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Math Help - solve the inequality

  1. #1
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    solve the inequality

    ok im not sure about this problem

    i am supposed to solve the inequality

    (x-6)^{2}(x+8)<0 the zero's are 6 and -8

    so the answer would be (\infty ,-8)?
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  2. #2
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    Re: solve the inequality

    Quote Originally Posted by slapmaxwell1 View Post
    ok im not sure about this problem

    i am supposed to solve the inequality

    (x-6)^{2}(x+8)<0 the zero's are 6 and -8

    so the answer would be (\infty ,-8)?
    use the "tex" tags ... and to answer your question, yes.
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  3. #3
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    Re: solve the inequality

    Quote Originally Posted by slapmaxwell1 View Post
    ok im not sure about this problem
    i am supposed to solve the inequality
    (x-6)^{2}(x+8)<0 the zero's are 6 and -8
    so the answer would be (\infty ,-8)?
    Use the LaTeX tags [TEX]...[/TEX]
    DO NOT USE [tex]...[/tex]
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  4. #4
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    Re: solve the inequality

    ok thanks....
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  5. #5
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    Re: solve the inequality

    Quote Originally Posted by slapmaxwell1 View Post
    ok im not sure about this problem

    i am supposed to solve the inequality

    (x-6)^{2}(x+8)<0 the zero's are 6 and -8

    so the answer would be (\infty ,-8)?
    First of all, you should write your solution regions from smallest to largest.

    Anyway, to solve \displaystyle \begin{align*} (x - 6)^2(x + 8) < 0 \end{align*} you are correct that you should check the roots of the equation \displaystyle \begin{align*} (x - 6)^2(x + 8) = 0 \end{align*} , which are \displaystyle \begin{align*} -8, 6 \end{align*} .

    The function will only ever stop being negative or positive at the roots. So we need to check each region that are separated by the roots.

    When \displaystyle \begin{align*} x = -9 , (x - 6)^2(x + 8) = -225 < 0 \end{align*}

    So \displaystyle \begin{align*} x \in (-\infty, -8) \end{align*} is acceptable.

    When \displaystyle \begin{align*} x = 0, (x - 6)^2(x + 8) = 288 > 0 \end{align*}

    So \displaystyle \begin{align*} x \in (-8, 6) \end{align*} is not acceptable.

    When \displaystyle \begin{align*} x = 7, (x - 6)^2(x + 8) = 15 > 0 \end{align*}

    So \displaystyle \begin{align*} x \in (6, \infty) \end{align*} is not acceptable.


    Therefore, the solution is \displaystyle \begin{align*} x \in (-\infty, 8) \end{align*} .
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