How do you solve #｜x-3｜=｜x^2-4x+3｜#?

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Answer 1 · 2016-10-07T14:23:41

Case 1: Both absolute values are positive

#x - 3 = x^2 - 4x + 3#

#0 = x^2 - 5x +6#

#0 = (x - 3)(x - 2)#

#x = 3 and 2#

Case 2: The absolute value on left is negative

#-(x- 3) = x^2 - 4x + 3#

#-x + 3 = x^2 - 4x + 3#

#0 = x^2 - 3x#

#0 = x(x - 3)#

#x = 0 and 3#

Case 3: The absolute value on right is negative

#x - 3 = -(x^2 - 4x + 3)#

#x - 3 =-x^2 + 4x - 3#

#x^2 - 3x = 0#

#x(x - 3) = 0#

#x = 0 and 3#

Hence, the solution set is #{0, 2, 3}#.

Hopefully this helps!