It is currently 18 Mar 2018, 06:42

### 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 square root of (-x|x|) is:

Author Message
TAGS:

### Hide Tags

Director
Joined: 03 Sep 2006
Posts: 850
If x<0, then square root of (-x|x|) is: [#permalink]

### Show Tags

06 May 2010, 12:33
7
KUDOS
20
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

57% (00:26) correct 43% (00:30) wrong based on 893 sessions

### HideShow timer Statistics

If x<0, then $$\sqrt{-x|x|}$$ is:

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

Attachments

PS1.PNG [ 3.17 KiB | Viewed 9087 times ]

Senior Manager
Joined: 19 Nov 2009
Posts: 280

### Show Tags

06 May 2010, 12:44
I think it's A.

Since x < 0 , let us consider x = -2

$$\sqrt{-x|x|}$$ = $$\sqrt{- (-2)(2)}$$ = $$\sqrt{4}$$ = 2

What's the OA ?
_________________

"Success is going from failure to failure without a loss of enthusiam." - Winston Churchill

As vs Like - Check this link : http://www.grammar-quizzes.com/like-as.html.

Math Expert
Joined: 02 Sep 2009
Posts: 44290
If x<0, then square root of (-x|x|) is: [#permalink]

### Show Tags

06 May 2010, 13:12
7
KUDOS
Expert's post
10
This post was
BOOKMARKED
LM wrote:
If x<0, then $$\sqrt{-x|x|}$$ is:

A. -x
B. -1
C. 1
D. x
E. $$\sqrt{x}$$

Given: $$x<0$$ Question: $$y=\sqrt{-x*|x|}$$?

Remember: $$\sqrt{x^2}=|x|$$.

As $$x<0$$, then $$|x|=-x$$ --> $$\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x$$.

_________________
Intern
Joined: 08 Apr 2010
Posts: 22

### Show Tags

18 May 2010, 15:36
Hey I'm sorry guys, this still does not make sense. Everyone's argument here is that the square root of 4 is 2, that is just not true! The square root of 4 is 2 OR -2. We're just accustomed to thinking that 2 is the "standard root" but -2 is just as correct. Therefore the square of -2 (which is x in this case) is 4, and the squareroot of that is 2 OR -2! So it could be x or -x.

This seems wrong and no one's explanation makes any sense.
Math Expert
Joined: 02 Sep 2009
Posts: 44290

### Show Tags

19 May 2010, 01:21
Expert's post
4
This post was
BOOKMARKED
shammokando wrote:
Hey I'm sorry guys, this still does not make sense. Everyone's argument here is that the square root of 4 is 2, that is just not true! The square root of 4 is 2 OR -2. We're just accustomed to thinking that 2 is the "standard root" but -2 is just as correct. Therefore the square of -2 (which is x in this case) is 4, and the squareroot of that is 2 OR -2! So it could be x or -x.

This seems wrong and no one's explanation makes any sense.

Red part is not correct.

THEORY:

GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root.

That is, $$\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. 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, $$\sqrt[3]{125} =5$$ and $$\sqrt[3]{-64} =-4$$.

Solution for the original question:

Given: $$x<0$$ Question: $$\sqrt{-x*|x|}=?$$.

Remember: $$\sqrt{x^2}=|x|$$.

As $$x<0$$, then $$|x|=-x$$ --> $$\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x$$.

Hope it helps.
_________________
Intern
Joined: 08 Apr 2010
Posts: 22

### Show Tags

19 May 2010, 07:49
Bunuel wrote:
Red part is not correct.

THEORY:

GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root.

[highlight]That is, $$\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.[/highlight]
Odd roots will have the same sign as the base of the root. For example, $$\sqrt[3]{125} =5$$ and $$\sqrt[3]{-64} =-4$$.

Solution for the original question:

Given: $$x<0$$ Question: $$\sqrt{-x*|x|}=?$$.

Remember: $$\sqrt{x^2}=|x|$$.

As $$x<0$$, then $$|x|=-x$$ --> $$\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x$$.

Hope it helps.

I'd really like an expert's explanation of this, you're explanation is not right.

You say the GMAT deals with real numbers, true... negative integers/nonintegers are real numbers, a non real number is 5i, or the
sqrrt of -1 (i). The square root of any positive number is real. period.

