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Re: Is a^2 + a > -b^2? [#permalink]
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.

Answer: B.



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]
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?
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Re: Is a^2 + a > -b^2? [#permalink]
Expert Reply
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.

Answer: B.



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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Re: Is a^2 + a > -b^2? [#permalink]
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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]
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Re: Is a^2 + a > -b^2? [#permalink]
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