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ParthKohli

  • 3 years ago

Denesting radicals. How to denest \(\sqrt{2 + \sqrt{2}}\)

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  1. ParthKohli
    • 3 years ago
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    \[\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2}}}}}\]I'm trying to do this problem.

  2. asnaseer
    • 3 years ago
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    one way is to write it as:\[x=\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2+...}}}}}\]and then square both sides to get:\[x^2=2+\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2+...}}}}}=2+x\]then just solve the quadratic

  3. asnaseer
    • 3 years ago
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    unless I have mis understood your question?

  4. ParthKohli
    • 3 years ago
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    Actually, I'd just want to know how to denest \(\sqrt{2 + \sqrt2}\)

  5. asnaseer
    • 3 years ago
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    I guess you could use the same technique, i.e:\[x=\sqrt{2+\sqrt{2}}\]\[x^2=2+\sqrt{2}\]\[x^2-2=\sqrt{2}\]then square both sides and solve the resulting quartic equation

  6. asnaseer
    • 3 years ago
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    which should end up being just a quadratic in \(x^2\)

  7. ParthKohli
    • 3 years ago
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    Ah,\[(x^2 - 2)^2 = 2\]\[x^4 -4x^2 +4 = 2 \]\[x^4 - 4x^2 - 2 = 0\]Assume \(y = x^2\).\[y^2 - 4y - 2 = 0\]\[y = 2\pm \sqrt{6}\]\[x = \sqrt{2 \pm \sqrt6}\]

  8. ParthKohli
    • 3 years ago
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    @asnaseer But that again brings up a nested radical.

  9. asnaseer
    • 3 years ago
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    I'm not sure what exactly you mean by "denest" here?

  10. ParthKohli
    • 3 years ago
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    Well, a radical expression containing another radical expression.

  11. asnaseer
    • 3 years ago
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    btw: you have a small sign error in your calculation above

  12. ParthKohli
    • 3 years ago
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    Oh, that was supposed to be a +... not -.

  13. asnaseer
    • 3 years ago
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    are you trying to simply \(\sqrt{2+\sqrt{2}}\) further?

  14. ParthKohli
    • 3 years ago
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    A second please.

  15. ParthKohli
    • 3 years ago
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    Yeah, I want to remove the inner radical.

  16. asnaseer
    • 3 years ago
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    why?

  17. ParthKohli
    • 3 years ago
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    http://math.stackexchange.com/questions/196155/strategies-to-denest-nested-radicals I can't get Bill Dubuque's technique to work here. The motivation behind this question is... I want to learn something new. :)

  18. asnaseer
    • 3 years ago
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    so you are trying to express this as:\[\sqrt{2+\sqrt{2}}=a\sqrt{b}\]and what to find a and b?

  19. ParthKohli
    • 3 years ago
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    Yeah, I want this in the form \(a + b\sqrt n \), not \(\sqrt{a + b\sqrt n }\)

  20. asnaseer
    • 3 years ago
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    ok, I see now - let me think over this. I am at work now so will probably get back to this after work - looks like an interesting problem. :)

  21. ParthKohli
    • 3 years ago
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    Thank you!\[2 + \sqrt 2\]Here a = 2, b = 1, n = 2. So,\[\sqrt{\text{parth}} = \sqrt{a^2 - nb^2 } = \sqrt{4 - 2} = \sqrt{2}\]Subtracting the above,\[2 + \sqrt{2} - \sqrt{2} = 2\]Now, find \(\sqrt{\text{kohli} } = \sqrt{2a} = \sqrt{4} = 2\). But that yields \(1\).

  22. mathslover
    • 3 years ago
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    1 Attachment
  23. mathslover
    • 3 years ago
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    answer = 1

  24. mathslover
    • 3 years ago
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    sorry ..

  25. ParthKohli
    • 3 years ago
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    Is 1 correct?

  26. Ishaan94
    • 3 years ago
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    Of course, NOT!

  27. asnaseer
    • 3 years ago
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    ok, after some research, I found out that this is actually quite a difficult problem. however, there are some special cases that can be solved for quite easily. see this wiki article for details: http://en.wikipedia.org/wiki/Nested_radical the article mentions: "In 1989 Susan Landau introduced the first algorithm for deciding which nested radicals can be denested. Earlier algorithms worked in some cases but not others."

  28. sauravshakya
    • 3 years ago
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    WELL:\[\sqrt{2+\sqrt{2}} \] cannot be expressed in the forms |dw:1348223973261:dw|

  29. ParthKohli
    • 3 years ago
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    Thank God! I was so confused.

  30. sauravshakya
    • 3 years ago
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    But I think it can be expressed in some form like|dw:1348224135337:dw|

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