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Math Help - Nonlinear Inequality problem?

  1. #1
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    Nonlinear Inequality problem?

    Ok what am I missing here?

    x2 + 7x + 6 > 0 I have this, to solve I factor,
    (x+1)(x+6)>0 then
    x+1>0 Solve for zero
    -1 -1
    x>-1 Answer 1
    (-1,Infinity) Interval Notation.

    Here's where im confused....then
    x+6>0
    -6 -6
    x>-6? right? Answer 2. What am I forgetting cause apparently this is wrong and the
    (-6,Infinity) is wrong?

    How is this true answer for interval notation (-Infinity, -6) U (-1,Infinity)?
    If x>-6, All answers should be going right on number line left.
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  2. #2
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    Re: Nonlinear Inequality problem?

    how about -6< x<-1
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  3. #3
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    Re: Nonlinear Inequality problem?

    x^2 + 7x + 6 = (x + 1)(x+6) .

    This will be larger than 0 if: (x + 1) and (x+6) are both > 0,

    or (x+1) and (x+6) are both < 0.

    x+1 > 0 when x > -1. x+6 > 0 when x > -6. So as long as x > -1, (x+1)(x+6) > 0.

    x + 1 < 0 when x < -1. x + 6 < 0 when x < -6. So (x+1)(x+6) > 0 as long as x < -6.
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  4. #4
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    Re: Nonlinear Inequality problem?

    Quote Originally Posted by sage1987 View Post
    Ok what am I missing here?

    x2 + 7x + 6 > 0 I have this, to solve I factor,
    (x+1)(x+6)>0 then
    a simple way ... think about the graph of the resulting parabola and where the graph is above the x-axis.
    Attached Thumbnails Attached Thumbnails Nonlinear Inequality problem?-parabola.jpg  
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  5. #5
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    Re: Nonlinear Inequality problem?

    Hello, sage1987!

    Ok, what am I missing here?

    I have this to solve: . x^2 + 7x + 6 \:>\: 0

    I factor: . (x+1)(x+6)\:>\:0

    We have: . (x + 1)(x+ 6) \:>\:0
    . . The product of two numbers is positive.

    There are two ways that this can happen.

    (1) Both factors are positive: . \begin{Bmatrix}x+1 \:>\: 0 & \Rightarrow & x \:>\: \text{-}1 \\ x+6 \:>\: 0 & \Rightarrow & x \:>\: \text{-}6 \end{Bmatrix}\;\;\Rightarrow\;\;x \:>\:-1

    (2) Both factors are negative: . \begin{Bmatrix}x+1\:<\:0 & \Rightarrow & x \:<\:\text{-}1 \\ x+6 \:<\:0 & \Rightarrow & x \:<\:\text{-}6 \end{Bmatrix} \;\;\Rightarrow\;\;x \:<\<-6


    Therefore: . (\text{-}\infty, \text{-}6) \cup (\text{-}1, \infty)

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