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If b is an even integer is b < 0?

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If b is an even integer is b < 0? [#permalink]

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If b is an even integer is b < 0?

(1) b^2 - 4b + 4 < 16
(2) b^2 > 9
[Reveal] Spoiler: OA

Last edited by Bunuel on 23 Nov 2013, 11:31, edited 1 time in total.
Renamed the topic and edited the question.

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Re: If b is an even integer is b < 0? [#permalink]

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If b is an even integer is b < 0?

(1) b^2 - 4b + 4 < 16 --> \((b-2)^2<16\). Take the square root: \(|b-2|<4\) --> \(-4<b-2<4\) --> \(-2<b<6\). Since given that b is an even integer, then b can be 0, 2, or 4. In either case b is not less than 0. Sufficient.

(2) b^2 > 9 --> \(|b|>3\) --> \(b<-3\) or \(b>3\). Not sufficient.

Answer: A.

Hope it's clear.
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Re: If b is an even integer is b < 0? [#permalink]

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New post 23 Nov 2013, 12:02
Bunuel wrote:
If b is an even integer is b < 0?

(1) b^2 - 4b + 4 < 16 --> \((b-2)^2<16\). Take the square root: \(|b-2|<4\) --> \(-4<b-2<4\) --> \(-2<b<6\). Since given that b is an even integer, then b can be 0, 2, or 4. In either case b is not less than 0. Sufficient.

(2) b^2 > 9 --> \(|b|>3\) --> \(b<-3\) or \(b>3\). Not sufficient.

Answer: A.

Hope it's clear.



thanks so much. It was so much simpler than i thought!! What would I do without this forum!! AWESOME RESOURCE!!

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Re: If b is an even integer is b < 0? [#permalink]

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New post 26 Nov 2013, 04:47
Bunuel wrote:
If b is an even integer is b < 0?

(1) b^2 - 4b + 4 < 16 --> \((b-2)^2<16\). Take the square root: \(|b-2|<4\) --> \(-4<b-2<4\) --> \(-2<b<6\). Since given that b is an even integer, then b can be 0, 2, or 4. In either case b is not less than 0. Sufficient.

(2) b^2 > 9 --> \(|b|>3\) --> \(b<-3\) or \(b>3\). Not sufficient.

Answer: A.

Hope it's clear.

Bunuel,

If we wanted to solve the quadratic we would have
b^2-4b-12<0
(b-6)(b+2)<0
then
b-6 <0 => b<6 and b+2<0 => b <-2
Is this wrong?

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Re: If b is an even integer is b < 0? [#permalink]

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New post 26 Nov 2013, 04:57
Skag55 wrote:

b-6 <0 => b<6 and b+2<0 => b <-2

Is this wrong?



Yes,but only the highlighted part. Note that if a*b<0, then a and b have to be of opposite signs.
If you apply that to the above inequality, you would get the same range as Bunuel got.
So yes, your method is also correct.
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Re: If b is an even integer is b < 0? [#permalink]

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New post 26 Nov 2013, 05:11
mau5 wrote:
Skag55 wrote:

b-6 <0 => b<6 and b+2<0 => b <-2

Is this wrong?



Yes,but only the highlighted part. Note that if a*b<0, then a and b have to be of opposite signs.
If you apply that to the above inequality, you would get the same range as Bunuel got.
So yes, your method is also correct.

Aha, so I need to elaborate more on the equation then, i.e. each term will have 2 solutions like:
(b+6) >0 or (b+6) < 0
and same for the other term.
Right?

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Re: If b is an even integer is b < 0? [#permalink]

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New post 21 Jul 2016, 09:19
Skag55 wrote:
mau5 wrote:
Skag55 wrote:

b-6 <0 => b<6 and b+2<0 => b <-2

Is this wrong?



Yes,but only the highlighted part. Note that if a*b<0, then a and b have to be of opposite signs.
If you apply that to the above inequality, you would get the same range as Bunuel got.
So yes, your method is also correct.

Aha, so I need to elaborate more on the equation then, i.e. each term will have 2 solutions like:
(b+6) >0 or (b+6) < 0
and same for the other term.
Right?


I am not completely sure but I don't think you need to do that because if you add 2 to b then that MUST be positive given the constraint explained. In other words (b-6)(b+2) < 0, if one of them must be negative I assume the parenthesis with b-6 would be it because both of them have b. If b was greater than 6 then BOTH would be positive and the equation would not be true.
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Re: If b is an even integer is b < 0? [#permalink]

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Re: If b is an even integer is b < 0?   [#permalink] 01 Aug 2017, 07:59
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