Find Solutions for |x-1|*|x+1|=0 with Step-by-Step Guide

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To solve the equation |x-1|*|x+1|=0, the solutions are found by setting each absolute value expression to zero, resulting in x=1 and x=-1. The discussion then shifts to a more complex equation, |x-1|*|x+2|=3, where the initial approach is flawed. The correct method involves recognizing that |(x-1)(x+2)|=3, which leads to a different set of solutions. The importance of correctly applying algebraic principles, especially when dealing with absolute values, is emphasized throughout the discussion. Proper categorization of the problem in the appropriate mathematical forum is also highlighted.
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Homework Statement


find all the numbers x for which |x-1|*|x+1|=0

Homework Equations





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!
 
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It's completely correct!
 
micromass said:
It's completely correct!

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...
please help, Thanks!
 
Last edited:
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.
 
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.
 
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