SOLUTION: The area of a rectangle is 165 ft (2)squared. The length of the rectangle is 4 feet longer than the width. Find the length and width. I need to know the equation and how to solve.

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: The area of a rectangle is 165 ft (2)squared. The length of the rectangle is 4 feet longer than the width. Find the length and width. I need to know the equation and how to solve.      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 280270: The area of a rectangle is 165 ft (2)squared. The length of the rectangle is 4 feet longer than the width. Find the length and width. I need to know the equation and how to solve.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The area of a rectangle is 165 ft (2)squared. The length of the rectangle is 4 feet longer than the width. Find the length and width. I need to know the equation and how to solve.
----
Let the width be "x":
Then the length is "x+4"
--------------
Area = x(x+4)
165 = x^2+4x
---
x^2 + 4x - 165 = 0
Quadratic formula:
x = [-4 +- sqrt(16-4*-165)]/2
x = [-4 +- sqrt(676)]/2
x = [-4 +- 26]/2
Positive solution:
x = [22/2]
x = 11 ft. (width)
x+4 = 15 ft (length)
===========================
Cheers,
Stan H.