VJesus12 wrote:If \(\frac{\left(ab\right)^2+3ab-18}{\left(a-1\right)\left(a+2\right)}=0\) where \(a\) and \(b\) are integers, which of the following could be the value of \(b?\)
I. 1
II. 2
III. 3
(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II and III only
[spoiler]OA=C[/spoiler]
Source: Manhattan GMAT
Given that \(\frac{\left(ab\right)^2+3ab-18}{\left(a-1\right)\left(a+2\right)}=0\), we have (ab)^2 + 3ab - 18 = 0
=> (ab)^2 + 6ab - 3ab - 18 = 0
ab(ab + 6) - 3(ab + 6) = 0
(ab + 6)*(ab - 3) = 0
=> ab = -6 or 3
Let's take each option one by one.
I. 1 : If b = 1, then with ab = -6, we have a = -6 (valid value for a); similarly, with ab = 3, we have a = 3 (valid value for a); so b can be 1.
II. 2: If b = 2, then with ab = -6, we have a = -3 (valid value for a); however, with ab = 3, we cannot have b = 2, else a would not be an integer; so b can be 2.
III. 3: If b = 3, then with ab = -6, we have a = -2 (invalid value for a since at a = -2, the denominator of the expression would turn 0, and the value of the expression would be indeterminable. Similarly, with ab = 3, we have a = 1 (invalid value for a since at a = 1, again, the denominator of the expression would turn 0, and the value of the expression would be indeterminable. Thus, b can't be 3.
The correct answer:
C
Hope this helps!
-Jay
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