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# M26-06

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Math Expert
Joined: 02 Sep 2009
Posts: 47983
M26-06  [#permalink]

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16 Sep 2014, 01:24
1
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Difficulty:

35% (medium)

Question Stats:

58% (00:20) correct 42% (00:38) wrong based on 24 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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Math Expert
Joined: 02 Sep 2009
Posts: 47983
Re M26-06  [#permalink]

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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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31 Aug 2016, 04:41
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 25 Aug 2007
Posts: 38
Re: M26-06  [#permalink]

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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?
Intern
Joined: 25 Aug 2007
Posts: 38
Re: M26-06  [#permalink]

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06 Jan 2017, 19:01
Any takers?
Senior CR Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1394
Location: Viet Nam
Re: M26-06  [#permalink]

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06 Jan 2017, 22:51
1
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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07 Jan 2017, 13:05
Thanks - for some reason I didn't sum it up before taking root.
Manager
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Posts: 82
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Concentration: Entrepreneurship
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Re: M26-06  [#permalink]

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14 Jul 2017, 07:09
1
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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Joined: 20 Apr 2018
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Re: M26-06  [#permalink]

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

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