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Bunuel
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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Skag55

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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mau5
Skag55

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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Skag55
mau5
Skag55

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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