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Math Help - Problem with an algebra exercice

  1. #1
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    Problem with an algebra exercice

    Hello!!
    I need your help with this exercise is about algebra i couldn't solve it
    thank you so much

    the answers are:
    Problem with an algebra exercice-nine2.gif
    and Problem with an algebra exercice-nine3.gif
    Attached Thumbnails Attached Thumbnails Problem with an algebra exercice-ninex.gif  
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  2. #2
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    Hastings
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    Expand and simplify the forumla and then use apply Quadratic formula to solve it.

    3(x^2 - 4 \sqrt{3}) - 8x(\sqrt{3} - 1) + 19 = 0
    3x^2 - 12 \sqrt{3} - 8x\sqrt{3} - 8x + 19 = 0
    3x^2 - 8x\sqrt{3}-8x + 19 - 12\sqrt{3} = 0

    Quadratic formula:

     a = 3
     b = -8\sqrt{3} - 8
     c = -12\sqrt{3} + 19

    x = \dfrac{-b-\sqrt{b^2-4ac}}{2a}

    x = \dfrac{-b+\sqrt{b^2-4ac}}{2a}

    Substitute in the values, simplify, and you get your answer.

    (I may have got it wrong... it might have been moved to the calculus section for a reason)
    Last edited by DavidM; August 14th 2010 at 02:06 AM.
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  3. #3
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    Completing the square

    Are you familiar with how to complete the square? If so, then we rearrange your original equation,

    3(x^2 - 4\sqrt{3}) - 8x(\sqrt{3} - 1) + 19 = 0

    into:
    3x^2 - 8(\sqrt{3} - 1)x + 19 - 12\sqrt{3} = 0

    Factor out the 3:
    3\left[x^2 - \frac{8(\sqrt{3} - 1)x}{3}\right] + 19 - 12\sqrt{3} = 0

    And complete the square:
    3\left[\left (x-\frac{4(\sqrt{3}-1)}{3}\right )^2 - \left (\frac{4(\sqrt{3} - 1)}{3}\right )^2\right] + 19 - 12\sqrt{3} = 0


    Then some algebraic manipulation gets you (incoming wall-of-step-by-step-text):

    \left (x-\frac{4(\sqrt{3}-1)}{3}\right )^2 - \left (\frac{4(\sqrt{3} - 1)}{3}\right )^2 = \frac{12\sqrt{3} - 19}{3}
    \left (x-\frac{4(\sqrt{3}-1)}{3}\right )^2 = \left (\frac{4(\sqrt{3} - 1)}{3}\right )^2 + \frac{12\sqrt{3} - 19}{3}
    \left (x-\frac{4(\sqrt{3}-1)}{3}\right )^2 = \frac{16(4-2\sqrt{3})}{9} + \frac{36\sqrt{3} - 57}{9}
    \left (x-\frac{4(\sqrt{3}-1)}{3}\right )^2 = \frac{64 - 32\sqrt{3} + 36\sqrt{3} - 57}{9}
    \left (x-\frac{4(\sqrt{3}-1)}{3}\right )^2 = \frac{7 + 4\sqrt{3}}{9}
    x-\frac{4(\sqrt{3}-1)}{3} = \pm\sqrt{\frac{7 + 4\sqrt{3}}{9}}
    x-\frac{4(\sqrt{3}-1)}{3} = \pm\frac{\sqrt{7 + 4\sqrt{3}}}{3}}


    From here, notice that:
    7+4\sqrt{3} = (2+\sqrt{3})^2

    Substituting that into the equation, we get:
    x-\frac{4(\sqrt{3}-1)}{3} = \pm\frac{\sqrt{(2+\sqrt{3})^2}}{3}}


    So,
    x=\frac{4(\sqrt{3}-1)}{3}\pm\frac{2+\sqrt{3}}{3}}


    Now, evaluate both solutions for x:
    x=\frac{4(\sqrt{3}-1)}{3}+\frac{2+\sqrt{3}}{3}}=\frac{5\sqrt{3}-2}{3}
    x=\frac{4(\sqrt{3}-1)}{3}-\frac{2+\sqrt{3}}{3}}=\frac{3\sqrt{3}-6}{3}=\sqrt{3}-2

    Voila! Done!

    If you're not sure how to complete the square, I'm sure there are other methods to doing this, such as using the quadratic equation as DavidM suggested. The "tricky" part is recognising that 7+4\sqrt{3} can be expressed as a square in itself.
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  4. #4
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    thank you for your help both methods have helped me.
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