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by canuckclint » Sat Oct 25, 2008 3:46 pm
Solve

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?

Answer choices

-4y^2/3


-2y^2


(2y^2+1)/3


2y^2


6y^2/3
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by Delpaiero » Sat Oct 25, 2008 6:32 pm
From 9x^4-4y^4=3x^2+2y^2
then (3x^2-2y^2)(3x^2+2y^2) = 3x^2+2y^2
(3x^2-2y^2) = 1
x^2 = (1+2y^2)/3


Good luck!!!!
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by canuckclint » Sat Oct 25, 2008 9:23 pm
Delpaiero wrote:From 9x^4-4y^4=3x^2+2y^2
then (3x^2-2y^2)(3x^2+2y^2) = 3x^2+2y^2
(3x^2-2y^2) = 1
x^2 = (1+2y^2)/3


Good luck!!!!
Nicely done, the trick here is to realize the difference of squares eqn can
be made.
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