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# is x<0, then sqrt(-x * |x|) 1. -x 2. -1 3. 1 4. x 5.

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VP
Joined: 29 Apr 2003
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is x<0, then sqrt(-x * |x|) 1. -x 2. -1 3. 1 4. x 5. [#permalink]

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05 May 2006, 23:02
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is x<0, then sqrt(-x * |x|)

1. -x
2. -1
3. 1
4. x
5. sqrt(x)

Kudos [?]: 30 [0], given: 0

Senior Manager
Joined: 29 Jun 2005
Posts: 403

Kudos [?]: 22 [0], given: 0

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05 May 2006, 23:45
tricky one. Thanks.

if x<0, then sqrt(-x * |x|)
-x=[-x] since x<0
the same is for [x]
sqrt([-x]^2)= -x

Kudos [?]: 22 [0], given: 0

VP
Joined: 29 Apr 2003
Posts: 1402

Kudos [?]: 30 [0], given: 0

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05 May 2006, 23:50
Dilshod wrote:
tricky one. Thanks.

if x<0, then sqrt(-x * |x|)
-x=[-x] since x<0
the same is for [x]
sqrt([-x]^2)= -x

sqrt([-x]^2)= -x

Should this be
[-x]^2 = x^2
and then sqrt (x^2) = x?

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VP
Joined: 29 Apr 2003
Posts: 1402

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06 May 2006, 00:04
Can anyone take at stab at explaining this?

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Senior Manager
Joined: 29 Jun 2005
Posts: 403

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06 May 2006, 06:14
sm176811 wrote:
Dilshod wrote:
tricky one. Thanks.

if x<0, then sqrt(-x * |x|)
-x=[-x] since x<0
the same is for [x]
sqrt([-x]^2)= -x

sqrt([-x]^2)= -x

Should this be
[-x]^2 = x^2
and then sqrt (x^2) = x?

Hi,
I’m not good in explanations, but I will try:

sqrt(x^2)=[x], which means that x in any case, whether it is positive or negative, should be +ve.
Because there is no such answer ([x]), we ought to choose the one which is always +ve. Since x<0, than –x is ALWAYS positive. That is the answer.

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Manager
Joined: 27 Mar 2006
Posts: 136

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06 May 2006, 10:44
So -x is positive since x is negative...hence -(-x) = postive

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06 May 2006, 10:44
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