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# If k≠0 and k-(3-2k^2)/k =x/k, then x= (A) -3-k^2 (B)

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Intern
Joined: 30 Mar 2006
Posts: 27

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Location: New Jersey
If k≠0 and k-(3-2k^2)/k =x/k, then x= (A) -3-k^2 (B) [#permalink]

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07 Apr 2006, 11:10
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45% (medium)

Question Stats:

49% (00:55) correct 51% (00:53) wrong based on 439 sessions

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If $$k\neq{0}$$ and $$k - \frac{3 -2k^2}{k} = \frac{x}{k}$$, then x =

(A) -3 - k^2
(B) k^2 -3
(C) 3k^2 - 3
(D) k - 3 - 2k^2
(E) k - 3 + 2k^2

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-k-0-and-k-3-2k-2-k-x-k-then-x-167913.html
[Reveal] Spoiler: OA

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Andre Crompton
andre.crompton@cit.com

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Re: If k≠0 and k-(3-2k^2)/k =x/k, then x= (A) -3-k^2 (B) [#permalink]

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07 Apr 2006, 11:20
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k - ( 3 - 2K^2)/k = x/k

k^2 - ( 3 - 2k^2) = x

k^2 - 3 + 2k^2 = x

x = (3k^2) - 3

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Intern
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Posts: 27

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Re: If k≠0 and k-(3-2k^2)/k =x/k, then x= (A) -3-k^2 (B) [#permalink]

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07 Apr 2006, 13:23
I dont understand how the -2k^2 became +2k^2

k^2 - ( 3 - 2k^2) = x

k^2 - 3 + 2k^2 = x
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Andre Crompton
andre.crompton@cit.com

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Re: If k≠0 and k-(3-2k^2)/k =x/k, then x= (A) -3-k^2 (B) [#permalink]

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07 Apr 2006, 15:10
andrecrompton wrote:
I dont understand how the -2k^2 became +2k^2

k^2 - ( 3 - 2k^2) = x

k^2 - 3 + 2k^2 = x

The brackets open and negative negative make a positive

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Re: If k≠0 and k-(3-2k^2)/k =x/k, then x= (A) -3-k^2 (B) [#permalink]

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23 Oct 2014, 18:22
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Re: If k≠0 and k-(3-2k^2)/k =x/k, then x= (A) -3-k^2 (B) [#permalink]

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24 Oct 2014, 02:34
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Expert's post
andrecrompton wrote:
If $$k\neq{0}$$ and $$k - \frac{3 -2k^2}{k} = \frac{x}{k}$$, then x =

(A) -3 - k^2
(B) k^2 -3
(C) 3k^2 - 3
(D) k - 3 - 2k^2
(E) k - 3 + 2k^2

$$k - \frac{3 -2k^2}{k} = \frac{x}{k}$$;

Multiply by k: $$k^2 - (3 -2k^2) = x$$;

$$x=3k^2-3$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-k-0-and-k-3-2k-2-k-x-k-then-x-167913.html

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Re: If k≠0 and k-(3-2k^2)/k =x/k, then x= (A) -3-k^2 (B)   [#permalink] 24 Oct 2014, 02:34
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# If k≠0 and k-(3-2k^2)/k =x/k, then x= (A) -3-k^2 (B)

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