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Math Expert V
Joined: 02 Sep 2009
Posts: 55714

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Difficulty:   25% (medium)

Question Stats: 71% (00:48) correct 29% (01:15) wrong based on 21 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 V
Joined: 02 Sep 2009
Posts: 55714

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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: $$x=3$$.

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I think this is a high-quality question and I agree with explanation.
Intern  B
Joined: 25 Aug 2007
Posts: 45

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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  B
Joined: 25 Aug 2007
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Any takers?
Retired Moderator V
Status: Long way to go!
Joined: 10 Oct 2016
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Location: Viet Nam

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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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Intern  B
Joined: 25 Aug 2007
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Thanks - for some reason I didn't sum it up before taking root.
Manager  S
Joined: 02 May 2016
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Concentration: Entrepreneurship
GRE 1: Q163 V154 WE: Information Technology (Computer Software)

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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 $$x$$
Intern  B
Joined: 25 Jun 2017
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GMAT 1: 660 Q45 V36 GPA: 3.5
WE: Human Resources (Consumer Products)

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Bunuel wrote:
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: $$x=3$$.

Can you explain why the expression in brackets would equal to x? I don't conceptually understand why if the expression under the square root extends infinitely, it equals to x.
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Intern  B
Joined: 11 Nov 2018
Posts: 2

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Hi rodri102,

Since x is that infinite number given to us in the question. what we are doing is replacing that infinite value in the question with X to get to a result.

Thanks. Re: M26-06   [#permalink] 12 Jun 2019, 01:17
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