If x and y are non-zero integers, and 9x^4 – 4y^4 = 3x^2 + 2

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If x and y are non-zero integers, and 9x^4 – 4y^4 = 3x^2 + 2 [#permalink]

25 May 2012, 15:14

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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

When a problem involves variables raised to the fourth power, it is often useful to represent them as a square of another square, since this approach will allow us to apply manipulations of squares. Also note that since we are dealing with high exponents, the approach of plugging numbers would prove time-consuming and prone to error in this case.

9x4 – 4y4 = (3x2)2 – (2y2)2 = (3x2 + 2y2)(3x2 – 2y2).

(3x2 + 2y2)(3x2 – 2y2) = (3x2 + 2y2)

I'm not getting how (3x2 + 2y2)(3x2 – 2y2) = (3x2 + 2y2). Can someone please explain?

[Reveal] Spoiler: OA

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Last edited by Bunuel on 25 May 2012, 15:22, edited 1 time in total.

Edited the question and moved to PS subforum

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Re: If x and y are non-zero integers, and 9x^4 – 4y^4 = 3x^2 + 2 [#permalink]

25 May 2012, 15:32

manimgoindowndown wrote:

Welcome to GMAT Club. Below is an answer to your question.

9x^4-4y^4=3x^2+2y^2 --> (3x^2)^2-(2y^2)^2=3x^2+2y^2 --> apply a^2-b^2=(a-b)(a+b): (3x^2-2y^2)(3x^2+2y^2)=3x^2+2y^2 --> reduce by 3x^2+2y^2: 3x^2-2y^2=1 --> x^2=\frac{2y^2+1}{3}.

Answer: C.

Hope it's clear.

P.S. Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/ and DS questions in the DS subforum: gmat-data-sufficiency-ds-141/ No posting of PS/DS questions is allowed in the main Math forum.
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Re: If x and y are non-zero integers, and 9x^4 – 4y^4 = 3x^2 + 2 [#permalink]

25 May 2012, 16:32

Not clear still
I break down the 9x^4 expression using a difference of squares to (3x^2 -2y^2)(3x^2+2y^2) I don't get it from there on especially how the later statment equals 3x^2+2y^2 all by itself
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Re: If x and y are non-zero integers, and 9x^4 – 4y^4 = 3x^2 + 2 [#permalink]

25 May 2012, 16:37

manimgoindowndown wrote:

Not clear still
I break down the 9x^4 expression using a difference of squares to (3x^2 -2y^2)(3x^2+2y^2) I don't get it from there on especially how the later statment equals 3x^2+2y^2 all by itself

From (3x^2-2y^2)(3x^2+2y^2)=3x^2+2y^2 divide both parts of the equation by 3x^2+2y^2 to get 3x^2-2y^2=1 --> 3x^2=2y^2+1--> x^2=\frac{2y^2+1}{3}.
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Re: If x and y are non-zero integers, and 9x^4 – 4y^4 = 3x^2 + 2 [#permalink]

26 May 2012, 13:33

Ok let's try this one more time since I dont' think you understood which part I didn't get

how do you get from/how does (3x^2-2y^2)(2x^2+2y^2) to/equal 3x^2+2y^2
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Re: If x and y are non-zero integers, and 9x^4 – 4y^4 = 3x^2 + 2 [#permalink]

26 May 2012, 14:13

(3x^2-2y^2)(3x^2+2y^2)=3x^2+2y^2

(3x^2-2y^2)(3x^2+2y^2) - [3x^2+2y^2 ]=0

[3x^2+2y^2 ] [(3x^2-2y^2) - (1) ] = 0
so , either [3x^2+2y^2 ] =0 or [(3x^2-2y^2) - (1) ]=0

[3x^2+2y^2 ] =0 can never be zero as question stem says that x,y are non-zero integers , so 3x^2+2y^2 will always be greater than zero .

[(3x^2-2y^2) - (1) ]=0
3x^2= 2y^2 + 1

x^2= (2y^2 + 1)/3

hence C is the answer.
Hope, it is clear now

Re: If x and y are non-zero integers, and 9x^4 – 4y^4 = 3x^2 + 2 [#permalink] 26 May 2012, 14:13

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If x and y are non-zero integers, and 9x^4 – 4y^4 = 3x^2 + 2

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If x and y are non-zero integers, and 9x^4 – 4y^4 = 3x^2 + 2

If x and y are non-zero integers, and 9x^4 – 4y^4 = 3x^2 + 2

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