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If a and b are non-zero integers and a^8*b^4 - a^4*b^2 = 12. Which of

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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, 23:53
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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

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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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14 Sep 2017, 00:10
=> 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.

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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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14 Sep 2017, 00:30
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Substitute the value for a^2 as 2/b, - 2/b and 3/b in given equation

LHS = RHS only for I and II
So option C is correct

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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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14 Sep 2017, 08:36
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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

Well working through options method has already been explained, so I'll go the Algebraic way

Given $$a^8*b^4 - a^4*b^2 = 12$$. it is evident that $$a^4b^2$$ is common in LHS. So let's assume $$a^4b^2 = x$$

so our equation becomes $$x^2-x-12=0$$. Now we can easily calculate the value of $$x$$ and substitute

Solving the equation we get $$x= 4$$ & $$-3$$. $$-3$$ is not possible as $$x$$ is a square. Hence $$x= 4$$.

or $$a^4b^2=4$$. Taking square root of both the sides we get

$$a^2b =2$$ or $$-2$$

So $$a^2 = \frac{2}{b}$$ or $$\frac{-2}{b}$$

Option C
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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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14 Sep 2017, 22:08
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(a⁸xb⁴)-(a⁴xb²)=12
or
{(a²)⁴xb⁴}-{(a²)²xb²}=12

Let a²=$$\frac{-2}{b}$$

{($$\frac{-2}{b}$$)⁴*b⁴}-{($$\frac{-2}{b}$$)²*b²}=12

{($$\frac{16}{b⁴}$$*b⁴)}-{($$\frac{4}{b²}$$*b²)}=12

16-4=12
12=12
Now all the powers in the above equation are even hence positive $$\frac{2}{b}$$ will also give the same result.
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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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15 Sep 2017, 03:00
Various methods to solve such problems.
1. Substitute appropriate nos to satisfy the equality
2. Algebraic method
3. Check for the option values.

Substituting a^2 as 2/b, -2/b and 3/b in equation , we can find that only I and II satisfies the equation.
Hence option C
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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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15 Sep 2017, 11:30
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.

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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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16 Sep 2017, 05:00
Can a^2 take a negative value? I figured it cannot which is why i eliminated C and chose A.
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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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16 Sep 2017, 05:26
mehrotrayash wrote:
Can a^2 take a negative value? I figured it cannot which is why i eliminated C and chose A.

Yes a^2 cannot take a negative value

so for a^2 = -2/b will only be true if b is -ve.

=> a^2 = -2/-2 =1 which is positive.

So condition I: a^2 =2/b will be true when b is +ve and condition II: a^2 =-2/b will be true when b is -ve.

And questions is asking for possible options so as we have 2 options for b +ve and -ve condition I and II are possible.
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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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16 Sep 2017, 06:16
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.

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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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16 Sep 2017, 06:18
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

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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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16 Sep 2017, 07:29
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.

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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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16 Sep 2017, 16:02
The above equation can be simplified to:
$$a^4b^2 (a^2b+1) (a^2b-1) = 12$$
or $$(a^2b)^2 (a^2b+1) (a^2b-1) =12$$

With some due diligence, we can find out that $$a^2b = 2$$ or $$-2$$.

Hence only option C implies.
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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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17 Sep 2017, 02:19
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a^8*b^4 - a^4*b^2 = 12
(a^4*b^2) (a^4*b^2) - (a^4*b^2) = 12
(a^4*b^2) (a^4*b^2 - 1) = 12

This says that they are consecutive numbers... only solution possible is 4*3. So a^4*b^2 = 4

factoring will give us a^4 = 4/B^2.

so a^2 = + or - 2/b

Kudos doesn't hurt anyone.
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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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18 Sep 2017, 20:51
Let a^4b^2 = x

X^2-x-12=0

X=4 or -3

Which is

a^4*b^2 = 4 or -3
a^2 = 2/b or -2/b or i(3^0.5)

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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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02 Oct 2017, 15:37
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

hi

the misdoings that I did to this problem are as under

a^4 x b^2 ( a^4 x b^2 - 1 ) = 12
the important thing to notice here is

n(n-1) = 4 x 3

so, a^4 x b^2 = 4

a^2 = 2/b or -2/b

cheers through the kudos button if this helps
thanks
Re: If a and b are non-zero integers and a^8*b^4 - a^4*b^2 = 12. Which of &nbs [#permalink] 02 Oct 2017, 15:37
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