Solving a quadratic equation with perturbation on the exponent. This is for analysis of chemical rate equations.   I have the equation $$ A(1-x)^2 = Bx^{2 + \epsilon} $$
which for the trivial case $\epsilon = 0 $, and ignoring the negative root, has the solution, $$ x = \frac{\sqrt A}{\sqrt A + \sqrt B}$$
I would like to vary $\epsilon$ from say -1 to 0, and have at least an approximate exporession for $x$ in the form
$$ x = \frac{A^\alpha}{A^\alpha + B^\beta}$$
if that is at all possible (variations for example including cross-terms are acceptable too).
With the first equation, I can plug in values for $A$, $B$, and $\epsilon$ and find a numerical solution for x, so I know it can exist, but I cannot come further than this.  I can prove the trivial solution is correct but cannot derive a method for a general solution involving the perturbation using this method.  
 A: Put $u = \sqrt{\dfrac{B}{A}}$, then we have:
$$1-x = u x^{1+\frac{\epsilon}{2}}= u x\exp\left[\frac{\epsilon}{2}\log(x)\right] = u x \left[1 + \frac{\epsilon}{2}\log(x) + \frac{\epsilon^2}{8}\log^2(x)+\cdots\right]$$
We then substitute the formal expansion:
$$x = x_0 + \epsilon x_1 + \epsilon^2 x_2 +\cdots$$
and expand the equation in powers of $\epsilon$ and equate equal powers of $\epsilon$. You then find the unperturbed solution:
$$x_0 = \frac{1}{1+u}$$
The coefficient of $\epsilon$ of the equation yields:
$$(1+u)x_1 + \frac{u}{2} x_0 \log(x_0) = 0$$
therefore:
$$x_1 = \frac{u\log(1+u)}{2(1+u)^2}$$
Extracting the coefficient of $\epsilon^2$ of the equation yields:
$$(1+u)x_2 + \frac{u}{2} x_1 \left[1+\log(x_0)\right] +\frac{u}{8}x_0\log^2(x_0)=0  $$
Here we've used the expansion:
$$\log(x) = \log(x_0 + \epsilon x_1 + \cdots) = \log(x_0) + \log(1+ \epsilon \frac{x_1}{x_0}+\cdots) = \log(x_0) + \epsilon \frac{x_1}{x_0}+\cdots$$
So, this way we get an expression for $x_2$ in terms of $u$ and it's easy to proceed in this way to find the higher order terms. A problem you may encounter when proceeding to higher and higher orders is that the perturbation series may not converge for the desired value of $\epsilon$. You can then resort to resummation methods.
