SOLUTION: The length of a rectangle is 3 feet less than twice its width and area of rectangle is 27 ft2. Find the dimension Lenth= Width=

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Question 949571: The length of a rectangle is 3 feet less than twice its width and area of rectangle is 27 ft2. Find the dimension
Lenth=
Width=
Answer by macston(5194)   (Show Source): You can put this solution on YOUR website!
W=width; L=length=2W-3 ft; A=L*W=27 sq ft
A=L*W Substitute for L
27 sq ft=(2W-3)(W)
Subtract 27 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=225 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 4.5, -3. Here's your graph:

ANSWER 1: W=4.5 ft The width is 4.5 feet.
L=2W-3 ft=2(4.5)-3 ft=9 feet-3 feet=6 feet ANSWER 2: The length is 6 feet.
CHECK:
A=L*W
27 sq ft=(6 ft)(4.5 ft)
27 sq ft=27 sq ft
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