## ParthKohli 3 years ago Denesting radicals. How to denest $$\sqrt{2 + \sqrt{2}}$$

1. ParthKohli

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

2. 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

3. asnaseer

unless I have mis understood your question?

4. ParthKohli

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

5. 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

6. asnaseer

which should end up being just a quadratic in $$x^2$$

7. 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}$

8. ParthKohli

@asnaseer But that again brings up a nested radical.

9. asnaseer

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

10. ParthKohli

11. asnaseer

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

12. ParthKohli

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

13. asnaseer

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

14. ParthKohli

15. ParthKohli

Yeah, I want to remove the inner radical.

16. asnaseer

why?

17. 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. :)

18. asnaseer

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

19. ParthKohli

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

20. 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. :)

21. 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 $$1$$.

22. mathslover

23. mathslover

24. mathslover

sorry ..

25. ParthKohli

Is 1 correct?

26. Ishaan94

Of course, NOT!

27. 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."

28. sauravshakya

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

29. ParthKohli

Thank God! I was so confused.

30. sauravshakya

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