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If x < 0, then (-x|x|)^(1/2) is:

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If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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New post 06 May 2010, 12:33
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If \(x < 0\), then \(\sqrt{-x|x|}\) is:


A. -x

B. -1

C. 1

D. x

E. \(\sqrt{x}\)


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Re: If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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New post 06 May 2010, 12:44
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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 ?
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Re: If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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New post 06 May 2010, 13:12
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Re: If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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New post 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.
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Re: If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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New post 19 May 2010, 01:21
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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 \(|x|=-x\) --> \(\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x\).

Answer: A.

Hope it helps.
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Re: If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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New post 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 \(|x|=-x\) --> \(\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x\).

Answer: A.

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...
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Re: If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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New post 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]
Answer: A.

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

See this link http://en.wikipedia.org/wiki/Square_root

I hope this helps.
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Re: If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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New post 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.
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Re: If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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New post 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)
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Re: If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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Re: If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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New post 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.
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Re: If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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New post 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 \(|x|=-x\) --> \(\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x\).

Answer: A.
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Re: If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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New post 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 \(|x|=-x\) --> \(\sqrt{-x*|x|}=\sqrt{(-x)*(-x)}=\sqrt{x^2}=|x|=-x\).

Answer: A.


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.
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Re: If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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New post 20 May 2015, 18:28
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.
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Re: If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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New post 20 May 2015, 19:53
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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

Final Answer:

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Re: If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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New post 23 May 2015, 10:55
1
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

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Re: If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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New post 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.
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Re: If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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New post 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.
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Re: If x < 0, then (-x|x|)^(1/2) is: [#permalink]

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New post 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 (-x|x|)^(1/2) is:   [#permalink] 11 Mar 2018, 08:35
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