How do you solve |x-3|=|x^2-4x+3|?

1 Answer
Oct 7, 2016

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!