GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 10 Dec 2018, 16:58

### 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 December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### Free lesson on number properties

December 10, 2018

December 10, 2018

10:00 PM PST

11:00 PM PST

Practice the one most important Quant section - Integer properties, and rapidly improve your skills.
• ### Free GMAT Prep Hour

December 11, 2018

December 11, 2018

09:00 PM EST

10:00 PM EST

Strategies and techniques for approaching featured GMAT topics. December 11 at 9 PM EST.

# If x < 0, then root({-x} •|x|) is

Author Message
TAGS:

### Hide Tags

Retired Moderator
Joined: 20 Dec 2010
Posts: 1820
Re: modes and square root  [#permalink]

### Show Tags

11 Feb 2011, 00:42
1
the question is: If x < 0, what is the value for $$sqrt{-x*\mid x \mid}$$:

$$sqrt{(+x)*(+x)}$$
$$sqrt{x^2}$$
$$\pm x$$

Reasoning is:

Since x < 0; only -x is the valid value. +x can be ignored.

Ans: "A"
_________________
Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 742
Re: Square root and Mod Problem  [#permalink]

### Show Tags

07 Mar 2011, 22:32
sqrt(x^2) = |x|
if x <0 |x| = -x
x^2 = -x|x|

Hence A
VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1033
Re: Square root and Modulus  [#permalink]

### Show Tags

14 Jun 2011, 01:07
A as root will always give positive value.
Intern
Joined: 25 Jun 2012
Posts: 35
Re: If x < 0, then root({-x} •|x|) is  [#permalink]

### Show Tags

29 Nov 2012, 13:10
x is < 0.

This makes sense, if it was positive, then it would be the square root of (-)(+)(|+|)= -. We can't have the square of a negative, not a real #

so x is -. Let's pick x= -5. (-1)(-5)(|-5|)= √25 = 5. answer is 5 which is -x.
Board of Directors
Joined: 01 Sep 2010
Posts: 3294
Re: Square root and Modulus  [#permalink]

### Show Tags

29 Nov 2012, 13:51
Bunuel wrote:
udaymathapati wrote:
If x < 0, then \sqrt{-x} •|x|) is
A. -x
B. -1
C. 1
D. x
E. \sqrt{x}

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

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

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

The point here is that square root function can not give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.

So $$\sqrt{x^2}\geq{0}$$. But what does $$\sqrt{x^2}$$ equal to?

Let's consider following examples:
If $$x=5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=x=positive$$;
If $$x=-5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=-x=positive$$.

So we got that:
$$\sqrt{x^2}=x$$, if $$x\geq{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<0$$.

What function does exactly the same thing? The absolute value function! That is why $$\sqrt{x^2}=|x|$$

Back to the original question:
$$\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x$$

Or just substitute the value let $$x=-5<0$$ --> $$\sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5=-(-5)=-x$$.

Hope it's clear.

Sorry Bunuel but the approach :$$\sqrt{-x* |x|}$$ if we put this one ^2 then we have simply $$-x * |x|$$. The latter is positive, so we have a quantity straight negative $$- X$$.

or is wrong this simple way ))

thanks
_________________
Board of Directors
Joined: 01 Sep 2010
Posts: 3294
Re: modes and square root  [#permalink]

### Show Tags

29 Nov 2012, 13:55
fluke wrote:
the question is: If x < 0, what is the value for $$sqrt{-x*\mid x \mid}$$:

$$sqrt{(+x)*(+x)}$$
$$sqrt{x^2}$$
$$\pm x$$

Reasoning is:

Since x < 0; only -x is the valid value. +x can be ignored.

Ans: "A"

Sorry fluke but the quantity $$\sqrt{x^2}$$ is not + or - X but only + X, if we have $$x^2= something$$ then we have $$+ or -$$.........
_________________
Senior Manager
Joined: 13 Aug 2012
Posts: 429
Concentration: Marketing, Finance
GPA: 3.23
Re: If x < 0, then root({-x} •|x|) is  [#permalink]

### Show Tags

03 Dec 2012, 19:59
Since $$x < 0$$,
$$\sqrt{-x|x|}$$
$$\sqrt{x^2}=|x|$$

Answer: -x since x is of negative value.
_________________

Impossible is nothing to God.

Manager
Joined: 29 Mar 2010
Posts: 120
Location: United States
GMAT 1: 590 Q28 V38
GPA: 2.54
WE: Accounting (Hospitality and Tourism)
Re: If x < 0, then root({-x} •|x|) is  [#permalink]

### Show Tags

08 May 2013, 20:43
I said X before I read the very important point that x<-1

I need to stop that
_________________

4/28 GMATPrep 42Q 36V 640

Senior Manager
Joined: 13 May 2013
Posts: 425
Re: Square root and Modulus  [#permalink]

### Show Tags

15 Jun 2013, 08:19
1
How can the solution be negative if we're taking the square root of a positive number?

Bunuel wrote:
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 $$\sqrt{-x*|x|}=\sqrt{positive*positive}=\sqrt{positive}$$.

