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Math Help - interval

  1. #1
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    interval

    1) How do i write the answer in interval notation for (x-5)(x-11)>0 I got (5,infinity)U(11,infinity) webwork said it is incorrect too, so can anyone help me out what mistake i made for those two problems?
    2)how will the problem be solved for the inequality in interval notation for -2x^2<18
    3)also for the problem -x^2+4x(equal or > to)0 ?
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  2. #2
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    Hello, Girlaaaaaaaa!

    With quadratic inequalities, we must be careful . . .


    1)\;(x-5)(x-11)\:>\:0

    It says: The product of two quantities is positive.

    This is true if: [1] both factors are positive, or [2] both factors are negative.
    . . And we must consider both cases.


    Both positive: . \begin{array}{ccc}x - 5 \:>\:0 & \Rightarrow & x \:> \:5 \\ x-11\:>\>0 & \Rightarrow & x \:>\:11\end{array}

    So x must be greater than 5 and greater than 11.
    . . Then: . x \:>\:11 covers both of them.


    Both negative: . \begin{array}{ccc}x - 5 \:<\:0 & \Rightarrow & x \:<\:5 \\ x-11 \:<\:0 & \Rightarrow & x \:<\:11 \end{array}

    So x must be less than 5 and less than 11.
    . . Then: .  x \:< \:5 covers both of them.


    Solution: . x \:<\:5\:\text{ or }\:x\:>\:11

    Interval notation: . 11,\,\infty) " alt="(-\infty,\,5)\:\cup\11,\,\infty) " />


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Here's a rather primitive approach to the problem.

    Consider the function: . x-5)(x-11)" alt="y \:=\x-5)(x-11)" />
    . . When is this function positive?

    We have a parabola (y \:=\:x^2 - 16x + 55)
    . . When is it above the x-axis?

    The parabola opens upward and it has x-intercepts 5 and 11.
    The graph is below the x-axis between the intercepts
    . . so it is above the x-axis to the left of 5 and to the right of 11.

    See it?

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  3. #3
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    I am not really understanding that. Well for example i have a somewhat similar problem of -x^2+4x>=0
    for that i then did -x^2+4x-0 then i got (-x+2)(x+2) and my answe in interval was (-infinity,-2]U[-2,infinity) it says that was wrong, so would it be somewhat what you showed?
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  4. #4
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    Quote Originally Posted by Girlaaaaaaaa View Post
    1) How do i write the answer in interval notation for (x-5)(x-11)>0
    x^2  - 16x + 55 > 0 \implies (x - 8)^2  > 9 \implies \left| {x - 8} \right| > 3.

    Finally - 3 > x - 8 > 3\,\therefore \,5 > x > 11.
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  5. #5
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    Quote Originally Posted by Krizalid View Post
    Finally - 3 > x - 8 > 3\,\therefore \,5 > x > 11.
    Excuse me? -3>x-8>3???
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