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: What is sqrt of x^2? [#permalink]
14 Dec 2009, 17:32

The problem is we can't simply assume that because \sqrt{x} can't be negative it isn't. Instead, we have to make sure x isn't negative through use of absolute value.

Reading further in the OG, "Every positive number n has two square roots, one positive and the other negative...The two square roots of 9 are 3 and -3."

In \(x^2\), x could be positive or negative for the equation to make sense, but to adhere to the rule you correctly stated, we use absolute value to ensure x is positive.

If x#0, then root(x^2)/x= [#permalink]
14 Dec 2009, 17:47

11

This post received KUDOS

Expert's post

14

This post was BOOKMARKED

If \(x\neq{0}\), then \(\frac{\sqrt{x^2}}{x}=\)

A. -1 B. 0 C. 1 D. x E. \(\frac{|x|}{x}\)

General rule: \(\sqrt{x^2}=|x|\).

When we see the equation of a type: \(y=\sqrt{x^2}\) then \(y=|x|\), which means that \(y\) can not be negative but \(x\) can.

\(y\)can not be negative as \(y=\sqrt{some \ expression}\), and even root from the expression (some value) is never negative (as for GMAT we are dealing only with real numbers).

When the GMAT provides the square root sign for an even root, such as a square root, then the only accepted answer is the positive root.

That is, \(\sqrt{16} = 4\), NOT +4 or -4. In contrast, the equation x^2 = 16 has TWO solutions, +4 and -4.

Even roots have only a positive value on the GMAT.

Odd roots will have the same sign as the base of the root. For example, CBRT(64) = 4, CBRT(-27) = -3.

In our original question we have: \(\frac{\sqrt{x^2}}{x}=\frac{|x|}{x}\), if we knew that \(x\) is positive then the answer would be 1, if we knew that \(x\) is negative the answer would be -1. BUT we don't know the sign of x, hence we cannot simplify expression \(\frac{|x|}{x}\) further.

Re: What is sqrt of x^2? [#permalink]
14 Dec 2009, 18:41

Bunuel wrote:

General rule: \(\sqrt{x^2}=|x|\).

When we see the equation of a type: \(y=\sqrt{x^2}\) then \(y=|x|\), which means that \(y\) can not be negative but \(x\) can.

\(y\)can not be negative as \(y=\sqrt{some expression}\), and even root from the expression (some value) is never negative (as for GMAT we are dealing only with real numbers).

When the GMAT provides the square root sign for an even root, such as a square root, then the only accepted answer is the positive root.

That is, \(\sqrt{16} = 4\), NOT +4 or -4. In contrast, the equation x^2 = 16 has TWO solutions, +4 and -4.

Even roots have only a positive value on the GMAT.

Odd roots will have the same sign as the base of the root. For example, CBRT(64) = 4, CBRT(-27) = -3.

In our original question we have: \(\frac{\sqrt{x^2}}{x}=\frac{|x|}{x}\), if we knew that x is positive then the answer would be 1, if we knew that x is negative the answer would be -1. BUT we don't know the sign of x, hence we can not simplify expression \(\frac{|x|}{x}\) further.

Hope it's clear.

Thank you. Much clear now. Also, Since x is Not equal to zero, is there a value of x for which this equation \(\frac{|x|}{x}\) WILL NOT simplify to \(1\)?

Re: What is sqrt of x^2? [#permalink]
14 Dec 2009, 19:50

Fremontian wrote:

Bunuel wrote:

General rule: \(\sqrt{x^2}=|x|\).

When we see the equation of a type: \(y=\sqrt{x^2}\) then \(y=|x|\), which means that \(y\) can not be negative but \(x\) can.

\(y\)can not be negative as \(y=\sqrt{some expression}\), and even root from the expression (some value) is never negative (as for GMAT we are dealing only with real numbers).

When the GMAT provides the square root sign for an even root, such as a square root, then the only accepted answer is the positive root.

