If x < 0, then root({-x} •|x|) is

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If x < 0, then root({-x} •|x|) is [#permalink]

02 Sep 2010, 13:20

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Question Stats:

46% (01:24) correct

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If x < 0, then \sqrt{-x*|x|}) is

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

[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Dec 2012, 02:22, edited 2 times in total.

Edited the question

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Re: Square root and Modulus [#permalink]

02 Sep 2010, 13:47

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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.

Answer: A.

Hope it's clear.
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Re: Square root and Modulus [#permalink]

02 Sep 2010, 14:07

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.
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Re: Square root and Modulus [#permalink]

02 Sep 2010, 14:14

Thanks, the way you posted the question clarified the solution. The original representation was quite confusing

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Re: Square root and Modulus [#permalink]

06 Sep 2010, 16:36

Thanks bunuel, what I would do without your explanations!!!

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Re: Square root and Modulus [#permalink]

21 Sep 2010, 04:14

Excellent post Bunuel +1

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Re: Square root and Modulus [#permalink]

22 Sep 2010, 10:56

Thanks to Bunuel. Key thing here is
Remember:
Square root of a number on gmat is positive number

I had seen similar problem earlier on gmatclub where I learnt this and got this one answer correct

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Re: Square root and Modulus [#permalink]

03 Nov 2010, 03:01

Thanks Bunuel for nice explanation.
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Re: Square root and Modulus [#permalink]

08 Dec 2010, 12:34

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shrive555 wrote:

Bunel: i've already seen all the explanation just one more question.

as If x<0, then \sqrt{-x*|x|}
Ans is -x
is Answer of the question depends on the condition x<0 or it depends on the sqrt (even root)

lets keep the condition same i.e x<0 and take odd root say cube root. i.e
\sqrt[3]{-x*|x}| . what would be the answer, would it be x then ?

Thanks

About the odd roots: odd roots will have the same sign as the base of the root. For example, \sqrt[3]{125} =5 and \sqrt[3]{-64} =-4.

So, if given that x<0 then |x|=-x and -x*|x|=(-x)*(-x)=positive*positive=x^2, thus odd root from positive x^2 will be positive.

But \sqrt[3]{-x*|x}| will equal neither to x nor to -x: \sqrt[3]{-x*|x|}=\sqrt[3]{x^2}=x^{\frac{2}{3}}, for example if x=-8<0 then \sqrt[3]{-x*|x|}=\sqrt[3]{x^2}=\sqrt[3]{64}=4.

If it were x<0 and \sqrt[3]{-x^2*|x}|=?, then \sqrt[3]{-x^2*|x}|=\sqrt[3]{-x^2*(-x)}=\sqrt[3]{x^3}=x<0. Or substitute the value let x=-5<0 --> \sqrt[3]{-x^2*|x}|=\sqrt[3]{-25*5}=\sqrt[3]{-125}=-5=x<0.

Now, back to the original question:

If x<0, then \sqrt{-x*|x|} equals:
A. -x
B. -1
C. 1
D. x
E. \sqrt{x}

As square root function can not give negative result, then options -1 (B) and x (D) can not be the answers as they are negative. Also \sqrt{x} (E) can not be the answer as even root from negative number is undefined for GMAT. 1 (C) also can not be the answer as for different values of x the answer will be different, so it can not be some specific value. So we are left with A.

Now, if it were x>0 instead of x<0 then the question would be flawed as in this case the expression under the even root would be negative (-x*|x|=negative*positive=negative).

Hope it's clear.
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Re: Square root and Modulus [#permalink]

08 Dec 2010, 23:10

Excellent !!! THanks B
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Re: Square root and Modulus [#permalink]

09 Dec 2010, 05:36

Friends,

In GMAT, to avoid the confusion with the negative #s, i use my own and simple technique that is as below.

PLEASE NOTE THAT: DEALING WITH THE INEQUALITIES/SQUARE-ROOTS/MODULUS THAT CONTAIN +VE #s IS EASIER THAN DEALING WITH THOSE CONTAIN -VE#S.

Given: x is a -ve
I take: y as a +ve #.
now i write -x = y

FROM NOW ON I AM GONNA DEAL WITH THE +VE # (i.e.y) ONLY. I CAN NOT BE TRAPPED BY THE TESTMAKER.

qtn: \sqrt{-x*|x|} = ?
==> \sqrt{y*|-y|}
==> \sqrt{y*y} as |-#|=#
==> \sqrt{y^2}
==> y (but not -y as Y is a positive #)
==> -x
Answer "A"

Refer can-anyone-answer-this-ds-question-94854.html for q qtn that can be solved using this technique.

