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Math Help - I have absolutely no idea how to solve this question? Please help!

  1. #1
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    I have absolutely no idea how to solve this question? Please help!

    Solve the nonlinear inequality: (
    x 1)(x+3)(x 7)2 0 Solve by marking zeros and undefined points on a number line and checking signs in the intervals created as in the examples in the narrative. Express your answer in interval notation.

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  2. #2
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    Quote Originally Posted by cwaddy View Post
    Solve the nonlinear inequality: (
    x 1)(x+3)(x 7)2 0 Solve by marking zeros and undefined points on a number line and checking signs in the intervals created as in the examples in the narrative. Express your answer in interval notation.

    A product is greater zero if the number of negative signs is even.

    Since (x-7)^2 > 0~\implies~x \in \mathbb{R} \setminus \{7\} you only have to examine the remaining 2 factors:

    \underbrace{x-1\geq 0\wedge x+3 \geq 0}_{\text{both factors positive}}~\vee~ \underbrace{x-1< 0\wedge x+3 < 0}_{\text{both factors negative}}

    After some simplification you'll get:

    x \in (-\infty,-3] \cup  [1,\infty)
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  3. #3
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    Thank you - can you help with this???

    Solve the inequality:
    x
    (x + 2)(x 5)(x + 7) 0 Solve by marking zeros and undefined points
    on a number line and checking signs in the intervals created as in the examples in the narrative. Express your answer in interval notation.

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  4. #4
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    Hello, cwaddy!

    Solve the inequality: . x(x+ 2)(x - 5) (x + 7)\:\leq\: 0

    Solve by marking zeros and undened points on a number line
    and checking signs in the intervals created.
    Express your answer in interval notation.
    The zeros of the inequality are: . -7,\:-2,\:0,\:5


    These four values divide the number line into five intervals.

    . . \begin{array}{cccccccccc}<br />
\underbrace{\;-----} & o & \underbrace{----- }& o & \underbrace{-----} & o & \underbrace{-----} & o & \underbrace{-----\;} \\<br />
& \text{-}7 && \text{-}2 && 0 && 5 \end{array}


    Now test a value from each interval:

    On (\text{-}\infty, \text{-}7), try x = \text{-}8:\;\;(\text{-}8)(\text{-}8+2)(\text{-}8-5)(\text{-}8+7) \:=\:(\text{-}8)(\text{-}6)(\text{-}13)(\text{-}1) \:=\:+182 . . . no

    On (\text{-}7,\text{-}2), try x = \text{-}3\!:;\;(\text{-}3)(\text{-}3+2)(\text{-}3-5)(\text{-}3+7) \:=\:(\text{-}3)(\text{-}1)(\text{-}8)(4) \:=\:-96 . . . yes

    On (\text{-}2,0), try x = \text{-}1\!:\;\;(\text{-}1)(\text{-}1+2)(\text{-}1-5)(\text{-}1+7) \:=\:(\text{-}1)(1)(\text{-}6)(6) \:=\:+36 . . . no

    On (0,5), try x = 1\!:\;\;(1)(1+7)(1-5)(1+7) \:=\:(1)(8)(\text{-}6)(6) \:=\:-288 . . . yes

    On (5,\infty), try x = 6\!:\;\;(6)(6+2)(6-5)(6+7) \:=\:(6)(8)(1)(13) \:=\:+624 . . . no


    The inequality is satisfied on: . (\text{-}7,\text{-}2) \,\cup\, (0,5)

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