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Math Help - How to solve x^2 + 2x - 8 > 0 algebraically

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    How to solve x^2 + 2x - 8 > 0 algebraically

    x^2 + 2x - 8 > 0

    (x + 1)^2 - 9 > 0

    (x + 1)^2 > 9

    x + 1 > 3 But what to do about the -3. Eg Do I say:

    x + 1 > -3 or x + 1 < -3

    It seems that that it has to be x + 1 < -3 - works if look at curve graphically.

    But then what if problem is x^2 + 2x - 8 < 0

    (x + 1)^2 < 9

    x + 1 < 3 and x + 1 < -3 which is wrong. Should be
    x + 1 < 3 and x + 1 > -3.
    -4 < x < 2 - can see graphically that is under x axis and so between -4 and 2.

    So what is the rule for bit when get square root of either side???
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  2. #2
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    Re: How to solve x^2 + 2x - 8 > 0 algebraically

    Quote Originally Posted by angypangy View Post
    x^2 + 2x - 8 > 0

    (x + 1)^2 - 9 > 0

    (x + 1)^2 > 9

    x + 1 > 3 But what to do about the -3. Eg Do I say:

    x + 1 > -3 or x + 1 < -3

    It seems that that it has to be x + 1 < -3 - works if look at curve graphically.

    But then what if problem is x^2 + 2x - 8 < 0

    (x + 1)^2 < 9

    x + 1 < 3 and x + 1 < -3 which is wrong. Should be
    x + 1 < 3 and x + 1 > -3.
    -4 < x < 2 - can see graphically that is under x axis and so between -4 and 2.

    So what is the rule for bit when get square root of either side???
    You need to say that \pm(x+1) > 3

    Then either x+1 > 3 \text{  or  } -(x+1) > 3 \Leftrightarrow x+1 < -3
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  3. #3
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    Re: How to solve x^2 + 2x - 8 > 0 algebraically

    Hello, angypangy!

    x^2 + 2x - 8 \,>\, 0

    (x + 1)^2 - 9 \,>\, 0

    (x + 1)^2 \,>\, 9

    x + 1 \,>\, 3 . . . . Not quite

    When we take the square root, we get: . |x+1| \:>\:3

    This means: . \begin{Bmatrix}x+1 \:>\:3 & \Rightarrow & x \:>\:2 \\ & \text{or} \\ x+1 \:<\:\text{-}3 & \Rightarrow & x \:<\:\text{-}4 \end{Bmatrix}

    The solution is: . (\text{-}\infty,\,-4)\,\cup\,(2,\,\infty)


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


    Graphically, we have a parabola: . y \:=\:x^2+2x-8

    The question is: when is the graph above the x-axis?

    Since the parabola opens upward,
    . . it is positive outside of its x-intercepts.

    The x-intercepts are: . (x+4)(x-2) \:=\:0 \quad\Rightarrow\quad x \:=\:\text{-}4,\,2

    And we can "see" the solution . . .
    Code:
                    |
         ♥          |      ♥
                    |
                    |
          ♥         |     ♥
                    |
           ♥        |    ♥
        - - * - - - + - * - - -
           -4 *     | * 2
                  * |
                    |
    If the problem were: . x^2 + 2x - 8 \;\;{\color{red}<}\;\;0
    . . the solution is between the x-intercepts.
    . . . . . \text{-}4 \;<\;x\,<\;2

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