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

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

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27 Jun 2017, 00:38
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Difficulty:

65% (hard)

Question Stats:

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

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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

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

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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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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

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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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

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

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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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05 Jul 2017, 06:33
1
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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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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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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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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Posts: 55715
Re: If x<0, which of the following must be true?  [#permalink]

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