cmath>\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3</cmath> | <cmath>\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3</cmath> | ||
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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>. | 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>. | ||
− | ==Solution 2== | + | ==Solution 2 (Similar) == |
+ | |||
+ | This is quite trivially solved, as <math>3x = \dfrac{9x}{3}</math>, so <math>P(3x) = P(9x/3) = 81x^2 + 9x + 1 = 7</math>. <math>81x^2+9x-6 = 0</math> has solutions <math>-\frac{1}{3}</math> and <math>\frac{2}{9}</math>. Adding these yields a solution of <math>\boxed{\textbf{(B) }-\frac19}</math>. | ||
+ | |||
+ | ~ icecreamrolls8 | ||
+ | |||
+ | ==Solution 3== | ||
Similar to Solution 1, we have <math>=81z^2+9z-6=0.</math> The answer is the sum of the roots, which by [[Vieta's Formulas]] is <math>-\frac{b}{a}=-\frac{9}{81}=\boxed{\textbf{(B) }-\frac19}</math>. | Similar to Solution 1, we have <math>=81z^2+9z-6=0.</math> The answer is the sum of the roots, which by [[Vieta's Formulas]] is <math>-\frac{b}{a}=-\frac{9}{81}=\boxed{\textbf{(B) }-\frac19}</math>. | ||
~dolphin7 | ~dolphin7 | ||
− | ==Solution | + | ==Solution 4== |
− | Set <math>f(\frac{x}{3}) = x^2+x+1=7</math> to get <math>x^2+x-6=0.</math> From either finding the roots (-3 and 2), or using Vieta's formulas, we find the sum of these roots to be <math>-1.</math> Each root of this equation is <math>9</math> times greater than a corresponding root of <math>f(3z) = 7</math> (because <math>\frac{x}{3} = 3z</math> gives <math>x = 9z</math>), thus the sum of the roots in the equation <math>f(3z)=7</math> is <math>-\frac{1}{9}</math> or <math>\boxed{\textbf{(B) }-\frac19}</math>. | + | Set <math>f\left(\frac{x}{3} \right) = x^2+x+1=7</math> to get <math>x^2+x-6=0.</math> From either finding the roots (-3 and 2), or using Vieta's formulas, we find the sum of these roots to be <math>-1.</math> Each root of this equation is <math>9</math> times greater than a corresponding root of <math>f(3z) = 7</math> (because <math>\frac{x}{3} = 3z</math> gives <math>x = 9z</math>), thus the sum of the roots in the equation <math>f(3z)=7</math> is <math>-\frac{1}{9}</math> or <math>\boxed{\textbf{(B) }-\frac19}</math>. |
− | ==Solution | + | ==Solution 5== |
Since we have <math>f(x/3)</math>, <math>f(3z)</math> occurs at <math>x=9z.</math> Thus, <math>f(9z/3) = f(3z) = (9z)^2 + 9z + 1</math>. We set this equal to 7: | Since we have <math>f(x/3)</math>, <math>f(3z)</math> occurs at <math>x=9z.</math> Thus, <math>f(9z/3) = f(3z) = (9z)^2 + 9z + 1</math>. We set this equal to 7: | ||
<math>81z^2 + 9z +1 = 7 \Longrightarrow 81z^2 + 9z - 6 = 0</math>. For any quadratic <math>ax^2 + bx +c = 0</math>, the sum of the roots is <math>-\frac{b}{a}</math>. Thus, the sum of the roots of this equation is <math>-\frac{9}{81} = \boxed{\textbf{(B) }-\frac19}</math>. | <math>81z^2 + 9z +1 = 7 \Longrightarrow 81z^2 + 9z - 6 = 0</math>. For any quadratic <math>ax^2 + bx +c = 0</math>, the sum of the roots is <math>-\frac{b}{a}</math>. Thus, the sum of the roots of this equation is <math>-\frac{9}{81} = \boxed{\textbf{(B) }-\frac19}</math>. | ||
+ | |||
+ | ==Note== | ||
+ | All solutions that apply Vieta must check if the discriminant is zero, which in this case it isn't. | ||
== Video Solutions == | == Video Solutions == | ||
https://youtu.be/qR85EBnpWV8 | https://youtu.be/qR85EBnpWV8 | ||
+ | |||
https://m.youtube.com/watch?v=NyoLydoc3j8&feature=youtu.be | https://m.youtube.com/watch?v=NyoLydoc3j8&feature=youtu.be | ||
+ | |||
+ | https://www.youtube.com/watch?v=ZbcJ0ja5TJ8 ~David | ||
== Video Solution 2== | == Video Solution 2== |
Latest revision as of 21:11, 29 April 2023
- The following problem is from both the 2000 AMC 12 #15 and 2000 AMC 10 #24, so both problems redirect to this page.
Contents
Problem
Let be a function for which
. Find the sum of all values of
for which
.
Solution 1
Let ; then
. Thus
, and
. These sum up to
.
Solution 2 (Similar)
This is quite trivially solved, as , so
.
has solutions
and
. Adding these yields a solution of
.
~ icecreamrolls8
Solution 3
Similar to Solution 1, we have The answer is the sum of the roots, which by Vieta's Formulas is
.
~dolphin7
Solution 4
Set to get
From either finding the roots (-3 and 2), or using Vieta's formulas, we find the sum of these roots to be
Each root of this equation is
times greater than a corresponding root of
(because
gives
), thus the sum of the roots in the equation
is
or
.
Solution 5
Since we have ,
occurs at
Thus,
. We set this equal to 7:
. For any quadratic
, the sum of the roots is
. Thus, the sum of the roots of this equation is
.
Note
All solutions that apply Vieta must check if the discriminant is zero, which in this case it isn't.
Video Solutions
https://m.youtube.com/watch?v=NyoLydoc3j8&feature=youtu.be
https://www.youtube.com/watch?v=ZbcJ0ja5TJ8 ~David
Video Solution 2
https://youtu.be/3dfbWzOfJAI?t=1300
~ pi_is_3.14
See also
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.