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What is X^2??

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by HeyArnold » Thu Sep 15, 2011 6:51 pm
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 [spoiler]C

Really complicated... steps would be helpful![/spoiler]
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by GMATGuruNY » Thu Sep 15, 2011 7:09 pm
HeyArnold 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

OA [spoiler]C

Really complicated... steps would be helpful![/spoiler]
9x^4 - 4y^4= 3x^2 + 2y^2

(3x² + 2y²)(3x² - 2y²) = 3x² + 2y²

3x² - 2y² = 1

3x² = 2y² + 1

x² = (2y² + 1)/3.

The correct answer is C.
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by Anurag@Gurome » Thu Sep 15, 2011 7:17 pm
HeyArnold 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

OA [spoiler]C




Really complicated... steps would be helpful![/spoiler]
9x^4 - 4y^4 = (3x^2 + 2y^2)(3x^2 - 2y^2) = 3x^2 + 2y^2
Or (3x^2 + 2y^2)(3x^2 - 2y^2 - 1) = 0
Now, (3x^2 + 2y^2) cannot be zero because both x and y are non zero.
Hence, 3x^2 - 2y^2 - 1 = 0 or x^2 = (2y^2 + 1)/3
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by HeyArnold » Sat Sep 24, 2011 1:15 pm
Ah... Thank you both.

In this case what I should have recognized is it's essentially a special product, X^2 - Y^2 = (x+y)(x-y)

It was just hidden with the ^4.
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by Cheese12 » Sun Sep 25, 2011 4:26 am
Can pluggin in work in this question ?
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