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If x<0, which of the following must be true?

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If x<0, which of the following must be true?  [#permalink]

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New post 27 Jun 2017, 00:38
1
10
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

52% (01:38) correct 48% (01:40) wrong based on 356 sessions

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Re: If x<0, which of the following must be true?  [#permalink]

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New post 27 Jun 2017, 01:12
1
Bunuel wrote:
If x < 0, which of the following must be true?

I. \(x<\sqrt{−x}\)

II. \(x^2>\sqrt{x^2}\)

III. \(x=−\sqrt{x^2}\)


A. I only
B. I and II only
C. I and III only
D. I, II, and III
E. None of the above


I
\(x<\sqrt{-x}\)

Take x=-2 and it does not fail
Take x= -1/2 and it does not fail


II
\(x^2>\sqrt{x^2}\)

Take x = -1/2 and it fails

III
\(x= -\sqrt{x^2}\)
Take x=-2 and it does not fail
Take x= -1/2 and it does not fail

Hence C
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If x<0, which of the following must be true?  [#permalink]

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New post Updated on: 27 Jun 2017, 05:15
1
Bunuel wrote:
If x < 0, which of the following must be true?

I. \(x<\sqrt{−x}\)

II. \(x^2>\sqrt{x^2}\)

III. \(x=−\sqrt{x^2}\)


A. I only
B. I and II only
C. I and III only
D. I, II, and III
E. None of the above


Another approach :-D

I \(x<\sqrt{−x}\), squaring both sides

\(x^2<-x\). \(x^2 > 0\) and cannot be less than a negative number,

So \(–x>0\) or \(x<0\). Hence must be true

II on squaring both sides we get \(x^4>x^2\)

or \(x^2>1\) or \(x^2-1>0\). Solving the inequality will give you a range \(-1>x\) and \(x>1\) so clearly our question stem must not be true

III\(x=−\sqrt{x^2}\)

square root is always positive, this directly implies that x = - (Some positive No). Hence x<0. Must be true

Option C

Originally posted by niks18 on 27 Jun 2017, 02:08.
Last edited by niks18 on 27 Jun 2017, 05:15, edited 1 time in total.
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Re: If x<0, which of the following must be true?  [#permalink]

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New post 27 Jun 2017, 03:04
Bunuel wrote:
If x < 0, which of the following must be true?

I. \(x<\sqrt{−x}\)

II. \(x^2>\sqrt{x^2}\)

III. \(x=−\sqrt{x^2}\)


A. I only
B. I and II only
C. I and III only
D. I, II, and III
E. None of the above



x < 0 which means x i s-ve.

I. \(x<\sqrt{−x}\)
Since x is -ve and \(\sqrt{−x}\) is +ve.
So it is CORRECT.

II. \(x^2>\sqrt{x^2}\) this is true for x<1
while for 0<x<1, \(x^2<\sqrt{x^2}\)

It is not correct for all values of x<0.
Henc INCORRECT.


III. \(x=−\sqrt{x^2}\)
x=-|x|

Since x<0 , this is CORRECT


Answer C.
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Re: If x<0, which of the following must be true?  [#permalink]

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New post 04 Jul 2017, 01:45
This problem is testing a very important concept of sign of square root
Even roots have only a positive value on the GMAT.
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Re: If x<0, which of the following must be true?  [#permalink]

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New post 05 Jul 2017, 06:33
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arvind910619 wrote:
This problem is testing a very important concept of sign of square root
Even roots have only a positive value on the GMAT.


Yes roots have only a positive value not only on GMAT but everywhere in the world.
\(\sqrt{x^2} = |x|\)

And this concept is more likely to be tested on GMAT in Data sufficiency questions.
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If x<0, which of the following must be true?  [#permalink]

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New post 27 Dec 2017, 08:20
Bunuel wrote:
If x < 0, which of the following must be true?

I. \(x<\sqrt{−x}\)

II. \(x^2>\sqrt{x^2}\)

III. \(x=−\sqrt{x^2}\)


A. I only
B. I and II only
C. I and III only
D. I, II, and III
E. None of the above


Bunuel please correct me if i am wrong
isn't \sqrt{x} =|x|

and if it is then by that logic how is this statement correct \(x<\sqrt{−x}\)

Bunuel VeritasPrepKarishma I am pretty sure i have seen you guys using \sqrt{x} =|x|
in answers . Then please tell me how can statement 1 be correct
thank you
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If x<0, which of the following must be true?  [#permalink]

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New post 27 Dec 2017, 08:26
Ravindra.here wrote:
Bunuel wrote:
If x < 0, which of the following must be true?

