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Re: If a and b are non-zero integers and a^8*b^4 - a^4*b^2 = 12. Which of [#permalink]

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13 Sep 2017, 23:10

=> a and b are non-zero integers \(a^8∗b^4−a^4∗b^2=12\) Find: a^2 in terms of b

\(a^8∗b^4−a^4∗b^2=12\) => \(a^4∗b^2(a^4∗b^2-1)=12\) => \(a^4∗b^2(a^2∗b-1)(a^2∗b+1)=12\) here given a and b are integers => by putting value a= 1 and b=-2 or b=2 satisfy the equation Now we have to find a^2 in terms of be... could be conditions

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

So 1^2 = (2/2) => when a=1 and b=2 . I satisfy So 1^2 = (-2/-2) => when a=1 and b=-2 . II satisfy II doesnot satisfy for both b=-2 and b=2.

Re: If a and b are non-zero integers and a^8*b^4 - a^4*b^2 = 12. Which of [#permalink]

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15 Sep 2017, 10:30

a^8*b^4-a^4*b^2=12 a^4*b^2(a^4*b^2-1)=12 lets say a^4*b^2 = x therefore x(x-1) =12 That will give us x^2-x-12=0 x=4 or x =-3

but any positive integer power of any integer cant be negative. so x = 4. i.e a^4*b^2=4 since a and b are non zero integers a can be 1,-1 and b can be 2,-2.

now a^1 = 1 for -1 and 1.

So 1) 2/b will give us 1 since b can be 2. 2) -2/b can give us 1 if we take b as -2. 3) 3/b is not possible for any of the values of b.

Re: If a and b are non-zero integers and a^8*b^4 - a^4*b^2 = 12. Which of [#permalink]

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16 Sep 2017, 05:16

The equation states a^8b^4-a^4b^2=12. Now assuming a^4b^2=x so we can solve the equation by substituting the assumption and the equation now becomes x^2-x-12=0 and on solving the equation, either x=4 or -3. Since the value is non negative, -3 cannot be accepted. So the accepted value is 4. Now putting the same against the assumption, a^4b^2=4, so (a^2b)^2=2^2. So further solving the same becomes (a^2b)^2- 4=0. Which is also a^2-b^2 and can be expressed as (a-b)*(a+b). So using the same formula in the above equation, the simlified equation becomes (a^2b-2)*(a^2b+2)=0 which makes a^2= 2/b and a^2=-2/b. Hence answer is C.

Re: If a and b are non-zero integers and a^8*b^4 - a^4*b^2 = 12. Which of [#permalink]

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16 Sep 2017, 05:18

Hope this helps mehrotrayash to understand how it can take negative values. Not required to substitute values and check. Simplifying the equation to its lowest form as per the question also clears how one can arrive at the values. Hope this helps

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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