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ParthKohliBest ResponseYou've already chosen the best response.0
\[\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2}}}}}\]I'm trying to do this problem.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.3
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
 one year ago

asnaseerBest ResponseYou've already chosen the best response.3
unless I have mis understood your question?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Actually, I'd just want to know how to denest \(\sqrt{2 + \sqrt2}\)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.3
I guess you could use the same technique, i.e:\[x=\sqrt{2+\sqrt{2}}\]\[x^2=2+\sqrt{2}\]\[x^22=\sqrt{2}\]then square both sides and solve the resulting quartic equation
 one year ago

asnaseerBest ResponseYou've already chosen the best response.3
which should end up being just a quadratic in \(x^2\)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Ah,\[(x^2  2)^2 = 2\]\[x^4 4x^2 +4 = 2 \]\[x^4  4x^2  2 = 0\]Assume \(y = x^2\).\[y^2  4y  2 = 0\]\[y = 2\pm \sqrt{6}\]\[x = \sqrt{2 \pm \sqrt6}\]
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
@asnaseer But that again brings up a nested radical.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.3
I'm not sure what exactly you mean by "denest" here?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Well, a radical expression containing another radical expression.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.3
btw: you have a small sign error in your calculation above
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Oh, that was supposed to be a +... not .
 one year ago

asnaseerBest ResponseYou've already chosen the best response.3
are you trying to simply \(\sqrt{2+\sqrt{2}}\) further?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Yeah, I want to remove the inner radical.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
http://math.stackexchange.com/questions/196155/strategiestodenestnestedradicals I can't get Bill Dubuque's technique to work here. The motivation behind this question is... I want to learn something new. :)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.3
so you are trying to express this as:\[\sqrt{2+\sqrt{2}}=a\sqrt{b}\]and what to find a and b?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Yeah, I want this in the form \(a + b\sqrt n \), not \(\sqrt{a + b\sqrt n }\)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.3
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. :)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
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 \(1\).
 one year ago

asnaseerBest ResponseYou've already chosen the best response.3
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."
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
WELL:\[\sqrt{2+\sqrt{2}} \] cannot be expressed in the forms dw:1348223973261:dw
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Thank God! I was so confused.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
But I think it can be expressed in some form likedw:1348224135337:dw
 one year ago
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