If x <0 , then \sqrt{-x*|x|} equals 1. -x 2. -1 3. 1 4. : GMAT Problem Solving (PS) - Page 3
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 19 Jan 2017, 22:50

### 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 \sqrt{-x*|x|} equals 1. -x 2. -1 3. 1 4.

Author Message
TAGS:

### Hide Tags

VP
Joined: 29 Dec 2005
Posts: 1348
Followers: 10

Kudos [?]: 60 [0], given: 0

### Show Tags

21 Mar 2006, 06:56
HongHu wrote:
Square root of any positive number is a positive number. When we say x^2=a, we know that x=+/-sqrt(a). You can see that sqrt(a) itself is positive, but x could be positive sqrt(a) or negative sqrt(a).

Using an example, say x^2=25. We know that x=+/-sqrt(25), where sqrt(25)=5.

Hope this helps.

i just want to be/make clear on why sqrt (25) = 5.

suppose x^2 = 25
x = sqrt (25) or - sqrt (25)
and we are here dealing with only "sqrt (25)".

therefore, sqrt (25) = 5

thank you honghu.
Director
Joined: 04 Jan 2006
Posts: 923
Followers: 1

Kudos [?]: 31 [0], given: 0

### Show Tags

21 Mar 2006, 08:33
I still dont understand..

isnt |x| just the positive value of the number x?
VP
Joined: 21 Sep 2003
Posts: 1065
Location: USA
Followers: 3

Kudos [?]: 73 [0], given: 0

### Show Tags

21 Mar 2006, 08:41
willget800 wrote:
I still dont understand..

isnt |x| just the positive value of the number x?

Yes. But what if the x is -ve (say x = -2), then the absolute value of x is 2
x = -2
|x| = 2 = -(-2) = -x

For more gyan (Hindi word for knowledge) see..

http://www.purplemath.com/modules/absolute.htm
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

SVP
Joined: 07 Nov 2007
Posts: 1820
Location: New York
Followers: 34

Kudos [?]: 862 [0], given: 5

### Show Tags

25 Jan 2009, 06:50
study wrote:
If x<0, then [sqrt (-x absolute x )] is
(please note that both variables - minus x and absolute x - come under the root sign)

1
-1
sqrt x
x
-x

let say -x=y (y is positive)

sqrt (-x |x|) = sqrt (y|-y|) = sqrt (y^2) = y or -y

when y +ve --> answer =y = -x
wnen y -ve --> answer = -y=x

here y is +ve so answer =-x

E
_________________

Smiling wins more friends than frowning

Manager
Joined: 04 Jan 2009
Posts: 241
Followers: 1

Kudos [?]: 12 [0], given: 0

### Show Tags

25 Jan 2009, 08:44
yes sir. sqrt(f) has to be positive otherwise, we run into the following problem:
2 is sqrt(4)
if we say sqrt(4)=-2 or 2
in one situation we run into situation where 2+2=0.
So, unless we are asked to find (-sqrt(f)) sqrt(f) always yields positive number.

x2suresh wrote:
study wrote:
If x<0, then [sqrt (-x absolute x )] is
(please note that both variables - minus x and absolute x - come under the root sign)

1
-1
sqrt x
x
-x

let say -x=y (y is positive)

sqrt (-x |x|) = sqrt (y|-y|) = sqrt (y^2) = y or -y

when y +ve --> answer =y = -x
wnen y -ve --> answer = -y=x

here y is +ve so answer =-x

E

_________________

-----------------------
tusharvk

Senior Manager
Joined: 05 Oct 2008
Posts: 274
Followers: 3

Kudos [?]: 379 [0], given: 22

### Show Tags

25 Jan 2009, 11:05
Thanks Ian for the explanation and for posting the question in mathematical format.

+2
Senior Manager
Joined: 20 Mar 2008
Posts: 454
Followers: 1

Kudos [?]: 119 [0], given: 5

Root of the product of -x and |x| [#permalink]

### Show Tags

31 Oct 2009, 21:40
Root of the product of -x and |x|
Attachments

root x mod x.JPG [ 9.55 KiB | Viewed 4394 times ]

Senior Manager
Joined: 20 Mar 2008
Posts: 454
Followers: 1

Kudos [?]: 119 [0], given: 5

Re: Root of the product of -x and |x| [#permalink]

### Show Tags

31 Oct 2009, 21:59
+1 to you.

Thanks for the explanation. When I saw this problem, I had a feeling there was something I was missing.
Intern
Joined: 22 Sep 2009
Posts: 36
Followers: 0

Kudos [?]: 24 [0], given: 3

Re: Root of the product of -x and |x| [#permalink]

### Show Tags

01 Nov 2009, 04:38
x<0 ....... let a=-x. so a is +ve

hence y= sqrt{-x*|x|}=sqrt{a*|-a|}
y = sqrt{a*a}
y = +a or -a
y=-x or x

Intern
Joined: 28 Apr 2009
Posts: 48
Followers: 0

Kudos [?]: 7 [0], given: 2

Re: Root of the product of -x and |x| [#permalink]

### Show Tags

14 Nov 2009, 06:11
Bunuel wrote:
I say that $$y$$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).

Consider this $$y=\sqrt{16}$$ --> $$y=4$$ and not $$y=4$$ or $$y=-4$$.

