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If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers

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If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers  [#permalink]

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Updated on: 16 Dec 2012, 06:20
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Difficulty:

95% (hard)

Question Stats:

44% (02:44) correct 56% (02:26) wrong based on 215 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

Originally posted by mun23 on 15 Dec 2012, 11:48.
Last edited by Bunuel on 16 Dec 2012, 06:20, edited 1 time in total.
Renamed the topic and edited the question.
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Joined: 24 Apr 2012
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16 Dec 2012, 03:22
3
1
Ans:

from the fraction we get that “a” cannot be 1 or -2, so putting the value of a in numerator we get the the values of “b” to 3,-6, -3/2 , therefore b cannot be 3 so the answer is (C).
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Re: If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers  [#permalink]

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16 Dec 2012, 06:29
From the Q,

Numerator = 0 whereas Denominator <> (cannot be) 0

Numerator = 0 provides: ab = -6 or ab = 3
Denominator <> 0 provides: a <> 1 and a <> -2

Put values of b in to ab to check for a:

1) b = 1; a = -6 or a = 3
2) b = 2; a = -3 or a = 3/2
3) b = 3; a = -2 or a = 1

But a <> 1 and a <> -2,

Hence, b can not be 3 and could be either 1 or 2.
Ans: C
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Re: If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers  [#permalink]

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20 Dec 2012, 12:50
2
mun23 wrote:
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

For the $$\frac{(ab)^2+3ab-18}{(a-1)(a+2)}= 0$$ = 0

numerator has to be zero
and a # 1 and a # -2 (Since these two values would make denominator zero and fraction undefined)

Now $$(ab)^2+3ab-18$$

Put ab = x

$$x^2 + 3x - 18$$

factorized to

(x-3)(x+6)

Case I
x = ab = 3, => a =1 , b = 3 -- NOT POSSIBLE as we cannot have a = 1
x = ab = 3, => a =3 , b = 1 -- POSSIBLE

Case II
x = ab = -6, => a = -1, 2,+3,+6 & b = +2, -3, +6

Hence we can note that b can assume values 1 & 2 but not 3.

Hence C
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Re: If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers  [#permalink]

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20 Dec 2012, 20:25
mun23 wrote:
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

For $$\frac{(ab)^2+3ab-18}{(a-1)(a+2)}= 0$$ to hold, $$(ab)^2+3ab-18 = 0$$

You need 'a' to be an integer so put in the values of b to check whether you get integral values for 'a'
b = 1 => a^2 + 3a - 18 = 0 => (a + 6)(a - 3) = 0 => Integral values so acceptable
b = 2 => 4a^2 + 6a - 18 = 0 => (2a + 6)(2a - 3) = 0 => We get a = -3 (an integer) hence acceptable
b = 3 => 9a^2 + 9a - 18 = 0 => (a + 2)(a - 1) = 0 => We get a = -2 or 1. a can take neither of these values since they make the denominator 0. Not acceptable

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Re: If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers  [#permalink]

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27 Nov 2016, 08:29

((ab)^2+3ab-18)/(a-1)(a+2) = 0 --> (ab)^2+3ab-18 = 0 --> let ab be x --> x^2+3x-18=0 --> (x+6)(x-3) --> ab=3 or ab= -6 --> now a and b could be any of the factors of 3 or -6, that means all 3 statements could be true. Please let me know what I am missing, thank you.
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Re: If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers  [#permalink]

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19 Apr 2018, 21:58
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Re: If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers &nbs [#permalink] 19 Apr 2018, 21:58
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