SOLUTION: The length of a rectangle is 2 feet more that its width. Find the dimensions of the rectangle if its area is 63 square feet. Show your work using a quadratic equation to solve.
Question 626408: The length of a rectangle is 2 feet more that its width. Find the dimensions of the rectangle if its area is 63 square feet. Show your work using a quadratic equation to solve. Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! The length of a rectangle is 2 feet more that its width. Find the dimensions of the rectangle if its area is 63 square feet. Show your work using a quadratic equation to solve.
.
Let w = width
then
w+2 = length
.
w(w+2) = 63
w^2 + 2w = 63
w^2 + 2w - 63 = 0
(w+9)(w-7) = 0
w = {-9, 7}
throw out the negative solution leaving
w = 7 feet (width)
.
Length:
w+2 = 7+2 = 9 feet
.
Answer: 7 feet by 9 feet
