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Math Help - Quadratic inequality confusion

  1. #1
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    Quadratic inequality confusion

    Ok so here goes,

    Find the set of values for which the graph of y=2x^{2}+x-10 is above the x-axis.

    Well that gives us the inequality
    2x^{2}+x-10>0
    \implies {(2x+5)(x-2)>0}
    \implies x<\frac {-5}{2}, x>2

    Here is where the confusion is; how do i write the set values because there is an infitesmal amount? I wrote it:

    x=3,4,5,6,7,8........ \infty
     <br />
x=-3,-4,-5,-6,-7,-8........ -\infty<br />

    However is this correct?

    Thanks in advance,

    Sean
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Consider the case when

    2x+5>0\,\wedge\,x-2>0 & 2x+5<0\,\wedge\,x-2<0
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  3. #3
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    I dont quite understand, but if
    x-2>0
    2x+5<0
    The same as in my first post is true?
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  4. #4
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    Hello, Sean!

    Krizalid is absolutely correct . . .


    Find the set of values for which the graph of y\:=\:2x^2+x-10 is above the x-axis.

    Well, that gives us the inequality: . 2x^{2}+x-10\:>\:0\quad\Rightarrow\quad(2x+5)(x-2)>0 . Right!

    If the product of those two factors is positive:
    . . (1) both factors are positive, or
    . . (2) both factors are negative.

    Case (1): . \begin{array}{ccccccc}2x+5 & > & 0 & \Rightarrow & x & > & \text{-}\frac{5}{2} \\<br />
x - 2 & > & 0 & \Rightarrow & x & > & 2\end{array} . The "stronger" inequality is: . x \:>\:2

    Case (2): . \begin{array}{ccccccc}2x + 5 & < & 0 & \Rightarrow & x & < &\text{-}\frac{5}{2} \\<br />
x - 2 & < & 0 & \Rightarrow & x & < & 2\end{array} . The "stronger" inequality is: . x \:<\:\text{-}\frac{5}{2}


    Therefore, the graph is above the x-axis on the intervals: . \left(\text{-}\infty,\,\text{-}\frac{5}{2}\right),\;(2,\,\infty)


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

    Of course, if we were allowed to graph the function,
    . . we could have "eyeballed" the answers.
    Code:
                      |
         *            |        *
                      |
          *           |       *
           *          |      *
        -----o--------+----o-----
            -2 *     | *  2
                    * |
                      |
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  5. #5
    MHF Contributor red_dog's Avatar
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    Let f:\mathbf{R}\to\mathbf{R}, \ f(x)=ax^2+bx+c a quadratic function, with real roots x_1,x_2, \ x_1<x_2.
    Then:
    a) If a>0 then
    f(x)>0, \ \forall x\in(-\infty,x_1)\cup(x_2,\infty)
    f(x)<0, \ \forall x\in(x_1,x_2)

    b) If a<0 then
    f(x)<0, \ \forall x\in(-\infty,x_1)\cup(x_2,\infty)
    f(x)>0, \ \forall x\in(x_1,x_2)
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  6. #6
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    Thanks guys,

    BTW i "eyeballed" the answer by drawing a quick sketch
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Sean12345 View Post
    BTW i "eyeballed" the answer by drawing a quick sketch
    no shame in that
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