# Zeros of real exponential polynominals

How can I solve an equation like $$x^a + bx + c = 0$$ ?

I figured that I can write this equation in a polynominal form as $$e^{wz} + b e^{z} + c$$ or more generic as $$\sum{v_i e^{w_i z}}$$ and that these are called exponential polynominals.

There are some papers about the zeros of these functions, but those are mostly concerned with problems where the exponent is imaginary and z is complex. In this case it seems, the exact positions of the zeros cannot be analytically calculated. (See "Polynomials" with non-integer exponents for a related question)

But I am only interested in the case of a real exponent ( $$z, w \in \mathbb{R}$$ ).

(The underlying problem: I want to get the parameters of the equation $$f(x) = a + b e^{c x}$$ defined by $$f(0) = y_0$$ , $$f(x_1) = y_1$$ , $$f(x_2) = y_2$$. I can solve this if $$x_1 = \frac{x_2}{2}$$, because then the problem becomes a simple quadratic equation. But if $$x_1 \neq \frac{x_2}{2}$$, then I get stuck with the problem at the top.)

There are no analytical solutions for the roots of such "polynomials", and usually the function is undefined for $$x<0$$.
Anyway, you can consider the intersections of the pseudo-parabola $$y=x^a$$ and the straight line $$y=-bx-c$$. If $$b>0$$, you must have $$c<0$$ for a real solution, which is unique.
Otherwise, you can easily obtain the position of the tangent of slope $$-b$$ to the curve, and from this check if the line lies below or above the tangent. This tells you the number of roots ($$0, 1$$ or $$2$$) and if there are two of them, it gives you a separation.