## Inequalities with Exponents

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### Inequalities with Exponents

by tonebeeze » Tue Apr 12, 2011 5:21 pm
Can someone please walk me through the work on statement 1. Thanks

If b is an even integer, is b<0?

1. b^2-4b+4<16

2. b^2>9

OA = A

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by Ian Stewart » Tue Apr 12, 2011 5:59 pm
tonebeeze wrote:Can someone please walk me through the work on statement 1. Thanks

If b is an even integer, is b<0?

1. b^2-4b+4<16

2. b^2>9

OA = A
From Statement 1:

b^2 - 4b + 4 < 16
b^2 - 4b - 12 < 0
(b - 6)(b + 2) < 0

We now have a product which is negative. So one factor must be negative, the other positive. Since b-6 is certainly smaller than b+2, it must be that b-6 is the negative factor, and b+2 is the positive factor. So b-6 < 0, and b < 6, and b+2 > 0, so b > -2. Combining these we know -2 < b < 6. Since b must be an even integer, b can only be 0, 2 or 4. So it is certainly *not* true that b<0, and we have a definite answer to the question, so Statement 1 is sufficient.

Statement 2 is not sufficient; b^2 > 9 is true whenever b > 3 or b < -3.
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by ldoolitt » Tue Apr 12, 2011 6:06 pm
From the stem

b={...-4, -2, 0, 2, 4...}

Question

is b = -2 or -4 or -6...

Attack statement 2 first because, off the bat it seems like it would not work
(2)

b^2 > 9

b=...-6 or -4 or 4 or 6...

Since we get both positives and negatives in our set, INSUFFICIENT

(1)
b^2 - 4b + 4 < 16
b^2 - 4b + -12 < 0
(b - 6)(b + 2) < 0

Since we must have a positive multiplied by a negative to get a value less than zero, one of b-6 or b+2 must be less than zero

If b>6, both parenthetical expressions are positive. If b < -2, both parenthetical expressions are negative. So we must have -2 < b < 6. But note that leaves us with the set b= {0, 2, 4} which is always positive or 0. Therefore the answer to the target question is no and this is SUFFICIENT.

That was kind of a tough one. There may be a more straightforward approach...

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