When working with inequalities what do you do when its x squared > 16 and vice versa does the sign flip.

Question

When working with inequalities what do you do when its x squared > 16 and vice versa does the sign flip.

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Quinn 3 weeks 2021-11-08T16:19:05+00:00 2 Answers 0 views 0

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    0
    2021-11-08T16:20:33+00:00

    No it doesn’t flip. You split into two possible cases with two separate inequalities.

    x^2 > 16

    | x | > 4

    And since now you are working with an absolute value, you split into 2 possible cases.

    if x >= 0: then x > 4

    if x < 0: then -x > 4

    x < -4

    FOR THE FIRST:

    x ∈ (4, + ∞)

    FOR THE SECOND:

    x ∈ (-∞, -4)

    Their UNITY is: x ∈ (-∞, -4) U (4, +∞)

    And that’s how you go about solving these. Hope I helped! 🙂

    0
    2021-11-08T16:20:55+00:00

    Answer:

    x<-4\quad \mathrm{or}\quad \:x>4

    \{x\in \mathbb{R}|  \:\left(-\infty \:,\:-4\right)\cup \left(4,\:\infty \:\right)    \}

    Step-by-step explanation:

    x^2>16

    x<-\sqrt{16}\quad \mathrm{or}\quad \:x>\sqrt{16}

    x<-4\quad \mathrm{or}\quad \:x>4

    The interval notation will be:

    \{x\in \mathbb{R}|  \:\left(-\infty \:,\:-4\right)\cup \left(4,\:\infty \:\right)    \}

    The contrary,

    x^2<16

    is -4<x<4

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