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Absolute value inequality

  1. Feb 9, 2008 #1

    abs(x^2 - 4) < 1

    implies that:
    x^2 - 4 < 1
    4 - x^2 < 1

    solving first equation for x gives:
    -sqrt(5) < x < sqrt(5)

    solving second equation for x gives:
    -sqrt(3) < x < sqrt(3)

    Now, my question is, what does that mean??
    How do I give the solution set, without a graphing utility and, if possible, without trial and error?

    Thank you
  2. jcsd
  3. Feb 9, 2008 #2


    User Avatar
    Homework Helper

    Draw the inequalities on two separate number lines. Then just find the intersection of the two(the region which is common to both).
  4. Feb 9, 2008 #3


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    Science Advisor

    NO! if 4- x^2< 1, then 3< x^2 so either x> sqrt{3} or x< -sqrt{3}.

    In order that |x^2- 4|< 1, you must have both x< -sqrt(3) or x> sqrt(3) and -sqrt(5)< x< sqrt(5). As rockfreak667 said, that is the intersection of the two sets:
    -sqrt(5)< x< -sqrt(3) or sqrt(3)< x< sqrt(5).

    I would prefer to do this problem in quite a different way: look at the corresponding equation |x^2- 4|= 1. Then either x^2- 4= 1 which leads to x^2= 5 and so x= +/- sqrt(5) or x^2- 4= -1 which leads to x^2= 3 and so x= +/- sqrt(3). But since the function is continuous, the only places it can change from "less than 1" to "greater than 1" is at one of those 4 points where it is "equal to 1".

    Those four points divide the line into 5 parts. Check one value of x< -sqrt(5), say 3: If x= 3, |3^2-4|= 5> 1. Check one value of x between -sqrt(5) and -sqrt(3): x= 2. If x= 2 |4- 4|= 0< 1. Check one point between -sqrt(3) and sqrt(3): x= 0. If x= 0 |0- 4|= 4> 1. Check one point between sqrt(3) and sqrt(5): 2. if x= 2, |4- 4|= 0< 1. Finally, check one point larger than sqrt(5): 3. If x= 3, |9- 4|= 5> 1. |x^2- 4|< 1 is satisfied for -sqrt(5)< x< -sqrt(3) and for sqr(3)< x< sqrt(5).
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