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Math Help - Square root manipulation

  1. #1
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    Square root manipulation

    Hello. From my calculator, I know that this is a true equality:

    \sqrt{5+2\sqrt{6}} = \sqrt{3} + \sqrt{2}

    However, I simply cannot come up with the method used to "simplify" the first expression to arrive at the second expression (or vice versa). Could someone demonstrate or explain what I am missing?
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  2. #2
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    Quote Originally Posted by drumist View Post
    Hello. From my calculator, I know that this is a true equality:

    \sqrt{5+2\sqrt{6}} = \sqrt{3} + \sqrt{2}

    However, I simply cannot come up with the method used to "simplify" the first expression to arrive at the second expression (or vice versa). Could someone demonstrate or explain what I am missing?
    Ok, let's start from

    C = \sqrt{3} + \sqrt{2}

    Square both sides

    C^2 = (\sqrt{3} + \sqrt{2})^2 = (\sqrt{3})^2 + 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6}

    Now take the positive root to find C = \sqrt{5 + 2\sqrt{6}}
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  3. #3
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    Well, to be clear, seeing the equality, I can definitely prove that it is true, just like you showed.

    I see that you can generalize it into an identity like this

    \sqrt{a+b\pm2\sqrt{ab}}=\sqrt{a}\pm\sqrt{b}

    I guess my point is: are there any other similar identities like this that would be useful in simplifying stacked square root expressions? I don't remember ever being taught this particular identity, although I'm not sure if it really occurs very often.
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  4. #4
    Moo
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    5+2\sqrt{6}=3+2+2\cdot\sqrt{2}\cdot\sqrt{3}=(\sqrt  {2}+\sqrt{3})^2

    you've got to recognize the terms of the identity/ies (a\pm b)^2=a^2+b^2\pm 2ab
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