Stumped on Word Problem
- Thread starter krisolaw
- Start date
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.
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.
Thanks for the explanation. I think I finally get it.