Ho do you approximate $4 \epsilon x + x^2 -1=0 $ using $x=x_0+x_1 \epsilon + x_2 \epsilon^2 $? Consider the quadratic $4 \epsilon x + x^2 -1=0 $ and suppose we want to approximate the solution to the equation using 
$$x=x_0+x_1 \epsilon + x_2 \epsilon^2 $$
Why does substituting the line directly above into the quadratic yield $x_0^2 -1=0$?
I know that after substituting into the quadratic the term independent of epsilon is $x_0^2 -1$ but why must it be zero? thanks
 A: The substitution gives
$$4\epsilon(x_0+x_1 \epsilon + x_2 \epsilon^2)+(x_0+x_1 \epsilon + x_2 \epsilon^2)^2-1=0$$
or, by increasing powers of $\epsilon$
$$x_0^2-1+(4x_0+2x_0x_1)\epsilon+(4x_1+x_1^2+2x_0x_2)\epsilon^2+(4x_2+2x_1x_2)\epsilon^3+x_2^2\epsilon^4=0.$$
To obtain a good approximation, you will cancel the terms of the lowest degree, as they give the largest contribution to the sum. As there are three parameters, you can cancel three coefficients.
$$\begin{cases}x_0^2-1=0,\\4x_0+2x_0x_1=0,\\4x_1+x_1^2+2x_0x_2=0.\end{cases}$$
A: NOTE: answer valid before the editing by the OP
By binomial expansion for $\epsilon<<1$
$$4 \epsilon x + x^2 -1=0\implies x=\frac{-4\epsilon\pm\sqrt{16\epsilon^2+4}}{2}=-2\epsilon\pm\sqrt{1+4\epsilon^2}\approx-2\epsilon\pm(1+2\epsilon^2)$$
then
$$x\approx\pm1-2\epsilon\pm2\epsilon^2$$
A: (Variables renamed because lazy.)
$x=u+vc+ wc^2
$
in
$  4cx + x^2 -1=0
$
gives,
assuming that
$c$ is small,
$\begin{array}\\
0
&=4cx + x^2 -1\\
&=4c(u+vc+ wc^2) + (u+vc+ wc^2)^2 -1\\
&=4cu+4vc^2+ 4wc^3 + u^2+2uvc+v^2c^2+2uwc^2+O(c^3) -1\\
&=u^2-1+c(4u+2uv)+c^2(4v+v^2+2uw)+O(c^3)\\
\end{array}
$
For this to hold,
$u^2 = 1$,
$4u+2uv = 0$,
and
$4v+v^2+2uw = 0$.
From the first,
$u = \pm 1$.
From the second,
$0
= 4u+2uv
=2u(2+v)
$.
Since
$u \ne 0$,
$v = -2$.
From the third,
$0
= -8+4\pm 2w
$
so
$w
=\mp 2
$.
Therefore
$x
=\pm 1 -2c\mp 2c^2
$.
