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If $A, B$ are non-zero integers and $A^8 B^4-A^4 B^2=12$, which of the
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19 Mar 2025, 00:58
19 Mar 2025, 00:58
Expert Reply
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Question Stats:
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100% (05:32) wrong based on 1 sessions
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If $\(A, B\)$ 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. $\(\quad \frac{2}{B}\)$
II. $\(\quad-\frac{2}{B}\)$
III. $\(\quad \frac{3}{B}\)$
(A) I only
(B) II only
(C) I and II
(D) I and III
(E) I, II, and III
I. $\(\quad \frac{2}{B}\)$
II. $\(\quad-\frac{2}{B}\)$
III. $\(\quad \frac{3}{B}\)$
(A) I only
(B) II only
(C) I and II
(D) I and III
(E) I, II, and III
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If $A, B$ are non-zero integers and $A^8 B^4-A^4 B^2=12$, which of the
[#permalink]
20 Mar 2025, 12:06
20 Mar 2025, 12:06
1
"Plugging in" the 3 options as a strategy makes sense here.
First, you can re-write \(A^8 B^4-A^4 B^2\) by factoring out \(A^2\) as \((A^2)^4B^4 - (A^2)^2B^2\)
Then, we substitute the options for \(A^2\).
A) If \(A^2 =\frac{2}{B}\) :
\((\frac{2}{B})^4B^4 - (\frac{2}{B})^2B^2\)
\(\dfrac{16B^4}{B^4} - \dfrac{4B^2}{B^2}\)
\(16 - 4 = 12\)
Thus, A is a possible option
A) If \(A^2 = -\frac{2}{B}\) :
At a quick glance, any negative value raised to an even degree (in this case, to the 2nd degree/squared) is a positive value, since the product of 2 negative values is a positive.
Thus, this will give us the same answer at Option A.
\((-\frac{2}{B})^4B^4 - (-\frac{2}{B})^2B^2\)
\(\dfrac{16B^4}{B^4} - \dfrac{4B^2}{B^2}\)
\(16 - 4 = 12\)
Thus, B is a possible option
A) If \(A^2 = \frac{3}{B}\) :
\((\frac{3}{B})^4B^4 - (\frac{3}{B})^2B^2\)
\(\dfrac{81B^4}{B^4} - \dfrac{9B^2}{B^2}\)
\(81 - 9 = 72\)
Thus, C is NOT a possible option.
First, you can re-write \(A^8 B^4-A^4 B^2\) by factoring out \(A^2\) as \((A^2)^4B^4 - (A^2)^2B^2\)
Then, we substitute the options for \(A^2\).
A) If \(A^2 =\frac{2}{B}\) :
\((\frac{2}{B})^4B^4 - (\frac{2}{B})^2B^2\)
\(\dfrac{16B^4}{B^4} - \dfrac{4B^2}{B^2}\)
\(16 - 4 = 12\)
Thus, A is a possible option
A) If \(A^2 = -\frac{2}{B}\) :
At a quick glance, any negative value raised to an even degree (in this case, to the 2nd degree/squared) is a positive value, since the product of 2 negative values is a positive.
Thus, this will give us the same answer at Option A.
\((-\frac{2}{B})^4B^4 - (-\frac{2}{B})^2B^2\)
\(\dfrac{16B^4}{B^4} - \dfrac{4B^2}{B^2}\)
\(16 - 4 = 12\)
Thus, B is a possible option
A) If \(A^2 = \frac{3}{B}\) :
\((\frac{3}{B})^4B^4 - (\frac{3}{B})^2B^2\)
\(\dfrac{81B^4}{B^4} - \dfrac{9B^2}{B^2}\)
\(81 - 9 = 72\)
Thus, C is NOT a possible option.
Re: If $A, B$ are non-zero integers and $A^8 B^4-A^4 B^2=12$, which of the
[#permalink]
23 Mar 2025, 04:00
23 Mar 2025, 04:00
Expert Reply
OFFICIAL EXPLANATION
We know $\(A^8 B^4-A^4 B^2=12\)$, where $A \& B$ are non-zero integers; we need to find the value of $\(A^2\)$ in terms of $\(B\)$
The given equation $\(A^8 B^4-A^4 B^2=12\)$ is same as $\(\left(A^4 B^2\right)^2-\left(A^4 B^2\right)-12=0\)$ in which substituting $\(A^4 B^2=y\)$ say, we get $\(y^2-y-12=0\)$ which when factorized gives $\((y-4)(y+3)=0 \Rightarrow y=4$ or $y=-3\)$
But $\(A^4 B^2\)$ having even powers cannot be negative can take only 4 as the value.
So, we get $\(A^4 B^2=4 \Rightarrow\left(A^4 B^2\right)^{1 / 2}=4^{1 / 2} \Rightarrow A^2 B= \pm 2\)$.
Hence the value of $\(\mathrm{A}^2\)$ in terms of B can be $\(\frac{2}{\mathrm{~B\)$ or $\(-\frac{2}{\mathrm{~B\)$ only, so the answer is (C).
We know $\(A^8 B^4-A^4 B^2=12\)$, where $A \& B$ are non-zero integers; we need to find the value of $\(A^2\)$ in terms of $\(B\)$
The given equation $\(A^8 B^4-A^4 B^2=12\)$ is same as $\(\left(A^4 B^2\right)^2-\left(A^4 B^2\right)-12=0\)$ in which substituting $\(A^4 B^2=y\)$ say, we get $\(y^2-y-12=0\)$ which when factorized gives $\((y-4)(y+3)=0 \Rightarrow y=4$ or $y=-3\)$
But $\(A^4 B^2\)$ having even powers cannot be negative can take only 4 as the value.
So, we get $\(A^4 B^2=4 \Rightarrow\left(A^4 B^2\right)^{1 / 2}=4^{1 / 2} \Rightarrow A^2 B= \pm 2\)$.
Hence the value of $\(\mathrm{A}^2\)$ in terms of B can be $\(\frac{2}{\mathrm{~B\)$ or $\(-\frac{2}{\mathrm{~B\)$ only, so the answer is (C).
