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[b^3 + 8][(64/b) + 16] = 0?

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by sanju09 » Wed Feb 29, 2012 12:11 am
What is the minimum value of b for which [b^3 + 8][(64/b) + 16] = 0?
A. - 4
B. - 2
C. − ¼
D. ¼
E. 2
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by Anurag@Gurome » Wed Feb 29, 2012 12:21 am
sanju09 wrote:What is the minimum value of b for which [b^3 + 8][(64/b) + 16] = 0?
A. - 4
B. - 2
C. − ¼
D. ¼
E. 2
Let us check each answer choice:

(A) [(-4)^3 + 8][(64/-4) + 16] = [(-4)^3 + 8][-16 + 16] = [(-4)^3 + 8]0 = 0, true
Of all the answer choices option A has the minimum value so no need to check other options.

The correct answer is A.
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