How to solve this equation for x? $0 = (x+k)e^{-(x+k)^2}+(x-k)e^{-(x-k)^2}$

How can I solve this equation:

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

To find x in terms of k?

The only analytical way to solve this is with the generalization of Lambert's Omega function:

Your equation is:

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

Multiplying both sides for $e^{x^2-2xk+k^2}$ we got:

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

Dividing both sides for $(x-k)$ we have:

$e^{-4kx}\frac {x+k} {x-k} =-1$

Whose solutions are (in terms of generalized Lambert's function):

$x=-k-\frac 1 {4k}W_{e^{-4k^2}}(-8k^2e^{-4k^2})$

I don't know if we can do any simplification here (notice that the pedix of the GLF is the same as the the term which multiplies $-8k^2$ in the argument of the function and that $-8k^2$ is double the exponent we just talked about). Hope this will help ;)

This answer can be considered an expansion of the @Renato Faraone answer; we start from point reached by Renato :

$$e^{-4kx}\frac {x+k} {x-k} =-1$$

we can rewrite:

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

Appliying Lagrange Inversion: $$x=k+\sum_{n=1}\frac{(-1)^{n}D^{n-1}\left(e^{-4nkx}(x+k)^n\right)|_{x=k}}{n!}$$ Remembering the Rodrigues formula of the Laguerre associated polynomials (http://en.wikipedia.org/wiki/Laguerre_polynomials):

$$L_n^{(\alpha)}(x) = {x^{-\alpha} e^x \over n!}{d^n \over dx^n} \left(e^{-x} x^{n+\alpha}\right)$$

we can write the general term of Lagrange expansion:

$$L_{n-1}^{(1)}(4nk(x+k)) = {e^{4knx} \over (x+k)(n-1)!}{d^{n-1} \over dx^{n-1}} \left(e^{-4knx} {(x+k)}^{n}\right)$$

So the Lagrange expansion becomes: $$x=k\left(1+2\sum_{n=1}^\infty\frac{(-1)^nL_{n-1}^{(1)}(8nk)e^{-4nk^2x}}{n}\right)$$

This is nothing else but the Lagrange expansion of the above seen generalization of Lambert W function.