Hope it's clear.
Math Expert
Joined: 02 Sep 2009
Posts: 51072
Re: Square root and Modulus  [#permalink]

### Show Tags

15 Jun 2013, 08:23
WholeLottaLove wrote:
How can the solution be negative if we're taking the square root of a positive number?

The answer is A, which is $$-x$$, since $$x$$ is negative then $$-x=-(negative)=positive$$.
_________________
Senior Manager
Joined: 13 May 2013
Posts: 425
Re: Square root and Modulus  [#permalink]

### Show Tags

15 Jun 2013, 08:25
Ha! I got it just before I read your response. That is a very tricky problem - it's -(-x). Thanks!

Bunuel wrote:
WholeLottaLove wrote:
How can the solution be negative if we're taking the square root of a positive number?

The answer is A, which is $$-x$$, since $$x$$ is negative then $$-x=-(negative)=positive$$.
Manager
Joined: 01 Nov 2016
Posts: 66
Concentration: Technology, Operations
If x < 0, then root({-x} •|x|) is  [#permalink]

### Show Tags

03 Apr 2017, 09:42
Bunuel wrote:
Square root function can not give negative result

Bunuel, I do not understand this. When you graph $$x^2 - 9 = 0$$ you will get a parabola with roots at 3 and -3. That is because $$\sqrt{9}$$ is 3 or -3, not just 3. When you square root a number, the answer will always be plus/minus because when you square a negative number, it becomes positive. For example, $$\sqrt{25}$$ is 5 or -5, not just 5. This is because $$5^2$$ and $$(-5)^2$$ both equal 25. Why is this different for this question?

I think what fluke wrote is correct:

fluke wrote:
the question is: If x < 0, what is the value for $$sqrt{-x*\mid x \mid}$$:

$$sqrt{(+x)*(+x)}$$
$$sqrt{x^2}$$
$$\pm x$$

Reasoning is:

Since x < 0; only -x is the valid value. +x can be ignored.

Ans: "A"
Math Expert
Joined: 02 Sep 2009
Posts: 51072
Re: If x < 0, then root({-x} •|x|) is  [#permalink]

### Show Tags

03 Apr 2017, 09:49
joondez wrote:
Bunuel wrote:
Square root function can not give negative result

Bunuel, I do not understand this. When you graph $$x^2 - 9 = 0$$ you will get a parabola with roots at 3 and -3. That is because $$\sqrt{9}$$ is 3 or -3, not just 3. When you square root a number, the answer will always be plus/minus because when you square a negative number, it becomes positive. For example, $$\sqrt{25}$$ is 5 or -5, not just 5. This is because $$5^2$$ and $$(-5)^2$$ both equal 25. Why is this different for this question?

I think what fluke wrote is correct:

fluke wrote:
the question is: If x < 0, what is the value for $$sqrt{-x*\mid x \mid}$$:

$$sqrt{(+x)*(+x)}$$
$$sqrt{x^2}$$
$$\pm x$$

Reasoning is:

Since x < 0; only -x is the valid value. +x can be ignored.

Ans: "A"

What I said there is a fact.

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, $$\sqrt{16}=4$$, NOT +4 or -4. Even roots have only a positive value on the GMAT.

In contrast, the equation $$x^2=16$$ has TWO solutions, +4 and -4.

Odd roots have the same sign as the base of the root. For example, $$\sqrt[3]{125} =5$$ and $$\sqrt[3]{-64} =-4$$.
_________________
Intern
Joined: 25 Jul 2018
Posts: 2
Re: If x < 0, then root({-x} •|x|) is  [#permalink]

### Show Tags

05 Aug 2018, 12:50
Bunuel wrote:
udaymathapati wrote:
If x < 0, then \sqrt{-x} •|x|) is
A. -x
B. -1
C. 1
D. x
E. \sqrt{x}

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

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

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

The point here is that square root function can not give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.

So $$\sqrt{x^2}\geq{0}$$. But what does $$\sqrt{x^2}$$ equal to?

Let's consider following examples:
If $$x=5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=x=positive$$;
If $$x=-5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=-x=positive$$.

So we got that:
$$\sqrt{x^2}=x$$, if $$x\geq{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<0$$.

What function does exactly the same thing? The absolute value function! That is why $$\sqrt{x^2}=|x|$$

Back to the original question:
$$\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x$$

Or just substitute the value let $$x=-5<0$$ --> $$\sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5=-(-5)=-x$$.

