fourier transform - Edge states of Kitaev chain

I am reading paper about Kitaev chain of electrons, which can exhibit famous Majorana fermions at ends of wire. The Hamiltonian (his Eq. (6)) reads

$H = \frac{i}{2} \sum_j - \mu c_{2j-1}c_{2j} +(w+|\Delta|)c_{2j}c_{2j+1} +(-w+|\Delta|)c_{2j-1}c_{2j+2}$

in terms of the Majorana operators $c_{2j-1}$ and $c_{2j}$ and constants $\mu, w, |\Delta|$ can be thought of as parameters controlling Fermi level, hopping and gap.

Kitaev shows there are solutions of Hamiltonian $H$ with zero-energy. He gives the operators associated to zero-energy solutions in ansatz form (his Eq. (14)):

$b' = \sum_j \left( \alpha_{+}' x_{+}^{j} + \alpha_{-}' x_{-}^{j}\right) c_{2j-1}$

$b'' = \sum_j \left( \alpha_{+}'' x_{+}^{-j} + \alpha_{-}'' x_{-}^{-j}\right) c_{2j}$

where all $\alpha$ are constants and $x_{\pm}$ are unknowns to be found.

Q1) How to find $x_{\pm}$?

Q2) How to show $x_{\pm} = \frac{-\mu \pm \sqrt{\mu^2 - 4 w^2 + 4 |\Delta|^2}}{2 (w+|\Delta|)}$?

I attempted to show it by computing $[b', H]=[b'', H]=0$, but it did not give correct answer.

Any help appreciated.