# Absolute Values (a non-urgent but roundabout question)

• Sep 10th 2009, 02:02 AM
Felix
Absolute Values (a non-urgent but roundabout question)
I have a question which is apparently worth 4 marks (of 10!) which states that there is a possible 4 answers for the following:

Solve $|x|(x-1)=2$ to which I simply worked out by expanding the LHS to $2x-x=2$ and thus $x=2$

I have no idea what other way I can look at this, so what am I missing?
• Sep 10th 2009, 02:12 AM
red_dog
If $x\geq 0$ then $x(x-1)=2\Rightarrow x^2-x-2=0\Rightarrow x_1=-1, \ x_2=2$.

But we have to choose the positive root, so $x=2$

If $x<0$ then $-x(x-1)=2\Rightarrow -x^2+x-2=0$ which has no real roots because the discriminant is negative.