Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Square root and Modulus [#permalink]
02 Sep 2010, 13:07

1

This post received KUDOS

Expert's post

mbafall2011 wrote:

udaymathapati wrote:

If x < 0, then \sqrt{-x} •|x|) is A. -x B. -1 C. 1 D. x E. \sqrt{x}

what is the source of this question. I havent seen any gmat question testing imaginary numbers

GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers. So you won't see any question involving imaginary numbers.

This question also does not involve imaginary numbers as expression under the square root is non-negative (actually it's positive): we have \(\sqrt{-x*|x|}\) --> as \(x<0\) then \(-x=positive\) and \(|x|=positive\), so \(\sqrt{-x*|x|}=\sqrt{positive*positive}=\sqrt{positive}\).

Re: Square root and Modulus [#permalink]
08 Dec 2010, 11:34

1

This post received KUDOS

Expert's post

shrive555 wrote:

Bunel: i've already seen all the explanation just one more question.

as If \(x<0\), then \(\sqrt{-x*|x|}\) Ans is -x is Answer of the question depends on the condition x<0 or it depends on the sqrt (even root)

lets keep the condition same i.e x<0 and take odd root say cube root. i.e \(\sqrt[3]{-x*|x}|\) . what would be the answer, would it be x then ?

Thanks

About the odd roots: odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).

So, if given that \(x<0\) then \(|x|=-x\) and \(-x*|x|=(-x)*(-x)=positive*positive=x^2\), thus odd root from positive \(x^2\) will be positive.

But \(\sqrt[3]{-x*|x}|\) will equal neither to x nor to -x: \(\sqrt[3]{-x*|x|}=\sqrt[3]{x^2}=x^{\frac{2}{3}}\), for example if \(x=-8<0\) then \(\sqrt[3]{-x*|x|}=\sqrt[3]{x^2}=\sqrt[3]{64}=4\).

If it were \(x<0\) and \(\sqrt[3]{-x^2*|x}|=?\), then \(\sqrt[3]{-x^2*|x}|=\sqrt[3]{-x^2*(-x)}=\sqrt[3]{x^3}=x<0\). Or substitute the value let \(x=-5<0\) --> \(\sqrt[3]{-x^2*|x}|=\sqrt[3]{-25*5}=\sqrt[3]{-125}=-5=x<0\).

Now, back to the original question:

If x<0, then \(\sqrt{-x*|x|}\) equals: A. \(-x\) B. \(-1\) C. \(1\) D. \(x\) E. \(\sqrt{x}\)

As square root function can not give negative result, then options -1 (B) and x (D) can not be the answers as they are negative. Also \(\sqrt{x}\) (E) can not be the answer as even root from negative number is undefined for GMAT. 1 (C) also can not be the answer as for different values of x the answer will be different, so it can not be some specific value. So we are left with A.

Now, if it were x>0 instead of x<0 then the question would be flawed as in this case the expression under the even root would be negative (-x*|x|=negative*positive=negative).

Re: Square root and Modulus [#permalink]
09 Dec 2010, 04:36

1

This post was BOOKMARKED

Friends,

In GMAT, to avoid the confusion with the negative #s, i use my own and simple technique that is as below.

PLEASE NOTE THAT: DEALING WITH THE INEQUALITIES/SQUARE-ROOTS/MODULUS THAT CONTAIN +VE #s IS EASIER THAN DEALING WITH THOSE CONTAIN -VE#S.

Given: x is a -ve I take: y as a +ve #. now i write -x = y

FROM NOW ON I AM GONNA DEAL WITH THE +VE # (i.e.y) ONLY. I CAN NOT BE TRAPPED BY THE TESTMAKER.

qtn: \(\sqrt{-x*|x|}\) = ? ==> \(\sqrt{y*|-y|}\) ==> \(\sqrt{y*y}\) as |-#|=# ==> \(\sqrt{y^2}\) ==> \(y\) (but not -y as Y is a positive #) ==> \(-x\) Answer "A"

Re: Square root and Modulus [#permalink]
30 Dec 2010, 14:20

Expert's post

marijose wrote:

I still do not understend why the x is negative. Is it because of the condition at the begining or because of the absolute value that is in the question?

here is whay I see. SQ ( -(-x)) *|-x|

you get... sq(x*x) therefore the answer is x, do you add the negative sign because the problem starts out saying that x is negative or is there a rule I am not seeing?

I was under the impression that the square root of any number cannot by negative or is this incorrect?

regards

x is negative because it's given in the stem that it's negative (if x<0...). Next, knowing that x is negative doesn't mean that you should write -x instead of x.

