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Modulus problem

  1. Jul 19, 2005 #1
    ( l 2-x l ^2 ) - 2 l 2-x l = 15

    ((2-x)^2) = 15 + 2 l 2-x l ------1
    ((2-x)^2) = -15 - 2 l 2-x l------2

    Am I doing correctly?
     
    Last edited: Jul 19, 2005
  2. jcsd
  3. Jul 19, 2005 #2

    HallsofIvy

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    How did you get 2?

    I think that you intended to drop the absolute value signs.
    Yes, |2-x|2l= (2-x)2. Either |2-x|= (2- x) (if 2-x>= 0) or |2-x|= -(2-x)

    Change the absolute values to parentheses and you will correct.
     
  4. Jul 19, 2005 #3
    The correct answer should be -3 and 7
    From 1, I get the answer of -1,-3,5,7

    Some more, how to do this question actually?
     
  5. Jul 19, 2005 #4

    Gokul43201

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    You also made the mistake of flipping the sign of the constant term : you accidentally made 15 into -15.

    PS : It might make things easier to first make a substitution, 2-x = y
     
  6. Jul 19, 2005 #5

    Gokul43201

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    The way you solve this problem is by using the definition of the modulus function.

    [tex]f(x) = |x| = x~,~~if~x \geq 0 [/tex]
    [tex]f(x) = |x| = -x~,~~if~x < 0 [/tex]

    Apply this definition to your problem and convert the equation to a pair of equations over two domains, one in which 2-x is negative and the other, where 2-x is non-negative.
     
  7. Jul 19, 2005 #6
    The first equation is correct and it is wrong for the second one. So, what should be the second equation?
    Thanks.
     
  8. Jul 19, 2005 #7
    ( ( 2-x ) ^2 ) - 2 l 2-x l = 15-----1
    -( ( 2-x ) ^2 ) - 2 l 2-x l = 15----2

    Do you mean this?
     
  9. Jul 19, 2005 #8
    Divide the above moduli into two parts , one being x>2 and second being x<2 ,

    Solve for both seperately and then check final value of x you get after solving under x>2 and x<2 , if it satisfies for both categories of x.

    BJ
     
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