SOLUTION: Arectangle 6 feet by 8 feet has a border of uniform width around the inside of all four edges. If the border has an area of 33 square feet, how wide is the border?

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Arectangle 6 feet by 8 feet has a border of uniform width around the inside of all four edges. If the border has an area of 33 square feet, how wide is the border?      Log On


   

Question 179283: Arectangle 6 feet by 8 feet has a border of uniform width around the inside of all four edges. If the border has an area of 33 square feet, how wide is the border?
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
6*8=48 is the area of the rectangle.
(6+2x)(8+2x)=48+33
48+16x+12x+4x^2=81
4x^2+28x+48-81=0
4x^2+28x-33=0
Using them quadratic equationx+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ we get:
x=(-28+-sqrt[28^2-4*4*-33])/2*4
x=(-28+-sqrt[784+528)/8
x=(-28+-sqrt1312)/8
x=(-28+-36.22)/8
x=(-28+36.22)/8
x=(8.22)/8
x=1.0277
Proof:
(6+2*1.0277)(8+2*1.0277)=81
(6+2.055)(8+2.055)=81
8.055*10.055=81
81~81