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| − | ==Problem==
| + | #REDIRECT [[2000 AMC 12 Problems/Problem 15]] |
| − | Let <math>f</math> be a function for which <math>f\left(\frac{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>.
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| − | <math>\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}</math>
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| − | ==Solution==
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| − | === Solution 1 ===
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| − | 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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| − | One can now either explicitly compute the roots, or use [[Vieta's formulas]]. According to them, the sum of the roots of <math>ax^2+bx+c=0</math> is <math>-\frac ba</math>. In our case this is <math>-\frac{9}{81}=\boxed{-\frac 19}</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>.)
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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>.
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| − | ==See Also==
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| − | {{AMC10 box|year=2000|num-b=23|num-a=25}}
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