SOLUTION: I can't come up with an answer to this problem, i was thinking that it has to be more than 15 square yards but if someone could help me that would be wonderful: Dan is building a r

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Question 129623: I can't come up with an answer to this problem, i was thinking that it has to be more than 15 square yards but if someone could help me that would be wonderful: Dan is building a rectangular dog pen that he wants to enclose. The width is 2 yards less than the length. If the area of the dog pen is 15 square yards, many yards of fencing would he need to completely enclose the pen?

Answer by Earlsdon(6103) About Me  (Show Source):
You can put this solution on YOUR website!
Start with the formula for the area of a rectangle:
A+=+L%2AW The problem states that the width (W) is 2 yards less than the length (L), so you can write:
W+=+L-2
It also states that the area (A) is 15 sq.yds. So, putting this altogether, you have:
A+=+L%2AW Substitute A = 15 and W = L-2 to get:
15+=+L%28L-2%29 Simplify this.
15+=+L%5E2-2L Subtract 15 from both sides.
L%5E2-2L-15+=+0 Factor this quadratic equation.
%28L%2B3%29%28L-5%29+=+0 Apply the zero products rule:
L%2B3+=+0 or L-5+=+0 so that...
L+=+-3 or L+=+5 Discard the negative solution as lengths are positive quantities.
So, The length, L, is 5 yards.
The width, W, is L-2 = 5-2 = 3 yards.
Now you can find the perimeter, or the number of yards of fencing needed to enclose the yard.
P+=2L%2B2W Substitute L = 5 and W = 3.
P+=+2%285%29%2B2%283%29
P+=+10%2B6
P+=+16
He would need 16 yards (not square yards) of fencing.