If |x^2 - x + 1| = 2x -1, then what is the sum of all possible values

There are 3 steps to solving equations involving ABSOLUTE VALUE:
1. Apply the rule that says: If |x| = k, then x = k or x = -k
2. Solve the resulting equations
3. Plug solutions into original equation to check for extraneous roots

So, we get: \(x^2-x+1=2x-1\) and \(x^2-x+1=-(2x-1)\)
Let's solve each equation separately

First equation: \(x^2-x+1=2x-1\)
Set equal to zero: \(x^2-3x+2=0\)
Factor: \((x-2)(x-1)=0\)
This gives us two possible solutions: \(x = 2\) and \(x = 1\)
Determine whether \(x = 2\) is an extraneous root by plugging it into the original equation to get: \(|2^2-2+1|=2(2)-1\). WORKS!
Determine whether \(x = 1\) is an extraneous root by plugging it into the original equation to get: \(|1^2-1+1|=2(1)-1\). WORKS!

Second equation: \(x^2-x+1=-(2x-1)\)
Simplify: \(x^2-x+1=-2x+1\)
Set equal to zero: \(x^2+x=0\)
Factor: \(x(x+1)=0\)
This gives us two possible solutions: \(x = 0\) and \(x = -1\)
Determine whether \(x = 0\) is an extraneous root by plugging it into the original equation to get: \(|0^2-0+1|=2(0)-1\). DOESN'T WORK
Determine whether \(x = -1\) is an extraneous root by plugging it into the original equation to get: \(|(-1)^2-(-1)+1|=2(-1)-1\). DOESN'T WORK

So, the only two solutions that work are \(x = 2\) and \(x = 1\)
The SUM \(= 2 + 1 = 3\)

Answer: D