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Solve for modulus inequality.

  1. Aug 27, 2011 #1
    1. The problem statement, all variables and given/known data
    Solve [tex]\frac{|x^2-5x+4|}{|x^2-4|}\le1[/tex]


    2. Relevant equations



    3. The attempt at a solution
    as

    [tex]|x^2-4|[/tex]will be positive always

    cross multiply and take 1 to other side of equation
    solve by taking LCM
    we get
    [tex]|x^2-5x+4|-(x^2-4)\le0[/tex]
    on solving we get

    [tex](x^2-5x+4)-(x^2-4)\le0[/tex] and [tex]-(x^2-5x+4)-(x^2-4)\le0[/tex]

    the other method I know is to square to remove the modulus function
    [tex](x^2-5x+4)^2-(x^2-4)^2\le0[/tex]


    among these which method is correct?
    the second method becomes equation of degree 4 i.e.[tex]x^4....[/tex]


    please provide hints.
     
  2. jcsd
  3. Aug 28, 2011 #2

    rock.freak667

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    Homework Helper


    Actually for x=1, 'x2-4' is negative, but you can use the fact that |a|/|b| = |a/b| iff b≠0.

    Just use the fact that |X|<A ⇒ -A<X<A and then just take each inequality separately and take the union of the sets.
     
  4. Aug 28, 2011 #3

    ehild

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    Homework Helper
    Gold Member

    Do not omit the modulus of x^2-4. Your equation has to be: [tex]|x^2-5x+4|-|x^2-4|\le0[/tex]
    The other method (squaring both the numerator and the denominator) is OK.

    ehild
     
  5. Aug 28, 2011 #4
    Thank you very much i got the answer:)
    is it easy to put random values before, between and after the zeros to check the sign
    or to make the sign table(attached)??
     

    Attached Files:

  6. Aug 28, 2011 #5

    HallsofIvy

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    How did [itex]|x^2- 4|[/itex] suddenly become [itex]x^2- 4[/itex]?

     
  7. Aug 29, 2011 #6
    as it is in modulus it will be positive for any real value of x

    if i am wrong please explain me?
     
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