That is, \(\sqrt{16} = 4\), NOT +4 or -4. In contrast, the equation x^2 = 16 has TWO solutions, +4 and -4.

Even roots have only a positive value on the GMAT.

Odd roots will have the same sign as the base of the root. For example, CBRT(64) = 4, CBRT(-27) = -3.

In our original question we have: \(\frac{\sqrt{x^2}}{x}=\frac{|x|}{x}\), if we knew that x is positive then the answer would be 1, if we knew that x is negative the answer would be -1. BUT we don't know the sign of x, hence we can not simplify expression \(\frac{|x|}{x}\) further.

Hope it's clear.

Thank you. Much clear now. Also, Since x is Not equal to zero, is there a value of x for which this equation \(\frac{|x|}{x}\) WILL NOT simplify to \(1\)?

if x=-1 then the exp will -1 _________________

GMAT is not a game for losers , and the moment u decide to appear for it u are no more a loser........ITS A BRAIN GAME

Re: What is sqrt of x^2? [#permalink]
14 Dec 2009, 21:06

xcusemeplz2009 wrote:

Fremontian wrote:

Bunuel wrote:

General rule: \(\sqrt{x^2}=|x|\).

When we see the equation of a type: \(y=\sqrt{x^2}\) then \(y=|x|\), which means that \(y\) can not be negative but \(x\) can.

\(y\)can not be negative as \(y=\sqrt{some expression}\), and even root from the expression (some value) is never negative (as for GMAT we are dealing only with real numbers).

When the GMAT provides the square root sign for an even root, such as a square root, then the only accepted answer is the positive root.

That is, \(\sqrt{16} = 4\), NOT +4 or -4. In contrast, the equation x^2 = 16 has TWO solutions, +4 and -4.

Even roots have only a positive value on the GMAT.

Odd roots will have the same sign as the base of the root. For example, CBRT(64) = 4, CBRT(-27) = -3.

In our original question we have: \(\frac{\sqrt{x^2}}{x}=\frac{|x|}{x}\), if we knew that x is positive then the answer would be 1, if we knew that x is negative the answer would be -1. BUT we don't know the sign of x, hence we can not simplify expression \(\frac{|x|}{x}\) further.

Hope it's clear.

Thank you. Much clear now. Also, Since x is Not equal to zero, is there a value of x for which this equation \(\frac{|x|}{x}\) WILL NOT simplify to \(1\)?

if x=-1 then the exp will -1

duh! that was a stupid question from me. Thanks for clarifying!

Re: What is sqrt of x^2? [#permalink]
15 Dec 2009, 03:28

1

This post received KUDOS

Expert's post

Fremontian wrote:

Thank you. Much clear now. Also, Since x is Not equal to zero, is there a value of x for which this equation \(\frac{|x|}{x}\) WILL NOT simplify to \(1\)?

For ANY value of \(x>0\) --> \(\frac{|x|}{x}=1\) For ANY value of \(x<0\) --> \(\frac{|x|}{x}=-1\) _________________

Re: So easy but couldnt do that explain me plz [#permalink]
18 Jun 2010, 05:39

Expert's post

You are asked to find the value of \(\frac {\sqrt{x^2}}{x}\)

Now, for the numerator term:

\(\sqrt{x^2} = x\) or \(-x\)

And the denominator is x.

So if you take the positive value of x as the numerator, the answer is 1, and if you take the negative value, the answer is -1. However, the root symbol specified is used only for positive answers.

Hence the only way to account for this is by using the mod sign. Hence the answer is E.

D is wrong, because either way you look at it \(\frac {\sqrt{x^2}}{x}\) can only be + or - 1

Re: Problem Solving [#permalink]
05 Jul 2010, 17:19

Lets try for say x=5; sqroot(5^2)/5 = +5/5 or -5/5 = 1 or -1

so, "a", "b" and "c" is not right.

|x|/x = |5|/5 = 1 Also, for negative value of x, (say -5), |x|/x = |-5|/-5 = 5/-5 = -1 It looks ok, however;

|x|/x = 1 only for x>0; |x|/x = -1 only for x<0; which not the case with sqroot(x^2)/x (as, we just saw even for positive value of x, we can have it's value as -1).

