SOLUTION: . The original plans for Jennifer’s house called for a square foundation. After increasing one side by 30ft and decreasing the other by 10ft, the area of the rectangular foundation

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: . The original plans for Jennifer’s house called for a square foundation. After increasing one side by 30ft and decreasing the other by 10ft, the area of the rectangular foundation      Log On


   



Question 999099: . The original plans for Jennifer’s house called for a square foundation. After increasing one side by 30ft and decreasing the other by 10ft, the area of the rectangular foundation was 2100 sqft. What was the area of the original square foundation? Can someone please solve this for me and show all of the work? thanks so much!
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
let x = the length of the original side of the square plot.

one side is increased by 30 feet and becomes x + 30.
the other side is decreased by 10 feet and becomes x - 10

the area of the rectangular foundation is 2100 square feet.
length * width = area
(x+30) * (x-10) = 2100
multiply out the factors to get:
x^2 - 10x + 30x - 300 = 2100
combine like terms to get:
x^2 + 20x - 300 = 2100
subtract 2100 from both sides of this equation to get:
x^2 + + 20x - 2400 = 0
factor this quadratic equation to get:
(x+60) * (x-40) = 0
solve for x to get:
x = -60 or x = 40
x can't be negative, so x = 40

the area of the original square foundation was x*x which becomes 40*40 = 1600 square feet.

the original foundation was 40 by 40.
the new foundation became 70 by 30

40 * 40 = 1600 square feet.
70 * 30 = 2100 square feet.

the original side length was 40.
the rectangular side lengths were x+30 and x-10.
x+30 = 70
x - 10 = 30

everything checks out ok.