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Solving quadratic inequalities and absolute values

  1. Oct 17, 2008 #1
    1. The problem statement, all variables and given/known data

    lxl <2

    The question is asking to solve this

    2. Relevant equations

    3. The attempt at a solution

    Ive tried bringint the 2 over which leads me to l-x-4l over lx +2l < 0 but then the absolute value confuses the heck out of me on where to go from here. Do I break it up into x -4 >0 and x+2 < 0 to solve it?

    Also if there is some literature on the site to help me with this subject further I would appreciate it.

  2. jcsd
  3. Oct 17, 2008 #2
    How did you get [tex]\frac{|-x-4|}{|x+2|} < 0[/tex] from [tex]\frac{|x|}{|x+2|} < 2[/tex]? You have assumed that x and x+2 have the same sign. You must test cases for where the stuff inside the absolute values are positive or negative independently of each other.
    Last edited: Oct 17, 2008
  4. Oct 17, 2008 #3
    Not necessary to consider cases where either inner function is positive/negative - that's more complicated in this case.

    You need to recognize that both sides of the inequality |x|/|x+2| < 2 are positive for all real x.

    Might be more apparent if you rewrite |x|/|x+2| = |x/(x+2)| < 2

  5. Oct 17, 2008 #4
    That is so far out of my league its sick.

    But im still fairly confused here but bare with me, id like to piece this together.

    Ive been doing some research on it and I found this example that may apply.

    http://www.intmath.com/Inequalities/4_Inequalities-Absolute-Values.php in this link down near the bottom there is excerise 2 solve

    http://http://www.intmath.com/Inequalities/Image2820.gif" [Broken]

    The answer being

    http://http://www.intmath.com/Inequalities/Image2821.gif" [Broken]

    So using that example I expanded this question to

    -2x-4 < x < 2x+4

    Am I on the right track? From there can I test the cases to find the solution?
    Last edited by a moderator: May 3, 2017
  6. Oct 17, 2008 #5
    I don't think that's correct. You can't multiply both sides by x+2, reason being that x+2 may be negative (reversal of inequality signs). Doing so would cause you to lose one set of values (x < -4 if I'm not wrong)

    But you can still use that method like this:

    1. But remember that you're solving the intersection of the two inequalities in this case. Let me give you an example. Say, suppose we have:

    -2x + 3 < x < 2x + 3
    => -2x + 3 < x AND x < 2x + 3
    => x > 1 AND x > -3
    => Taking the intersection of {x > 1} and {x > -3}, required set = { x ϵ R : x > 1 }

    2. Also, take careful note that |f(x)| < 2 and |f(x)| > 2 are VASTLY different, and that you can't use this (shortcut) method when the other side of the inequality is not a real constant.

    It's a matter of preference. You can also solve the problem by a graphical method, the testing of cases for inner functions, and in this case, squaring and the one you've highlighted. I preferred squaring in this case because it was the fastest for me. But there are instances where you have no other choice but to test the inner functions (the 'sureproof' method, but the slowest too), e.g.:

    |x+2| < 2x - |x+1|

    because both sides are not necessarily positive.
  7. Oct 20, 2008 #6
    Would you be able to explain how you got this..

    x < -4

    For the equation x/x+2 <2
    I get x/x+2 - 2 < 0
    expanded we get x-2x+4 < 0
    then -x + 4 < 0
    -x < -4
    x > -4

    If you could let me know where we differ that would be awesome!

    Thank you!
  8. Oct 21, 2008 #7
    Your expansion is wrong.

    x/(x+2) - 2(x+2)/(x+2) < 0

    (x - 2x - 4)/(x+2) < 0

    (-x-4)/(x+2) < 0

    Also, it's wrong to remove the denominator just like that, and moreover, doing so loses you the asymptotic value of x (it is just coincidental that in this case your asymptote is in the "negative" region).
  9. Oct 21, 2008 #8
    "(x - 2x - 4)/(x+2) < 0"

    this would mean that x > 4 then. Cause the top half of that is -x < 4 which leads to x > 4 correct?

    and then x < -2 so..... we have x > 4, x < -2, and then x > -4/3?
  10. Oct 21, 2008 #9
    I don't know how you got your numbers, perhaps you need to review your basic inequalities involving polynomials before you attempt absolute-valued functions. You can use number lines to visualize the solution. I hope this helps:

    (-x-4)/(x+2) < 0
    (x+4)/(x+2) > 0

    ----o--------o-----> x

    {x < -4 or x > -2}

    [x/(x+2)] + 2 > 0
    (3x+4)/(x+2) > 0

    ----o--------o-----> x

    {x < -2 or x > -4/3}

    Intersection of the 2 sets (you might want to draw a 3rd number line if you can't see this) gives you required set = { x ϵ R : x < -4 OR x > -4/3 }

    (Edit; PS: sorry for the strange lines, had to use those because the encoding prevented me from using more than one spacing between characters)
    Last edited: Oct 21, 2008
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