SOLUTION: The length of a rectangle is 11ft more then twice the width, and the area of the rectangle is 63 ft^2. Find the dimensions of the rectangle.

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Question 949487: The length of a rectangle is 11ft more then twice the width, and the area of the rectangle is 63 ft^2. Find the dimensions of the rectangle.
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
W=width; L=length=2W+11 ft; A=area=L*W=63 sq ft
A=L*W Substitute for L
63+sq+ft=%282W%2B11%29%28W%29
63+sq+ft=2W%5E2%2B11W Subtract 63 sq ft from each side
0=2W%5E2%2B11W-63+sq+ft
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aW%5E2%2BbW%2Bc=0 (in our case 2W%5E2%2B11W%2B-63+=+0) has the following solutons:

W%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=%2811%29%5E2-4%2A2%2A-63=625.

Discriminant d=625 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-11%2B-sqrt%28+625+%29%29%2F2%5Ca.

W%5B1%5D+=+%28-%2811%29%2Bsqrt%28+625+%29%29%2F2%5C2+=+3.5
W%5B2%5D+=+%28-%2811%29-sqrt%28+625+%29%29%2F2%5C2+=+-9

Quadratic expression 2W%5E2%2B11W%2B-63 can be factored:
2W%5E2%2B11W%2B-63+=+2%28W-3.5%29%2A%28W--9%29
Again, the answer is: 3.5, -9. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B11%2Ax%2B-63+%29

W=3.5 ft ANSWER the width is 3.5 feet.
L=2W+11 feet=2(3.5 feet)+11 ft=18 feet ANSWER The length is 18 feet.
CHECK
A=L*W
63 sq ft=18 ft*3.5 ft
63 sq ft=63 sq ft