Solving |x-1|*|x+1|=0

  • #1

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!
 

Answers and Replies

  • #2
22,129
3,297
It's completely correct!!
 
  • #3
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:
  • #4
22,129
3,297
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:

[tex]|x-1||x+1|=3[/tex]

thus

[tex]|(x-1)(x+1)|=3[/tex]

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
529
28
In both cases you are following the procedure
[itex]a*b=c \Rightarrow a=\frac{c}{b}[/itex]
which works as long as [itex]b\neq0[/itex]. When you do it withe [itex]c=0[/itex] then [itex]\frac{c}{b}=0[/itex], but this is true for any [itex]b[/itex] (again, I'm assuming b is not zero). So you just skip right to [itex]a=0[/itex] or [itex]b=0[/itex].

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

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