For what k values are there more than one solution to the following equation? Suppose we have the equation, where k is some constant:

$0 = (x+k)e^{-(x+k)^2} + (x-k)e^{-(x-k)^2} + xe^{-x^2}$

A trivial solution exists, where x = 0.

How can I figure out what range of values for k enable additional solutions for x to be found in addition to the trivial x = 0 solution?



Context:

If $y =e^{-(x+k)^2} + e^{-(x-k)^2} + e^{-x^2}$

Then y is the sum of three normal distributions, with means 0, k and -k.  If k is small, y will have one maximum at x=0.  If y is larger, there will be 3 maximums.  Therefore if we set y' = 0, we should find what k values enable more than one solution, and thus find the value of k required for three peaks to be distinguishable.
The process to solve this equation is similar to the other one you posted:

$(x+k)e^{-x^2-2kx-k^2}+(x-k)e^{-x^2+2kx-k^2}+xe^{-x^2}=0$

Multyplying both sides for $e^{x^2+2kx+k^2}$ we get:

$(x+k)+(x-k)e^{4kx}+xe^{4kx+k^2}=0$

$x+k+e^{4kx}(x(1+e^{k^2})-k)=0$

Dividing first for $x+k$ and then for $1+e^{k^2}$ leads us to:

$e^{4kx}\frac {(x+ak)} {x+k}=a$ where $a=-\frac 1 {1+e^{k^2}}$

Then the solution(s) is/are (in terms of generalized Lambert's function):

$x=-ak+\frac 1 {4k}W_{-ae^{4ak^2}}(4kae^{4ak^2}(k-ak))$

Now if you want to know how many solutions there are you must check the value of $-a$

1) iff $-a\ge\frac 1 {e^2}$ you have only one solution;

2) iff $0\lt-a\lt\frac 1 {e^2}$ you have three solutions (one positive and two negatives);

3) iff $-a\lt0$ you have two solutions (one positive and the other negative).
Numerically it's pretty simple -- and if there's a computational tool available to you (specifically Matlab) I could advise on finding a solution. But numerically I'm not sure if a sufficiently efficient method exists as not to be a complete waste of your time.