Bunuel wrote:

If a and b are non-zero integers and \(a^8*b^4 - a^4*b^2 = 12\). Which of the following could be a^2 in terms of b?

I. 2/b

II. -2/b

III. 3/b

A. I only

B. II only

C. I and II only

D. I and III only

E. I, II and III

Let’s factor (a^4)(b^2) from the left side:

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

The arithmetic fact that 4 x 3 = 12 now comes into play for approaching the solution to this equation. We see that (a^4)(b^2) - 1 is exactly one less than (a^4)(b^2), so (a^4)(b^2) = 4 and (a^4)(b^2) - 1 = 3 OR (a^4)(b^2) = -3 and (a^4)(b^2) - 1 = -4. However, the latter can’t be true since (a^4)(b^2) is a nonnegative quantity. Thus, only (a^4)(b^2) = 4 and (a^4)(b^2) - 1 = 3 can be true.

Taking the square root of (a^4)(b^2) = 4, we have a^2 * b = 2 or a^2 * b = -2. Thus a^2 = 2/b or a^2 = -2/b.

Alternate Solution:

Let’s let (a^4)(b^2) = x. Then, the given equation becomes x^2 - x = 12, or x^2 - x - 12 = 0. We can factor this equation as (x - 4)(x + 3) = 0; therefore x = 4 or x = -3. Then, (a^4)(b^2) = 4 or (a^4)(b^2) = -3. Since (a^4)(b^2) is nonnegative, it can’t equal -3; thus, it must be true that (a^4)(b^2) = 4.

Taking the square root of (a^4)(b^2) = 4, we have a^2 * b = 2 or a^2 * b = -2. Thus, a^2 = 2/b or a^2 = -2/b.

Answer: C

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Jeffery Miller

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