is x<0, then sqrt(-x * |x|) 1. -x 2. -1 3. 1 4. x 5. : DS Archive
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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
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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, 22:02
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is x<0, then sqrt(-x * |x|)

1. -x
2. -1
3. 1
4. x
5. sqrt(x)
Senior Manager
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05 May 2006, 22:45
tricky one. Thanks.

Answer is -X (1 choice)
if x<0, then sqrt(-x * |x|)
-x=[-x] since x<0
the same is for [x]
sqrt([-x]^2)= -x
VP
Joined: 29 Apr 2003
Posts: 1403
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05 May 2006, 22:50
Dilshod wrote:
tricky one. Thanks.

Answer is -X (1 choice)
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?
VP
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05 May 2006, 23:04
Can anyone take at stab at explaining this?
Senior Manager
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06 May 2006, 05:14
sm176811 wrote:
Dilshod wrote:
tricky one. Thanks.

Answer is -X (1 choice)
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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06 May 2006, 09:44
So -x is positive since x is negative...hence -(-x) = postive
06 May 2006, 09:44
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# is x<0, then sqrt(-x * |x|) 1. -x 2. -1 3. 1 4. x 5.

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