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

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.

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.

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

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.

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

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

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.

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).

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