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# If b<0, is 5b(a+b)>-a^2-b^2? 1) a+2b>0 2) a+3b>0

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Joined: 14 Nov 2016
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Location: Malaysia
If b<0, is 5b(a+b)>-a^2-b^2? 1) a+2b>0 2) a+3b>0  [#permalink]

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09 Mar 2017, 21:30
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Difficulty:

95% (hard)

Question Stats:

33% (03:00) correct 67% (02:48) wrong based on 63 sessions

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If $$b<0$$, is $$5b(a+b)>-a^2-b^2$$?

1) $$a+2b>0$$

2) $$a+3b>0$$

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Senior SC Moderator
Joined: 14 Nov 2016
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Location: Malaysia
Re: If b<0, is 5b(a+b)>-a^2-b^2? 1) a+2b>0 2) a+3b>0  [#permalink]

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03 Apr 2017, 15:48
1
2
ziyuen wrote:
If $$b<0$$, is $$5b(a+b)>-a^2-b^2$$?

1) $$a+2b>0$$

2) $$a+3b>0$$

OFFICIAL EXPLANATION

You need preliminary knowledge to solve this question.

If b<0, you get -3b>-2b>-b. If you take the 1st step of the variable approach and modify the original condition and the question,
Que: is 5b(a+b) > -a^2-b^2 → 5ab+5b2+a^2+b^2>0, a^2+5ab+6b^2>0, or (a+3b)(a+2b)>0

In the original condition, there are 2 variables (a, b) and 1 equation (b<0), and in order to match the number of variables to the number of equations, there must be 1 more equation. Therefore, D is most likely to be the answer.

In the case of con 1), a>-2b, so (a, b) = (3,-1) no, and (a, b) = (5,-1) yes, hence it is not sufficient.

In the case of con 2), a>-3b>-2b. This is because b<0, and it is shown in the preliminary knowledge above. If so, a>-3b and a>-2b, then a+3b>0 and a+2b>0. Since (a+3b)(a+2b)>0 is always yes, it is sufficient. The answer is B.
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Re: If b<0, is 5b(a+b)>-a^2-b^2? 1) a+2b>0 2) a+3b>0  [#permalink]

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04 Apr 2017, 00:29
Can someone explain this more easily? Don't get it at all.
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Joined: 25 Apr 2016
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If b<0, is 5b(a+b)>-a^2-b^2? 1) a+2b>0 2) a+3b>0  [#permalink]

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04 Apr 2017, 00:51
after factorizing the inequality would boils down to (2b+a)(a+3b)>0 and as "b"<0, the solution for the inequality would be a>-2b or a<-3b. Try consider b=-1, now, the 1st statement says a>-2 and that to me is sufficient and not to mention that 2nd statement would clearly be ambiguous as the solution suggest. -> for me it's A
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Re: If b<0, is 5b(a+b)>-a^2-b^2? 1) a+2b>0 2) a+3b>0  [#permalink]

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26 Jul 2018, 05:52
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Re: If b<0, is 5b(a+b)>-a^2-b^2? 1) a+2b>0 2) a+3b>0 &nbs [#permalink] 26 Jul 2018, 05:52
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