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Thread: Surds question

  1. #1
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    Surds question

    I need help with this question.
    Attached Thumbnails Attached Thumbnails Surds question-img_3125.jpg  
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  2. #2
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    Re: Surds question

    Give it a go.
    Work out 1+c by getting a common denominator.
    Then do the same for 1-c.
    Then do the division.
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  3. #3
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    Re: Surds question

    Quote Originally Posted by Britt00 View Post
    I need help with this question.
    $\dfrac{\sqrt3+1}{\sqrt3-1}=\dfrac{3+2\sqrt3+1}{2}$
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  4. #4
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    Re: Surds question

    Here is the method step by step.

    $\dfrac{1 + \sqrt{3}}{1 - \sqrt{3}} = \dfrac{1 + \sqrt{3}}{1 - \sqrt{3}} * 1 = \dfrac{1 + \sqrt{3}}{1 - \sqrt{3}} * \dfrac{1 + \sqrt{3}}{1 + \sqrt{3}} =$

    $ \dfrac{1^2 + \sqrt{3} + \sqrt{3} + (\sqrt{3})^2}{1^2 + \sqrt{3} - \sqrt{3} - (\sqrt{3})^2} = \dfrac{1 + 2\sqrt{3} + 3}{1 - 3} = -\ \dfrac{2(2 + \sqrt{3})}{2}.$

    EDIT: I believe Plato read the question incorrectly so he and I disagree in terms of sign only. Of course it is always possible I misread the problem, which was hard to parse on a tablet.
    Last edited by JeffM; Mar 5th 2017 at 07:24 PM.
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  5. #5
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    Re: Surds question

    Please post your future algebra questions in the algebra forum. Thank you for your cooperation.
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  6. #6
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    Re: Surds question

    Jeff, why are you not using 1/sqrt(3)?

    I get same as Sir Plato.
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  7. #7
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    Re: Surds question

    Because I did manage to misread the question. I hate it when people put those tiny jpegs in instead of writing out the question.

    Yes of course Plato is correct.

    $\dfrac{1 + \dfrac{1}{\sqrt{3}}}{1 - \dfrac{1}{\sqrt{3}}} = \dfrac{\dfrac{\sqrt{3} + 1}{\cancel{\sqrt{3}}}}{\dfrac{\sqrt{3} - 1}{\cancel{\sqrt{3}}}} = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}$.
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