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Homework Help: Fixing domains with mods

  1. Jun 20, 2012 #1
    mod(x-6) > mod(x^2 - 5x + 9)

    Can anyone tell me about domain fixtures with mods using the above inequation?

    A real beginner, so have mercy, be elementry .:D
     
  2. jcsd
  3. Jun 20, 2012 #2

    Curious3141

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    The word "mods" can have different meanings in Maths. I think you're using it in the sense of "modulus", or absolute value here.

    So just to be clear, are you asking how to solve this inequality?

    [tex]|x-6| > |x^2 - 5x + 9|[/tex]

    You should start by making a careful sketch of both curves. Remember to consider the discriminant of the quadratic expression, its minimum point and the fact that [itex]|x-6|[/itex] is actually represented by two lines at right angles to each other.
     
  4. Jun 20, 2012 #3

    HallsofIvy

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    I am going to assume you are using "mod" to mean "absolute value" (unfortunately "mod" has many different meanings in different circumstances.)

    To find values of x for which [itex]|x- 6|> |x^2- 5x+ 9|[/itex] consider the possible signs for the quantities inside the absolute values. Solve x- 6> 0 and x^2- 5x+ 9> 0 to determine where those are positive or negative.

    Where x- 6 and [itex]x^2- 5x+ 9[/itex] are both positive, the absolute values can be taken off to give [itex]x- 6> x^2- 5x+ 9[/itex] or [itex]0> x^2- 5x+ 9[/itex]. Solve that inequality and take the values of x that satisfy it as well as [itex]x- 6> 0[/itex] and [itex]x^2- 5x+ 9> 0[/itex].

    Where x- 6 is positive and [itex]x^2- 5x+ 9[/itex] is negative, the absolute values can be taken off to give [itex]x- 6> -(x^2- 6x+ 9)[/itex] which is the same as [itex]x^2- 5x+ 3> 0[/itex].

    Where x- 6 is negative and [itex]x^2- 5x+ 9[/itex] is positive, the absolute values can be taken off to give [itex]-(x- 6)> x^2- 6x+ 9[/itex] which is the same as [itex]0>x^2- 5x+ 3> 0[/itex].

    Where both x- 6 and [itex]x^2- 5x+ 9[/itex] are negative, the absolute values can be taken off to give [itex]-(x- 6)> -(x^2- 6x+ 9)[/itex] which is the same as [itex]x^2- 7x+ 15> 0[/itex].
     
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