# How do you solve ｜x-3｜=｜x^2-4x+3｜?

Oct 7, 2016

Case 1: Both absolute values are positive

$x - 3 = {x}^{2} - 4 x + 3$

$0 = {x}^{2} - 5 x + 6$

$0 = \left(x - 3\right) \left(x - 2\right)$

$x = 3 \mathmr{and} 2$

Case 2: The absolute value on left is negative

$- \left(x - 3\right) = {x}^{2} - 4 x + 3$

$- x + 3 = {x}^{2} - 4 x + 3$

$0 = {x}^{2} - 3 x$

$0 = x \left(x - 3\right)$

$x = 0 \mathmr{and} 3$

Case 3: The absolute value on right is negative

$x - 3 = - \left({x}^{2} - 4 x + 3\right)$

$x - 3 = - {x}^{2} + 4 x - 3$

${x}^{2} - 3 x = 0$

$x \left(x - 3\right) = 0$

$x = 0 \mathmr{and} 3$

Hence, the solution set is $\left\{0 , 2 , 3\right\}$.

Hopefully this helps!