# Homework Help: Solving |x-1|*|x+1|=0

1. Jan 17, 2012

### solar nebula

1. The problem statement, all variables and given/known data
find all the numbers x for which |x-1|*|x+1|=0

2. Relevant equations

3. The attempt at a solution
|x-1|*|x+1|=0

|x-1|=0 or |x+1|=0

x-1=0 or x+1=0

x=1 or x=-1

but I think there's a flaw some in my attempt solution...or the whole thing is a flaw..
any help will be much appreciated!!!
Thanks!

2. Jan 17, 2012

### micromass

It's completely correct!!

3. Jan 17, 2012

### solar nebula

But when I try to do the same thing for |x-1|*|x+2|=3 which is:

|x-1|*|x+2|=3

|x-1|=3 or |x+2|=3

x-1=3 or x+2=3

x=4 or x=1

But this solution does not make sense...

Last edited: Jan 17, 2012
4. Jan 17, 2012

### micromass

Of course not. You basically did "xy=3 thus x=3 or y=3". This is obviously false.

To solve that, first do the following:

$$|x-1||x+1|=3$$

thus

$$|(x-1)(x+1)|=3$$

So the absolute value of the number (x-1)(x+1) equal 3. Then what can the number (x-1)(x+1) equal?

BTW I moved this to precalculus, because it obviously does not belong in calculus and beyond. Please post in the correct forum next time.

5. Jan 17, 2012

### DrewD

In both cases you are following the procedure
$a*b=c \Rightarrow a=\frac{c}{b}$
which works as long as $b\neq0$. When you do it withe $c=0$ then $\frac{c}{b}=0$, but this is true for any $b$ (again, I'm assuming b is not zero). So you just skip right to $a=0$ or $b=0$.

micromass's explanation is probably more intuitive, but I like to think of things algorithmically.