Hope it's clear.

I have a different theory here. They explicitly state that x is negative. Then they give us an eq. Your solution neglects the fact that x is negative.
$$\sqrt{-x*|x|}=\sqrt{-(negative)*|negative|)}=\sqrt{positive*positive}=|x|=|x|$$
There are two solutions for |x| and both of them are in the answers. Some thing is off with this question imho
Math Expert
Joined: 02 Sep 2009
Posts: 51072
Re: If x < 0, then root({-x} •|x|) is  [#permalink]

### Show Tags

05 Aug 2018, 21:27
nobilisrex wrote:
Bunuel wrote:
udaymathapati wrote:
If x < 0, then \sqrt{-x} •|x|) is
A. -x
B. -1
C. 1
D. x
E. \sqrt{x}

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

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

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

The point here is that square root function can not give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.

So $$\sqrt{x^2}\geq{0}$$. But what does $$\sqrt{x^2}$$ equal to?

Let's consider following examples:
If $$x=5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=x=positive$$;
If $$x=-5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=-x=positive$$.

So we got that:
$$\sqrt{x^2}=x$$, if $$x\geq{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<0$$.

What function does exactly the same thing? The absolute value function! That is why $$\sqrt{x^2}=|x|$$

Back to the original question:
$$\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x$$

Or just substitute the value let $$x=-5<0$$ --> $$\sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5=-(-5)=-x$$.

Hope it's clear.

I have a different theory here. They explicitly state that x is negative. Then they give us an eq. Your solution neglects the fact that x is negative.
$$\sqrt{-x*|x|}=\sqrt{-(negative)*|negative|)}=\sqrt{positive*positive}=|x|=|x|$$
There are two solutions for |x| and both of them are in the answers. Some thing is off with this question imho

To check your theory, I'd suggest to plug some negative number and check what you'd get.

P.S. The question as well as the solution is fine.
_________________
Manager
Joined: 29 May 2017
Posts: 127
Location: Pakistan
Concentration: Social Entrepreneurship, Sustainability
Re: If x < 0, then root({-x} •|x|) is  [#permalink]

### Show Tags

19 Sep 2018, 17:38
Bunuel wrote:
udaymathapati wrote:
If x < 0, then \sqrt{-x} •|x|) is
A. -x
B. -1
C. 1
D. x
E. \sqrt{x}

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

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

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

The point here is that square root function can not give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.

So $$\sqrt{x^2}\geq{0}$$. But what does $$\sqrt{x^2}$$ equal to?

Let's consider following examples:
If $$x=5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=x=positive$$;
If $$x=-5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=-x=positive$$.

So we got that:
$$\sqrt{x^2}=x$$, if $$x\geq{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<0$$.

What function does exactly the same thing? The absolute value function! That is why $$\sqrt{x^2}=|x|$$

Back to the original question:
$$\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x$$

Or just substitute the value let $$x=-5<0$$ --> $$\sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5=-(-5)=-x$$.

Hope it's clear.

since we are given x<0, this means x is -ve. therefore:
1. -x--> -(-x) --> x

so,
2. sqrt(-x . |x|) --> sqrt(x . |x|)

can you explain what is wrong with stmt 1?
i lose the plot on stmt 2....can you explain why |x|--> -x? it should be +x

thanks
Intern
Joined: 07 Jul 2018
Posts: 24
If x < 0, then root({-x} •|x|) is  [#permalink]

### Show Tags

20 Sep 2018, 08:35
Bunuel wrote:
udaymathapati wrote:
If x < 0, then \sqrt{-x} •|x|) is
A. -x
B. -1
C. 1
D. x
E. \sqrt{x}

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

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

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

The point here is that square root function can not give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.

So $$\sqrt{x^2}\geq{0}$$. But what does $$\sqrt{x^2}$$ equal to?

Let's consider following examples:
If $$x=5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=x=positive$$;
If $$x=-5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=-x=positive$$.

So we got that:
$$\sqrt{x^2}=x$$, if $$x\geq{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<0$$.

What function does exactly the same thing? The absolute value function! That is why $$\sqrt{x^2}=|x|$$

Back to the original question:
$$\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x$$

Or just substitute the value let $$x=-5<0$$ --> $$\sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5=-(-5)=-x$$.

Hope it's clear.

I understood the concept, However I don't understand how you did this $$\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}$$
Inside mod X can be either positive or negative, how can you assume it to be negative?
If x < 0, then root({-x} •|x|) is &nbs [#permalink] 20 Sep 2018, 08:35

Go to page   Previous    1   2   [ 37 posts ]

Display posts from previous: Sort by