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If z^2 = (2+(2+(2+(2+...)^(1/2))^(1/2))^(1/2))^(1/2), where the given

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niteshwaghray
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If z^2 = (2+(2+(2+(2+...)^(1/2))^(1/2))^(1/2))^(1/2), where the given [#permalink] New post  Updated on: 26 Aug 2019, 02:23
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If \(z^2=\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}\), where the given expression extends till infinity, which of the following statements must be true ?

I. Two values are possible for z
II. 4 - z^2 = 2
III. z^8 = 16


A. I only
B. II only
C. III only
D. I, II and III
E. None of the above
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Originally posted by niteshwaghray on 31 May 2017, 23:14.
Last edited by Bunuel on 26 Aug 2019, 02:23, edited 3 times in total.
Renamed the topic and edited the question.
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Bunuel
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Re: If z^2 = (2+(2+(2+(2+...)^(1/2))^(1/2))^(1/2))^(1/2), where the given [#permalink] New post  31 May 2017, 23:33
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niteshwaghray wrote:
If \(z^2=\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}\), where the given expression extends till infinity, which of the following statements must be true ?

I. Two values are possible for z
II. 4 - z^2 = 2
III. z^8 = 16


A. I only
B. II only
C. III only
D. I, II and III
E. None of the above


\(z^2=\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}\);

\(z^2=\sqrt{2+({\sqrt{2+\sqrt{2+\sqrt{2+...}})}}}\), as the expression under square root extends infinitely then expression in brackets would equal to z^2 itself so we can rewrite given expression as \(z^2=\sqrt{2+z^2}\). Square both sides \(z^4=2+z^2\) --> \(z=\sqrt{2}\) or \(z=-\sqrt{2}\).

We have two possible values, so I is true. Also, both of them satisfy II and III.

Answer: D.
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If z^2 = (2+(2+(2+(2+...)^(1/2))^(1/2))^(1/2))^(1/2), where the given [#permalink] New post  31 May 2017, 23:34
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Bunuel wrote:
niteshwaghray wrote:
If \(z^2=\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}\), where the given expression extends till infinity, which of the following statements must be true ?

I. Two values are possible for z
II. 4 - z^2 = 2
III. z^8 = 16


A. I only
B. II only
C. III only
D. I, II and III
E. None of the above


\(z^2=\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}\);

\(z^2=\sqrt{2+({\sqrt{2+\sqrt{2+\sqrt{2+...}})}}}\), as the expression under square root extends infinitely then expression in brackets would equal to z^2 itself so we can rewrite given expression as \(z^2=\sqrt{2+z^2}\). Square both sides \(z^4=2+z^2\) --> \(z=\sqrt{2}\) or \(z=-\sqrt{2}\).

We have two possible values, so I is true. Also, both of them satisfy II and III.

Answer: D.


Similar questions to practice:
https://gmatclub.com/forum/tough-and-tri ... l#p1029228
https://gmatclub.com/forum/find-the-valu ... 38049.html
https://gmatclub.com/forum/find-the-valu ... 75403.html
https://gmatclub.com/forum/if-the-expre ... 32547.html
https://gmatclub.com/forum/if-the-expre ... 98647.html

Hope it helps.
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philipssonicare
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Re: If z^2 = (2+(2+(2+(2+...)^(1/2))^(1/2))^(1/2))^(1/2), where the given [#permalink] New post  08 Sep 2019, 03:59
niteshwaghray
What is the question source?

Bunuel
"as the expression under square root extends infinitely then expression in brackets would equal to z^2 itself"
I don't know why this is the case. Is there a name for this theory so I can learn about it?
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Re: If z^2 = (2+(2+(2+(2+...)^(1/2))^(1/2))^(1/2))^(1/2), where the given [#permalink] New post  09 Sep 2019, 00:08
Bunuel wrote:
Bunuel wrote:
niteshwaghray wrote:
If \(z^2=\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}\), where the given expression extends till infinity, which of the following statements must be true ?

I. Two values are possible for z
II. 4 - z^2 = 2
III. z^8 = 16


A. I only
B. II only
C. III only
D. I, II and III
E. None of the above


\(z^2=\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}\);

\(z^2=\sqrt{2+({\sqrt{2+\sqrt{2+\sqrt{2+...}})}}}\), as the expression under square root extends infinitely then expression in brackets would equal to z^2 itself so we can rewrite given expression as \(z^2=\sqrt{2+z^2}\). Square both sides \(z^4=2+z^2\) --> \(z=\sqrt{2}\) or \(z=-\sqrt{2}\).

We have two possible values, so I is true. Also, both of them satisfy II and III.

Answer: D.


Similar questions to practice:
https://gmatclub.com/forum/tough-and-tri ... l#p1029228
https://gmatclub.com/forum/find-the-valu ... 38049.html
https://gmatclub.com/forum/find-the-valu ... 75403.html
https://gmatclub.com/forum/if-the-expre ... 32547.html
https://gmatclub.com/forum/if-the-expre ... 98647.html

Hope it helps.



Thanks a lot. this helped me understand the various scenarios relating to such a question.
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Crytiocanalyst
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Re: If z^2 = (2+(2+(2+(2+...)^(1/2))^(1/2))^(1/2))^(1/2), where the given [#permalink] New post  03 Aug 2021, 02:57
niteshwaghray wrote:
If \(z^2=\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}\), where the given expression extends till infinity, which of the following statements must be true ?

I. Two values are possible for z
II. 4 - z^2 = 2
III. z^8 = 16


A. I only
B. II only
C. III only
D. I, II and III
E. None of the above


The key to sloving the equation is understanding the fact that the equation tending to infinity doesn't change if just one part is eleminated

Therefore z^4 =2+z^2
=>z^4 -z^2 -2 =0

Gives us 2 values +-(2)^1/2
4-z^2
4-2=2
and finally
2^4= 16

All the three conditions are satisfied therefore IMO D
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Re: If z^2 = (2+(2+(2+(2+...)^(1/2))^(1/2))^(1/2))^(1/2), where the given [#permalink]
03 Aug 2021, 02:57
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