Stumped on Word Problem

krisolaw

New member
Joined
Jun 5, 2005
Messages
45
The length of a rectangular ice-skating rink is 20 meters greater than twice its width. Find the width. (Additional information in pic form~ Area=6,000m^2 and w m(located on width side of pic) Please help and explain. Thanks in advance. :?
 

dagr8est

Junior Member
Joined
Nov 2, 2004
Messages
128
Use variables for the width and length:
Let w=width
Let l=length

It says that the length is 20m greater than twice the width, so:
l=2w+20

The area of a rectangle is A=wl, remember l=2w+20:

A=lw
6000=w(2w+20)
6000=2w^2+20w
0=2w^2+20w-6000
0=2(w^2+10w-3000)
0=2(w-50)(w+60)
w=50m, -60m

Note: If you haven't learned how to factor a quadratic, the above might not make sense.

I leave it to you from there. Remember we are talking about a physical object so the width and length cannot be negative.
 

krisolaw

New member
Joined
Jun 5, 2005
Messages
45
:D Thanks for the explanation. I think I finally get it.
 
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