SOLUTION: The area of a rectangular ping pong table is {{{45 feet^2}}}. The length is 4 feet longer than the width. Let x = the width. Find the length in feet. So far I have {{{45x^2+4x=0

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: The area of a rectangular ping pong table is {{{45 feet^2}}}. The length is 4 feet longer than the width. Let x = the width. Find the length in feet. So far I have {{{45x^2+4x=0      Log On


   

Question 825846: The area of a rectangular ping pong table is 45+feet%5E2. The length is 4 feet longer than the width. Let x = the width. Find the length in feet.
So far I have 45x%5E2%2B4x=0
Found 2 solutions by htmentor, josgarithmetic:
Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
The area of a rectangle is length*width
Given: width = x, length, l = x+4
The area = x*(x+4) = 45
x^2+4x-45 = 0
This can be factored as (x-5)(x+9) = 0
Taking the positive solution, we have x = 5
Therefore, the length = 5+4 = 9 ft.

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
L=length
L=x+4
x=width

Area, xL=45
x%28x%2B4%29=45
highlight_green%28x%28x%2B4%29-45=0%29