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

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

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Updated on: 24 Apr 2019, 13:45
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Question Stats:

45% (02:40) correct 55% (02:27) wrong based on 141 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, 12:48.
Last edited by Bunuel on 24 Apr 2019, 13:45, edited 2 times in total.
Renamed the topic and edited the question.
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Re: If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers,which of  [#permalink]

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16 Dec 2012, 04:22
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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,which of  [#permalink]

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20 Dec 2012, 13:50
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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,which of  [#permalink]

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20 Dec 2012, 21:25
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,which of  [#permalink]

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02 Dec 2018, 23:54
topper97 wrote:
If $$\frac{(ab)^2+3ab-18}{(a-1)(a+2)}=0$$, where a and b are integers and a does not equal 1 or −2, 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

Here's what I've done.
$$\frac{(ab)^2+3ab-18}{(a-1)(a+2)=0}$$
$$(ab)^2$$+3ab-18=0
ab(ab+3)=18
ab=18 or ab+3=18
ab=18
6*3
9*2
18*1
or ab=15
5*3
So b can be 1,2 or 3 but the OA says b can only be 1 and 2. Where am I going wrong?

For $$\frac{(ab)^2+3ab-18}{(a-1)(a+2)}$$ to equal to 0, the numerator must be 0. Thus, $$(ab)^2+3ab-18 =0$$.

Factor to get: (ab - 3)(ab + 6) = 0. So, ab = 3 or ab = -6.

Check options:

I. If b = 1, then a = 3 or -6. No problem there, a could be either of those values.

II. If b = 2, then a = 3/2 or -3. No problem there, a could be -3.

III. If b = 3, then a = 1 or -2. PROBLEM, a cannot take those values becasue in this case the denominator becomes 0, and we cannot divide by 0.

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

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03 Dec 2018, 00:08
Bunuel wrote:
For $$\frac{(ab)^2+3ab-18}{(a-1)(a+2)}$$ to equal to 0, the numerator must be 0. Thus, $$(ab)^2+3ab-18 =0$$.

Factor to get: (ab - 3)(ab + 6) = 0. So, ab = 3 or ab = -6.

Excuse me if my question is silly but I still didn't get how the highlighted part was derived from $$(ab)^2+3ab-18 =0$$
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Re: If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers,which of  [#permalink]

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03 Dec 2018, 00:11
topper97 wrote:
Bunuel wrote:
For $$\frac{(ab)^2+3ab-18}{(a-1)(a+2)}$$ to equal to 0, the numerator must be 0. Thus, $$(ab)^2+3ab-18 =0$$.

Factor to get: (ab - 3)(ab + 6) = 0. So, ab = 3 or ab = -6.

Excuse me if my question is silly but I still didn't get how the highlighted part was derived from $$(ab)^2+3ab-18 =0$$

By factoring for ab. Or denote ab as x, and factor x^2 + 3x - 18 = 0.

7. Algebra

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Re: If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers,which of  [#permalink]

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03 Dec 2018, 00:18
"By factoring for ab. Or denote ab as x, and factor x^2 + 3x - 18 = 0"

Now it's clear. I was confused on how we can factorize with variable ab. Thanks a lot Bunuel!
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Re: If ((ab)^2+3ab-18)((a-1)(a+2))=0 where a and b are integers,which of   [#permalink] 03 Dec 2018, 00:18
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