SOLUTION: The student center contain 4800 square ft. and its length is 20 ft. longer than its width. What are the dimensions?

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: The student center contain 4800 square ft. and its length is 20 ft. longer than its width. What are the dimensions?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 201096: The student center contain 4800 square ft. and its length is 20 ft. longer than its width. What are the dimensions?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let the length = L and the width = W.
The area of a rectangle is given by:
A+=+L%2AW Substitute L = W+20 and A = 4800.
4800+=+%28W%2B20%29W Simplify.
4800+=+W%5E2%2B20W Subtract 4800 from both sides.
W%5E2%2B20W-4800+=+0 Solve by factoring.
%28W-60%29%28W%2B80%29+=+0Apply the zero product rule.
w-60+=+0 or W%2B80+=+0 so that...
W+=+60 or W+=+-80 Discard the negative solution as width can only be positive.
The width is 60 feet and the length is 60+20 = 80 feet.