# SOLUTION: This word problem involves quadratic equations which I am having a very difficult understanding. The width of a rectangle is 4 ft less than the length. The area is 12ft^2. Find the

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Geometry -> SOLUTION: This word problem involves quadratic equations which I am having a very difficult understanding. The width of a rectangle is 4 ft less than the length. The area is 12ft^2. Find the      Log On

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 Question 167832: This word problem involves quadratic equations which I am having a very difficult understanding. The width of a rectangle is 4 ft less than the length. The area is 12ft^2. Find the length and width. I start out translating the equation like this: A=LW L=length 4-L=width 12ft^2=L(4-L) then I get lost from that point. Answer by nerdybill(6948)   (Show Source): You can put this solution on YOUR website!This word problem involves quadratic equations which I am having a very difficult understanding. The width of a rectangle is 4 ft less than the length. The area is 12ft^2. Find the length and width. I start out translating the equation like this: A=LW L=length 4-L=width <<--SHOULD BE L-4 . Let x = length then x-4 = width . 12 = x(x-4) 12 = x^2-4x 0 = x^2-4x-12 Factoring: 0 = (x-6)(x+2) x = {-2, 6} . A negative solution doesn't make sense -- so, toss it out. x = 6 feet (length) . Width: x-4 = 6-4 = 2 feet (width)