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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: \(x=5\).

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 \(\sqrt[2]{20+\sqrt[2]{20+5}\)=\(\sqrt[2]{20+5}\)=\(5\)

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.

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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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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