emdorsheimer
New member
- Joined
- Oct 6, 2006
- Messages
- 17
Algebra and word problems. I can't seem to get the right answer.
The length of a rectangle is 3 cm more than 2 times its width. If the area of the rectangle is 99 cm2, find the dimensions of the rectangle to the nearest thousandth.
This is what I did:
Let the length be denoted by "L" and the width be denoted by "w". Then the equations are:
L = 2w + 3
L - 2w = 3………(1)
Lw = 99…………………(2)
From (2) I get:
w=99/L
Substituting this in (1) I get:
L - 2(99/L) = 3
Multiplying throughout by L I get:
L^2 - 198 = 3L
L^2 - 3L - 198 = 0
L = [3+/-sqrt(9 + 792)]/21 = 3+/-sqrt(801)/21 = 3 + 28.3/21 = 15.65
Is this right?
______________________________
Edited by stapel -- Reason for edit: formatting
The length of a rectangle is 3 cm more than 2 times its width. If the area of the rectangle is 99 cm2, find the dimensions of the rectangle to the nearest thousandth.
This is what I did:
Let the length be denoted by "L" and the width be denoted by "w". Then the equations are:
L = 2w + 3
L - 2w = 3………(1)
Lw = 99…………………(2)
From (2) I get:
w=99/L
Substituting this in (1) I get:
L - 2(99/L) = 3
Multiplying throughout by L I get:
L^2 - 198 = 3L
L^2 - 3L - 198 = 0
L = [3+/-sqrt(9 + 792)]/21 = 3+/-sqrt(801)/21 = 3 + 28.3/21 = 15.65
Is this right?
______________________________
Edited by stapel -- Reason for edit: formatting