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Square root within a square root

  1. Sep 7, 2012 #1
    Hoping someone can push me in the right direction with this one. Plume snookered.

    It's to simplify:
    2√3(3+√3)

    Guessing first calculate (a^2 - b^2*c) in the square, though the 2 is throwing this an i'm not sure how the answer is 6√3 + 6, an not 18√3 ?
     
  2. jcsd
  3. Sep 7, 2012 #2
    It's just the Distributive property:
    [itex](2)(\sqrt{3})(3 + \sqrt{3})[/itex]
    [itex](2)(\sqrt{3})(3 ) + (2)(\sqrt{3})(\sqrt{3})[/itex]
    [itex](2)(3 )(\sqrt{3}) + (2)(\sqrt{3})(\sqrt{3})[/itex]
     
  4. Sep 7, 2012 #3
    Thanks coolul007, havent really been shown the Distributive way of working with squares like that, has helped shed some light on another tricky simplication.
     
  5. Sep 7, 2012 #4

    haruspex

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    The title said 'square root within a square root', which implies you meant 2√(3(3+√3)), but the answer you say is correct matches (2√3)(3+√3).
     
  6. Sep 8, 2012 #5

    Mentallic

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    There is no square root within a square root as your title suggested :tongue:

    Unless you meant that 2√3(3+√3) as [tex]2\sqrt{3(3+\sqrt{3})}[/tex] but that doesn't follow from the required answer, so yep, this problem can simply be solved with the distributive property.

    edit: beaten to it.
     
  7. Sep 8, 2012 #6

    Mark44

    Staff: Mentor

    Mod note: Moving this thread to the Precalculus section under Homework & Coursework, which is where the OP should have started this thread.
     
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