Is there a simple criterion for when $e^{x}+ ax^{2}+ bx+ c$ has a zero?

Question

Is there a simple criterion for when $$e^{x}+ ax^{2}+ bx+ c$$ has a zero ?
Source: MathOverflow/@MattF.

one simpler function $e^{x}+ ax+ b$ has a zero iff either or $- a+ a\!\ln\!\left ( -a \right )+ b\!\leq 0>\!a,$ or $a\!= 0>\!b,$ or $a> 0,$ where the last clause comes from using calculus to minimize the function.
First edit. It might be possible to extract something from Wilkie's approach in his
paper "Model completeness results for expansions of the ordered field of real numbers by restricted Pfaffian functions and the exponential function," which implies, among other things, that there is a system of exponential equations and inequalities that has a solution iff $e^{x}+ ax^{2}+ bx+ c$ doesn't.
Second edit. There is an algorithm to determine whether $e^{x}+ ax^{2}+ bx+ c$ has a zero :
simultaneously check for each positive rational $q$ whether $e^{q}+ aq^{2}+ bq+ c< 0,$ for each $n$ whether for all the positives $x$ so that $ax^{2}+ bx+ c+ \sum_{k= 0}^{n}\frac{x^{k}}{k!}\geq 0,$ and something similar for the negatives $x.$ The first piece comes from the proof of the transcendence of $e,$ like_ https://mathoverflow.net/a/378064/164469, and the second is standard Tarski-Seidenberg quantifier elimination. But this algorithm does not provide a time bound for the algorithm or a single set of discriminants which can be used for all $a, b, c.$

Answer 1

It's one of my first answers, so my MathJax and my answer won't be perfect. I appreciate every help.

I am not sure if my answer is correct, because the first answer has many likes and an easier criterium. However, when graphing the function, it doesn't seem to work for me.

I've gone the same way. It is the question for which $a, b, c$ are the functions $f(x)=e^x$ and $g(x)=-ax^2 -bx -c$ intersect. $a<0$ is trivial, so we will assume $a>0.$

The function $g(x)$ is concave and the function $f(x)$ is convex, so we will look at the argument, where both functions have the same derivative. If $f=g$ at that point, there is exactly one intersection. If $f>g,$ there is no intersection and when $f<g$ there are two intersections.

Set $f'(k)=g'(k)$ to get the desired argument. If you solve $$e^k = -2ak-b$$ you get $$k=-W(\frac{e^{- \frac{b}{2a} }}{2a})- \frac {b}{2a}$$

There are zeroes, when $e^k \leq -ak^2 -bk-c.\,k$ is the root of the above, so we can substitute $e^k.$

We get : $$-2ak-b\leq -ak^2-bk-c$$

So the answer to your question is $:\quad h(x)=e^x+ax^2+bx+c$ has zeroes, iff $$ak^2+(b-2a)k-b+c\leq 0$$ or $$a<0$$ with $k$ being the term at the top. One zero with equality and two zeroes with strict inequality.

Answer 2

You basically ask when

$f(x)=e^x+c$ and $g(x)=-ax^2-bx=-x(ax+b)$ cross each other.

Always when $a < 0$.

For $a \geq 0$ it is as follows.

The reason we have chosen the above two functions is that it is easier to visualize what is going on. $g(x)$ has one of the legs through the origin and $e^x$ is elevated by $c$. For such arrangement, there is one critical value $c_0$ when $f(x)$ and $g(x)$ are touching each other. Then, if $c > c_0$, $f(x)$ cannot cross $g(x)$ or if $c < c_0$ they will have two intersections.

When $f(x)$ is touching $g(x)$ it is at that point $f'(x_0)=g'(x_0)$ giving us the condition:

$$e^{x_0}=-2 a x_0 - b$$

which we can effectively solve and find $x_0$. Another condition we are getting from replacing $e^{x_0}$ into $f(x_0)=g(x_0)$ which is a common point.

$$-2 a x_0 - b+c_0=-ax_0^2-bx_0$$

giving

$$c_0=-ax_0^2-bx_0+2 a x_0 + b$$

So the condition is

$$c < c_0=-ax_0^2 +(2 a-b) x_0 + b \text{ for } x_0 = -W(\frac{e^{\frac{-b}{2a}}}{2a}) - \frac{b}{2 a}$$

If $a = 0$ we have always the solution if $b>0$ and otherwise

$$e^{x_0}= - b, x_0=\ln(-b)$$

$$c_0=b-bx_0$$

or $$c < b(1-\ln(-b)), b<0$$

If $a = b = 0$ we have the solution only if $c < 0$.