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Math Help - Solve inequality

  1. #1
    Junior Member
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    Solve inequality

    task is x-4 / x^2 + 3x >= 0 solve inequality

    x-4 >= 0
    x >= 4

    x (x+3) , x=0 & x>= -3

    thus, (-3,0] and (4, infinity)

    is this correct?
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  2. #2
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    By looking at your working I see you mean to solve \frac{x-4}{x^2+3x} \geq 0. You should use brackets to be more clear in future please...

    It should be x \in \left(-3,0\right) \cup \left[\,4, \infty \right)

    0 is not included in the first interval as that would cause division by zero. 4 is included in the second interval as we are solving for greater than or equal to 0.

    EDIT: sorry fixed a mistake.
    Last edited by nzmathman; April 11th 2009 at 05:40 PM.
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  3. #3
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    Hello, makaveli89!

    Solve: . \frac{x-4}{x^2 + 3x}\:\geq \:0

    x-4 \:\geq\:0 \quad\Rightarrow\quad x \:\geq\:4

    x(x+3)\:\geq\:0 \quad\Rightarrow\quad x\:\geq\:0\:\text{ and }\:x\:\geq\:-3

    Thus: . (-3,0]\:\text{ and }\:(4, \infty)

    Is this correct? . . . . no

    You have: . \begin{array}{c}x \geq 4 \\ x \geq 0 \\ x \geq \text{-}3 \end{array}\quad \hdots which means: . x \geq 4\:\text{ or }\:[4,\infty) . . . and that's all.



    That fraction is positive if both the numerator and denominator are negtive.


    Numerator is negative: . x-4 \:\leq \:0 \quad\Rightarrow\quad x \:\leq\:4


    Denominator is negative: . x(x+3) \:< \:0
    There are two cases to consider:

    [1]\;\;\begin{array}{ccccccc}x > 0 & \Rightarrow & x\text{ positive} \\ x+3 < 0 & \Rightarrow & x < -3 \end{array}\;\;\hdots\;\text{ impossible}

    [2]\;\;\begin{array}{cccccc}x < 0 & \Rightarrow & x\text{ negative} \\<br />
x + 3 > 0 & \Rightarrow& x > \text{-}3 \end{array}


    We have: . \begin{array}{c}x \:\leq \:4 \\ x \:<\:0 \\ x \:> \:\text{-}3 \end{array}\quad\hdots which means: . -3 < x < 0 \:\text{ or }\;(-3,0)


    The answer is: . (\text{-}3,0) \,\cup\, [4,\infty)

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