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
The equation is valid only if (ab)²+3ab-18 = 0.
I: b=1
Plugging b=1 into (ab)²+3ab-18 = 0, we get:
a²+3a-18 = 0
(a+6)(a-3) = 0
a=-6 or a=3
This works.
Since the correct answer must include I, eliminate B.
II: b=2
Plugging b=2 into (ab)²+3ab-18 = 0, we get:
4a²+6a-18 = 0
2a²+3a-9 = 0
(2a-3)(a+3) = 0
Since a must be an integer, a=-3.
This works.
Since the correct answer must include II, eliminate A and D.
III: b=3
Plugging b=3 into (ab)²+3ab-18 = 0, we get:
9a²+9a-18 = 0
a²+a-2 = 0
(a+2)(a-1) = 0
The expression in red constitutes the denominator of the equation in the prompt.
Since division by 0 is not allowed, this expression cannot be equal to 0.
Thus, b=3 is not viable.
Since the correct answer cannot include III, eliminate E.
The correct answer is
C.
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