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Find the value of x

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Find the value of x [#permalink] New post 04 Feb 2009, 01:04
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Find the value of x

x= \sqrt[2]{20+\sqrt[2]{20+\sqrt[2]{20}}}

1. 20
2. 5
3. 2
4. 8
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Re: square root [#permalink] New post 04 Feb 2009, 02:51
B. 5 (by process of elimination).
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Re: square root [#permalink] New post 04 Feb 2009, 03:05
How did u get 5?
quote="scthakur"]B. 5 (by process of elimination).[/quote]
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Re: square root [#permalink] New post 04 Feb 2009, 04:41
ritula wrote:
Find the value of x

x= \sqrt[2]{20+\sqrt[2]{20+\sqrt[2]{20}}}

1. 20
2. 5
3. 2
4. 8


Since the answer choices are dissimilar, we can estimate the answer choice here. The \sqrt[2]{20} is somewhere between 4 and 5. Suppose it's 5, then we'll get
\sqrt[2]{20+\sqrt[2]{20+5}=\sqrt[2]{20+5}=5
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Re: square root [#permalink] New post 04 Feb 2009, 07:29
agree with 5..

i estimated it to be 5..

now if they had a 4 in the ans choices..that would have been tough..
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Re: square root [#permalink] New post 04 Feb 2009, 07:32
FN wrote:
agree with 5..

i estimated it to be 5..

now if they had a 4 in the ans choices..that would have been tough..


even if you have 4.. its not tough.. sqrt(20) clearly.. >4

I agree if answer choice has options like 4.9 or 4.8...
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Re: square root [#permalink] New post 04 Feb 2009, 07:38
ritula wrote:
How did u get 5?
quote="scthakur"]B. 5 (by process of elimination).
[/quote]

actual value of x= 4.994690378 ~5

only way to do is process of elimination.
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Re: square root [#permalink] New post 04 Feb 2009, 07:59
ritula wrote:
Find the value of x

x= \sqrt[2]{20+\sqrt[2]{20+\sqrt[2]{20}}}

1. 20
2. 5
3. 2
4. 8



1: x= \sqrt{20+\sqrt{20+\sqrt{20}}}
x= \sqrt{20+\sqrt{20+4.47}}
x= \sqrt{20+\sqrt{24.47}}
x= \sqrt{20+4.95}
x= \sqrt{24.95}
x= 4.995 = approx. 5.00

2: From third sqrt: sqrt 20 = 4 and a fraction of 1.
From second sqrt: sqrt (20+4.00 = 24) = 4 and a fraction of 1.
From first sqrt: sqrt (20+4.00 = 24) = 4 and a fraction of 1. which is definitely close to none other than 5.

3: Using POE:

A: it cannot be 20 cuz for 20, the value under root must be 400, which is impossible. so A is ruled out.
B: It could be 5 as done above in method 2.
3. 2 is also not possible because even if we consider first 20 under root, the value must not be smaller than 4.
4. 8 is not possible because for 8, the value under root must be 64. Even if we add up all three 20s, the sum would not be more than 60. so it is also not possible. So left with 5.

So B make sense.
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Re: square root [#permalink] New post 04 Feb 2009, 11:39
If so, shouldn't the question be "What is the approximate value of x?" Just curious :wink:
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Re: square root [#permalink] New post 02 Oct 2009, 08:49
x=root of 20+x
x^2=20+x
x^2-x=20
solving we get x=5,-4
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Re: square root [#permalink] New post 02 Oct 2009, 11:53
boiled it down to roughly 5.5ish. 8 is too high. must be 5.
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Re: square root [#permalink] New post 29 Sep 2010, 05:15
Bunuel, would you be so kind and look at this question. Is there any other way to solve it rather than elimination? Can you describe elimination in greater detail? Thank you.
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Re: square root [#permalink] New post 29 Sep 2010, 06:45
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nonameee wrote:
Bunuel, would you be so kind and look at this question. Is there any other way to solve it rather than elimination? Can you describe elimination in greater detail? Thank you.


Find the value of x

x= \sqrt{20+\sqrt{20+\sqrt{20}}}

1. 20
2. 5
3. 2
4. 8

Question should be what is the approximate value of x.

Obviously answer choice C (2) is out as \sqrt{20+some \ #}>4.

Now, 4<\sqrt{20}<5:
x= \sqrt{20+\sqrt{20+\sqrt{20}}}= \sqrt{20+\sqrt{20+(# \ less \ than \ 5)}}= \sqrt{20+\sqrt{# \ less \ than \ 25}}= \sqrt{20+(# \ less \ than \ 5)}=\sqrt{# \ less \ than \ 25}=# \ less \ than \ 5\approx{5}.

Answer: B.

Next, exactly 5 to be the correct answer question should be:

If the expression x=\sqrt{20+{\sqrt{20+\sqrt{20+\sqrt{20+...}}}}} extends to an infinite number of roots and converges to a positive number x, what is x?

x=\sqrt{20+{\sqrt{20+\sqrt{20+\sqrt{20+...}}}}} --> x=\sqrt{20+({\sqrt{20+\sqrt{20+\sqrt{20+...})}}}}, as the expression under square root extends infinitely, then expression in brackets would equal to x itself so we can rewrite given expression as x=\sqrt{20+x}. Square both sides x^2=20+x --> x=5 or x=-4. As given that x>0 then only one solution is valid: x=5.

Hope it helps.
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Re: square root [#permalink] New post 29 Sep 2010, 08:09
+1 B
I think the best way to solve this question is by watching the options, don't you think?
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Re: square root [#permalink] New post 04 Oct 2010, 02:31
Bunuel, thank you very much.
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Re: Find the value of x [#permalink] New post 22 Sep 2013, 02:55
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Re: Find the value of x [#permalink] New post 22 Sep 2013, 03:02
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Re: Find the value of x   [#permalink] 22 Sep 2013, 03:02
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