SOLUTION: a rectangle is 6 feet long and 4 feet wide. if each dimension is increased by the same number of feet, a new rectangle is formed whose area is 39 square feet more then the area of

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: a rectangle is 6 feet long and 4 feet wide. if each dimension is increased by the same number of feet, a new rectangle is formed whose area is 39 square feet more then the area of       Log On


   

Question 249377: a rectangle is 6 feet long and 4 feet wide. if each dimension is increased by the same number of feet, a new rectangle is formed whose area is 39 square feet more then the area of the original rectangele. by how many feet was each dimension increased?
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
AREA=6*4=24 FT^2
39+24=(6+X)(4+X)
63=24+10X+X^2
X62+10X+24-63
X^2+10X-39=0
(X+13)(X-3)=0
X-3=0
X=3 FT. ANS. FOR THE INCREASE IN LENGTH & WIDTH.
pROOF:
63=(6+3)(4+3)
63=9*7
63=63