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Re: If \frac{(ab)^2+3ab-18}{(a-1)(a+2)}=0 [#permalink]

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16 Dec 2012, 03:22

3

This post received KUDOS

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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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

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

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25 Sep 2015, 05:06

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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 Nov 2016, 08:23

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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

Bunuel, need your insights -

((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.

gmatclubot

Re: If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers
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27 Nov 2016, 08:29

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