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Algebra Difficulty 3.9 AMC 10/12 Find the answer China

Suppose the domain of function f(x)f(x) is D=(−∞,0)∪(0,+∞)D = (-\infty, 0) \cup (0, +\infty) and there is f(x)=f(1)⋅x2+f(2)⋅x−1xf(x) = \frac{f(1) \cdot x^2 + f(2) \cdot x - 1}{x} for any x∈Dx \in D. Then the sum of all the zeros of f(x)f(x) is ______.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let x1,x2x_1, x_2 and we get
f(1)=f(1)+f(2)−1,f(2)=2f(1)+f(2)−12, \begin{align*} f(1) &= f(1) + f(2) - 1, \\ f(2) &= 2f(1) + f(2) - \frac{1}{2}, \end{align*}
and the solutions are f(2)=1f(2) = 1, f(1)=14f(1) = \frac{1}{4}. Therefore,
f(x)=1x⋅(14x2+x−1)(x≠0). f(x) = \frac{1}{x} \cdot \left( \frac{1}{4}x^2 + x - 1 \right) \quad (x \neq 0).
Let f(x)=0f(x) = 0 and we get x=−2±22x = -2 \pm 2\sqrt{2}, so the sum of all the zeros of f(x)=0f(x) = 0 is −4-4. □\square

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