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# F = a^2 - b^2 where a and b are different non zero integers. Is f>0?

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Math Expert
Joined: 02 Sep 2009
Posts: 52344
F = a^2 - b^2 where a and b are different non zero integers. Is f>0?  [#permalink]

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10 Jun 2018, 04:16
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Difficulty:

55% (hard)

Question Stats:

32% (01:50) correct 68% (01:23) wrong based on 37 sessions

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F = a^2 - b^2 where a and b are different non zero integers. Is f > 0?

(1) a > b
(2) a/b > 0

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Joined: 10 Jun 2018
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Re: F = a^2 - b^2 where a and b are different non zero integers. Is f>0?  [#permalink]

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10 Jun 2018, 04:29
This question seems to work best by testing a few numbers.

For (1), I started out by testing two options: when both numbers are positive and when both numbers are negative. Here, if we take a = 2 and b = 1, then f = 2^2 - 1^2 = 4 - 1 = 3, and thus > 0. However, if we take a = -1 and b = -2, then we get f = (-1)^2 - (-2)^2 = 1 - 4 = -3, and thus < 0. So (1) is not sufficient.

For (2), what a/b > 0 means is that a and b shares a sign, and are non-zero. We just need to test two positive and two negative values for a and b. Hey... didn't we already do that for (1)? Yeah we did. (2) is not sufficient.

Even when we put (1) and (2) together, those examples still work. So E is our answer.
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Re: F = a^2 - b^2 where a and b are different non zero integers. Is f>0?  [#permalink]

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10 Jun 2018, 17:06
$$F = a^2 - b^2$$
This can be written as F = (a+b)(a-b)

Statement 1: a > b => (a-b) > 0
We cannot comment on (a+b) with this knowledge. So, Statement 1 is insufficient.

Statement 2: $$\frac{a}{b}$$ > 0
Add 1 on both sides, $$\frac{a}{b} + 1 > 1$$
This can be further written as $$\frac{(a+b)}{b} > 1$$
We have 2 cases here.
Case 1: b > 0
(a+b) > b
a > 0.
Case 2: b < 0
(a+b) < b
a < 0
So, b > 0 a > 0 or b < 0 a < 0
So, a + b > 0 or a + b < 0.
So, Statement 2 is insufficient.
From Statement 1 and Statement 2:
(a - b) > 0 but (a + b) can be > 0 or < 0.
So, Statement 1 and Statement 2 are insufficient.
Re: F = a^2 - b^2 where a and b are different non zero integers. Is f>0? &nbs [#permalink] 10 Jun 2018, 17:06
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