The number of real solutions of the equation |x|2-3|x|+2=0 are

Given |x|2-|x|+2=0Here we consider two cases viz. x<0 and x>0Case I: x<0 This gives x2+3x+2=0⇒(x+2)(x+1)=0⇒x=-2, -1Also x=-1, -2 satisfy x<0, so x=-1, -2 is solution in this case.Case II: x>0. This gives x2-3x+2=0⇒(x-2)(x-1)=0⇒x=2, 1, so x=2, 1 is solutions are four i.e., x=-1, 1, 2, -2