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!

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