SOLUTION: the perimeter of a garden is 27 ft, and the area is 35 ft^2 . find the length and the width of the garden

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Question 903032: the perimeter of a garden is 27 ft, and the area is 35 ft^2 . find the length and the width of
the garden



Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
2(35/w) + 2w = 27ft
2w^2 - 27w + 70 = 0
dimensions are 10ft by 3.5ft
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-27x%2B70+=+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=%28-27%29%5E2-4%2A2%2A70=169.

Discriminant d=169 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--27%2B-sqrt%28+169+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-27%29%2Bsqrt%28+169+%29%29%2F2%5C2+=+10
x%5B2%5D+=+%28-%28-27%29-sqrt%28+169+%29%29%2F2%5C2+=+3.5

Quadratic expression 2x%5E2%2B-27x%2B70 can be factored:
2x%5E2%2B-27x%2B70+=+2%28x-10%29%2A%28x-3.5%29
Again, the answer is: 10, 3.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-27%2Ax%2B70+%29