Question 151133
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) &#8805 12
<br>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
<br>Solving for w:
w<sup>2</sup> + w = 12
w<sup>2</sup> + w - 12 = 0
(w+4)(w-3) = 0
<br>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.
<br>Therefore:
l = w+1
l = 3+1 
l = 4
<br> According to the original question, any integer above 3 would also be an acceptable width (at <b>least</b> 12 square feet), as long as the corresponding length were one foot longer.