SOLUTION: The perimeter of a rectangula town is 24m.if the are of the town is 35cm^2,how wide is the town?

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: The perimeter of a rectangula town is 24m.if the are of the town is 35cm^2,how wide is the town?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 912085: The perimeter of a rectangula town is 24m.if the are of the town is 35cm^2,how wide is the town?
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The perimeter of a rectangula town is 24m.if the are of the town is 35cm^2,how wide is the town?
---------------
P = 2W + 2L = 24m = 2400 cm
W + L = 1200
W*L = 35
-----
W + L = 1200 --> L = 1200 - W
---
W*L = 35
W*(1200 - W) = 35
W^2 - 1200W + 35 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-1200x%2B35+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-1200%29%5E2-4%2A1%2A35=1439860.

Discriminant d=1439860 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--1200%2B-sqrt%28+1439860+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-1200%29%2Bsqrt%28+1439860+%29%29%2F2%5C1+=+1199.97083262439
x%5B2%5D+=+%28-%28-1200%29-sqrt%28+1439860+%29%29%2F2%5C1+=+0.0291673756131559

Quadratic expression 1x%5E2%2B-1200x%2B35 can be factored:
1x%5E2%2B-1200x%2B35+=+%28x-1199.97083262439%29%2A%28x-0.0291673756131559%29
Again, the answer is: 1199.97083262439, 0.0291673756131559. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1200%2Ax%2B35+%29

===========================
W = 0.029167 cm