SOLUTION: Kay wants a rectangular area in greenhouse reserved for African violets.She says the widths of the area must be one foot less then the length she also tells me the area must be at

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Question 151133: Kay wants a rectangular area in greenhouse reserved for African violets.She says the widths of the area must be one foot less then the length she also tells me the area must be at least 12 square feet. What widths satisfy Kay's requirements?
Answer by mducky2(62) About Me  (Show Source):
You can put this solution on YOUR website!
If w = width and w+1 = length, then the area would be the width times the length. If this amount has to be at least 12 square feet, then the real equation is:
w(w+1) ≥ 12

However, this will come up with an infinite number of answers, since the area can be anywhere between 12 and infinity. Therefore, it's better to look for the minimum width, which could be represented by:
w(w+1) = 12

Solving for w:
w2 + w = 12
w2 + w - 12 = 0
(w+4)(w-3) = 0

In order for this whole left side to be zero, only one of these, w+3 or w-4 has to be zero.
w + 4 = 0
w = -4
Since the width can't be negative, that can't be the right answer. It must be the other part.
w - 3 = 0
w = 3
This is an acceptable answer.

Therefore:
l = w+1
l = 3+1
l = 4

According to the original question, any integer above 3 would also be an acceptable width (at least 12 square feet), as long as the corresponding length were one foot longer.