SOLUTION: The length of a rectangle is 3 m more than twice the width and the area of the rectangle is 35 m2 . Find the length and width of the rectangle.

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Question 1043016: The length of a rectangle is
3 m
more than twice the width and the area of the rectangle is
35 m2
. Find the length and width of the rectangle.
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is
3 m
more than twice the width and the area of the rectangle is
35 m2
--------
L = 2W + 3
L*W = 35
---
sub for L
W*(2W+3) = 35
2W^2 + 3W - 35 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B3x%2B-35+=+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=%283%29%5E2-4%2A2%2A-35=289.

Discriminant d=289 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-3%2B-sqrt%28+289+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%283%29%2Bsqrt%28+289+%29%29%2F2%5C2+=+3.5
x%5B2%5D+=+%28-%283%29-sqrt%28+289+%29%29%2F2%5C2+=+-5

Quadratic expression 2x%5E2%2B3x%2B-35 can be factored:
2x%5E2%2B3x%2B-35+=+%28x-3.5%29%2A%28x--5%29
Again, the answer is: 3.5, -5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B3%2Ax%2B-35+%29

============
W = 3.5 meters
L = 10