I. \(x<\sqrt{−x}\)

II. \(x^2>\sqrt{x^2}\)

III. \(x=−\sqrt{x^2}\)


A. I only
B. I and II only
C. I and III only
D. I, II, and III
E. None of the above


Bunuel please correct me if i am wrong
isn't \sqrt{x} =|x|

and if it is then by that logic how is this statement correct \(x<\sqrt{−x}\)

Bunuel VeritasPrepKarishma I am pretty sure i have seen you guys using \sqrt{x} =|x|
in answers . Then please tell me how can statement 1 be correct
thank you


Hi Ravindra.here

The question mentions that \(x<0\) i.e negative so \(-x\) will be positive

if \(x=-3\), then \(-x=-(-3)=3\)

so \(x<\sqrt{-x}\) is possible
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If x<0, which of the following must be true?  [#permalink]

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New post 27 Dec 2017, 09:16
niks18 wrote:
Ravindra.here wrote:
Bunuel wrote:
If x < 0, which of the following must be true?

I. \(x<\sqrt{−x}\)

II. \(x^2>\sqrt{x^2}\)

III. \(x=−\sqrt{x^2}\)


A. I only
B. I and II only
C. I and III only
D. I, II, and III
E. None of the above


Bunuel please correct me if i am wrong
isn't \sqrt{x} =|x|

and if it is then by that logic how is this statement correct \(x<\sqrt{−x}\)

Bunuel VeritasPrepKarishma I am pretty sure i have seen you guys using \sqrt{x} =|x|
in answers . Then please tell me how can statement 1 be correct
thank you


Hi Ravindra.here

The question mentions that \(x<0\) i.e negative so \(-x\) will be positive

if \(x=-3\), then \(-x=-(-3)=3\)

so \(x<\sqrt{-x}\) is possible


Hi niks18 my confusion is this one
\sqrt{-x} is positive here and as \sqrt{x}=|x|
here \sqrt{-x} = |-x|=|x|
now lets take x=-9
Therefore \sqrt{-x}= |3|
this means now two values 3 and -3
Here for both values \(x<\sqrt{−x}\)
but what if i take x=-1/4
now \sqrt{-x}= |-1/2|=|1/2|
this yeilds two values again

for 1/2
\(x<\sqrt{−x}\) holds

for -1/2 x will be greater than \sqrt{−x}

Please correct me if i am wrong
thank you
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Re: If x<0, which of the following must be true?  [#permalink]

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New post 27 Dec 2017, 09:26
Ravindra.here wrote:
Bunuel wrote:
If x < 0, which of the following must be true?

I. \(x<\sqrt{−x}\)

II. \(x^2>\sqrt{x^2}\)

III. \(x=−\sqrt{x^2}\)


A. I only
B. I and II only
C. I and III only
D. I, II, and III
E. None of the above


Bunuel please correct me if i am wrong
isn't \sqrt{x} =|x|

and if it is then by that logic how is this statement correct \(x<\sqrt{−x}\)

Bunuel VeritasPrepKarishma I am pretty sure i have seen you guys using \sqrt{x} =|x|
in answers . Then please tell me how can statement 1 be correct
thank you


The red part is not correct.

\(\sqrt{x^2}=|x|\), NOT \(\sqrt{x}=|x|\).

Next, \(x<\sqrt{−x}\) is correct because \((x=negative)<(\sqrt{−x}=\sqrt{positive}=positive)\)

Hope it's clear.
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Re: If x<0, which of the following must be true?  [#permalink]

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New post 27 Dec 2017, 09:33
Ravindra.here wrote:
Hi niks18 my confusion is this one
\sqrt{-x} is positive here and as \sqrt{x}=|x|
here \sqrt{-x} = |-x|=|x|
now lets take x=-9
Therefore \sqrt{-x}= |3|
this means now two values 3 and -3
Here for both values \(x<\sqrt{−x}\)
but what if i take x=-1/4
now \sqrt{-x}= |-1/2|=|1/2|
this yeilds two values again

for 1/2
\(x<\sqrt{−x}\) holds

for -1/2 x will be greater than \sqrt{−x}

Please correct me if i am wrong
thank you


Hi Ravindra.here

\(\sqrt{x^2}=|x|\), note here x is a variable and you do not know whether it is positive or negative, hence both values are possible

for eg. if \(x^2=4\), then \(x=2\) or \(-2\)

Also for GMAT square root is always positive. concept of imaginary number i is not tested in GMAT

so \(\sqrt{9}=3\) (square root of a positive number will be positive) and not \(-3\)

Now coming to this question. Here we know that x is negative. So if \(x=-\frac{1}{2}\), then \(-x=-(-\frac{1}{2})=\frac{1}{2}\)

\(\sqrt{1/2}\) will be a positive number. Here negative number is not possible because square root of positive number will be positive

so here \(\sqrt{-x}\) is a positive number
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Re: If x<0, which of the following must be true?  [#permalink]

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New post 27 Dec 2017, 09:46
niks18 Bunuel ty for your suggestions
That's really kind of you both .
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Re: If x<0, which of the following must be true?   [#permalink] 27 Dec 2017, 09:46
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