Solve :- $|x^2 - 1| < 3x + 3|x-2|$ I was trying to solve this inequality with two absolute values (see below), I don't know why I got the wrong answer, because this is the standard method for this type of exercises.
I've got four different systems of inequalities, but two (those in the middle) of them don't exist in real set of solutions.
The result must be $-\sqrt7<x<5$, but I got another result.
Before doing this one, I've done other similar exercises, therefore I think I've got an error in this exercise because of frivolous errors I didn't take into account, but let me know.
My attempt to solve it:
$$|x^2 - 1| < 3x + 3|x-2|$$
$$\rightarrow|x^2 - 1| \rightarrow (x \leq -1) /(x \geq 1)$$
$$\rightarrow |x-2| \rightarrow x \geq 2$$
If ($x < -1$), then I've got:-=
$$x^2 - 1 < 3x -3x + 6$$
$$\rightarrow -\sqrt7 < x < \sqrt7$$
If ($-1 < x < 2$), then both of them are negative and in this interval I've got no set of solutions.
If ($x>2$), then I got:-
$x^2 - 1 < 3x + 3x -6$
$\rightarrow 1<x<5$.
Now, I've put on the number line in these interval in order to check all the possible solutions. I got:- $$\rightarrow 2<x<\sqrt7$$
But it's not the correct result. In fact, it must be :- $$\rightarrow-\sqrt7<x<5$$
Can anyone help me?
 A: If I wasn't wrong, I think what you did was let $x^2 - 1 > 0$ and $x - 2 \geq 0$, then take the solution of those two inequalities as intervals to check the sign in the absolute and solve the original inequality, and that would be wrong.
Let look at the inequality again. At first we have:
$$|x^2-1| =
\begin{cases}
x^2 - 1  & \text{if $x^2-1 \geq 0 \implies x \in (-\infty, -1] \cup [1, +\infty)$} \\
1 - x^2 & \text{if $x^2-1 < 0 \implies x \in (-1, 1)$}
\end{cases}
$$
$$|x - 2| =
\begin{cases}
x - 2  & \text{if $x \geq 2$} \\
2 - x & \text{if $x < 2$}
\end{cases}
$$
Within that, we should consider $4$ different cases:

*

*Case $1 \,(x \leq -1) : $
$$\begin{align} 
& x^2 - 1 < 3x + 3(2-x) \\
\implies & x^2 < 7 \\
\implies & -\sqrt{7} < x < \sqrt{7}
\end{align}$$
With $x \leq -1 \implies -\sqrt{7} < x \leq -1 \, (1)$

*

*Case $2 \,(-1 < x < 1) : $
$$\begin{align} 
& 1 - x^2 < 3x + 3(2-x) \\
\implies & x^2 + 5 > 0 \,\, \text{(always true)}
\end{align}$$
Therefore $-1 < x < 1$ is a solution for case $2$ $(2)$

*

*Case $3 \,(1 \leq x < 2) : $
$$\begin{align} 
& x^2 - 1 < 3x + 3(2-x) \\
\implies & x^2 < 7 \\
\implies & -\sqrt{7} < x < \sqrt{7}
\end{align}$$
With $1 \leq x < 2 \implies 1 \leq x < 2 \, (3)$

*

*Case $4 \,(x \geq 2) : $
$$\begin{align} 
& x^2 - 1 < 3x + 3(x - 2) \\
\implies & x^2 - 6x + 5 < 0 \\
\implies & 1 < x < 5
\end{align}$$
With $x \geq 2 \implies 2 \leq x < 5 \, (4)$
From $(1), (2), (3), (4)$, you get the sotultion to the inequality:
$$-\sqrt{7} < x < 5$$
