Quadratic 6

$x^2=|x|^2$

$|x|=p$

$p^2-3p+2=0\\\Rightarrow (p-2)(p-1)=0\\\Rightarrow p=1,2$

$|x|=1,2\Rightarrow x=\pm 1,\pm 2$

First, this problem is too overrated :D

Second, we divide this equation into 2 cases:

Case 1: If $x \geq 0\Leftrightarrow \left| x \right| =x \Leftrightarrow { x }^{ 2 }-3x+2=0 \Leftrightarrow x\in \{ 1;2\}$ , which gives us $2$ solutions of $x$

Case 2: If $x<0\Leftrightarrow \left| x \right| =-x \Leftrightarrow { x }^{ 2 }+3x+2=0 \Leftrightarrow x\in \{ -1;-2\}$ , which gives us another $2$ solutions of $x$

In total, there are $2+2=\boxed{4}$ solutions of $x$ satisfy the equations.

Number of reals solutions will be 4 because it is mod x

So there will be two equations

so number of real solutions comes out to be 4

Notice that the function is even: we can restrict our study to $x^2-3x+2=(x-2)\cdot(x-1)$ which has two positive roots. By simmetry (with respect to the y axis), we can conclude that the number of solutions of the starting equation is therefore $2 \cdot2=4$ .