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# For which of the following values of x is

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Intern
Joined: 06 Aug 2007
Posts: 30
Location: Montreal
For which of the following values of x is [#permalink]

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Updated on: 16 May 2012, 03:53
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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

Originally posted by Safiya on 24 Aug 2010, 12:41.
Last edited by Bunuel on 16 May 2012, 03:53, edited 1 time in total.
Edited the question
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Joined: 02 Sep 2009
Posts: 46328
Re: NOT defined as a real number [#permalink]

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24 Aug 2010, 12:49
2
8
Safiya wrote:
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.

Hope it helps.
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Re: NOT defined as a real number [#permalink]

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24 Aug 2010, 13:04
It helped indeed, thank you very much!
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Re: NOT defined as a real number [#permalink]

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20 Dec 2010, 01:18
Bunuel -

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.
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Re: NOT defined as a real number [#permalink]

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20 Dec 2010, 01:35
6
3
tonebeeze wrote:
Bunuel -

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 $$4-\sqrt{x}$$: $$1-\sqrt{4-\sqrt{x_{min}}}=1-\sqrt{4-\sqrt{4}}=1-1.41=-0.41<0$$.

Hope it's clear.
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Re: NOT defined as a real number [#permalink]

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20 Dec 2010, 02:02
Great visual explanation. I now understand. Thanks for taking the time to help.
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Joined: 29 Jan 2011
Posts: 317
Re: NOT defined as a real number [#permalink]

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11 Jul 2011, 04:04
Bunuel wrote:
Safiya wrote:
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.

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?
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Re: NOT defined as a real number [#permalink]

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11 Jul 2011, 05:36
@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.
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Re: For which of the following values of x is [#permalink]

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16 May 2012, 04:41
1
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.

Hope it's clear.
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Re: For which of the following values of x is [#permalink]

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16 May 2012, 05:58
O yes,yes,yes... thanks! Sometimes I just look at numbers and don't see anything no matter how obvious it is.
I need a break!
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Re: For which of the following values of x is [#permalink]

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05 Jun 2013, 04:28
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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05 Jun 2013, 10:09
1
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.
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Re: For which of the following values of x is [#permalink]

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15 Jul 2015, 21:29
Safiya wrote:
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

CONCEPT: Any Number is considered NON-REAL if the number under the square root is negative (i.e.√(-ve)) .

Such questions require us to check the smallest or highest values among option because at the extremes only the number will result in Real or NON-Real

Here, We need to check the highest value because x is being subtracted from other numbers

@x=5 (Option E)

√[1−√(2−√x)] = √[1−√(2−√5)] = √[1−√(2−2.2)] = √[1−√(-ve)] i.e. NON-REAL Number
Hence, Correct Option

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

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11 Jun 2016, 10:48
Nwsmith11 wrote:
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.

How would you solve this (the question mentioned by buneul above):
Consider another example: For which of the following values of x 1−4−x√‾‾‾‾‾‾‾√‾‾‾‾‾‾‾‾‾‾‾‾‾√1−4−x is NOT defined as a real number?

A. 16
B. 12
C.10
D. 9
E. 4
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Joined: 09 Mar 2016
Posts: 615
For which of the following values of x is [#permalink]

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09 Jun 2018, 04:46
Bunuel wrote:
tonebeeze wrote:
Bunuel -

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 $$4-\sqrt{x}$$: $$1-\sqrt{4-\sqrt{x_{min}}}=1-\sqrt{4-\sqrt{4}}=1-1.41=-0.41<0$$.

Hope it's clear.

hi pushpitkc

can you please explain the technique of taking square root, when one radical is under another radical sign ? i am kinda confused:-)

do we take square root $$\sqrt{1}$$ by multiplying by exponent 2 ? ?

then we get $$1-\sqrt{4-\sqrt{x}}$$ and after we get this

$$1-2-\sqrt{x}$$ and then again multiply by exponent 2 and get $$1-2-x$$
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For which of the following values of x is [#permalink]

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09 Jun 2018, 05:00
1
dave13 wrote:
Bunuel wrote:
tonebeeze wrote:
Bunuel -

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 $$4-\sqrt{x}$$: $$1-\sqrt{4-\sqrt{x_{min}}}=1-\sqrt{4-\sqrt{4}}=1-1.41=-0.41<0$$.

Hope it's clear.

hi pushpitkc

can you please explain the technique of taking square root, when one radical is under another radical sign ? i am kinda confused:-)

do we take square root $$\sqrt{1}$$ by multiplying by exponent 2 ? ?

then we get $$1-\sqrt{4-\sqrt{x}}$$ and after we get this

$$1-2-\sqrt{x}$$ and then again multiply by exponent 2 and get $$1-2-x$$

Hi dave13

Whenever we have a square root under another square root, for example $$\sqrt{7 + \sqrt{81}}$$

First step will be to do the square root of the number underneath.
The expression now becomes $$\sqrt{7 + 9}$$ - because $$\sqrt{81} = 9$$

Now, after adding the two integers(in this case) under the radical sign,
we need to perform the second square root operation ($$\sqrt{16} = 4$$)

Hope this helps you!
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For which of the following values of x is [#permalink]

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Updated on: 09 Jun 2018, 07:56
pushpitkc thanks for explanation

if you say "First step will be to do the square root of the number underneath."

then why Bunuel after this $$\sqrt{1-\sqrt{4-\sqrt{x}}}$$ gets this $$1-\sqrt{4-\sqrt{x}}$$ why he takes square root of 1 and not x which is underneath

hello generis, i ve decided to have rest and relax by switching to number properties questions today

I thought perhaps you can explain ?:) I still dont get, though pushpitkc tried to explain

pushpitkc says "First step will be to do the square root of the number underneath."

this is how i tackled question

$$(\sqrt{1-\sqrt{4-\sqrt{x}}})^2$$ ( square to get rid of the root )

we get this $$(1-\sqrt{4-\sqrt{x}})^2$$ (square again)

we get this $$1-4-\sqrt{x}$$ ---> $$(-3-\sqrt{x} )^2$$ square again

we get $$9-x$$:? now what ? ...

that`s strange when we square (as you see above) first we get rid of external radical signs for example, Bunuel after this $$\sqrt{1-\sqrt{4-\sqrt{x}}}$$ get this $$1-\sqrt{4-\sqrt{x}}$$ this means he squared and got rid of bigger radical sign... on the other hand pushpitkc says First step will be to do the square root of the number underneath which is $$x$$

can you please investigate this case generis007

Originally posted by dave13 on 09 Jun 2018, 05:22.
Last edited by dave13 on 09 Jun 2018, 07:56, edited 1 time in total.
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Re: For which of the following values of x is [#permalink]

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09 Jun 2018, 05:37
dave13 wrote:
pushpitkc thanks for explanation

if you say "First step will be to do the square root of the number underneath."

then why Bunuel after this $$\sqrt{1-\sqrt{4-\sqrt{x}}}$$ gets this $$1-\sqrt{4-\sqrt{x}}$$ why he takes square root of 1 and not x which is underneath

Hi dave13

If you read the solution that Bunuel has written, it is also saying exactly what I said.

Read this post for more clarity:
https://gmatclub.com/forum/for-which-of ... ml#p838260

Hope this helps you!
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Re: For which of the following values of x is [#permalink]

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12 Jun 2018, 06:10
pushpitkc
what if it were like this $$\sqrt{7 + \sqrt{11}}$$ how would you solve it ?
thanks!:)
Re: For which of the following values of x is   [#permalink] 12 Jun 2018, 06:10

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