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
Value of b
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rijul007
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rijul007 wrote: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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[email protected]
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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...
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...
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pemdas
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'a' need not to be 3/2, as it can be -3
[email protected] wrote: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...
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GMATGuruNY
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The numerator must be equal to 0.((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 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.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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