And, on the highlighted part, you have no justification for your delineation, Those are literally identical equations, you just chose to give one two roots and the other 1...
I looked it up in OFFICIAL GMAT Literature... EVERYONE go look at pg 114 of OG 12 it says, $$\sqrt{9}=3 OR -3$$. It defines ALL EVEN ROOTS AS HAVING 2 solutions, plus and minus.

It is just incorrect to claim the square root of 4 is 2. We learned this in middle school... And I absolutely know the GMAT has tested negative values of the root before...
Senior Manager
Joined: 25 Jun 2009
Posts: 293

### Show Tags

19 May 2010, 08:04
shammokando wrote:
Bunuel wrote:
Red part is not correct.

THEORY:

GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root.

[highlight]That is, $$\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.[/highlight]
Odd roots will have the same sign as the base of the root. For example, $$\sqrt[3]{125} =5$$ and $$\sqrt[3]{-64} =-4$$.

Solution for the original question:

Given: $$x<0$$ Question: $$\sqrt{-x*|x|}=?$$.

Remember: $$\sqrt{x^2}=|x|$$.

[highlight]As $$x<0$$, then $$|x|=-x$$ --> $$\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x$$.[/highlight]

Hope it helps.

I'd really like an expert's explanation of this, you're explanation is not right.

You say the GMAT deals with real numbers, true... negative integers/nonintegers are real numbers, a non real number is 5i, or the
sqrrt of -1 (i). The square root of any positive number is real. period.

And, on the highlighted part, you have no justification for your delineation, Those are literally identical equations, you just chose to give one two roots and the other 1...
I looked it up in OFFICIAL GMAT Literature... EVERYONE go look at pg 114 of OG 12 it says, $$\sqrt{9}=3 OR -3$$. It defines ALL EVEN ROOTS AS HAVING 2 solutions, plus and minus.

It is just incorrect to claim the square root of 4 is 2. We learned this in middle school... And I absolutely know the GMAT has tested negative values of the root before...

Read that last part of the Bunuel Explanation which is highlighted here

And trust me Bunuel is one of the best guys on this forum.

For all real numbers x

$$\sqrt{x^2} = |x| = x ....... if ...... x>= 0$$
$$= -x ........if ....... x <0$$
And here in the question its mentioned that x<0 and thats why the answer is -x

I hope this helps.
Intern
Joined: 08 Apr 2010
Posts: 22

### Show Tags

19 May 2010, 08:16
I mean absolutely no disrespect, I'm very impressed and grateful of the help Bruenel has personally given me. I'm just still not getting it...

From Wiki:
Quote:
Although the principal square root of a positive number is only one of its two square roots, the designation "the square root" is often used to refer to the principal square root.

That's pretty arbitrary and would HAVE to be explicitly stated by the GMAT somewhere, otherwise that's insanely controversial. I pointed out the official GMAC Literature, Pg 114 of the OG 12 states that
Quote:
Every positive number n has 2 square roots

It says N has 2 square roots! Replace N with X^2, and you have that X^2 has 2 square roots, +/- X. Therefore it could be -(+x) or it could be +(-x) and be the right answer.

I just don't see where you have seen the claim that x^2 is the absolute value of x... The wiki page doesn't even state that. You're claiming the root of x^2 is ONLY x. -x*-x is also x^2.

Sorry to be difficult.
Senior Manager
Joined: 25 Jun 2009
Posts: 293

### Show Tags

19 May 2010, 08:28
shammokando wrote:
I mean absolutely no disrespect, I'm very impressed and grateful of the help Bruenel has personally given me. I'm just still not getting it...

From Wiki:
Quote:
Although the principal square root of a positive number is only one of its two square roots, the designation "the square root" is often used to refer to the principal square root.

That's pretty arbitrary and would HAVE to be explicitly stated by the GMAT somewhere, otherwise that's insanely controversial. I pointed out the official GMAC Literature, Pg 114 of the OG 12 states that
Quote:
Every positive number n has 2 square roots

It says N has 2 square roots! Replace N with X^2, and you have that X^2 has 2 square roots, +/- X. Therefore it could be -(+x) or it could be +(-x) and be the right answer.

I just don't see where you have seen the claim that x^2 is the absolute value of x... The wiki page doesn't even state that. You're claiming the root of x^2 is ONLY x. -x*-x is also x^2.

