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

Given |x|2−3|x|+2=0 Here we consider two cases viz,x<0  and x>0 .Case I: x<0 . This gives x2+3x+2=0 ⇒       (x+2)(x+1)=0⇒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 x2−3x+2=0 ⇒(x−2)(x−1)=0⇒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