Perturbation theory - Algebraic equations (Repeated roots) Obtain a three term approximation (for $E\to 0$) of the roots of the following equation:
$$x^2 + (4+E)x + (4-E) = 0$$
I understand what to do in basic cases but I've no idea what to do in this situation as I get a repeated root and obviously need two solutions. This is as far as I've got: 
Look for solution 
$$ x(E) = x_0 + Ex_1 + E^2 x_2 + \cdots$$
Substitute into equation:
$$\left(x_0 + Ex_1 + E^2 x_2\right)^2 + (4+E)x + (4-E) = 0$$
$$x_0^2 + 2E x_0 x_1 + E^2 (x_1^2 + 2x_0 x_2) + 4x_0 + 4Ex_1 + 4E^2 x_2 + 4Ex_0 + 4E^2 x_1 = E-4$$
$$ O(E^0) : X_0^2 + 4X_0 + 4 = 0 \implies X_0 = -2$$ 
(REPEATED ROOT)
Trying to continue on anyway:
$$ O(E) : 2x_0 x_1 + 4x_1 + 4x_0 = 1 \implies -4x_1 + 4x + 4x_0 = 1 \implies -7=0$$
which is clearly wrong.
Can someone please tell me the correct method for repeated roots? Any and all help is appreciated :)
 A: This is actually quit subtle. At leading order you have $x_0=-2,-2$, but it all goes wrong at $O(\epsilon)$, and you get an inconsistent equation (even if you fix your slight error in the expansion). This is a sign that the ansatz you chose is not right, and we should look for a different expansion. Let's consider $x=x_0+\epsilon^\alpha x_1$. Writing out the equation gives
$$ x_0^2+2\epsilon^\alpha x_0x_1+\epsilon^{2\alpha}x_1^2+4x_0+\epsilon x_0+4\epsilon^\alpha x_1+\epsilon^{\alpha+1}x_1+4-\epsilon=0. $$
Now, we must have $\epsilon>0$ by assumption (since the $x^2$ term has an $O(1)$ coefficient the perturbation is regular), so the largest terms are $O(1)$, giving
$$ x_0^2+4x_0+4=0 $$
so $x_0=-2,-2$. Now substituting back into the expansion something very interesting happens,
$$ \epsilon^{2\alpha}x_1^2+\epsilon x_0+\epsilon^{\alpha+1}x_1-\epsilon=0 $$
the $\epsilon^{\alpha}$ terms cancel out! Now do dominant balance on this equation, we can choose to set $\alpha=1/2$, $\alpha=1$ or $\alpha=-1$. If you look at those choices, the only one we can use is $\alpha=1/2$, which gives
$$ x_1^2+x_0-1=0. $$
If you solve this you get $x_1=\pm\sqrt 3$. So, the right expansion is $x=x_0+\epsilon^{1/2}x_1+\epsilon x_2+\epsilon^{3/2}x_3+\ldots$, and continuing on you get
$$ x=-2\pm\sqrt3\,\epsilon^{1/2}-\frac{\epsilon}{2}\pm\frac{\sqrt3}{24}\,\epsilon^{3/2}+O(\epsilon^2) $$
which does converge pretty well to the right roots.
If the $\epsilon^\alpha$ terms hadn't of cancelled out, your expansion would have worked (dominant balance would choose $\alpha=1$).
Actually, the exact solution is
$$ x=-2-\frac{\epsilon}{2}\pm\sqrt{\epsilon}\sqrt{1+\frac{\epsilon}{2}} $$
and $\sqrt{1+\frac{\epsilon}{2}}$ can be expressed as a Taylor series of integer powers of $\epsilon$, so you will only get the $\epsilon^0$ and $\epsilon^1$ term in the expansion of $x$, all the other powers will be $\epsilon^{k/2}$ for integers $k$.
