Rational Inequality...4

Discussion in 'Other Pre-University Math' started by nycmathguy, Oct 3, 2021.

  1. nycmathguy

    nycmathguy

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    Section 2.7
    Question 50

    20210930_232249.jpg

    (x^2 + x - 6)/x ≥ 0

    Immediately I see that x cannot be zero. To show this fact, we draw a hole on the real number line at x = 0.

    Factoring the numerator, I get (x - 2)(x + 3).
    Setting to zero, I get x = 2 and x = -3.

    See number line below.

    When x = -4, we get -3/2 ≥ 0. False statement.
    When x = -3, we get 0 ≥ 0. True statement. The hole is filled at x = -3.
    When x = -2, we get 2 ≥ 0. True statement.
    When x = 1, we get -4 ≥ 0. False statement.
    When x = 2, we get 0 ≥ 0. True statement. The hole is filled at x = 2.
    When x = 3, we get 2 ≥ 0.

    The following intervals satisfy the original inequality:

    [-3, 0) U (0, -2] U [2, infinity)

    20211003_095750.jpg
     
    nycmathguy, Oct 3, 2021
    #1
  2. nycmathguy

    MathLover1

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    perfect

    here is the graph:

    [​IMG]
     
    MathLover1, Oct 3, 2021
    #2
    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    I got it right? Really? Ha!
     
    nycmathguy, Oct 3, 2021
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