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If x<0, which of the following must be true?
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27 Jun 2017, 00:38
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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?
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27 Jun 2017, 01:12
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?
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Updated on: 27 Jun 2017, 05:15
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 \(x^21>0\). 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?
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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 sve. 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?
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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?
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05 Jul 2017, 06:33
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?
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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?
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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.hereThe 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?
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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.hereThe 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?
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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?
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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?
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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?
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