The number of real solutions of the equation |x|2−3|x|+2=0 are
4
Given |x|2−3|x|+2=0
Here we consider two cases viz.x < 0 and x > 0
Case 1: 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 is solution in this case
Case 2: x > 0. This gives x2−3x+2=0
⇒ (x-2)(x-1) = 0 ⇒ x =2,1. so, x=2,1 is solution in this case. Hence the number5 of solutions are four i.e. x = -1, -2, 1, 2
Aliter : |x|2−3|x|+2=0
⇒ (|x|-1)(|x|-2)=0
⇒ |x| = 1 and |x| = 2 ⇒ x = ±1, x = ±2.