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M26-06

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M26-06  [#permalink]

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New post 16 Sep 2014, 01:24
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60% (00:22) correct 40% (00:38) wrong based on 25 sessions

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If \(x\) is a positive number and equals to \(\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}\), where the given expression extends to an infinite number of roots, then what is the value of \(x\)?

A. \(\sqrt{6}\)
B. 3
C. \(1+\sqrt{6}\)
D. \(2\sqrt{3}\)
E. 6

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Re M26-06  [#permalink]

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New post 16 Sep 2014, 01:24
Official Solution:

If \(x\) is a positive number and equals to \(\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}\), where the given expression extends to an infinite number of roots, then what is the value of \(x\)?

A. \(\sqrt{6}\)
B. 3
C. \(1+\sqrt{6}\)
D. \(2\sqrt{3}\)
E. 6

Given: \(x \gt 0\) and \(x=\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}\). Re-write it as: \(x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}\), as the expression under the square root extends infinitely then expression in brackets would equal to \(x\) itself and we can safely replace it with \(x\) and rewrite the given expression as \(x=\sqrt{6+x}\). Square both sides: \(x^2=6+x\). Now, re-arrange and factorize \((x+2)(x-3)=0\), so \(x=-2\) or \(x=3\), but since \(x \gt 0\) then: \(x=3\).

Answer: B
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Re M26-06  [#permalink]

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New post 31 Aug 2016, 04:41
I think this is a high-quality question and I agree with explanation.
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Re: M26-06  [#permalink]

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New post 01 Jan 2017, 12:10
I am missing something....
Square root of 6 is about 2.45
Double square root of 6 is about 1.57
Triple square root of 6 is about 1.25
Sum of these three numbers is greater than 5. How come the answer is 3?
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Re: M26-06  [#permalink]

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New post 06 Jan 2017, 19:01
Any takers?
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Re: M26-06  [#permalink]

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New post 06 Jan 2017, 22:51
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inter wrote:
I am missing something....
Square root of 6 is about 2.45
Double square root of 6 is about 1.57
Triple square root of 6 is about 1.25
Sum of these three numbers is greater than 5. How come the answer is 3?



\(\sqrt{6} \approx 2.4495\)
\(\sqrt{6+\sqrt{6}} \approx 2.9068\)
\(\sqrt{6+\sqrt{6+\sqrt{6}}} \approx 2.98443\)
\(\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}} \approx 2.9974\)

You could see that the value is getting nearer to 3
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Re: M26-06  [#permalink]

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New post 07 Jan 2017, 13:05
Thanks - for some reason I didn't sum it up before taking root.
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Re: M26-06  [#permalink]

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New post 14 Jul 2017, 07:09
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I do understand this:
nguyendinhtuong wrote:
You could see that the value is getting nearer to 3


Is this some kind of rule or something? :
Bunuel wrote:
\(x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}\), as the expression under the square root extends infinitely then expression in brackets would equal to \(x\) itself and we can safely replace it with \(x\)
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Re: M26-06  [#permalink]

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New post 09 Jun 2018, 07:05
Hi Bunuel,

Can you expand upon why we can assume the infinite addition of square root 6 is equal to x? Is this because the infinite sum can never actually surpass x?

Thanks in advance.
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Re: M26-06 &nbs [#permalink] 09 Jun 2018, 07:05
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M26-06

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