But why ?? is this a standard formula ?
Math Expert
Joined: 02 Sep 2009
Posts: 36568
Followers: 7081

Kudos [?]: 93225 [0], given: 10553

Re: Root of the product of -x and |x| [#permalink]

### Show Tags

14 Nov 2009, 06:24
mbaquestionmark wrote:
Bunuel wrote:
I say that $$y$$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).

Consider this $$y=\sqrt{16}$$ --> $$y=4$$ and not $$y=4$$ or $$y=-4$$.

But why ?? is this a standard formula ?

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.

Hope it's clear.
_________________
Intern
Joined: 23 Sep 2009
Posts: 11
Location: uzbekistan
Schools: Chicago booth,Dartmouth,Duke
Followers: 0

Kudos [?]: 1 [0], given: 0

Re: Square root - positive or negative? [#permalink]

### Show Tags

31 Dec 2009, 03:22
I think the answer is 4. X

Coz: given that x<0 , -(-x) is positive and module x is also positive so we get x^2 , and end result is x

If you do not think so please correct me
Math Expert
Joined: 02 Sep 2009
Posts: 36568
Followers: 7081

Kudos [?]: 93225 [0], given: 10553

Re: Square root - positive or negative? [#permalink]

### Show Tags

31 Dec 2009, 06:57
sher1978 wrote:
I think the answer is 4. X

Coz: given that x<0 , -(-x) is positive and module x is also positive so we get x^2 , and end result is x

If you do not think so please correct me

Answer to this question is A (-x). Please refer to the solutions above. Do ask if anything remains unclear.
_________________
Senior Manager
Joined: 22 Dec 2009
Posts: 362
Followers: 11

Kudos [?]: 375 [0], given: 47

Re: Square root - positive or negative? [#permalink]

### Show Tags

31 Dec 2009, 11:46
I guess the best way to solve this question is to plug in values and see how the result features.

Before we do the same, explanation by HongHu is perfect. (Cant post the link to his reply as I am new to the club )

i.e Square root of any positive number is a positive number

Now getting back to the actual question:

If x<0, then $$\sqrt{-x*|x|}$$ equals:
1. -x
2. -1
3. 1
4. x
5. $$\sqrt{x}$$

Approach:
Since x <0, let use plug in a negative integer as x, for e.g.

Let x = -2 ---- [1]

Hence $$\sqrt{-x*|x|}= \sqrt{-(-2)*|-2|} = \sqrt{2*2} = \sqrt{4}$$ ---- [2]

Now as per the discussion going on, the point of disagreement is that $$\sqrt{4}$$ = 2 or -2. But as stated by HongHu, we would consider $$\sqrt{4}$$ = 2 only! The reason for this conclusion is that the radical sign used in $$sqrt{x}$$ is used to show the positive square root of a number x.

Therefore $$\sqrt{4} = 2$$ ---- [3]

Now as per Equation [1] marked above, x = -2 which can also be written as -x = 2 ---- [4]

Hence if we subsitute the values of $$\sqrt{4}$$ from equation 2 ( i.e. $$\sqrt{-x*|x|}$$ ) and value of 2 from equation [4] ( i.e. -x )... in equation [3], we get the following result:

$$\sqrt{-x*|x|} = -x$$ and hence the correct answer is 1

This approach can be verified by another question which featured in the GMAT prep exams...

If $$x \neq 0$$, then $$\frac{\sqrt{x^2}}{x}$$ equals:
1. -1
2. 0
3. 1
4. x
5. $$\frac{|x|}{x}$$

Approach:
In this question, we don't know whether x is negative or positive. Hence $$\sqrt{x^2}$$ is always equal to |x|. Reason for the same is: x is a variable here and we dont know the exact value. The absolute value sign is needed when we are taking the square root of a square of a variable, which may be positive or negative

Therefore in this situation, the answer would be 5

Hope this helps!

Cheers!
JT
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

Joined: 20 Aug 2009
Posts: 311
Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)
Followers: 18

Kudos [?]: 140 [0], given: 69

### Show Tags

08 Jan 2010, 03:17
No it's (A): -x

$$\sqrt{(-x)|x|}=\sqrt{(-x)(-x)}=\sqrt{x^2}=|x| = -x$$ (because x<0)
Re: if x<0   [#permalink] 08 Jan 2010, 03:17

Go to page   Previous    1   2   3   [ 55 posts ]

Similar topics Replies Last post
Similar
Topics:
4 If x is a positive integer such that (x-1)(x-3)(x-5)....(x-93) < 0, ho 2 29 Jan 2016, 02:14
5 The value of 1/(1-x) + 1/(1+x) + 2/(1+x^2) + 4/(1+x^4) 6 10 Oct 2015, 20:40
4 If x = -1, then -(x^4 + x^3 + x^2 + x) = 7 03 Feb 2014, 00:23
1 If x = -1, then (x^4 - x^3 + x^2)/(x - 1) = 3 19 Dec 2012, 05:17
33 For x < 0. Simplify {-(x + 1) |x-1| + 1}^1/2? 12 26 Jul 2012, 04:35
Display posts from previous: Sort by