If x#0, then root(x^2)/x= : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 24 Feb 2017, 14:50

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If x#0, then root(x^2)/x=

Author Message
TAGS:

### Hide Tags

Intern
Status: Applying
Joined: 14 Oct 2009
Posts: 31
Location: California
Schools: Cornell AMBA, Kellogg, Oxford, Cambridge
Followers: 0

Kudos [?]: 64 [4] , given: 4

### Show Tags

14 Dec 2009, 16:58
4
KUDOS
25
This post was
BOOKMARKED
00:00

Difficulty:

5% (low)

Question Stats:

71% (01:15) correct 29% (00:15) wrong based on 1037 sessions

### HideShow timer Statistics

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

A. -1
B. 0
C. 1
D. x
E. $$\frac{|x|}{x}$$
[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Dec 2012, 02:14, edited 1 time in total.
Renamed the topic and edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 37108
Followers: 7253

Kudos [?]: 96530 [14] , given: 10753

### Show Tags

14 Dec 2009, 17:47
14
KUDOS
Expert's post
37
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.

Hope it's clear.
_________________
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 560
Followers: 155

Kudos [?]: 1242 [2] , given: 100

Re: If x does not equal 0, then root x^2/x = [#permalink]

### Show Tags

28 May 2015, 22:59
2
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?

Hi naeln,

$$\sqrt{x^2} = |x|$$ i.e. square root function will not give a negative value. Let me explain this to with you an example.

Assume $$x = 2$$, so $$x^2 = 4$$ and hence $$\sqrt{x^2} = \sqrt{4} = 2$$ $$= x$$ when $$x => 0$$

Similarly if $$x = -2, x^2 = 4$$ and hence $$\sqrt{x^2} = \sqrt{4} = 2$$ $$= - x$$ when $$x < 0$$

So the square root function has the same behavior as the modulus function. Hence we need to represent $$\sqrt{x^2} = |x|$$.

Hope it's clear

Regards
Harsh
_________________

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Math Expert
Joined: 02 Sep 2009
Posts: 37108
Followers: 7253

Kudos [?]: 96530 [1] , given: 10753

Re: What is sqrt of x^2? [#permalink]

### Show Tags

15 Dec 2009, 03:28
1
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 $$x<0$$ --> $$\frac{|x|}{x}=-1$$
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 37108
Followers: 7253

Kudos [?]: 96530 [1] , given: 10753

### Show Tags

19 Oct 2010, 13:16
1
KUDOS
Expert's post
Other questions about the same concept:
if-x-81600.html
square-root-and-modulus-100303.html

Hope it helps.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 37108
Followers: 7253

Kudos [?]: 96530 [1] , given: 10753

Re: If x#0, then root(x^2)/x= [#permalink]

### Show Tags

15 Jun 2015, 01:54
1
KUDOS
Expert's post
bgpower wrote:
Another conceptual question:

If we have in a question square root of any variable (e.g. $$\sqrt{x^2}$$) then the answer would modulus x, provided no information is provided in the question stem.
If we have square root of any number (e.g. $$\sqrt{25}$$) then we take only the positive root, viz. 5 (even without any further information), because that's what the GMAT does. Nevertheless, normally it could be both 5 and negative 5.

Thanks for the clarification!

Everything is correct, except the red text.
_________________
Manager
Status: Getting ready for the internship summer
Joined: 07 Jun 2009
Posts: 51
Location: Rochester, NY
Schools: Simon
WE 1: JPM - Treasury
Followers: 0

Kudos [?]: 25 [0], given: 23

Re: What is sqrt of x^2? [#permalink]

### Show Tags

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.
Intern
Status: Applying
Joined: 14 Oct 2009
Posts: 31
Location: California
Schools: Cornell AMBA, Kellogg, Oxford, Cambridge
Followers: 0

Kudos [?]: 64 [0], given: 4

Re: What is sqrt of x^2? [#permalink]

