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

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(Don't Vieta's Formulas work even when the discriminant isn't zero?)
 
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== Problem ==
 
== Problem ==
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Let <math>f</math> be a [[function]] for which <math>f(x/3) = x^2 + x + 1</math>. Find the sum of all values of <math>z</math> for which <math>f(3z) = 7</math>.
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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>.
  
−
<math>\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3</math>
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<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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== Solution ==
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==Solution 1==
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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{-\frac{1}{9}\ \mathrm{(B)}}</math>. (We can also use [[Vieta's formulas]] to find the sum more quickly.)
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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>.
  
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'''Alternative solution:''' 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 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>
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==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
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 +
==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>.
 +
 
 +
~dolphin7
 +
 
 +
==Solution 4==
 +
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>.
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 +
==Solution 5==
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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>.
 +
 
 +
== Video Solutions ==
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 +
https://youtu.be/qR85EBnpWV8
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 +
https://m.youtube.com/watch?v=NyoLydoc3j8&feature=youtu.be
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 +
https://www.youtube.com/watch?v=ZbcJ0ja5TJ8  ~David
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 +
== Video Solution 2==
 +
https://youtu.be/3dfbWzOfJAI?t=1300
 +
 
 +
~ pi_is_3.14
  
 
== See also ==
 
== See also ==

Latest revision as of 23:22, 15 October 2024

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 $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$. 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)

This is quite trivially solved, as $3x = \dfrac{9x}{3}$, so $P(3x) = P(9x/3) = 81x^2 + 9x + 1 = 7$. $81x^2+9x-6 = 0$ has solutions $-\frac{1}{3}$ and $\frac{2}{9}$. Adding these yields a solution of $\boxed{\textbf{(B) }-\frac19}$.

~ icecreamrolls8

Solution 3

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 4

Set $f\left(\frac{x}{3} \right) = x^2+x+1=7$ to get $x^2+x-6=0.$ From either finding the roots (-3 and 2), 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 5

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 Solutions

https://youtu.be/qR85EBnpWV8

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

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