Solving solution set inequalities problems I need some guidance with solving problems of the following nature:
$$|x^2-3x-1|<3$$
Instinctively, my way of solving it is splitting it into the following equations:
$$x^2-3x-1< 3$$          
$$x^2-3x-1> -3$$
Solving $x^2-3x-1< 3$
$$x^2-3x-4< 0$$ 
$$(x+1)\ (x-4)< 0$$
$$x < -1$$
$$x < 4$$
Solving $x^2-3x-1> -3$
$$x^2-3x-1> -3$$
$$x^2-3x+2> 0$$
$$(x-2)(x-1)> 0$$
$$x > 2$$
$$x > 1$$
So, here $x < -1$ is a valid solution and $x > 1$ is also valid.
In addition, $x < 4$ is valid and $ x > 2$ is also valid.
So we have $-1 > x > 1$ and $2<x<4$.
Firstly, is this possible? Does any $x$ have to satisfy $-1 > x > 1$ and $2<x<4$ simultaneously? Or are these the ranges of allowed inputs? Like, $x$ can be $-2$ despite $2<-2<4$ is not true. 
Secondly, WolframAlpha has the following numberline:

Which means I've not solved this correctly. So how do I solve these properly?
 A: $$(x+1)\ (x-4)< 0$$
$$x < -1$$
$$x < 4$$
Really? Did you try to plug in $x=-2$ to see if the inequality holds?

Remember, $a\cdot b<0$ if exactly one of the two numbers is negative. So, either $(x+1)$ is negative and $(x-4)$ is positive (which, since $x+1>x-4$, is impossible) or $x+1$ is positive and $x-4$ is negative.
You can also rewrite the inequality as
$$(x+1)\cdot(4-x)>0$$
and then use the fact that $a\cdot b>0$ if the sign of both numbers is equal.

You made the same mistake in the other inequality. 
$$(x-2)(x-1)> 0$$
$$x > 2$$
$$x > 1$$
This excludes a lot of options, for ecample, $x=-100$ is also a perfectly valid solution

Your main mistake is that you go from
$$f(x)\cdot g(x) > 0$$
to
$$f(x)>0\text{ and }g(x)>0$$
which is a pretty big mistake. Remember, the product of two negative numbers is also positive! So, once you get to $$f(x)g(x)>0$$, you have two options:


*

*$f(x)>0$ and $g(x)>0$

*$f(x)<0$ and $g(x)<0$.


similarly, if you get to $f(x)g(x)<0$, the two options are


*

*$f(x)>0$ and $g(x)<0$

*$f(x)<0$ and $g(x)>0$.

A: $$|x^2-3x-1|<3\Rightarrow -3<x^2-3x-1<3\\\Rightarrow -3<x^2-2(3/2)x+(3/2)^2-(3/2)^2-1<3\\\Rightarrow 1/4<(x-3/2)^2<25/4$$
Now you have two inequation:


*

*$1/4<(x-3/2)^2\Rightarrow x-3/2<-1/2\space \text{or}\space x-3/2>1/2\Rightarrow x<1\space \text{or}\space x>2$.

*$(x-3/2)^2<25/4\Rightarrow -5/2<x-3/2<5/2\Rightarrow x>-1\space \text{or} \space x<4$
Hence the final answer is $\{x\in\mathbb{R}: -1<x<1\space\text{and}\space 2<x<4\}\space\space\space\blacksquare$
A: It's $$(x^2-3x-1)^2<3^2$$ or
$$(x^2-3x+2)(x^2-3x-4)<0$$ or
$$(x+1)(x-1)(x-2)(x-4)<0,$$ which gives the answer:
$$(-1,1)\cup(2,4).$$