Re: Problem Solving [#permalink]
06 Jul 2010, 05:23

Expert's post

You are basically asked to find out what \(\frac{\sqrt{x^2}}{x}\) is.

If it had been given as \(\frac{\sqrt{x}}{x}\) then we can say that the numerator is simply x, and neglect the -x value since the square root sign considers only the positive radical.

But judging by the OA, I think the question was given in terms of what I had written in the first statement. In that case, the numerator is either a +x or a -x, depending on the original value of x. But since we don't know whether the original value was a + or a - number, we use the mod sign to indicate that we are taking the absolute value of the number, which is always positive. So your final answer will be \(\frac{|x|}{x}\)

As an example, let's consider one positive and one negative case.

x = 1 \(x^2\)= 1 \(\sqrt{x^2}\) = x = 1 So here, \(\frac{\sqrt{x^2}}{x}\) = 1

This poses no confusion since the original x value was a positive number by itself.

x = -1 \(x^2\) = 1 \(\sqrt{x^2}\) = (-x) = 1 [Note: The radical sign only indicates that the final result has to be a positive number. This doesn't necessarily mean that the answer is always 'x'] So here, we have \(\frac{\sqrt{x^2}}{x}\) = \(\frac{-x}{x}\) = -1

So, to combine both these results into one answer that fits both, we use \(\frac{|x|}{x}\)

Re: Square Root - Absolute Value [#permalink]
25 Aug 2010, 12:15

I think that A and C are out because \(\sqrt{x^2}/x\) could be +1 and -1. Remember that the square root of a number can have a positive and negative value.

E makes sense, but let's wait a better explanation. _________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

Square root of (x^2) ,can be +x or -x and this can be written as |x|.Because Modulus of anything should be always positive as we are just takking the magnitude.If x is +ve ,|x| will be x as its +ve.If x is -ve ,|x| will be -x as -ve of -ve number will be +ve.

Re: If x#0, then root(x^2)/x= [#permalink]
09 Mar 2014, 01:14

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: If x#0, then root(x^2)/x= [#permalink]
01 Apr 2015, 06:51

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

If x does not equal 0, then root x^2/x = [#permalink]
28 May 2015, 22:29

If x does not equal 0, then\(\sqrt{x^2}\)/x = a. -1 b. 0 c. 1 d. x e. lxl/x

Could anyone explain why C is wrong (why couldn't we take the square root of \(\sqrt{x^2}\) so it would equal x) and how come E is right?! Also, what is the level of difficulty for a question like this?

Re: If x does not equal 0, then root x^2/x = [#permalink]
28 May 2015, 22:59

2

This post received KUDOS

Expert's post

naeln wrote:

If x does not equal 0, then\(\sqrt{x^2}\)/x = a. -1 b. 0 c. 1 d. x e. lxl/x

Could anyone explain why C is wrong (why couldn't we take the square root of \(\sqrt{x^2}\) so it would equal x) and how come E is right?! Also, what is the level of difficulty for a question like this?

Re: If x#0, then root(x^2)/x= [#permalink]
29 May 2015, 03:48

Expert's post

naeln wrote:

If x does not equal 0, then\(\sqrt{x^2}\)/x = a. -1 b. 0 c. 1 d. x e. lxl/x

Could anyone explain why C is wrong (why couldn't we take the square root of \(\sqrt{x^2}\) so it would equal x) and how come E is right?! Also, what is the level of difficulty for a question like this?

Merging similar topics. Please refer to the solutions provided and ask if anything remains unclear. _________________

Hey, everyone. After a hectic orientation and a weeklong course, Managing Groups and Teams, I have finally settled into the core curriculum for Fall 1, and have thus found...

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

After I was accepted to Oxford I had an amazing opportunity to visit and meet a few fellow admitted students. We sat through a mock lecture, toured the business...