= Solution
A <regular perturbation> admits an <asymptotic expansion> about the limiting problem on the relevant scale without losing its solutions or conditions. A <singular perturbation> changes the limiting structure, so some solutions require a different scale: a differential equation may lose an order, or a <polynomial> may lose its degree.
The two finite roots approach $s=\pm\sqrt2$. Set $x=s+\epsilon a+O(\epsilon^2)$. Substitution into the <polynomial> gives $s^2-2=0$ at leading order and $s^3+2sa=0$ at the next order. Since $s^2=2$, $a=-1$ for either sign. Each branch is a <regular perturbation of a simple algebraic root>, with
$$
\boxed{x_+=\sqrt2-\epsilon+O(\epsilon^2),\qquad x_-=-\sqrt2-\epsilon+O(\epsilon^2).}
$$
The limiting quadratic has only two roots, whereas the original equation has three. The missing root is a <divergent algebraic root in a singular perturbation>. The <dominant balance for algebraic roots> sets $x=\epsilon^{-1}X$, giving $X^3+X^2=2\epsilon^2$. Its nonzero leading root is $X=-1$. Put $X=-1+a\epsilon^2+O(\epsilon^4)$; the left side is $a\epsilon^2+O(\epsilon^4)$, so $a=2$. Hence the second nonzero term occurs at order $\epsilon$, not order one:
$$
\boxed{x_3=-\epsilon^{-1}+2\epsilon+O(\epsilon^3).}
$$
For small nonzero real $\epsilon$ these three branches are real and distinct; the divergent branch changes sign with $\epsilon$. The two bounded branches are regular even though the full root problem is singular.
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