SOLUTION: A retangular badminton court has an original dinmesions of 14 feet wide by 18 feet long. It is determined that the court is not big enough and the length and width are increased, b

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Question 721228: A retangular badminton court has an original dinmesions of 14 feet wide by 18 feet long. It is determined that the court is not big enough and the length and width are increased, by the same amount, so that the new court is 437 square feet. By how much did the length and width increase?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Let's call the increased amount: "x". This makes the new dimensions 14+x and 18+x. The area of this rectangle would be:
(14+x)(18+x)
Simplifying (using FOIL) we get:
14%2A18%2B14%2Ax%2Bx%2A18%2Bx%2Ax
252%2B14x%2B18x%2Bx%5E2
x%5E2%2B+30x%2B252
Since we told that this new area is 437:
x%5E2%2B+30x%2B252=437

Now we solve. First we want a zero on one side. Subtracting 437:
x%5E2%2B+30x-185=0
This will not factor so we must use the quadratic formula:
x+=+%28-%2830%29+%2B-+sqrt%2830%5E2-4%281%29%28-185%29%29%29%2F2%281%29
Simplifying...
x+=+%28-%2830%29+%2B-+sqrt%28900-4%281%29%28-185%29%29%29%2F2%281%29
x+=+%28-%2830%29+%2B-+sqrt%28900%2B740%29%29%2F2%281%29
x+=+%28-%2830%29+%2B-+sqrt%281640%29%29%2F2%281%29
x+=+%28-30+%2B-+sqrt%281640%29%29%2F2
x+=+%28-30+%2B-+sqrt%284%2A410%29%29%2F2
x+=+%28-30+%2B-+sqrt%284%29%2Asqrt%28410%29%29%2F2
x+=+%28-30+%2B-+2%2Asqrt%28410%29%29%2F2
x+=+%282%28-15+%2B-+sqrt%28410%29%29%29%2F2
x+=+%28cross%282%29%28-15+%2B-+sqrt%28410%29%29%29%2Fcross%282%29
x+=+-15+%2B-+sqrt%28410%29
which is short for:
x+=+-15+%2B+sqrt%28410%29 or x+=+-15+-+sqrt%28410%29
The second solution is negative. It makes no sense for the amount we add to the sides to be negative so we reject that solution. So the only solution is:
x+=+-15+%2B+sqrt%28410%29
If you need/want a decimal approximation, get out your calculator.

P.S. The facts that we got an equation that did not factor and we got an answer with a square root suggest that maybe one or more numbers in the problem might have been wrong. Was the new area supposed to be 427 not 437? If so then the equation factors, leading to an easier solution. The quadratic formula will still work but you don't end up with a square root.

P.P.S. Thanks for finding my error! Here's the solution without the error:
x%5E2%2B+32x%2B252=437
x%5E2%2B+32x-185=0
Factor:
%28x-5%29%28x%2B37%29=0
From the Zero Product Property:
x-5 = 0 or x+37 = 0
x = 5 or x = -37
Since x represents the increase, we reject the negative solution because it doesn't make sense. So each of the dimensions increased by 5.