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A^8 x B ^4 - A^4x B^ 2= 12

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by neelgandham » Sun Dec 18, 2011 4:54 am
aditya.j wrote:If A,B are non zero integers and A^8xB^4 - A^4xB 2= 12, which of the following could be A^2 in terms of B?
I. 2/B
II. - 2/B
III. 3/B
:roll:
A^8xB^4 - A^4xB^2= 12
Option I - A^2 = 2/B =? A^2*B = 2
A^8xB^4 - A^4xB^2 = (A^2*B)^4 - (A^2*B)^2 = 2^4 - 2^2 = 16-4 = 12 Satisfies.

Option II - A^2 = -2/B =? A^2*B = -2
A^8xB^4 - A^4xB^2 = (A^2*B)^4 - (A^2*B)^2 = (-2)^4 - (-2)^2 = 16-4 = 12 Satisfies.

Option III - A^2 = 3/B =? A^2*B = 3
A^8xB^4 - A^4xB^2 = (A^2*B)^4 - (A^2*B)^2 = 3^4 - 3^2 = 27-9 = 18 Nopes!

Option I and II.

You can also solve this solving the equation
A^8xB^4 - A^4xB^2= 12, Let A^4xB^2 = K then
K^2 - K = 12
K(K-1) = 12 where K is an integer
K = 4. i.e. A^4xB^2 = 4,
Implies A^2*B = +2
A^2 = +2/B
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by Ian Stewart » Sun Dec 18, 2011 4:54 am
aditya.j wrote:If A,B are non zero integers and A^8xB^4 - A^4xB 2= 12, which of the following could be A^2 in terms of B?

I. 2/B
II. - 2/B
III. 3/B

:roll: [/spoiler]
While you don't need to here, I'd find it easiest to factor first. We can take out the common factor of a^4 * b^2 from each expression, then use a difference of squares:

(a^8)(b^4) - (a^4)(b^2) = 12
(a^4)(b^2) [ (a^4)(b^2) - 1] = 12
(a^2 * b)^2 [ (a^2)(b) + 1 ] [ (a^2)(b) - 1 ] = 12

You could use divisibility properties here (each term on the left side needs to be a divisor of 12, so there are almost no possibilities to consider), but I think the easiest thing to do is to simply plug in each answer choice. If you replace a^2 with 2/b or with -2/b, the left side is equal to 12. If you replace a^2 with 3/b, then the left side is much larger than 12. So the answer is I and II only.
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by aditya.j » Sun Dec 18, 2011 5:40 am
Thanks a lot for the quick replies!..

Although i did factorize A^8 x B ^4 - A^4x B^ 2= 12

I got confused when i read which of the following could be A^2 in terms of B. Didnt strike me that i could plug the values. In addition to that, the 2 minute deadline was something that froze my thought process.

Any suggestions on what approach could be used for such a question stem in the future?..and how to improve my speed in such sums?..

Thanks

aJ
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