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If x<0, then sqrt(-x|x|) is -x -1 1 x sqrt(x

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If x<0, then sqrt(-x|x|) is -x -1 1 x sqrt(x [#permalink]

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New post 18 Nov 2007, 17:02
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If x<0, then sqrt(-x|x|) is

-x
-1
1
x
sqrt(x)
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New post 18 Nov 2007, 17:27
Expert's post
A. -x

for x<0: |x|=-x

f=sqrt(-x|x|)=sqrt((-x)*(-x))=sqrt(x^2)=|x|=-x
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Re: DS absolute value [#permalink]

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New post 18 Nov 2007, 17:28
young_gun wrote:
If x<0, then sqrt(-x|x|) is

-x
-1
1
x
sqrt(x)


Quick guess - A -x?

if x < 0, then x is negative, thus -x is positive, and |x| as the absolute value is a positive value - so it's equivalent is positive positive X^2 the square root of which is positive x. If x is negative, then -x is positive.
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New post 18 Nov 2007, 17:49
I guess I have not paid much attention to abs value concepts...does anyone know of any websites/posts that cover everything from the basic to the advanced? thx
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New post 18 Nov 2007, 17:51
Clear A.

We know that the value is -x (from the question). So irrespective of any mathmatical calculation its value will remains the same.

Amar
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New post 18 Nov 2007, 18:29
Guys, does n't this have to be -X instead of X ?
If it were -X and let us say we pick a value of -3 for X (because X < 0), then the result would be -(-3) = 3. This does not agree with what was given in the question.

Please explain if my approach was wrong.
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New post 18 Nov 2007, 20:06
Is n't OA: D?
Please see my previous post above...
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New post 18 Nov 2007, 20:51
Actually, I'm the second one who also thinks that the answer should be X ..

"The square of any positive or negative number is positive, and the square of 0 is 0. Therefore, no negative number can have a real square root. "
http://en.wikipedia.org/wiki/Square_roo ... ex_numbers
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 [#permalink]

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New post 18 Nov 2007, 22:19
Expert's post
ok

definition:

f=|x| ==> 1. x>=0: f=x 2. x<0: f=-x

therefore on should use -x instead |x| in the case x<0
.......

let x=-10
sqrt(-x|x|)=sqrt(-(-10)*|-10|)=sqrt(-(-10)*(10))=sqrt(-(-100))=sqrt(100)=10=-x
  [#permalink] 18 Nov 2007, 22:19
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If x<0, then sqrt(-x|x|) is -x -1 1 x sqrt(x

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