SOLUTION: Planting windscreens: A farmer intends to construct a windscreen by planting trees in a quarter-mile row. His daughter points out that 44 fewer trees will be needed if they are

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Question 156035This question is from textbook College Algebra
: Planting windscreens:
A farmer intends to construct a windscreen by planting trees in a quarter-mile row. His daughter points out that 44 fewer trees will be needed if they are planted 1 foot farther apart. If her dad takes her advice, how many trees will be needed? A row starts and ends with a tree. (Hint: 1 mile = 5,280 feet)
I give up...Everything I have tried does not seem to work. This question is from textbook College Algebra

Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
A farmer intends to construct a windscreen by planting trees in a quarter-mile row. His daughter points out that 44 fewer trees will be needed if they are planted 1 foot father apart. If her dad takes her advice, how many trees will be needed. A row starts and ends with a tree. (1 mile = 5,280 feet)
:
Find ft in 1/4 mile; *5280 = 1320 ft
:
Not including the extra tree on the end
Let x = no of trees suggested
Then
(x+44) = original no. of trees proposed
:
Write expressions that calculate distance between the trees:(1320/no. of trees)
:
Original dist + 1 ft = suggested distance
+ 1 =
;
Multiply equation by common denominator x(x+44); results:
1320x + x(x+44) = 1320(x+44)
:
1320x + x^2 + 44x = 1320x + 58080
:
Arrange on the left as a quadratic equation:
x^2 + 44x + 5280x - 5280x - 58080= 0
;
x^2 + 44x - 58080 = 0
;
This will actually factor, but you can use the quadratic formula if wish:
(x-220)(x+264) = 0
The positive solution
x = 220 + 1 = 221 trees to save 44 trees
;
We added one because there is a tree on each end
;
:
Check solution by finding the distance between each tree
1320/220 = 6 ft
1320/264 = 5 ft (with 44 more trees)