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

- ParthKohli

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

- katieb

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- ParthKohli

\[\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2}}}}}\]I'm trying to do this problem.

- asnaseer

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

- asnaseer

unless I have mis understood your question?

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- ParthKohli

Actually, I'd just want to know how to denest \(\sqrt{2 + \sqrt2}\)

- asnaseer

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

- asnaseer

which should end up being just a quadratic in \(x^2\)

- ParthKohli

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

- ParthKohli

@asnaseer But that again brings up a nested radical.

- asnaseer

I'm not sure what exactly you mean by "denest" here?

- ParthKohli

Well, a radical expression containing another radical expression.

- asnaseer

btw: you have a small sign error in your calculation above

- ParthKohli

Oh, that was supposed to be a +... not -.

- asnaseer

are you trying to simply \(\sqrt{2+\sqrt{2}}\) further?

- ParthKohli

A second please.

- ParthKohli

Yeah, I want to remove the inner radical.

- asnaseer

why?

- ParthKohli

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. :)

- asnaseer

so you are trying to express this as:\[\sqrt{2+\sqrt{2}}=a\sqrt{b}\]and what to find a and b?

- ParthKohli

Yeah, I want this in the form \(a + b\sqrt n \), not \(\sqrt{a + b\sqrt n }\)

- asnaseer

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. :)

- ParthKohli

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 \(1\).

- mathslover

##### 1 Attachment

- mathslover

answer = 1

- mathslover

sorry ..

- ParthKohli

Is 1 correct?

- anonymous

Of course, NOT!

- asnaseer

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."

- anonymous

WELL:\[\sqrt{2+\sqrt{2}} \] cannot be expressed in the forms |dw:1348223973261:dw|

- ParthKohli

Thank God! I was so confused.

- anonymous

But I think it can be expressed in some form like|dw:1348224135337:dw|

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