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iit should be 4
x^2-3|x|+2=0
Case 1: when x > 0
x2 - 3x + 2 = 0
x2 - 2x - x + 2 = 0
x(x-2) -1(x-2) = 0
(x-1) (x-2) = 0
x = 1 and x = 2
Case 2: when x < 0
x2 + 3x + 2 = 0
x2 + 2x + x + 2 = 0
x(x+2) + 1(x+2) = 0
(x+1) (x+2) = 0
x = -1 and x = -2
From both cases x belongs to {-1, -2, 1, 2}
Hence it has four no. of solutions.
PLZ APPROVE!
Take two cases:
1.x>=0:
x2-3x+2
Soln: 2 solutions.(x=2,1)
2.x<0
x2+3x+2
Solution: 2 solutions(x=-2,-1)
Therefore 4 solutions.
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