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Absolute Value inequalities

  1. Sep 26, 2012 #1
    I was hoping someone could give a little more insight, or perhaps enlighten me to a better way of approaching solving these seemingly simple Algebra 2 inequalities.

    I did some google searching but I was not able to find the answers I seek.

    The problem came up when a friend of mine had an equation that looked like this:

    |x-3| > -2 + |2x + 8|

    Question 1: What is the generally accepted best way to solve this inequality.

    My method of solving involved setting up 4 equalities:

    1) x-3 = -2 + 2x + 8
    2) -(x-3) = -2 + 2x + 8
    3) x-3 = -2 - (2x + 8)
    4) -(x-3) = -2 - (2x + 8)

    These yielded the following values for x:

    1) x= -9
    2) x = -1
    3) x = -7/3
    4) x = -13

    With each of these x values I then went back and plugged into the original inequality. My goal was to check for equality.

    Only 2) and 4) resulted in equality so I threw the other answers out.

    Once I had these two solutions I tested the points 0, -5, and -20 in the original inequality. And since -5 was the only value that yielded a true statement, I declared the solution:

    (technically I had checked -7/3 and -9, checking -5 was unnecessary I realize now)

    -13 < x < -1

    Question 2: Can someone give me more information on why there are extraneous solutions?

    As I understand things, when you have:

    |x| = A

    You need to solve x = A and x = -A

    When you have:

    |x| = |y|

    You need to solve x = y, x = -y, -x = y and -x = -y, but because x = -y, -x = y are the same equation and x = y, -x = -y are the same equation you only need to solve two equations.

    When you have:

    |x| = A + |y|

    You need to solve x = A + y, x = A - y, -x = A + y, and -x = A - y, since these are all different, you need to solve all four. Essentially the prior case (|x| = |y|) is a special case of this, just two of the equations you need to solve happen to be the same.

    Question 3: Is there a more rigorous way of handling these equations? Should I be using |x| = Sqrt(x^2) here?

    Thank you in advance.
     
    Last edited: Sep 26, 2012
  2. jcsd
  3. Sep 26, 2012 #2

    haruspex

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    Your method does not scale well. For n abs value terms you would have 2n cases. Better to list out the critical x values (-4 and 3 in your first example), then consider the n+1 ranges these generate: <-4, -4 to 3, >3.
     
  4. Sep 26, 2012 #3
    Thank you very much for your response.

    I do not fully understand.

    I understand that my method will result in 2n cases. That is clear to me.

    I understand how to find what you call critical x values for each set of absolute value quantities. This is also clear.

    I do not understand what it means to "consider" these ranges. How would I go about solving once I have the ranges: <-4, -4 to 3, >3?

    Thanks.
     
  5. Sep 26, 2012 #4

    haruspex

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    Within each of those ranges, every abs value term has a known simplification. For x<-4, |x-3|=3-x and |2x+8|=-2x-8. For -4<x<3, |x-3|=3-x and |2x+8|=2x+8, etc.
     
  6. Sep 26, 2012 #5
    Thank you.

    Please allow me to continue for my own benefit so that I may understand completely.

    For x < -4 we have |x-3|=3-x and |2x+8|=-2x-8

    For -4 < x < 3 we have |x-3|=3-x and |2x+8|= 2x+8

    For x > 3 we have |x-3|= x - 3 and |2x+8|= 2x+8

    Thus I must solve 3 equations, with constraints, as you predicted n + 1 = 3 in this case.


    For x < -4 since |x-3|=3-x and |2x+8|=-2x-8
    We must solve 3-x = -2 -2x-8. This yields x = -13 which is a valid answer since we require x < -4.

    For -4 < x < 3 since|x-3|=3-x and |2x+8|= 2x+8
    We must solve 3-x = -2 + 2x+8

    This yields x = -1 which is a valid answer since we require -4 < x < 3.

    For x > 3 since |x-3|= x - 3 and |2x+8|= 2x+8
    We must solve x - 3 = -2 + 2x + 8

    This yields x = -9. Which is NOT valid because we require x > 3.

    Then by solving just 3 equations I can arrive at the critical values -1 and -13. I can then do my tests between -1 and -13 and beyond -1 and -13 to determine where the original inequality is true and where it is false.

    Is this correct?

    Thank you.
     
  7. Sep 26, 2012 #6

    haruspex

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    No, for this range (actually, x<=-4) you have 3-x > -2 + (-2x-8), i.e x>-13. So (-13,-4] is an entire interval of valid solutions. Proceed similarly with the other two cases.
     
  8. Sep 28, 2012 #7
    Ah yes. It took me a while to understand what you are saying here. Very clever.

    So for
    For -4 <= x <= 3 we have 3-x > -2 + 2x+8 ie x < -1, So [-4, -1) is valid.

    And for x>= 3 we have x - 3 > -2 + 2x + 8 ie x < -9 So no solutions added in this case.

    Thus (-13, -1) is our solution set. This way I don't have to do all that business with testing points in between....

    Thank you haruspex for your help.
     
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