Andrew M. answered 04/14/15
Tutor
New to Wyzant
Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors
Let the width of the pool be designated as W
The length of the pool L = W + 30
When one sketches out the rectangular pad along one side of the pool it adds
20 feet to the length so the length of the pool and pad together is W + 50
The dimensions of the pool and pad together is W by (W+50)
The total area is given by (length)(width) so the area of the pool and pad together is:
W(W+50) = 5000
W2 + 50W = 5000
W2 + 50W -5000 = 0
Lets factor this out... the factors of -5000 that will add to 50 are (100)(-50) so this gives us
(W+100)(W-50) = 0
This is only true if wither w + 100 = 0 or W-50 = 0
Thus, the solution of the equation is W = -100 ft or W = 50 ft
We can discard W = -100ft since the pool will have positive dimensions so the width of the pool is 50ft
From this we can then get the length of the pool as W + 30 .... So the length of the pool L = 80 ft
Thus the pool is 50ft x 80ft
The concrete pad is 50ft x 20ft