Quadratic equation with absolute value Prepping for the GMAT, I came across the following question:

What is the product of all solutions of:
$$x^2 - 4x + 6 = 3 - |x - 1|?$$


First, I set up two equations, ie:
$$x^2 - 4x + 6 = 3 - (x - 1),$$
and
$$x^2 - 4x + 6 = 3 - (-1) \times (x - 1).$$
These factor down to $3$ solutions: $1$, $2$ and $4$. And $8$ is correct solution in the back of the prep book.
However, when plugging $4$ back into the original equation, it reduces to $6 = 3$, so $4$ does not seem to be a solution. Also, when graphing both, they only intersect at $1$ and $2$.

What part of my process (and seemingly the practice books process) is wrong?

The solutions to
$$x^2-4x+6 = 3-(x-1)$$
are only valid when $x-1 \ge 0$, i.e. when $x \ge 1$. Likewise, the solutions to
$$x^2-4x+6 = 3+(x-1)$$
are only valid when $x-1 \le 0$, i.e. when $x \le 1$.
Solving the first equation gives $x=1,2$, both of which are valid. Solving the second gives $x=1,4$. Notice that in this latter case $4$ is not valid since $4 \nleq 1$, and so the only solutions to $x^2-4x+6=3-\left|x-1\right|$ are $x=1,2$. (The textbook is wrong!)
