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Difference between revisions of "2000 AMC 10 Problems/Problem 24"
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==Solution==
 
==Solution==
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=== Solution 1 ===
  
 
In the definition of <math>f</math>, let <math>x=9z</math>. We get: <math>f(3z)=(9z)^2+(9z)+1</math>. As we have <math>f(3z)=7</math>, we must have <math>f(3z)-7=0</math>, in other words <math>81z^2 + 9z - 6 = 0</math>.
 
In the definition of <math>f</math>, let <math>x=9z</math>. We get: <math>f(3z)=(9z)^2+(9z)+1</math>. As we have <math>f(3z)=7</math>, we must have <math>f(3z)-7=0</math>, in other words <math>81z^2 + 9z - 6 = 0</math>.
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(Note that for the above approach to be completely correct, we should additionally verify that there actually are two distinct real roots. This is, for example, obvious from the facts that <math>81>0</math> and <math>f(0)-7 < 0</math>.)
 
(Note that for the above approach to be completely correct, we should additionally verify that there actually are two distinct real roots. This is, for example, obvious from the facts that <math>81>0</math> and <math>f(0)-7 < 0</math>.)
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=== Solution 2 ===
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<math> f(\frac{x}{3}) = x^2 + x + 1 = [3(\frac{x}{3})]^2 + 3(\frac{x}{3}) + 1 </math> Multiplying by 9, we get
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<math> f(x) = 9x^2 + 3x + 1 </math>
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so <math> f(3z) = 9(3z)^2 + 3(3z) + 1 = 81z^2 + 9z + 1 </math>
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Now, if <math> f(3z) = 7 </math>, then <math> 81z^2 + 9z + 1 = 7 </math> The subtract 7 from both sides we get:
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<math> 81z^2 + 9z - 6 = 0 </math>. Divide by 3 and we get:
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<math> 27z^2 + 3z - 2 = 0 </math>
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The solutions for <math> z </math> are <math> \frac{2}{9} </math> and <math> - \frac{1}{3} </math>
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so the sum is <math> \frac{2}{9} - \frac{1}{3} = \boxed{- \frac{1}{9}} </math>.
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2000|num-b=23|num-a=25}}
 
{{AMC10 box|year=2000|num-b=23|num-a=25}}

Revision as of 11:04, 16 March 2011

Problem

Let $f$ be a function for which $f\left(\frac{x}{3}\right)=x^2+x+1$. Find the sum of all values of $z$ for which $f(3z)=7$.

$\mathrm{(A)}\ -\frac{1}{3} \qquad\mathrm{(B)}\ -\frac{1}{9} \qquad\mathrm{(C)}\ 0 \qquad\mathrm{(D)}\ \frac{5}{9} \qquad\mathrm{(E)}\ \frac{5}{3}$

Solution

Solution 1

In the definition of $f$, let $x=9z$. We get: $f(3z)=(9z)^2+(9z)+1$. As we have $f(3z)=7$, we must have $f(3z)-7=0$, in other words $81z^2 + 9z - 6 = 0$.

One can now either explicitly compute the roots, or use Vieta's formulas. According to them, the sum of the roots of $ax^2+bx+c=0$ is $-\frac ba$. In our case this is $-\frac{9}{81}=\boxed{-\frac 19}$.

(Note that for the above approach to be completely correct, we should additionally verify that there actually are two distinct real roots. This is, for example, obvious from the facts that $81>0$ and $f(0)-7 < 0$.)

Solution 2

$f(\frac{x}{3}) = x^2 + x + 1 = [3(\frac{x}{3})]^2 + 3(\frac{x}{3}) + 1$ Multiplying by 9, we get

$f(x) = 9x^2 + 3x + 1$

so $f(3z) = 9(3z)^2 + 3(3z) + 1 = 81z^2 + 9z + 1$

Now, if $f(3z) = 7$, then $81z^2 + 9z + 1 = 7$ The subtract 7 from both sides we get:

$81z^2 + 9z - 6 = 0$. Divide by 3 and we get:

$27z^2 + 3z - 2 = 0$

The solutions for $z$ are $\frac{2}{9}$ and $- \frac{1}{3}$

so the sum is $\frac{2}{9} - \frac{1}{3} = \boxed{- \frac{1}{9}}$.

See Also

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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