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So the approach to problems such as this one is to work your way from the most inner root, out toward the main root? Always keeping track of whether the underlying roots are (1) negative (2) are larger than the main root.

Consider another example: For which of the following values of x \(\sqrt{1-\sqrt{4-\sqrt{x}}}\) is NOT defined as a real number?

A. 16 B. 12 C.10 D. 9 E. 4

First see whether \(4-\sqrt{x}\) could be negative for some value of \(x\) so you should test max value of \(x\): \(4-\sqrt{x_{max}}=4-\sqrt{16}=0\). As it's not negative then see whether \(1-\sqrt{4-\sqrt{x}}\) can be negative for some value of \(x\), so you should test min value of \(x\) to maximize \(4-\sqrt{x}\): \(1-\sqrt{4-\sqrt{x_{min}}}=1-\sqrt{4-\sqrt{4}}=1-1.41=-0.41<0\).

I'd be glad if someone could explain the logic for this question;

For which of the following values of x is

\(\sqrt{1-}\)\(\sqrt{2-}\)\(\sqrt{x}\)

NOT defined as a real number?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

Real Numbers are: Integers, Fractions and Irrational Numbers. Non-real numbers are even roots (such as square roots) of negative numbers.

We have \(\sqrt{1-\sqrt{2-\sqrt{x}}}\). For \({x=5}\) expression becomes:\(\sqrt{1-\sqrt{2-\sqrt{5}}}\) and \(2-\sqrt{5}<0\), thus square root from this expression is not a real number.

Re: For which of the following values of x is [#permalink]

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16 May 2012, 04:41

1

This post received KUDOS

Expert's post

Stiv wrote:

I don't understand how did you came from \(\sqrt{1-\sqrt{2-\sqrt{5}}} to 2-\sqrt{5}<0\) ? I would appreciate if you explain, because I'm obviously missing something.

If \({x=5}\) then the expression becomes:\(\sqrt{1-\sqrt{2-\sqrt{5}}}\). The expression under the second square root is \(2-\sqrt{5}\). Now, since \(2-\sqrt{5}<0\) then the square root from this expression is not a real number.

Re: For which of the following values of x is [#permalink]

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05 Jun 2013, 10:09

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Can we not simply take each value (calculated or not) under a radical, inside to out, and test if it's less than zero?

\(x<0\) -- no answer choice satisfies

\(2-\sqrt{x}<0\) -- otherwise reads \(2<\sqrt{x}\) so \(x>2^2\)or 4....answer is E, no need to test [m]1-\sqrt{2-[square_root]x}[/square_root] as we have only one answer choice that meets this criteria. This approach is very similar to what Bunuel did but seems much quicker upfront in this case.

So the approach to problems such as this one is to work your way from the most inner root, out toward the main root? Always keeping track of whether the underlying roots are (1) negative (2) are larger than the main root.

I'd be glad if someone could explain the logic for this question;

For which of the following values of x is

\(\sqrt{1-}\)\(\sqrt{2-}\)\(\sqrt{x}\)

NOT defined as a real number?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

Real Numbers are: Integers, Fractions and Irrational Numbers. Non-real numbers are even roots (such as square roots) of negative numbers.

We have \(\sqrt{1-\sqrt{2-\sqrt{x}}}\). For \({x=5}\) expression becomes:\(\sqrt{1-\sqrt{2-\sqrt{5}}}\) and \(2-\sqrt{5}<0\), thus square root from this expression is not a real number.

Answer: E.

Hope it helps.

Why is 4 not the correct answer as it would yield root ( 2 - (root (4)) => root(2 - 2) = 0 .. Is root (0) not a real number?

@siddhans, root(0) = 0, which is a real number, and the question is asking for a value that is not a real number. So 4 is not a correct choice. _________________

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Re: For which of the following values of x is [#permalink]

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16 May 2012, 04:27

I don't understand how did you came from \(\sqrt{1-\sqrt{2-\sqrt{5}}} to 2-\sqrt{5}<0\) ? I would appreciate if you explain, because I'm obviously missing something. _________________

Re: For which of the following values of x is [#permalink]

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11 Jul 2014, 22:19

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Re: For which of the following values of x is [#permalink]

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