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Consider the following equation:

$$ax^2 + bx + c = f(x)$$

$a$, $b$, and $c$ are arbitrary real constants. $f(x)$ is not a polynomial.

Does there exist a condition on $f(x)$ such that the solutions are guaranteed to be real?

Update:

A fixed, more detailed version of the question can be found here Do nth degree polynomials derived using Least Squares Interpolation always have n+1 intersections with the function?

Community
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user120404
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  • lazy answer, $f(x) = ax^2 + b_1x + c_1$ – abel Feb 18 '15 at 20:31
  • Like $b^2 - 4ac \ge 0$ – Alice Ryhl Feb 18 '15 at 20:32
  • All functions f such that their graphs intersect the parabola $y=ax^2+bx+c$ in the plane $\mathbb R ^2$ – Fermat Feb 18 '15 at 20:51
  • This question is only slightly less vague than asking "Under what conditions will $f(x) = g(x)$?" with no constraints on either function. Which is to say, I don't think you're going to get anything very informative as an answer. – pjs36 Feb 18 '15 at 23:20
  • Here's a more detailed version. http://math.stackexchange.com/questions/1166392/do-nth-degree-polynomials-derived-using-least-squares-interpolation-always-have – user120404 Feb 26 '15 at 13:12

1 Answers1

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Plenty. For instance: $$\begin{align}f(x) &= c \\ f(x) &= bx + k\, , \quad k \geq |c| \\ f(x) &= ax^2 + bx + c\quad\text{(solution all of $\Bbb R$)} \\ f(x) &= g(x) + ax^2 + bx + c\, ,\quad g^{-1}(\{0\})\subset \Bbb R\end{align}$$

Maybe you want a specific type of function though? The last example is probably the most general condition in this case.

GPerez
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