As for the square root: yes, even roots can not give negative result, so when you say that \(\sqrt{-x*|x|}\) equals to x, you are basically saying that square root of an expression equals to the negative number (as x is negative) which can not be true. Now, the answer is -x or -(negative)=positive, so square root of an expression equals to positive number which can be true. Also \(\sqrt{x^2}\) equals to |x| not to x. It's all explained above, so please read the solutions again and ask if anything needs further clarification.

Re: Square root and Modulus [#permalink]
30 Dec 2010, 14:47

Expert's post

marijose wrote:

Ive read the explanations .. sorry to keep bugging but

lets say you plug in a a -5. by doing all the calculations you get that the SQ(-(-5)*|-5|) = sq(5*5) = 5 so because and only because the question states that x is negative we have to change the value to -5. I this the correct reasoning?

I'm not sure understood your question.

What do you mean "change the value to -5"? If \(x=-5<0\) then \(\sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5\) so the answer is 5 or \(-x\), as \(-x=-(-5)=5\).

Re: Square root and Modulus [#permalink]
31 Dec 2010, 08:06

Expert's post

shrive555 wrote:

marijose wrote:

Ive read the explanations .. sorry to keep bugging but

lets say you plug in a a -5. by doing all the calculations you get that the SQ(-(-5)*|-5|) = sq(5*5) = 5 so because and only because the question states that x is negative we have to change the value to -5. I this the correct reasoning?

you don't have to plug in -5. Just Plug in 5, keep in mind that x<0 so it will make 5 as -5 automatically. so when you get the final answer after taking square root of 25 = 5. due to given condition of x<0 . you have to pick -5 or (negative value)from the answer choices. The only trick is " x<0 "

i was stuck in the same point

What do you mean by "it will make 5 as -5 automatically"???

And again: given that \(x<0\). If you want to solve this question by number plugging method you should plug negative values of \(x\) --> if \(x=-5<0\) then \(\sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5\) so the answer is 5 or \(-x\), as \(-x=-(-5)=5\). _________________

Re: Square root and Modulus [#permalink]
31 Dec 2010, 10:33

Bunuel wrote:

shrive555 wrote:

marijose wrote:

Ive read the explanations .. sorry to keep bugging but

lets say you plug in a a -5. by doing all the calculations you get that the SQ(-(-5)*|-5|) = sq(5*5) = 5 so because and only because the question states that x is negative we have to change the value to -5. I this the correct reasoning?

you don't have to plug in -5. Just Plug in 5, keep in mind that x<0 so it will make 5 as -5 automatically. so when you get the final answer after taking square root of 25 = 5. due to given condition of x<0 . you have to pick -5 or (negative value)from the answer choices. The only trick is " x<0 "

i was stuck in the same point

What do you mean by "it will make 5 as -5 automatically"???

And again: given that \(x<0\). If you want to solve this question by number plugging method you should plug negative values of \(x\) --> if \(x=-5<0\) then \(\sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5\) so the answer is 5 or \(-x\), as \(-x=-(-5)=5\).

Since x=5 but x<0 so x=-5 probably marijose is confusing the negative sign - with negative x .... \(\sqrt{-(x)*|x|}\) _________________

Re: Square root and Modulus [#permalink]
31 Dec 2010, 11:31

Bunuel wrote:

marijose wrote:

I still do not understend why the x is negative. Is it because of the condition at the begining or because of the absolute value that is in the question?

here is whay I see. SQ ( -(-x)) *|-x|

you get... sq(x*x) therefore the answer is x, do you add the negative sign because the problem starts out saying that x is negative or is there a rule I am not seeing?

I was under the impression that the square root of any number cannot by negative or is this incorrect?

regards

x is negative because it's given in the stem that it's negative (if x<0...). Next, knowing that x is negative doesn't mean that you should write -x instead of x.

This is what i meant but messed up in mathematics lingo _________________

Re: Is the square root always positive? [#permalink]
01 Feb 2011, 05:23

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

Merging similar topics.

karenhipol wrote:

I encountered this question in gmat prep:

If x is < 0, what is the Square root of -x/x/

a. 0 b. 1 c. x d. -x e. -1

The official answer is -x but I answered x because I thought the answer should be positive.

Should I consider however, that since x <0 then I should multiply it by -1 to get a positive number so the answer is - x ( to get -(-x) = +x.

Please let me know If I reasoned this out correctly.

Thanks

Two things:

1. Square root function can not give negative result: \(\sqrt{some \ expression}\geq{0}\) so square root of a non-negative number is not always positive, it's never negative (always non-negative).

2. As \(x<0\) then \(-x=-negative=positive\) so the answer is still positive.

For the complete solution refer to the above posts. _________________

Type of Visa: You will be applying for a Non-Immigrant F-1 (Student) US Visa. Applying for a Visa: Create an account on: https://cgifederal.secure.force.com/?language=Englishcountry=India Complete...

I started running back in 2005. I finally conquered what seemed impossible. Not sure when I would be able to do full marathon, but this will do for now...