# 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?

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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!)

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Thanks Clive. Makes sense. Plotting the equation shows that as well: wolframalpha.com/input/… – jim_shook Aug 2 '12 at 1:38

You seem to be right: only intersections at 1 and 2. For 4 you get 16=0.

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Hint $\rm\,\ |x\!-\!1| = (x\!-\!1)(3\!-\!x).\:$ Thus either $\rm\: x=1\:$ or $\rm\: 3\!-\!x = |x\!-\!1|/(x\!-\!1) = sgn(x\!-\!1).\:$ Therefore $\rm\:x = 3-sgn(x\!-\!1)\:\Rightarrow\:x>1\:\Rightarrow\: x = 3\!-\!1 = 2.$

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May I suggest saying something to the effect of "The given equation is equivalent to $|x-1|=(x-1)(3-x)$" instead, so it doesn't look like you're stating an identity? I was confused for a moment. – Rahul Aug 2 '12 at 6:15