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sovlve in terms of y.

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Senior Manager
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sovlve in terms of y. [#permalink]

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New post 09 Aug 2009, 09:00
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C
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If x and y are non-zero integers, and 9x^4 – 4y^4 = 3x^2 + 2y^2, which of the following could be the value of x2 in terms of y?
A) -4y^2/3

B) -2y^2

C) (2y^2+1)/3

D) 2y^2

E) 6y^2/3

OA:
[Reveal] Spoiler:
C


Explain please...

Last edited by crejoc on 09 Aug 2009, 11:58, edited 1 time in total.

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Re: sovlve in terms of y. [#permalink]

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New post 09 Aug 2009, 11:31
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9x^4 – 4y^4 = 3x^2 + 2y^2

(3x^2)^2 – (2y^2)^2 = (3x^2 + 2y^2)

(3x^2 + 2y^2)(3x^2 - 2y^2)=(3x^2 + 2y^2)...this means (3x^2 - 2y^2)= 1

hence x^2 = (2y^2+1)/3...OA C
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Re: sovlve in terms of y. [#permalink]

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New post 10 Aug 2009, 01:18
Agree with Bhushan! Answer would be C.
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Re: sovlve in terms of y. [#permalink]

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New post 10 Aug 2009, 02:34
bhushan252 wrote:
9x^4 – 4y^4 = 3x^2 + 2y^2

(3x^2)^2 – (2y^2)^2 = (3x^2 + 2y^2)

(3x^2 + 2y^2)(3x^2 - 2y^2)=(3x^2 + 2y^2)...this means (3x^2 - 2y^2)= 1

hence x^2 = (2y^2+1)/3...OA C

Perfect ! Only one thing to note here, we can reduce to the expression marked in RED only if we are sure that the expression (3x^2 + 2y^2) is non-zero, which is evident from the fact that x and y are non-zero :)

Kudos [?]: 817 [0], given: 18

Re: sovlve in terms of y.   [#permalink] 10 Aug 2009, 02:34
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