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Math Help - Surds

  1. #1
    Junior Member
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    Surds

    Hi, could anyone help me solve this equation?

    3x = sqrt.5 (x + 2) ----- square root applies to 5 only
    in the form a + b sqrt.5

    Thanks.
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  2. #2
    Banned
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    Quote Originally Posted by Geometor View Post
    Hi, could anyone help me solve this equation?

    3x = sqrt.5 (x + 2) ----- square root applies to 5 only
    in the form a + b sqrt.5

    Thanks.
    3x = \sqrt{5} \;(x+2)

    3x = x\sqrt{5} \;+2\sqrt{5}

    3x - x\sqrt{5} =2\sqrt{5}

    x(3 - \sqrt{5}) =2\sqrt{5}

    x=\frac {2\sqrt{5}}{3 - \sqrt{5}}

    Now Rationalize the denominator,

    x=\frac {2\sqrt{5}}{3 - \sqrt{5}}\times \frac {3+ \sqrt{5}}{3 + \sqrt{5}}

    x=\frac {6\sqrt{5} - 2\times 5}{9-5}

    x=\frac {-10+6\sqrt{5}}{4}

    x=\frac {-10}{4}+ \frac{6}{4}\sqrt{5}

    x=\frac {-5}{2}+ \frac{3}{2}\sqrt{5}
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  3. #3
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    ah lol I see, thank you
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