SOLUTION: The length of a rectangle is 8 feet longer than its width. The area is 3500 sq ft. I have to find the width of the rectangle and just one length because this is for a fence that i

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Question 872843: The length of a rectangle is 8 feet longer than its width. The area is 3500 sq ft.
I have to find the width of the rectangle and just one length because this is for a fence that is connected to the back of a house. The back of the house serves as one fenced area so I just need the length to the part I have to fence.
Found 2 solutions by Seutip, lwsshak3:
Answer by Seutip(231) About Me  (Show Source):
You can put this solution on YOUR website!
Let:
x - be the width
x+8 - be the length
So since the area is 3500+ft%5E2
A=lw
3500=x%28x%2B8%29
3500=x%5E2%2B8x

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 8 feet longer than its width. The area is 3500 sq ft.
I have to find the width of the rectangle and just one length because this is for a fence that is connected to the back of a house. The back of the house serves as one fenced area so I just need the length to the part I have to fence.
***
let x=width
x+8=length
length*width=area
x(x+8)=3500
x^2+8x=3500
x^2+8x-3500=0
solve for x using quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a=1, b=8, c=-3500
ans:
x≈55.30
x+8≈63.30
Fencing required=twice the width+length=2x+(x+8)=3x+8≈174 ft