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Math Help - Various Surds Questions

  1. #1
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    Various Surds Questions

    These are the more difficult questions at the end of my chapter and I REALLY have no idea how to even approach them.

    Would any of you mind giving me a few hints for them and nudging me
    in the right direction?

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  2. #2
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    Hello prettiestfriend
    Quote Originally Posted by prettiestfriend View Post
    These are the more difficult questions at the end of my chapter and I REALLY have no idea how to even approach them.

    Would any of you mind giving me a few hints for them and nudging me
    in the right direction?

    (5) \sqrt{\frac{3+2\sqrt2}{3-2\sqrt2}} =\sqrt{\frac{(3+2\sqrt2)^2}{(3-2\sqrt2)(3+2\sqrt2)}}

    = \frac{3+2\sqrt2}{3^2 - (2\sqrt2)^2}

    =3 + 2\sqrt2

    (6) \frac{1}{\sqrt3}-\frac{1}{\sqrt2}=\frac{\sqrt2-\sqrt3}{\sqrt2\sqrt3}=p

    \Rightarrow \left(\frac{\sqrt2-\sqrt3}{\sqrt2\sqrt3}\right)^2=p^2

    \Rightarrow \frac{2 - 2\sqrt2\sqrt3 + 3}{6} = p^2

    Can you complete this now?

    (7) x = \sqrt5-\sqrt3

    \Rightarrow x^2 = 5 - 2\sqrt5\sqrt3 + 3 = 8 - 2\sqrt{15}

    \Rightarrow x^4 = 64 - 32\sqrt{15} + 4\times 15 = 124 - 32\sqrt{15}

    Can you complete it?

    (8) For n=1: u_1 = \frac{1}{\sqrt5}\left[\left(\frac{1+\sqrt5}{2}\right)^1-\left(\frac{1-\sqrt5}{2}\right)^1\right]

    = \frac{1}{\sqrt5}\left[\frac{1+\sqrt5 - 1 +\sqrt5}{2}\right]=\frac{1}{\sqrt5}\times \sqrt5=1

    For n = 2, note that \left(\frac{1+\sqrt5}{2}\right)^2 = \frac{1 + 2\sqrt5+ 5}{4}= \frac{3+\sqrt5}{2}. Use a similar expression for \left(\frac{1-\sqrt5}{2}\right)^2, and simplify the result.

    For n = 3, note that \left(\frac{1+\sqrt5}{2}\right)^3= \left(\frac{1+\sqrt5}{2}\right)^2\times\left(\frac  {1+\sqrt5}{2}\right)

    =\left(\frac{3+\sqrt5}{2}\right)\times\left(\frac{  1+\sqrt5}{2}\right)

    Multiply out the brackets, and simplify. Can you do this, and n = 4 now?

    Grandad
    Last edited by Grandad; June 8th 2009 at 11:22 PM. Reason: Corrected typo
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