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If x < 0, then sqrt(-x|x|) = a) -x b) -1 c) 1 d)

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If x < 0, then sqrt(-x|x|) = a) -x b) -1 c) 1 d) [#permalink]

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New post 28 Jul 2008, 10:56
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If \(x < 0\), then \(sqrt(-x|x|)\) =

a) \(-x\)
b) \(-1\)
c) \(1\)
d) \(x\)
e) \(sqrt(x)\)



I don't get it. I chose D, but the OA is A. How come??

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Re: PS: What is x? [#permalink]

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New post 28 Jul 2008, 11:06
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tarek99 wrote:
If \(x < 0\), then \(sqrt(-x|x|)\) =

a) \(-x\)
b) \(-1\)
c) \(1\)
d) \(x\)
e) \(sqrt(x)\)



I don't get it. I chose D, but the OA is A. How come??


OA as A) is right because if you replace the eqn with +ve x then it will become sqrt (-x^2)

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Re: PS: What is x? [#permalink]

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New post 28 Jul 2008, 11:11
nmohindru wrote:
tarek99 wrote:
If \(x < 0\), then \(sqrt(-x|x|)\) =

a) \(-x\)
b) \(-1\)
c) \(1\)
d) \(x\)
e) \(sqrt(x)\)



I don't get it. I chose D, but the OA is A. How come??


OA as A) is right because if you replace the eqn with +ve x then it will become sqrt (-x^2)


but when you look at |x|, the result here will always be positive. As for -x, I read somewhere that whenever you see such an expression, I should treat the negative sign attached to x separately. So since x should be negative, negative x times the attached negative sign should be positive, therefore \(sqrt(x^2)\) which is equal to |x|, therefore x. What's wrong with my approach?

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Re: PS: What is x? [#permalink]

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New post 28 Jul 2008, 12:07
You have sqrt(x^2) = |x| = -x ( since x<0)
The ans is A
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Re: PS: What is x? [#permalink]

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New post 28 Jul 2008, 13:15
tarek99 wrote:
If \(x < 0\), then \(sqrt(-x|x|)\) =

a) \(-x\)
b) \(-1\)
c) \(1\)
d) \(x\)
e) \(sqrt(x)\)



I don't get it. I chose D, but the OA is A. How come??



I solved by plugging in x= -3

\(sqrtx\) = +/-x, given x<0, so must be -x

i.e. \(sqrt9\) = +/-3

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Re: PS: What is x? [#permalink]

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New post 28 Jul 2008, 13:50
You are very correct when you say :

Sqrt (x^2) = |x|

But it is also given in the question stem that x < 0, Hence out ot the two possible values of |x|: +x and -x, the answer to be chosen would become -x.

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Re: PS: What is x? [#permalink]

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New post 28 Jul 2008, 14:11
tarek99 wrote:
If \(x < 0\), then \(sqrt(-x|x|)\) =

a) \(-x\)
b) \(-1\)
c) \(1\)
d) \(x\)
e) \(sqrt(x)\)

I don't get it. I chose D, but the OA is A. How come??


if we assume, it woud be better understandable: suppose x = -5.

\(sqrt(-x|x|)\) = \(sqrt[-(-5)|-5|]\) = \(sqrt(25)\) = 5

Since x = -5, -x = 5

therefore: \(sqrt(-x|x|)\) = \(-x\)

So its A.
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Re: PS: What is x? [#permalink]

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New post 28 Jul 2008, 21:32
I see.....thanks a lot everyone for your input!

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Re: PS: What is x?   [#permalink] 28 Jul 2008, 21:32
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If x < 0, then sqrt(-x|x|) = a) -x b) -1 c) 1 d)

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