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# M12-07

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Math Expert
Joined: 02 Sep 2009
Posts: 49300

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16 Sep 2014, 00:46
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Difficulty:

15% (low)

Question Stats:

83% (01:04) correct 17% (01:03) wrong based on 71 sessions

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What is the value of $$(a + b)^2$$?

(1) $$a^2-a-2=0$$.

(2) $$b^2+b-2=0$$.

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Math Expert
Joined: 02 Sep 2009
Posts: 49300

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16 Sep 2014, 00:46
Official Solution:

(1) $$a^2-a-2=0$$. From this statement it follows that $$a=-1$$ or $$a=2$$. Not sufficient.

(2) $$b^2+b-2=0$$. From this statement it follows that $$b=-2$$ or $$b=1$$. Not sufficient.

(1)+(2) If $$a=-1$$ and $$b=-2$$ then the answer is 9 but if $$a=-1$$ and $$b=1$$ then the answer is 0. Not sufficient.

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Joined: 12 Nov 2015
Posts: 59
Location: Uruguay
Concentration: General Management
Schools: Goizueta '19 (A)
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01 Jun 2016, 16:36
Hello,
why is it ok not to specify that "a" and b" are integers?
thank you
Math Expert
Joined: 02 Sep 2009
Posts: 49300

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01 Jun 2016, 22:48
Avigano wrote:
Hello,
why is it ok not to specify that "a" and b" are integers?
thank you

Why should it be specified? We can solve for a and b and get their values. It turns out that both of them are integers but the values could have been non-integers as well.
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Joined: 19 May 2016
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Location: United States
Concentration: Strategy, Human Resources
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08 Aug 2016, 08:07
Bunuel wrote:
Official Solution:

(1) $$a^2-a-2=0$$. From this statement it follows that $$a=-1$$ or $$a=2$$. Not sufficient.

(2) $$b^2+b-2=0$$. From this statement it follows that $$b=-2$$ or $$b=1$$. Not sufficient.

(1)+(2) If $$a=-1$$ and $$b=-2$$ then the answer is 9 but if $$a=-1$$ and $$b=1$$ then the answer is 0. Not sufficient.

Thanks for the elegant solution. I approached it in a different way, and arrived at the same answer.

I expanded the expression in the stem, and for (1), I saw that it gives no information about a, so insufficient. I did the same for (2), as it gives no info about a. I figured both statements together are insufficient as well, since both terms together cannot be manipulated either by addition, multiplyication, subtraction, or division in such a way that is helpul to solving the stem.

Is this a valid aprroach?

Thanks!
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Joined: 27 Jul 2016
Posts: 14

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09 Aug 2016, 16:43
I thought I could get away with not solving for #2, but upon thinking about it, you should solve for both a and b.

In this particular case, of (a+b)^2, what if #1 was the same but #2 was a different equation: b^2 + 5= 5 ? If b = 0, then #1 & #2 is sufficient.
Intern
Joined: 08 Aug 2017
Posts: 21
GMAT 1: 690 Q49 V35

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07 Jul 2018, 14:44
Hi Bunuel

Statement 1:
a^2-a-2=0
so a^-a=2
a(a-1)=2
so either a=2 or a-1=2; a=3 (this is where i am wrong, i get different value of a)

similarly
statement 2:
b^2+b-2=0
b(b+1)=2
so either b=2 or b+1=2; b=1 (again different value of b)

Thanks.
Math Expert
Joined: 02 Sep 2009
Posts: 49300

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10 Jul 2018, 02:02
pantera07 wrote:
Hi Bunuel

Statement 1:
a^2-a-2=0
so a^-a=2
a(a-1)=2
so either a=2 or a-1=2; a=3 (this is where i am wrong, i get different value of a)

similarly
statement 2:
b^2+b-2=0
b(b+1)=2
so either b=2 or b+1=2; b=1 (again different value of b)

Thanks.

a(a - 1) = 2 does not mean than a = 2 or a - 1 = 2. You can do this only when the LHS equals to 0. This is because the product is 0 if either of the multiples is 0. For example, a(a - 1) = 0 to be true either a or a - 1 must be true. But the roots of a^2 - a - 2 = 0 are a = -1 and a = 2.
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Re: M12-07 &nbs [#permalink] 10 Jul 2018, 02:02
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# M12-07

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