If x

Question

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

Luckisnoexcuse · Accepted Answer

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

niks18 · Answer

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

shashankism · Answer

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.

arvind910619 · Answer

This problem is testing a very important concept of sign of square root 
Even roots have only a positive value on the GMAT.

shashankism · Answer

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.

Ravindra.here · Answer

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

niks18 · Answer

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

Ravindra.here · Answer

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

Bunuel · Answer

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.

niks18 · Answer

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

Ravindra.here · Answer

niks18   Bunuel  ty for your suggestions 
That's really kind of you both .

cottonmouth · Answer

If -x>x^2 and say we take x= -2 (because x<0) then this statement doesn't hold. 
-(-2) is not greater than (-2)^2
2 is not greater than 4

Please help clear the confusion.

Bunuel · Answer

Ignore that solution. You cannot square  x<\sqrt{−x}  because x there is negative and we can only square an inequality when both sides are non-negative.

The reason  x<\sqrt{−x}  must be true is that the left hand side, x, is negative while, the right hand side,  \sqrt{−x}= \sqrt{positive} , which is positive. Therefore, (LHS = negative) < (RHS = positive).

Kinshook · Answer

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

I. \(x<\sqrt{−x}\)
\(x < 0 < \sqrt{−x}\)
MUST BE TRUE

II. \(x^2>\sqrt{x^2}\)
or
xˆ2 > |x|
Case 1: x=-.5; xˆ2 = .25; |x| = .5; xˆ2 < |x|
Case 2: x=-2; xˆ2 = 4; |x| = 2; xˆ2 > |x|
COULD BE TRUE

III. \(x=−\sqrt{x^2}\)
or
x = - |x|
Since x < 0
MUST BE TRUE

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

IMO C

ajdevgmat · Answer

I have a doubt here.
for the (iii) option,  x=−\sqrt{x^2}

\sqrt{x^2}  can be equal to   -x   or   x   .

This would mean   x = -x   or   x = x   ? Is my understanding incorrect?

Bunuel · Answer

\sqrt{x^2}=|x| , and since x < 0,  |x| = -x . Therefore, on the right-hand side we have:

−\sqrt{x^2}=-|x|=-(-x)=x

bumpbot · Answer

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

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