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# Is a^2 + a > -b^2?

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Intern
Joined: 05 Nov 2010
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Is a^2 + a > -b^2? [#permalink]

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07 Nov 2010, 13:39
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53% (02:00) correct 47% (01:15) wrong based on 156 sessions

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Is a^2 + a > -b^2?

(1) a^2 + b^2 = 1
(2) a > 0
[Reveal] Spoiler: OA

Kudos [?]: 26 [1], given: 0

Math Expert
Joined: 02 Sep 2009
Posts: 41873

Kudos [?]: 128604 [2], given: 12180

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07 Nov 2010, 14:54
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Expert's post
Creeper300 wrote:
a^2+a > -b^2?

1. a^2+ b^2=1
2. a>0

Is $$a^2+b^2>-a$$? or is $$a^2+a+b^2>0$$

(1) $$a^2+b^2=1$$ --> is $$1>-a$$ or is $$a>-1$$, not sufficient. For example if $$a=1$$ and $$b=0$$ then the answer will be YES but if $$a=-1$$ and $$b=0$$ then the answer will be NO.

(2) $$a>0$$ --> so $$a^2+a+b^2=(positive)+(positive)+(non-negative)$$ which is clearly positive, so $$a^2+a+b^2>0$$. Sufficient.

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Is a^2 + a > -b^2? [#permalink]

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23 Nov 2014, 08:45
Creeper300 wrote:
Is a^2 + a > -b^2?

(1) a^2 + b^2 = 1
(2) a > 0

did it in the following way:

question becomes:
is a^2 + b^2 + a > 0, as the minimum value of a^2 + b^2 is 0, the question is is a> 0?

from statement 1: a^2 + b^2 = 1, which reduces to a> -1, when a = b= 0 then a = 0 and not greater than 0, so NSF
statement 2 directly answers the question, sufficient

[B]

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Re: Is a^2 + a > -b^2? [#permalink]

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26 Nov 2014, 19:16
Bunuel wrote:
Creeper300 wrote:
a^2+a > -b^2?

1. a^2+ b^2=1
2. a>0

Is $$a^2+b^2>-a$$? or is $$a^2+a+b^2>0$$

(1) $$a^2+b^2=1$$ --> is $$1>-a$$ or is $$a>-1$$, not sufficient. For example if $$a=1$$ and $$b=0$$ then the answer will be YES but if $$a=-1$$ and $$b=0$$ then the answer will be NO.

(2) $$a>0$$ --> so $$a^2+a+b^2=(positive)+(positive)+(non-negative)$$ which is clearly positive, so $$a^2+a+b^2>0$$. Sufficient.

Hi Bunnel ,

Shouldnt this be D , since statement (1) gives a>-1 ,so a cannot be negative , and both a & b are not zero (a^2 + b^=1).

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Is a^2 + a > -b^2? [#permalink]

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26 Nov 2014, 20:28
Is a^2 + a > -b^2?

Is (a)(a+1) > -b*b?

(1) a^2 + b^2 = 1
a^2 - 1 = - b^2
(a-1)(a+1) = -b^2
Sufficient, because if (a-1)(a+1) = - b* then (a)(a+1) > -b^2

(2) a>0,
Sufficient, because if a>0, then both factors (a) and (a+1) will always be positive satisfying the equation (a)(a+1) > -b^2.

I would guess the answer D?

Kudos [?]: 75 [0], given: 98

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Re: Is a^2 + a > -b^2? [#permalink]

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27 Nov 2014, 03:14
mattapattu wrote:
Bunuel wrote:
Creeper300 wrote:
a^2+a > -b^2?

1. a^2+ b^2=1
2. a>0

Is $$a^2+b^2>-a$$? or is $$a^2+a+b^2>0$$

(1) $$a^2+b^2=1$$ --> is $$1>-a$$ or ", not sufficient. For example if $$a=1$$ and $$b=0$$ then the answer will be YES but if $$a=-1$$ and $$b=0$$ then the answer will be NO.

(2) $$a>0$$ --> so $$a^2+a+b^2=(positive)+(positive)+(non-negative)$$ which is clearly positive, so $$a^2+a+b^2>0$$. Sufficient.

Hi Bunnel ,

Shouldnt this be D , since statement (1) gives a>-1 ,so a cannot be negative , and both a & b are not zero (a^2 + b^=1).

You missed a point there.

(1) says that $$a^2+b^2=1$$. If we substitute this into the question (is $$a^2+a+b^2>0$$?), the question becomes "is $$a>-1$$?". As shown in the solution it can be more than -1 (a=1 and b=0) as well as equal to -1 (a=-1 and b=0). So, we cannot answer the question with a definite YES or NO. Thus the statement is not sufficient.

Hope it's clear.
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Kudos [?]: 128604 [0], given: 12180

Math Expert
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Posts: 41873

Kudos [?]: 128604 [0], given: 12180

Re: Is a^2 + a > -b^2? [#permalink]

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27 Nov 2014, 03:16
Expert's post
1
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viktorija wrote:
Is a^2 + a > -b^2?

Is (a)(a+1) > -b*b?

(1) a^2 + b^2 = 1
a^2 - 1 = - b^2
(a-1)(a+1) = -b^2
Sufficient, because if (a-1)(a+1) = - b* then (a)(a+1) > -b^2

(2) a>0,
Sufficient, because if a>0, then both factors (a) and (a+1) will always be positive satisfying the equation (a)(a+1) > -b^2.

I would guess the answer D?

The red part is not correct. Consider a=-1 and b=0. The correct answer is B.
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Re: Is a^2 + a > -b^2?   [#permalink] 27 Nov 2014, 03:16
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