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Math Help - Solving this inequation

  1. #1
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    Solving this inequation

    I did have a look at the sticky but I'm still confused.

    3x^2 + x - 2 > 0
    I got
    x > 2/3 and x > -1

    For the interval notation I got
    (-1, infinity) union (2/3, infinity)
    But the answer is:
    (negative infinity, -1) union (2/3, infinity)

    Why is that? :S
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  2. #2
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    Quote Originally Posted by Klutz View Post
    I did have a look at the sticky but I'm still confused.

    3x^2 + x - 2 > 0
    I got
    x > 2/3 and x > -1

    For the interval notation I got
    (-1, infinity) union (2/3, infinity)
    But the answer is:
    (negative infinity, -1) union (2/3, infinity)

    Why is that? :S
    3x^2 + x - 2 = (3x - 2)(x + 1) > 0

    the graph of the parabola crosses the x-axis at x = -1 and x = 2/3

    now, think about the graph of the parabola (opens upward, correct?) ... where is the graph below the x-axis? where is it above?
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  3. #3
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    oh I get it now thanks!

    is it required to sketch the graph for this everytime?
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  4. #4
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    Quote Originally Posted by Klutz View Post
    oh I get it now thanks!

    is it required to sketch the graph for this everytime?
    not really "required", but it sure helps to visualize the intervals.
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  5. #5
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    Hello, Klutz!


    You had: . 3x^2 + x - 2 \:>\:0

    You factored and got: . (x + 1)(3x-2) \:>\:0


    We have the product of two factors and it's suppose to be positive.
    . . Under what conditions is this true?


    Obviously, if both factors are positive, their product is positive.
    . . (You saw this one.) **


    Also, if both factors are negative, their product is positive.
    . . (You missed this one!)

    We have: . \begin{Bmatrix}x + 1 \:<\:0 & \Rightarrow & x\:<\:-1 \\ 3x-2 \:<\:0 & \Rightarrow & x \:<\:\frac{2}{3} \end{Bmatrix}

    These can be condensed into: . x \:<\:-1


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

    **

    You had: .  x \:>\:-1\,\text{ and }\,x \:>\:\frac{2}{3}

    Can you see that this is redundant?

    You want numbers greater than \text{-}1 and greater than \frac{2}{3}

    Any number greater than \frac{2}{3} satisfies both inequalities.

    They can be condensed: . x \:>\:\frac{2}{3}


    The answer is: .  \begin{Bmatrix}<br />
x \:<\:\text{-}1 & \text{or} & x \:>\:\frac{2}{3} \\ \\[-2mm]<br />
(\text{-}\infty,\:\text{-}1) & \cup &(\frac{2}{3},\:\infty) \end{Bmatrix}

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