Regards,
Murali.
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Re: Square root and Modulus [#permalink]

09 Dec 2010, 10:03

Murali: THe concept is same

\sqrt{-x*|x|} given x<0

=> \sqrt{x^2} = |x|

|x| = -x if x<0
|x| = x if x>0
since x<0 so -x
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Re: Square root and Modulus [#permalink]

23 Dec 2010, 19:04

great post Bunuel, thanks for the explanation.

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Re: Square root and Modulus [#permalink]

30 Dec 2010, 15:20

marijose wrote:

I still do not understend why the x is negative. Is it because of the condition at the begining or because of the absolute value that is in the question?

here is whay I see.
SQ ( -(-x)) *|-x|

you get... sq(x*x) therefore the answer is x, do you add the negative sign because the problem starts out saying that x is negative or is there a rule I am not seeing?

I was under the impression that the square root of any number cannot by negative or is this incorrect?

regards

x is negative because it's given in the stem that it's negative (if x<0...). Next, knowing that x is negative doesn't mean that you should write -x instead of x.

As for the square root: yes, even roots can not give negative result, so when you say that \sqrt{-x*|x|} equals to x, you are basically saying that square root of an expression equals to the negative number (as x is negative) which can not be true. Now, the answer is -x or -(negative)=positive, so square root of an expression equals to positive number which can be true. Also \sqrt{x^2} equals to |x| not to x. It's all explained above, so please read the solutions again and ask if anything needs further clarification.

You can also check Absolute Value (math-absolute-value-modulus-86462.html) and Number Theory (math-number-theory-88376.html) chapters of Math Book for fundamentals.
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Re: Square root and Modulus [#permalink]

30 Dec 2010, 15:47

marijose wrote:

Ive read the explanations .. sorry to keep bugging but

lets say you plug in a a -5.
by doing all the calculations you get that the SQ(-(-5)*|-5|) = sq(5*5) = 5 so because and only because the question states that x is negative we have to change the value to -5. I this the correct reasoning?

I'm not sure understood your question.

What do you mean "change the value to -5"? If x=-5<0 then \sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5 so the answer is 5 or -x, as -x=-(-5)=5.

Hope it's clear.
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Re: Square root and Modulus [#permalink]

31 Dec 2010, 09:06

shrive555 wrote:

marijose wrote:

you don't have to plug in -5.
Just Plug in 5, keep in mind that x<0 so it will make 5 as -5 automatically. so when you get the final answer after taking square root of 25 = 5. due to given condition of x<0 . you have to pick -5 or (negative value)from the answer choices. The only trick is " x<0 "

i was stuck in the same point :-D

What do you mean by "it will make 5 as -5 automatically"???

And again: given that x<0. If you want to solve this question by number plugging method you should plug negative values of x --> if x=-5<0 then \sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5 so the answer is 5 or -x, as -x=-(-5)=5.
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Re: Square root and Modulus [#permalink]

31 Dec 2010, 11:33

Bunuel wrote:

shrive555 wrote:

marijose wrote:

Since x=5 but x<0 so x=-5
probably marijose is confusing the negative sign - with negative x .... \sqrt{-(x)*|x|}
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Re: Square root and Modulus [#permalink]

31 Dec 2010, 11:44

shrive555 wrote:

Since x=5 but x<0 so x=-5
probably marijose is confusing the negative sign - with negative x .... \sqrt{-(x)*|x|}

The red part does not make any sense. Please refer to the solutions on the previous page.
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Re: Square root and Modulus [#permalink]

31 Dec 2010, 12:31

Bunuel wrote:

marijose wrote:

x is negative because it's given in the stem that it's negative (if x<0...). Next, knowing that x is negative doesn't mean that you should write -x instead of x.

This is what i meant but messed up in mathematics lingo :think:

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Re: Is the square root always positive? [#permalink]

01 Feb 2011, 06:23

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Merging similar topics.

karenhipol wrote:

I encountered this question in gmat prep:

If x is < 0, what is the Square root of -x/x/

a. 0
b. 1
c. x
d. -x
e. -1

The official answer is -x but I answered x because I thought the answer should be positive.

Should I consider however, that since x <0 then I should multiply it by -1 to get a positive number so the answer is - x ( to get -(-x) = +x.

Please let me know If I reasoned this out correctly.

Thanks

Two things:

1. Square root function can not give negative result: \sqrt{some \ expression}\geq{0} so square root of a non-negative number is not always positive, it's never negative (always non-negative).

2. As x<0 then -x=-negative=positive so the answer is still positive.

For the complete solution refer to the above posts.
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Re: Is the square root always positive? [#permalink] 01 Feb 2011, 06:23

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