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# If x < 0, then root({-x} •|x|) is

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

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02 Sep 2010, 13:20
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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]

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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 $$x=-5<0$$ --> $$\sqrt{-x*|x|}=\sqrt{-(-5)*|-5|}=\sqrt{25}=5=-(-5)=-x$$.

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

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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]

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02 Sep 2010, 14:07
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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: Is the square root always positive? [#permalink]

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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: Square root and Modulus [#permalink]

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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]

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06 Sep 2010, 16:36
Thanks bunuel, what I would do without your explanations!!!
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Re: Square root and Modulus [#permalink]

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21 Sep 2010, 04:14
Excellent post Bunuel +1
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Re: Square root and Modulus [#permalink]

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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]

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03 Nov 2010, 03:01
Thanks Bunuel for nice explanation.
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Re: Square root and Modulus [#permalink]

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08 Dec 2010, 23:10
Excellent !!! THanks B
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Re: Square root and Modulus [#permalink]

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09 Dec 2010, 05:36
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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 #)
==> $$-x$$

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]

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09 Dec 2010, 10:03
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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]

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23 Dec 2010, 19:04
great post Bunuel, thanks for the explanation.
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Re: Square root and Modulus [#permalink]

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30 Dec 2010, 15:20
Expert's post
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]

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30 Dec 2010, 15:47
Expert's post
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]

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31 Dec 2010, 09:06
Expert's post
shrive555 wrote:
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?

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

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]

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31 Dec 2010, 11:33
Bunuel wrote:
shrive555 wrote:
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?

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

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 $$-x=-(-5)=5$$.

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]

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31 Dec 2010, 11:44
Expert's post
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]

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31 Dec 2010, 12:31
Bunuel wrote:
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.

This is what i meant but messed up in mathematics lingo
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Re: Square root and Modulus   [#permalink] 31 Dec 2010, 12:31

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