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

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manimgoindown
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If x and y are non-zero integers, and 9x^4 – 4y^4 = 3x^2 + 2y^2, which [#permalink] New post   Updated on: 23 Jan 2022, 03:11
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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 x^2 in terms of y?


A. \(\frac{4y^2}{3}\)

B. \(2y^2\)

C. \(\frac{(2y^2+1)}{3}\)

D. \(2y^2\)

E. \(\frac{6y^2}{3}\)

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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?
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C

Originally posted by manimgoindown on 25 May 2012, 15:14.
Last edited by Bunuel on 23 Jan 2022, 03:11, edited 2 times in total.
Edited the question and moved to PS subforum
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If x and y are non-zero integers, and 9x^4 – 4y^4 = 3x^2 + 2y^2, which [#permalink] New post   25 May 2012, 15:32
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manimgoindowndown wrote:
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?


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 \(a^2-b^2=(a-b)(a+b)\) to get \((3x^2-2y^2)(3x^2+2y^2)=3x^2+2y^2\);

Reduce by \(3x^2+2y^2\) to get \(3x^2-2y^2=1\);

\(x^2=\frac{2y^2+1}{3}\).

Answer: C.

Hope it's clear.
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Re: If x and y are non-zero integers, and 9x^4 – 4y^4 = 3x^2 + 2y^2, which [#permalink] New post   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 + 2y^2, which [#permalink] New post   25 May 2012, 16:37
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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 + 2y^2, which [#permalink] New post   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 :P

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 + 2y^2, which [#permalink] New post   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
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Re: If x and y are non-zero integers, and 9x^4 – 4y^4 = 3x^2 + 2y^2, which [#permalink] New post   19 Oct 2013, 07:38
The way I solved it is the following:

9x^4 - 4y^4 = 3x^2 + 2y^2
9x^4 - 3x^2 = 4y^4 + 2y^2
3x^2 ( 3x^2 - 1) = 2y^2 (2y^2 + 1)

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

so now we can equate the numerators for example:

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

Answer: C
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Re: If x and y are non-zero integers, and 9x^4 – 4y^4 = 3x^2 + 2y^2, which [#permalink] New post   26 Feb 2014, 20:00
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9x^4 – 4y^4 = 3x^2 + 2y^2

Divide both sides by 3x^2 + 2y^2

3x^2 - 2y^2 = 1

x^2 = (2y^2 + 1) / 3 = Answer = C
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Re: If x and y are non-zero integers, and 9x^4 – 4y^4 = 3x^2 + 2y^2, which [#permalink] New post   03 Dec 2014, 04:36
9x^4 – 4y^4 = 3x^2 + 2y^2

P^2 - q^2 = (p+q) (p-q)

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

as x and y bot non zero integers
(3x^2 + 2y^2) > 0

Divide both sides by 3x^2 + 2y^2

3x^2 - 2y^2 = 1

x^2 = (2y^2 + 1) / 3 = Answer = C
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Re: If x and y are non-zero integers, and 9x^4 4y^4 = 3x^2 + 2y^2, which [#permalink] New post   20 Nov 2023, 19:53
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Re: If x and y are non-zero integers, and 9x^4 4y^4 = 3x^2 + 2y^2, which [#permalink]
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