abs(2*x*x)<1 inequation

Given the inequality:
$$\left|{x 2 x}\right| < 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\left|{x 2 x}\right| = 1$$
Solve:
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.

1.
$$x^{2} \geq 0$$
or
$$-\infty < x \wedge x < \infty$$
we get the equation
$$2 x^{2} - 1 = 0$$
after simplifying we get
$$2 x^{2} - 1 = 0$$
the solution in this interval:
$$x_{1} = - \frac{\sqrt{2}}{2}$$
$$x_{2} = \frac{\sqrt{2}}{2}$$

2.
$$x^{2} < 0$$
The inequality system has no solutions, see the next condition


$$x_{1} = - \frac{\sqrt{2}}{2}$$
$$x_{2} = \frac{\sqrt{2}}{2}$$
$$x_{1} = - \frac{\sqrt{2}}{2}$$
$$x_{2} = \frac{\sqrt{2}}{2}$$
This roots
$$x_{1} = - \frac{\sqrt{2}}{2}$$
$$x_{2} = \frac{\sqrt{2}}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{\sqrt{2}}{2} - \frac{1}{10}$$
=
$$- \frac{\sqrt{2}}{2} - \frac{1}{10}$$
substitute to the expression
$$\left|{x 2 x}\right| < 1$$
$$\left|{\left(- \frac{\sqrt{2}}{2} - \frac{1}{10}\right) 2 \left(- \frac{\sqrt{2}}{2} - \frac{1}{10}\right)}\right| < 1$$

            /       ___\    
/1     ___\ |1    \/ 2 |    
|- + \/ 2 |*|-- + -----| < 1
\5        / \10     2  /    
    

but

            /       ___\    
/1     ___\ |1    \/ 2 |    
|- + \/ 2 |*|-- + -----| > 1
\5        / \10     2  /    
    

Then
$$x < - \frac{\sqrt{2}}{2}$$
no execute
one of the solutions of our inequality is:
$$x > - \frac{\sqrt{2}}{2} \wedge x < \frac{\sqrt{2}}{2}$$

         _____  
        /     \  
-------ο-------ο-------
       x1      x2