Your analysis is correct.
Let's look at the original equation
$${\epsilon ^2}{x^4} - \epsilon {x^3} - 2{x^2} + 2 = 0$$
and plot the Newton-Kruskal diagram:
There are only two possible placements of straight lines passing through two or more points with all remaining points "above the line"; those two lines correspond exactly to
- $\delta^2 \sim 2 \Rightarrow \delta \sim 1$, and
- $\epsilon^2\delta^4 \sim \epsilon \delta^3 \sim \delta^2 \Rightarrow \delta \sim \epsilon^{-1}.$
Before we continue, let's rescale the original equation by making the change of variables $x=\delta y$ to arrive at
$${\epsilon ^2}{\delta ^4}{y^4} - \epsilon {\delta ^3}{y^3} - 2{\delta ^2}{y^2} + 2 = 0.$$
From the first balance, $\delta^2 \sim 2 \Rightarrow \delta \sim 1$, we immediately recover the original equation:
$$\tag{1} {\epsilon ^2}{y^4} - \epsilon {y^3} - 2{y^2} + 2 = 0.$$
Letting $\epsilon \rightarrow 0$, we get
$$2{y^2} + 2 = 0 \Rightarrow y = \pm 1$$
Let´s check if there are roots close to $y=-1$ and $y=1$.
Plugging an asymptotical expansion of the form $y \sim \pm 1 + {a_1}\epsilon + {a_2}{\epsilon ^2} + {a_3}{\epsilon ^3} + {a_4}{\epsilon ^4} + \ldots $ into (1), and then matching coefficients, we arrive at
$${x_a} \sim {y_a} \sim - 1 - \frac{\epsilon }{4} - \frac{{13}}{{32}}{\epsilon ^2} + O\left( {{\epsilon ^3}} \right),$$
$${x_b} \sim {y_b} \sim 1 - \frac{\epsilon }{4} + \frac{{13}}{{32}}{\epsilon ^2} + O\left( {{\epsilon ^3}} \right).$$
Regarding the second possibility of balancing, $\epsilon^2\delta^4 \sim \epsilon \delta^3 \sim \delta^2 \Rightarrow \delta \sim \epsilon^{-1}$, we get the following equation
$$\tag{2}{y^4} - {y^3} - 2{y^2} + 2{\epsilon ^2} = 0$$
Letting $\epsilon \rightarrow 0$, we get
$${y^4} - {y^3} - 2{y^2} = 0 \Rightarrow {y^2}\left( {{y^2} - y - 2} \right) = 0 \Rightarrow y = 0 \vee y = - 1 \vee y = 2.$$
Let´s check if there are roots close to $y=-1$ and $y=2$ ($y=0$ is of no interest, as we just recover one of the roots previously found). Plugging an asymptotical expansion of the form $y \sim {a_0} + {a_1}\epsilon + {a_2}{\epsilon ^2} + {a_3}{\epsilon ^3} + {a_4}{\epsilon ^4} + \ldots $ (with ${a_0} = - 1 \vee {a_0} = 2$) into (2), and then matching coefficients, we arrive at
$${x_c} \sim {\epsilon ^{ - 1}}{y_c} \sim - {\epsilon ^{ - 1}} + \frac{2}{3}\epsilon + O\left( {{\epsilon ^3}} \right),$$
$${x_d} \sim {\epsilon ^{ - 1}}{y_d} \sim 2{\epsilon ^{ - 1}} - \frac{1}{6}\epsilon + O\left( {{\epsilon ^3}} \right).$$
Thus, we've determined all the four roots of the original equation.
-EDIT-
Here's the Mathematica code I used to do my calculations, more specifically, the last one.
n = 4;
a[0] = 2;
x = a[0] + Sum[a[i] \[Epsilon]^i, {i, 1, n}] + O[\[Epsilon]]^(n + 1);
x^4 - x^3 - 2 x^2 + 2 \[Epsilon]^2 == 0;
LogicalExpand[%]
Solve[%]
You just need to shift by $x_0$ first. If you take $x = 1 + X \epsilon^\alpha$, the two lowest powers of $\epsilon$ will be $1$ and $2 \alpha$. Equating the lowest powers gives $\alpha$, and equating the coefficients gives $X$.
Alternatively, if $x = 1 + X$, the equation becomes
$$X^3 - \epsilon X^2 + 3 X^2 - 2 \epsilon X - \epsilon = 0.$$
The side of the Newton polygon which is closest to the origin goes through the vertices corresponding to $X^2$ and $\epsilon$, which gives $X = \pm \sqrt {\epsilon/3}$.
Best Answer
The dominant balance argument goes like this:
Let $x=\epsilon^\alpha x_0+\epsilon^\beta x_1+\ldots$ with $\alpha<\beta<\ldots$. We start by working out $\alpha$, so substitute $x=\epsilon^\alpha x_0$ (we can ignore the $\epsilon^\beta$ and smaller terms because they must be small than the $\epsilon$^\alpha$ terms).
$$\epsilon^{1+3\alpha}x_0^3-\epsilon^{2\alpha}x_0^2+2\epsilon^\alpha x_0-1=0$$
Dominant balance gives either
That's all the combinations, so there are two possible balances, $\alpha=-1$ and $\alpha=0$.
With $\alpha=0$ you get the regular expansion, $x=x_0+\epsilon^\beta x_1+\epsilon^\gamma x_2+\ldots$. The $O(1)$ equation is $$x_0^2+2x_0-1=0$$ with solutions $x_0=1,1$. Now you do the same thing again with $x=1+\epsilon^\beta x_1$, (with $\beta>0$), $$ \epsilon\left(1+3\epsilon^\beta x_1+3\epsilon^{2\beta}x_1^2+\epsilon^{3\beta}x_1^3\right)-\left(1+2\epsilon^\beta x_1+\epsilon^{2\beta}x_1^2\right)+2\left(1+\epsilon^\beta x_1\right)-1=0 $$ which simplifies to $$ \epsilon-\epsilon^{2\beta}x_1^2=0 $$ and so $\beta=1/2$ and $x_1^2=1$ so $x_1=\pm1$.
You can (usually) assume the pattern continues now, and let $x=1\pm\sqrt\epsilon+\epsilon x_2+\epsilon^{3/2}x_3+\ldots$.
For the singular root, you have $x=\epsilon^{-1}x_0+\epsilon^\beta x_1+\ldots$ with $\beta>-1$. With this, you get $x_0=0,0,1$ (the two zeros correspond to the regular roots we found before). For the actual singular root, you find $\beta=0$ from dominant balance and the $O(\epsilon^{-1})$ equation is $$3x_1-2x_1+2=0$$ so $x_1=-2$. Continuing on, you get $x=\epsilon^{-1}-2-3\epsilon+\ldots$
Putting it all together, we get that the roots are $$x=1+\sqrt\epsilon+O(\epsilon),\quad1-\sqrt\epsilon+O(\epsilon),\quad\frac{1}{\epsilon}-2-3\epsilon+O(\epsilon^2).$$