### Show Tags

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 $$1$$?
Manager
Joined: 09 May 2009
Posts: 203
Followers: 1

Kudos [?]: 227 [0], given: 13

Re: What is sqrt of x^2? [#permalink]

### Show Tags

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

Intern
Status: Applying
Joined: 14 Oct 2009
Posts: 31
Location: California
Schools: Cornell AMBA, Kellogg, Oxford, Cambridge
Followers: 0

Kudos [?]: 64 [0], given: 4

Re: What is sqrt of x^2? [#permalink]

### Show Tags

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!
Ms. Big Fat Panda
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1922
Concentration: General Management, Nonprofit
Followers: 452

Kudos [?]: 2001 [0], given: 210

Re: So easy but couldnt do that explain me plz [#permalink]

### Show Tags

18 Jun 2010, 05:39
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

Hope this helps!
Manager
Joined: 25 Jun 2010
Posts: 91
Followers: 1

Kudos [?]: 35 [0], given: 0

### Show Tags

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).

So, "e" should be the right option.

Thanks
Ms. Big Fat Panda
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1922
Concentration: General Management, Nonprofit
Followers: 452

Kudos [?]: 2001 [0], given: 210

### Show Tags

06 Jul 2010, 05:23
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 $$\frac{|x|}{x}$$

Hope this helps.
Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1713
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
Followers: 101

Kudos [?]: 940 [0], given: 109

Re: Square Root - Absolute Value [#permalink]

### Show Tags

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."

My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html

GMAT Club Premium Membership - big benefits and savings

Manager
Joined: 08 Sep 2010
Posts: 232
Location: India
WE 1: 6 Year, Telecom(GSM)
Followers: 4

Kudos [?]: 261 [0], given: 21

### Show Tags

19 Oct 2010, 01:16
satishreddy wrote:
need help

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.

Hence teh answer will be |x|/x

Consider KUDOS if it is helpful to u in some way.
Director
Joined: 01 Feb 2011
Posts: 755
Followers: 14

Kudos [?]: 119 [0], given: 42

Re: What is sqrt of x^2? [#permalink]

### Show Tags

19 Apr 2011, 18:17
if x>0 then sqrt(x^2) = x

if x<0 then sqrt(x^2) = -x

so we generalize the given expression sqrt(x^2)/x = |x|/x

Senior Manager
Joined: 13 Aug 2012
Posts: 464
Concentration: Marketing, Finance
GMAT 1: Q V0
GPA: 3.23
Followers: 26

Kudos [?]: 443 [0], given: 11

### Show Tags

03 Dec 2012, 19:40
I'll just remember that $$\sqrt{x^2}=|x|$$

_________________

Impossible is nothing to God.

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13955
Followers: 590

Kudos [?]: 167 [0], given: 0

Re: If x#0, then root(x^2)/x= [#permalink]

### Show Tags

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.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13955
Followers: 590

Kudos [?]: 167 [0], given: 0

Re: If x#0, then root(x^2)/x= [#permalink]

### Show Tags

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.
_________________
Intern
Joined: 05 Aug 2014
Posts: 36
Followers: 0

Kudos [?]: 52 [0], given: 24

If x does not equal 0, then root x^2/x = [#permalink]

### Show Tags

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

Go to page    1   2    Next  [ 26 posts ]

Similar topics Replies Last post
Similar
Topics:
If x ≠ 0 then x^-2/1/x^2 3 02 Jan 2017, 04:31
8 If 5 ≥ |x| ≥ 0, which of the following must be true? 4 02 Jan 2017, 04:26
1 What is the value of 3x^2 − 1.8x + 0.3 for x = 0.6? 3 26 Jul 2016, 08:37
36 Solve for x: 0<|x|-4x<5 = ? 23 15 Aug 2009, 23:44
12 If x#0, then root(x^2)/x= 3 18 May 2008, 12:29
Display posts from previous: Sort by