Value of b

[knight247]

No OA on this one.

[rijul007]

You need to plug in the value of b for each option and solve for a
If you get a as integer and not equal to 1 or -2, that option stands correct.

I. b=1

a^2 +3a-18 = 0
(a+6)(a-3) = 0
a = -6 or 3

This is correct
Eliminate B

II. b=2

4a^2 +6a-18 = 0
2a^2 +3a-9 = 0
2a^2 +6a-3a-9=0
(2a-3)(a+3)=0
a=3/2 or -3

b=2 is correct

Eliminate A and D

III. b=3
9a^2 +9a-18 = 0
a^2 +a-2=0
a^2 +2a-a-2=0
(a+2)(a-1)=0
a=-2 or 1

b=3 is possible


Option E is correct

[knight247]

Sorry the OA is C

[pemdas]

Opppppps


the expression is valid only if a=! (1;-2) Restriction put on a, as we can't divide by 0
there's no unique solution for b, and b can be only I and II

Indeed c

[rijul007]

You need to plug in the value of b for each option and solve for a
If you get a as integer and not equal to 1 or -2, that option stands correct.

I. b=1

a^2 +3a-18 = 0
(a+6)(a-3) = 0
a = -6 or 3

This is correct
Eliminate B

II. b=2

4a^2 +6a-18 = 0
2a^2 +3a-9 = 0
2a^2 +6a-3a-9=0
(2a-3)(a+3)=0
a=3/2 or -3

b=2 is correct

Eliminate A and D

III. b=3
9a^2 +9a-18 = 0
a^2 +a-2=0
a^2 +2a-a-2=0
(a+2)(a-1)=0
a=-2 or 1




Option E is correct

III. a= 1 or -2
hence b cannot be equal to 3

Option C

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II. b=2

4a^2 +6a-18 = 0
2a^2 +3a-9 = 0
2a^2 +6a-3a-9=0
(2a-3)(a+3)=0
a=3/2 ??????? or -3


How can the answer be c. According to me the answer should be A. Tell me guyzzz how can the value of a be 3/2 when it is specified that a and b are integers....

Plss get back to this sum...

[pemdas]

'a' need not to be 3/2, as it can be -3

II. b=2

4a^2 +6a-18 = 0
2a^2 +3a-9 = 0
2a^2 +6a-3a-9=0
(2a-3)(a+3)=0
a=3/2 ??????? or -3


How can the answer be c. According to me the answer should be A. Tell me guyzzz how can the value of a be 3/2 when it is specified that a and b are integers....

Plss get back to this sum...

[GMATGuruNY]

((ab)² + 3ab - 18) / (a-1)(a+2) = 0

Which of the following could be the value of b?

I: 1
II: 2
III: 3

A) I
B) II
C) I and II
D) I and III
E) I, II and III

The numerator must be equal to 0.
The denominator cannot be equal to 0, requiring that a≠1 and a≠-2.

(ab)² + 3ab - 18 = 0
(ab + 6)(ab - 3) = 0
ab = -6 or ab = 3.

I: b=1
If ab = -6, then a = -6.
This works.
Eliminate B, which does not include I.

II: b=2.
If ab = -6, then a = -3.
This works.
Eliminate A and D, which do not include II.

III: b=3
If ab = -6, then a = -2.
Not possible, since a≠-2.
If ab = 3, then a=1.
Not possible, since a≠1.
Eliminate E, which includes III.

The correct answer is C.