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I am dealing with an expression of the form $$\alpha f(x)^2 + \beta f(x) + \gamma = 0,$$ where $f: \mathbb{R} \to \mathbb{R}$ is smooth and not identically zero, and $\alpha$, $\beta$, and $\gamma$ are real coefficients.

I strongly suspect that for the above to always hold, the $x$-dependence on the LHS must be removed. This would entail that $\alpha = \beta = 0$ (and $\gamma = 0$ as a result). Is my suspicion correct? In case it is correct, is there an algebraic argument for why it holds?

Edit: I forgot to specify that $f$ is not constant.

FraGrechi
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2 Answers2

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If $p$ is any non-zero polynomial and $p(f(x))=0$ for all $x$ then the range of $f$ is contained in the set of roots of $p$, Hence $f$ is a constant. Since $f$ is assumed to be non-constant the conclusion is that $p$ is the zero polynomial. This makes all the coefficients $0$.

Kavi Rama Murthy
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  • Thank you very much! I figured that the correct approach was using properties of polynomials, but was unsure about precisely how. – FraGrechi Apr 17 '19 at 16:27
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Solve the equation as a quadratic for $f(x)$, and you get that either $$\alpha = \beta = \gamma = 0$$ or $$\alpha = 0, f(x) = -\frac \gamma \beta$$ or $$f(x) = \frac{-\beta \pm \sqrt{\beta^2 - 4\alpha\beta}}{2\alpha}$$

requiring $\beta^2 - 4\alpha\beta \ge 0$. Except the first case $f(x)$ is a constant.

DanielV
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