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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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Bunuel
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niteshwaghray
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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niteshwaghray
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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↧↧↧ Detailed Video Solution to the Problem Series ↧↧↧




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

Now the expression inside the square roots extends till infinity and if we note then the terms are repeating.
So, we can approximate the problem by writing terms after the first \(\sqrt{2}\) = \(z^2\)

=> \(z^2=\sqrt{2 + z^2}\)

Squaring both the sides
\(z^4 = 2 + z^2\)
=> \(z^4 - z^2 - 2 = 0\)
=> \(z^4 + z^2 - 2z^2 - 2 = 0\)
=> \(z^2(z^2 + 1) - 2 (z^2 + 1) = 0\)
=> \((z^2 + 1)(z^2 - 2) = 0\)
=> \(z^2\) = -1 or +2

But square a number cannot be negative
=> \(z^2\) = 2
=> z = + \(\sqrt{2}\) or -\(\sqrt{2}\)

I. Two values are possible for z
z = + \(\sqrt{2}\) or -\(\sqrt{2}\)
=> TRUE

II. 4 - \(z^2\) = 2
=>
\(z^2\) = 4 - 2
=> TRUE

III. \(z^8\) = 16
\(z^2\) = 2
=> \(z^4\) = \(2^2 \)= 4
=> \(z^8\) = \(4^2\) = 16
=> TRUE

So, Answer will be D
Hope it helps!

Watch the following video to MASTER Roots

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