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 150815: 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) (Show Source):
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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 w = width of field
2w+8 = length of field
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w(2w+8)= 540
2w^2+8w= 540
2w^2+8w-540 = 0
w^2+4w-270 = 0
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Since you can't factor you must use the quadratic equation. Doing so yields:
w={14.55, -18.55}
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Only the positive answer would be correct. Therefore:
w = 14.55 feet (width)
2w+8 = 37.1 feet (length)
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Below is the quadratic solution:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=1096 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 14.5529453572468, -18.5529453572468. Here's your graph: