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)   (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

Subtract 63 sq ft from each side
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=625 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 3.5, -9. Here's your graph:

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
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