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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] New post 23 Nov 2013, 10:28
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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, 10:31, 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] New post 23 Nov 2013, 10:35
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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] New post 23 Nov 2013, 11:02
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] New post 26 Nov 2013, 03:47
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] New post 26 Nov 2013, 03:57
Expert's post
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] New post 26 Nov 2013, 04:11
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?
Re: If b is an even integer is b < 0?   [#permalink] 26 Nov 2013, 04:11
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