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

  1. Jan 17, 2012 #1
    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. jcsd
  3. Jan 17, 2012 #2

    micromass

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    It's completely correct!!
     
  4. Jan 17, 2012 #3
    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: Jan 17, 2012
  5. Jan 17, 2012 #4

    micromass

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    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.
     
  6. Jan 17, 2012 #5
    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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