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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers, which of
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17 Aug 2021, 09:31
17 Aug 2021, 09:31
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Question Stats:
30% (03:29) correct
70% (02:23) wrong based on 10 sessions
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If \(\frac{(ab)^2+3ab-18}{(a-1)(a+2)}= 0\) 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
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
If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers, which of
[#permalink]
23 Aug 2021, 14:41
23 Aug 2021, 14:41
2
Expert Reply
The original equation states
\(\frac{(ab)^2 + 3ab - 18 }{ (a-1)(a+2)}\)
The first thing to note is the denominator. For the given expression to be true
\( \implies a \neq 1 \) and \(a \neq -2 \)
as these a =1 or a = -2 will result in an undefined expression irrespective of the value of b.
Since the numerator is a quadratic expression in the variables ab, lets replace ab with x.
Factorizing
Since the possible values of b include only 1, 2 and 3, therefore we will focus only on these factors of b.
For \(ab =3 \implies a =1 \ \ b = 3 \) or \(a = 3, b = 1\)
However since \(a \neq 1 \), only the second option i.e. b = 1 offers a valid choice. Thus I. is valid.
For
\(ab =-6 \implies a = -6, b = 1 \) or \( a = -3, b = 2 \) or \( a = -2,\ b = 3\)
The first choice for b is already valid i.e. b = 1 and does not invalidate the conditions for a. The second choice i.e b = 2 is also valid, thus II. is true. However, the third option is invalid since it requires a = -2 which we understand is an invalid choice for a
Option C states both I and II and thus Option C is the right answer.
\(\frac{(ab)^2 + 3ab - 18 }{ (a-1)(a+2)}\)
The first thing to note is the denominator. For the given expression to be true
\((a -1) \neq 0 \) and \( a+2 \neq 0 \)
\( \implies a \neq 1 \) and \(a \neq -2 \)
as these a =1 or a = -2 will result in an undefined expression irrespective of the value of b.
Since the numerator is a quadratic expression in the variables ab, lets replace ab with x.
\( \implies x^2 - 3x + 18 = 0 \)
Factorizing
\(\implies x^2 +6x - 3x + 18 = 0 \implies x(x-6) -3( x +6) = 0 \)
\(\implies (x-3)(x+6) = 0 \implies (ab - 3) (ab +6) = 0\)
\(\implies ab = 3 \) OR \(ab = -6 \)
Since the possible values of b include only 1, 2 and 3, therefore we will focus only on these factors of b.
For \(ab =3 \implies a =1 \ \ b = 3 \) or \(a = 3, b = 1\)
However since \(a \neq 1 \), only the second option i.e. b = 1 offers a valid choice. Thus I. is valid.
For
\(ab =-6 \implies a = -6, b = 1 \) or \( a = -3, b = 2 \) or \( a = -2,\ b = 3\)
The first choice for b is already valid i.e. b = 1 and does not invalidate the conditions for a. The second choice i.e b = 2 is also valid, thus II. is true. However, the third option is invalid since it requires a = -2 which we understand is an invalid choice for a
Option C states both I and II and thus Option C is the right answer.
gmatclubot
If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers, which of [#permalink]
23 Aug 2021, 14:41
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