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Math Help - Solving Absolute value quadratic inequalities

  1. #1
    Senior Member sakonpure6's Avatar
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    Solving Absolute value quadratic inequalities

    Solve | x2 - 4| < 8
    I have not done any absolute value quadratic inequalities, here is my attempt. Please check my work, thank you!

    |x2 - 4| < 8
    |x2| +4 < 8
    |x2| - 4 < 0
    (|x2| -2 ) ( |x2| + 2) < 0

    Do we stop here? What do we do next?
    (Not sure about this...)
    So, a. -x2 - 2 < 0 and b. x2 -2 < 0 & c. - x2 +2 < 0 and d. x2 + 2 < 0
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    Re: Solving Absolute value quadratic inequalities

    Quote Originally Posted by sakonpure6 View Post
    Solve | x2 - 4| < 8
    I have not done any absolute value quadratic inequalities, here is my attempt. Please check my work, thank you!
    None of that is correct.
    \\|x^2-4|<8\\-8<x^2-4<8\\-4<x^2<12\\-\sqrt{12}<x<\sqrt{12}

    Have a look at this
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    Senior Member sakonpure6's Avatar
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    Re: Solving Absolute value quadratic inequalities

    how would you solve |x^2 - 4| > 8 ?
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    Re: Solving Absolute value quadratic inequalities

    Quote Originally Posted by sakonpure6 View Post
    how would you solve |x^2 - 4| > 8 ?
    Can you solve x^2>12\text{ or }x^2<-4~?
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    Re: Solving Absolute value quadratic inequalities

    Quote Originally Posted by sakonpure6 View Post
    how would you solve |x^2 - 4| > 8 ?
    Copy what Plato did, but change the < sign to a > sign.

    |x^2-4|>8

    (x^2-4)<-8 or 8<(x^2-4)

    x^2<-4 or 12 < x^2

    There are no real values of x with x^2<-4, so
    x< -\sqrt{12} or \sqrt{12}< x.
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  6. #6
    Senior Member sakonpure6's Avatar
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    Re: Solving Absolute value quadratic inequalities

    Oh okay thank you guys!
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