Sorry to be difficult.

I got your point, and I understand how irritating it can be when there is some thing which you are not able to understand,

Anyway, about the wiki link if you scroll a bit you will that ... I am attaching a screen shot for reference.

Second, about the 2 roots, yes you are right that there are 2 roots ( -+x) but that depends upon the value of x and thats why we always say that roots are in the form of $$\sqrt{x^2}=|x| = +-x$$ ( depending where x lies on the number line as in whether its -ve or +ve)
Attachments

untitled.JPG [ 31.34 KiB | Viewed 8703 times ]

Math Expert
Joined: 02 Sep 2009
Posts: 44290

### Show Tags

19 May 2010, 08:55
First of all: EVERYTHING in my previous post is correct.

Second: square root symbol "√" gives only one, non-negative value, non-negative square root.

So $$\sqrt{9}=3$$. NOT 3 or/and -3.
_________________
Intern
Joined: 08 Apr 2010
Posts: 22

### Show Tags

19 May 2010, 09:03
I see. Again apologies for the difficulties. I was just trying to fully understand it. These are subtleties I was never taught

Thanks again.
Intern
Joined: 23 Jul 2012
Posts: 4
Re: If x<0, then square root of (-x|x|) is: [#permalink]

### Show Tags

01 Jul 2014, 03:34
Hi Bunuel,

I really did not understand why you wrote , "As x < 0, then |x| = -x"

As I see, even if x is < 0 still since its inside mod sign, its value will always be positive. Also, we cannot have value of absolute structure as a negative number. The whole stuff inside "...." is confusing.

Can you please clarify on this.

Kind Regards,
DK

Bunuel wrote:
LM wrote:
If x<0, then $$\sqrt{-x|x|}$$ is:

A. -x
B. -1
C. 1
D. x
E. $$\sqrt{x}$$

Given: $$x<0$$ Question: $$y=\sqrt{-x*|x|}$$?

Remember: $$\sqrt{x^2}=|x|$$.

As $$x<0$$, then $$|x|=-x$$ --> $$\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x$$.

Math Expert
Joined: 02 Sep 2009
Posts: 44290
Re: If x<0, then square root of (-x|x|) is: [#permalink]

### Show Tags

01 Jul 2014, 03:55
dk19761976 wrote:
Hi Bunuel,

I really did not understand why you wrote , "As x < 0, then |x| = -x"

As I see, even if x is < 0 still since its inside mod sign, its value will always be positive. Also, we cannot have value of absolute structure as a negative number. The whole stuff inside "...." is confusing.

Can you please clarify on this.

Kind Regards,
DK

Bunuel wrote:
LM wrote:
If x<0, then $$\sqrt{-x|x|}$$ is:

A. -x
B. -1
C. 1
D. x
E. $$\sqrt{x}$$

Given: $$x<0$$ Question: $$y=\sqrt{-x*|x|}$$?

Remember: $$\sqrt{x^2}=|x|$$.

As $$x<0$$, then $$|x|=-x$$ --> $$\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x$$.

If x is negative, then |x| = -x = -negative = positive. So, as you can see the result is still positive. For example, if x=-5, then |-5| = -(-5) = 5.

Absolute value properties:

When $$x\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \ expression\leq{0}$$ then $$|some \ expression|={-(some \ expression)}$$. For example: $$|-5|=5=-(-5)$$;

When $$x\geq{0}$$ then $$|x|=x$$, or more generally when $$some \ expression\geq{0}$$ then $$|some \ expression|={some \ expression}$$. For example: $$|5|=5$$.

Theory on Abolute Values: math-absolute-value-modulus-86462.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope this helps.
_________________
GMAT Tutor
Joined: 24 Jun 2008
Posts: 1346
If x < 0, then √((-x)·|x|) is [#permalink]

### Show Tags

20 May 2015, 18:28
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
If x < 0, then |x| = -x. So by substituting, we have:

$$\sqrt{ (-x) ( |x| )} = \sqrt{ (-x)(-x)} = \sqrt{x^2}$$

Now it's important to understand that √(x^2) is not necessarily equal to x. That is only true when x is positive (or zero). You can see, if you plug in any negative number here, say x = -3, that √(x^2) = √9 = 3, which is not equal to x because the sign changed; it's actually equal to -x. In general, √(x^2) is always equal to |x|. Since x < 0 in this question, √(x^2) = |x| = -x.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 11255
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Re: If x < 0, then √((-x)·|x|) is [#permalink]

### Show Tags

20 May 2015, 19:53
1
KUDOS
Expert's post
Hi jchae90,

This question is perfect for TESTing VALUES.

