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If the expression x=sqrt(2+sqrt(2+sqrt(2+sqrt(2+...) extends [#permalink]

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06 Aug 2010, 01:38

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If the expression \(x=\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}\) extends to an infinite number of roots and converges to a positive number x, what is x?

This question is out of scope for the GMAT, but there's an interesting trick to questions like this. We know that:

x=sqrt(2+sqrt(2+sqrt(2+sqrt(2+...)

Now, notice that the part I've highlighted in red is actually equal to x itself. So we can replace it with x, to get the much simpler equation:

x = sqrt(2 + x) x^2 = 2 + x x^2 - x - 2 = 0 (x - 2)(x + 1) = 0 x = 2 or -1

and since x cannot be negative, x = 2. You won't see anything like this on the GMAT though, so it's for interest only.
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\(x=\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}\) --> \(x=\sqrt{2+({\sqrt{2+\sqrt{2+\sqrt{2+...})}}}}\), as the expression under square root extends infinitely then expression in brackets would equal to \(x\) itself so we can rewrite given expression as \(x=\sqrt{2+x}\). Square both sides \(x^2=2+x\) --> \(x=2\) or \(x=-1\). As given that \(x>1\) then only one solution is valid: \(x=2\).

Re: If the expression x=sqrt(2+sqrt(2+sqrt(2+sqrt(2+...) extends [#permalink]

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06 Aug 2010, 22:45

That's a clever solution, putting x into the equation itself. Sort of recursive, and very elegant. These type of problems give me a hard time.
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Re: If the expression x=sqrt(2+sqrt(2+sqrt(2+sqrt(2+...) extends [#permalink]

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26 Aug 2012, 12:29

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Please let me know if you get the formula. I have to find the sum of the square root of 2 plus Sq root of sq root of 2 plus.... (I have pasted a pic)...

Attachments

File comment: This series goes on forever. I could draw only three twos, but the no is infinite

Sum of infinite root sequence.png [ 7.4 KiB | Viewed 80299 times ]

Re: If the expression x=sqrt(2+sqrt(2+sqrt(2+sqrt(2+...) extends [#permalink]

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27 Feb 2013, 11:12

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The expression sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+.....extends to an infinite number of roots.Which of the following choices most closely approximates the value of this expression?

The expression sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+.....extends to an infinite number of roots.Which of the following choices most closely approximates the value of this expression?

Dear mun23, The trick of this question is to give the entire expression a name --- I am going to call it S, and then I am going to square it.

Attachment:

nested roots.JPG [ 18.63 KiB | Viewed 79592 times ]

Square the expression produces 2 plus a copy of itself --- that's why we can replace it on the other side with S, and then solve for S algebraically: S^2 = 2 + S S^2 - S - 2 = 0 (S - 2)(S + 1) = 1 S = 2 or S = -1 The negative root makes no sense in this context, so S = 2, and the answer = B

Does all this make sense? Mike
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Hi mikemcgarry I am not understanding why the entire expression is given a named?

Dear mun23, This is a standard trick in mathematics. We give the entire expression a name, the name S, a variable, because that allows us to manipulate it algebraically. We want to know the value of the entire expression, so we set the entire expression equal to a variable, then ultimately all we have to do is solve for the value of this variable. Because the variable equals the whole expression, when we know the value of the variable, we also know the value of the whole expression.

This is an extension of the fundamental power of algebra --- when we assign a variable to any unknown quantity, then the whole panoply of algebraic techniques comes to bear on the problem.

Keep in mind that material like this ---- infinitely recursive expressions --- is exceedingly unlike to appear on the GMAT. I have never seen anything like this. If it did appear at all, it would only appear to someone getting virtually everything else correct on the Quant section. Folks in the Q < 45 range will NEVER see a question about this stuff, and even folks in the high 50s would only see it less than 1% of the time.

Does all this make sense?

Mike
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Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

The expression sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+.....extends to an infinite number of roots.Which of the following choices most closely approximates the value of this expression?

\(x=\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}\) --> \(x=\sqrt{2+({\sqrt{2+\sqrt{2+\sqrt{2+...})}}}}\), as the expression under square root extends infinitely then expression in brackets would equal to \(x\) itself so we can rewrite given expression as \(x=\sqrt{2+x}\). Square both sides \(x^2=2+x\) --> \(x=2\) or \(x=-1\). As given that \(x>1\) then only one solution is valid: \(x=2\).

Answer: B.

Shouldn't the answer to this be infinity....I have been looking at this one for about 45 minutes, and I can't figure it out. We start with 2, and then add to that \(\sqrt{2}\), which is about 1.4, then we add to that the square root of the square root of 2, or about 1.18, and then add the square root of the square root of the square root of 2, which is 1.09. The numbers CAN NOT ever be below 1. Just taking it out through 10 cycles, the total is almost 11. And this is an infinite sequence, so the answer is whatever the square root of infinity is. I've looked at the solutions above and they don't make sense to me, at all. The way the problem is written, the answer CANT be any of the options listed

\(x=\sqrt{2+{\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}\) --> \(x=\sqrt{2+({\sqrt{2+\sqrt{2+\sqrt{2+...})}}}}\), as the expression under square root extends infinitely then expression in brackets would equal to \(x\) itself so we can rewrite given expression as \(x=\sqrt{2+x}\). Square both sides \(x^2=2+x\) --> \(x=2\) or \(x=-1\). As given that \(x>1\) then only one solution is valid: \(x=2\).

Answer: B.

Shouldn't the answer to this be infinity....I have been looking at this one for about 45 minutes, and I can't figure it out. We start with 2, and then add to that \(\sqrt{2}\), which is about 1.4, then we add to that the square root of the square root of 2, or about 1.18, and then add the square root of the square root of the square root of 2, which is 1.09. The numbers CAN NOT ever be below 1. Just taking it out through 10 cycles, the total is almost 11. And this is an infinite sequence, so the answer is whatever the square root of infinity is. I've looked at the solutions above and they don't make sense to me, at all. The way the problem is written, the answer CANT be any of the options listed

Consider the examples below: \(\sqrt{2}\approx{1.4}\);

As you can see the result approaches to 2 by decreasing pace. If we extend that to an infinite number of roots the result will be exactly 2.
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Re: If the expression x=sqrt(2+sqrt(2+sqrt(2+sqrt(2+...) extends [#permalink]

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04 Dec 2013, 17:41

jakolik wrote:

If the expression x=sqrt(2+sqrt(2+sqrt(2+sqrt(2+...) extends to an infinite number of roots and converges to a positive number x, what is x?

A. Sqrt(3) B. 2 C. 1+sqrt(2) D. 1+sqrt(3) E. 2*sqrt(3)

Sqrt of 2 is approx 1.4

1.4+2 = 3.4 close enough to 4. Square root of 4 is 2 and then +2 = square root of 4 again, so this will happen indefinetely Hence x= Square root of 4 = 2

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