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2000 AMC 10 Problems/Problem 24

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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\math...

Solution

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.)

See Also

2000 AMC 10 (ProblemsResources)
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
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