Solving $|x^2 - 1| - 1 = 3x - 2$ without graphing The book I'm working though has just showed how to find the intersection points of functions containing moduli by graphing them, visually spotting the intersection points, and then solving algebraically. I've done the exercises and am happy with that. Now it's showing another method to solve without graph sketching. I've taken a picture of the text:

The first exercise is asking to solve the equation $|x^2 - 1| - 1 = 3x - 2$, so I start by adding 1 to both sides, squaring, and simplifying to get $x^4 - 11x^2 + 6x = 0$. I now have no idea what to do! Can someone give he any hints on how to approach this please?
 A: $|x^2-1|=3x-1$
Case 1: $|x|<1$
$1-x^2=3x-1$
$x^2+3x-2=0$
$x=\frac{-3\pm \sqrt{17}}{2}$
Case 2: $|x|\geq 1$
$x^2-1=3x-1$
$x(x-3)=0$
$x=0$ or $x=3$
Therefore, $x=\frac{\sqrt{17}-3}{2}$or $x=3$.
A: Answer based on did's comment. In this case you can use your quartic equation
$$\begin{equation*}
x^{4}-11x^{2}+6x=0
\end{equation*}$$
and factor it
$$\begin{eqnarray*}
0=x^{4}-11x^{2}+6x &=&x(x^{3}-11x+6)=x\left( x-3\right) \left(
x^{2}+3x-2\right)  \\
&=&x\left( x-3\right) (x-\frac{-3+\sqrt{17}}{2})(x-\frac{-3-\sqrt{17}}{2}).
\end{eqnarray*}$$
The first factor, $x$, is easy 
$$\begin{equation*}
x^{4}-11x^{2}+6x=x(x^{3}-11x+6).
\end{equation*}$$
As for the second, $x-3$, you need to find a root of $x^{3}-11x+6=0$. By the
rational root theorem you should try factors of the constant term $6$, i.e $
x=\pm 1,\pm 2,\pm 3,\pm 6$. From these $x=3$ is a root. By polynomial long
division you get the 2nd degree polynomial
$$\begin{equation*}
\frac{x^{3}-11x+6}{x-3}=x^{2}+3x-2,
\end{equation*}$$
whose roots are $\frac{-3\pm \sqrt{17}}{2}$. From the four possible roots
$$\begin{equation*}
x_{1}=0,\quad x_{2}=3,\quad x_{3}=\frac{-3+\sqrt{17}}{2},\quad x_{4}=\frac{-3-\sqrt{17}}{2},
\end{equation*}$$
you can check that only $x_{2}$ and $x_{3}$ are roots of the original
equation
$$\begin{equation*}
\left\vert x^{2}-1\right\vert -1=3x-2.
\end{equation*}$$
A: An easier, though perhaps lengthier method. Divide by cases:
$$(1)\;\;x^2-1\geq 0\Longleftrightarrow x\leq -1\,\,\vee\,\,x\geq 1:\Longrightarrow |x^2-1|=x^2-1\Longrightarrow$$
$$x^2-1-1=3x-2\Longrightarrow x^2-3x=0\Longleftrightarrow x=0\,\,\vee\,\,\color{red}{x=3}$$
$$(2)\;\;x^2-1<0\Longleftrightarrow -1<x<1:\Longrightarrow |x^2-1|=1-x^2\Longrightarrow$$
$$1-x^2-1=3x-2\Longrightarrow x^2+3x-2=0\Longrightarrow x_{1,2}=\frac{-3\pm\sqrt{17}}{2}\Longrightarrow \color{red}{x=\frac{-3+\sqrt{17}}{2}}$$
The solutions are the red ones (why did we ban the other apparent solutions?)
A: Let's go back for a moment. For an arbitrary example, notice that $|x| = 2 \Longleftrightarrow x = \pm 2$. Similarly, we have two cases:
THE FIRST CASE: $$\begin{aligned}x^2 - 1 &= 3x - 1 \\ x(x - 3)& = 0 \\ \Longleftrightarrow \boxed {x= 0 \ {\rm or} \ x = 3}\end{aligned}$$
*THE SECOND CASE:*$$\begin{aligned}x^2 - 1 &= -3x + 1 \\ \Longleftrightarrow \boxed{x = {-3 \pm \sqrt{17} \over 2}} \end{aligned}$$We have four solutions. 
Notice that two solutions will not be taken in the real answer. Guess!
