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

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

Plugin
(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.
Answer I and II

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

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.

Answer is C

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.

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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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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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

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.

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

Answer C.

Kudos doesn't hurt anyone.

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)

Hence answer is C


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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

Answer: C

cheers through the kudos button if this helps
thanks

\(a^8*b^4 - a^4*b^2 = 12\)
\(a^8*b^4 - a^4*b^2 -12 = 0\)
\(a^8*b^4 - 4 a^4*b^2 + 3 a^4*b^2 -12 = 0\)

\(a^4*b^2( a^4*b^2 - 4) + 3 (a^4*b^2 - 4) = 0\)

Only \((a^4*b^2 - 4)\) will satisfy

\(a^4*b^2 = 4\)

a^4 = b^2/4

We will get 2 values

i and ii

C

\(a^8*b^4 - a^4*b^2 = 12\)
\((a^4b^2)^2 - (a^2b)^2=12\)
\(special:x^2 - y^2=(x+y)(x-y)\)
\((a^4b^2+a^2b)(a^4b^2-a^2b)=12\)
\((a^4b^2+a^2b)(a^4b^2-a^2b)=[12*1];[6*2];[4*3]\)
\(test[12*1]:(a^4b^2+a^2b)=12…(a^4b^2-a^2b)=1…combined:2(a^4b^2)=13…13/2≠integer\)
\(test[4*3]:(a^4b^2+a^2b)=4…(a^4b^2-a^2b)=3…combined:2(a^4b^2)=7…7/2≠integer\)
\(test[6*2]:(a^4b^2+a^2b)=6…(a^4b^2-a^2b)=2…combined:2(a^4b^2)=8…8/2=integer\)
\(2(a^4b^2)=8…(a^4b^2)=8/2…(a^4b^2)=4…|a^2b|=2…a^2=2/|b|…a^2=(2/b,-2/b)\)

Answer (C)

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