We're told that X < 0, so let's TEST X = -2

We're asked to determine the value of..... √((-x)·|x|)

√((-(-2))·|-2|) = √(2)·|2|) = √4 = 2

So we're looking for an answer that equals 2 when X = -2

Answer A: –X = -(-2) = 2 This IS a match
Answer B: -1 NOT a match
Answer C: 1 NOT a match
Answer D: X = -2 NOT a match
Answer E: √X = √2 NOT a match

[Reveal] Spoiler:
A

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Manager
Joined: 28 Dec 2012
Posts: 110
Location: India
Concentration: Strategy, Finance
WE: Engineering (Energy and Utilities)
Re: If x<0, then square root of (-x|x|) is: [#permalink]

### Show Tags

23 May 2015, 10:55
1
KUDOS
LM wrote:
If x<0, then $$\sqrt{-x|x|}$$ is:

A. -x
B. -1
C. 1
D. x
E. $$\sqrt{x}$$

Clearly B, C and E are not the answers. Now square root always returns a non negative ( 0 or positive) value... Since x < 0 ... We -x >=0. Thus Option A

Kudos... If this helps
_________________

Impossibility is a relative concept!!

Senior Manager
Joined: 25 Mar 2013
Posts: 271
Location: United States
Concentration: Entrepreneurship, Marketing
GPA: 3.5
If x<0, then square root of (-x|x|) is: [#permalink]

### Show Tags

01 Jan 2017, 15:44
Test It process was x < 0, so is negative number
x = -4
$$\sqrt{-(-4 ) I -4 I}$$
$$\sqrt{16}$$

4

Substitute x in options
-(x) = -(-4) = 4
A

Thanks Rich.
_________________

I welcome analysis on my posts and kudo +1 if helpful. It helps me to improve my craft.Thank you

Senior Manager
Joined: 13 Apr 2013
Posts: 347
Location: India
GMAT 1: 480 Q38 V22
GPA: 3.01
WE: Engineering (Consulting)
Re: If x<0, then square root of (-x|x|) is: [#permalink]

### Show Tags

05 Mar 2018, 21:32
IanStewart wrote:
If x < 0, then |x| = -x. So by substituting, we have:

$$\sqrt{ (-x) ( |x| )} = \sqrt{ (-x)(-x)} = \sqrt{x^2}$$

Now it's important to understand that √(x^2) is not necessarily equal to x. That is only true when x is positive (or zero). You can see, if you plug in any negative number here, say x = -3, that √(x^2) = √9 = 3, which is not equal to x because the sign changed; it's actually equal to -x. In general, √(x^2) is always equal to |x|. Since x < 0 in this question, √(x^2) = |x| = -x.

A very very important point discussed by you. That's the way I did, and found answer as "x" not "-x". I cancelled root with power of x.
_________________

"Success is not as glamorous as people tell you. It's a lot of hours spent in the darkness."

Intern
Joined: 30 May 2017
Posts: 1
Re: If x<0, then square root of (-x|x|) is: [#permalink]

### Show Tags

11 Mar 2018, 08:35
shammokando wrote:
Hey I'm sorry guys, this still does not make sense. Everyone's argument here is that the square root of 4 is 2, that is just not true! The square root of 4 is 2 OR -2. We're just accustomed to thinking that 2 is the "standard root" but -2 is just as correct. Therefore the square of -2 (which is x in this case) is 4, and the squareroot of that is 2 OR -2! So it could be x or -x.

This seems wrong and no one's explanation makes any sense.

Since x<0, The mod of x inside the square root will be -x. So, Sqrt of(-x)(-x) = mod (x). This is the answer and now since x <0 is already the condition , therefore mod of x must be -x.
Re: If x<0, then square root of (-x|x|) is:   [#permalink] 11 Mar 2018, 08:35
Display posts from previous: Sort by