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

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SVP
Joined: 21 Jul 2006
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If x < 0, then sqrt(-x|x|) = a) -x b) -1 c) 1 d) [#permalink]

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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??

Kudos [?]: 1010 [0], given: 1

Senior Manager
Joined: 06 Apr 2008
Posts: 429

Kudos [?]: 165 [1], given: 1

Re: PS: What is x? [#permalink]

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28 Jul 2008, 11:06
1
KUDOS
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)

Kudos [?]: 165 [1], given: 1

SVP
Joined: 21 Jul 2006
Posts: 1512

Kudos [?]: 1010 [0], given: 1

Re: PS: What is x? [#permalink]

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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?

Kudos [?]: 1010 [0], given: 1

Manager
Joined: 18 Aug 2006
Posts: 61

Kudos [?]: 9 [0], given: 2

Location: Houston TX
Re: PS: What is x? [#permalink]

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28 Jul 2008, 12:07
You have sqrt(x^2) = |x| = -x ( since x<0)
The ans is A
_________________

haveaniceday

Kudos [?]: 9 [0], given: 2

Manager
Joined: 27 Mar 2008
Posts: 81

Kudos [?]: 122 [0], given: 0

Re: PS: What is x? [#permalink]

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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

Kudos [?]: 122 [0], given: 0

Intern
Joined: 06 Jul 2008
Posts: 30

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

Re: PS: What is x? [#permalink]

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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.

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

SVP
Joined: 29 Aug 2007
Posts: 2472

Kudos [?]: 843 [0], given: 19

Re: PS: What is x? [#permalink]

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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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GT

Kudos [?]: 843 [0], given: 19

SVP
Joined: 21 Jul 2006
Posts: 1512

Kudos [?]: 1010 [0], given: 1

Re: PS: What is x? [#permalink]

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

Kudos [?]: 1010 [0], given: 1

Re: PS: What is x?   [#permalink] 28 Jul 2008, 21:32
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