Verifying a computation in a basic asymptotic expansion Just want to check that I'm not going crazy because I don't have someone to work with on this. For the perturbed quadratic
$$ x^2 - \epsilon x - 1 = 0 \hspace{5em} (*) $$
I want to insert the ansatz
$$ x = \epsilon^{\alpha_0}(x_0 + \epsilon^{\alpha_1}x_1 + \dots ) $$
where without loss $x_0≠0≠x_1$, $\alpha_0 \in \mathbb R$ and $0<\alpha_1<\alpha_2<\dots$ are to be determined: I will be happy with determining $\alpha_0$ only here. This PDF draft I found$^1$
 claims that on substituting the ansatz into the equation, I should see that (after correcting for the assumed typo of $x\leftrightarrow \epsilon$...?)
$$ 0 = 
\epsilon^{2\alpha_0}x_0^2 
+2\epsilon^{2\alpha_0+\alpha_1}x_0x_1 
+\epsilon^{2(\alpha_0+\alpha_1)}x_1^2 + \dots \\ - \epsilon^{\alpha_0}x_0 - \epsilon^{\alpha_0+\alpha_1}x_1  + \dots  \\ -1$$
where I have split the lines into what I assume are from each term of the quadratic $(*)$. However this is what I get if I substitute the ansatz into $x^2 - x - 1 = 0$. Should they have used $(*)$ instead, or am I misunderstanding the method? 
The explanation is also a little terse; assume the expansion is correct, and say $\alpha_0 > 0$. How should one conclude that $x_0=0$(in contradiction with assumptions)? Is it not enough to take a (formal) limit $\epsilon \to 0$ to see that this would imply $0=-1$?

$^1$ I know I should not be relying on a draft, but this should be a simple example. I have had a cursory glance at some text books but they seem to assume $\alpha_0=0$ without explanation or work immediately with more difficult examples. If any, good references are very welcome.
 A: First, the unperturbed quadratic $x^2 -1$ has two roots $\pm 1$. You choose to perturb one of these roots, say $+1$. Thus, the ansatz is
$$
x = 1 + \epsilon^{\alpha_1} x_1 + \ldots,
$$
i.e. $\alpha_0 = 0$ (this is obviously also true for the other root). Substitution into the perturbed quadratic yields
$$
0 = - \epsilon + 2 x_1 \epsilon^{\alpha_1} + x_1^2 \epsilon^{2 \alpha_1} - x_1 \epsilon^{1+\alpha_1} + \ldots
$$
It is clear that, since $2 \alpha_1 > \alpha_1$ and $\alpha_1 + 1 > \alpha_1$. the last two terms are of higher order than the second term. Therefore, we have to leading order
$$
0= - \epsilon + 2 x_1 \epsilon^{\alpha_1},
$$
which implies that $\alpha_1 = 1$ and $x_1 = \frac{1}{2}$. 
There are a couple of good resources out there if you want to learn more about perturbation theory. To name one, you could look at Chapter 9 in
F. Verhulst, Nonlinear Differential Equations and Dynamical Systems, Springer, 1996
or try
M. Holmes, Introduction to Perturbation Methods, Springer, 2013.
