SOLUTION: If the length of a rectangular field is 8 feet more than double of its width. Find the dimensions of the field if its area is 540 square feet?

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Question 165010: If the length of a rectangular field is 8 feet more than double of its width. Find the dimensions of the field if its area is 540 square feet?
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
If the length of a rectangular field is 8 feet more than double of its width. Find the dimensions of the field if its area is 540 square feet?
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Let x = width
then from "length of a rectangular field is 8 feet more than double of its width"
2x+8 = length
.
x(2x+8) = 540
2x^2+8x = 540
x^2+4x = 270
x^2+4x-270= 0
.
Since you can't factor you must use the quadratic equation.
This will result in:
x = {14.55, -18.55}
.
Since width can't be negative:
x = 14.55 feet
.
Length:
2x+8 = 2(14.55)+8 = 37.1 feet
.
Dimensions: 14.55 by 37.1 feet
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Details of quadratic:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B4x%2B-270+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%284%29%5E2-4%2A1%2A-270=1096.

Discriminant d=1096 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-4%2B-sqrt%28+1096+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%284%29%2Bsqrt%28+1096+%29%29%2F2%5C1+=+14.5529453572468
x%5B2%5D+=+%28-%284%29-sqrt%28+1096+%29%29%2F2%5C1+=+-18.5529453572468

Quadratic expression 1x%5E2%2B4x%2B-270 can be factored:
1x%5E2%2B4x%2B-270+=+1%28x-14.5529453572468%29%2A%28x--18.5529453572468%29
Again, the answer is: 14.5529453572468, -18.5529453572468. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B4%2Ax%2B-270+%29