The number of real solutions of the equation ${\left|x\right|}^2-3\left|x\right|+2=0$  are:

Given ${\left|x\right|}^2-3\left|x\right|+2=0$

Here we consider two cases $viz,x<0$  and $x>0$ .

Case I: $x<0$ . This gives $x^2+3x+2=0$

$\Rightarrow$       $\left(x+2\right)\left(x+1\right)=0\Rightarrow x=-2,-1$

Also $x=-1,-2$  satisfy $x<0$ , so $x=-1,-2$  are solutions in this case.

Case II:  $x>0$ .this gives $x^2-3x+2=0$

$\Rightarrow \left(x-2\right)\left(x-1\right)=0\Rightarrow x=2,1,$  so $x=2,1$  are solutions in this case.

Hence, the number of solutions are four i.e., $x=-1,1,2,-2$