The length of a rectangle is 2ft. longer than the width. If the area is 16ft^, then what are the length and width?
Please show your work.
Thanks so much!
2007-01-30 05:59:31 · 7 answers · asked by SweetnSpiceyBrown 2 in Science & Mathematics ➔ Mathematics
W: x
L: x+2
A= wl
16= x(x+2)
16= x^2+2x
x^2+2x-16
use quadradic fomula and you'll get x= 3.123ft
Width=3.123ft
Length= 5.123ft
2007-01-30 06:05:37 · answer #1 · answered by 7 · 3⤊ 1⤋
L = 2 + w
L * w = 16
(2 + w)*w = 16
w^2 +2w - 16 = 0
w = [-b +/- sqrt(b^2 - 4ac)]/(2a)
{only use the + from the +/- above since w cannot be negative}
w = [-2 + sqrt( 4 - (-64)]/2
w = [-2 + sqrt(68)]/2
w = -1 + sqrt(17)
L = 1 + sqrt(17)
2007-01-30 06:09:23 · answer #2 · answered by jimvalentinojr 6 · 1⤊ 1⤋
given length L = W+2
Area = L * W = 16 given
Therefore, (W+2) * W = 16
W^2 +2 W -16 +0
W = [- 2 ± Sqrt {(-2*-2 - 4*1*(-16)}] /2*1
W = [- 2 ± Sqrt {4 +64}] /2
W = [- 2 ± Sqrt {68}] /2 = [-2 ± 8.246]/2
> W = 3.123 (+ value) or -5.123 ft (ignore)
> L = W +2 = 5.123 (+ value) or -3.123 ft (ignore)
2007-01-30 06:19:06 · answer #3 · answered by anil bakshi 7 · 0⤊ 1⤋
You are going to use two key pieces of information.
First:
l=2+w.
Second:
A=l*w
So you know that A=16, or
16=l*w,
but l=2+w, so
16=(2+w)*w, or
w^2+2w-16=0;
Solve for w,
w=-5.12 or w=3.12.
Since w cannot be negative, the length is 5.12, width is 3.12 ft
2007-01-30 06:09:32 · answer #4 · answered by Daniel W 3 · 0⤊ 1⤋
length = x+2
width = x
x*(x+2) = 16
x^2 +2x = 16
x^2 + 2x - 16 = 0
now use the quadratice equation to solve that.
2007-01-30 06:09:13 · answer #5 · answered by Michael Dino C 4 · 1⤊ 2⤋
Do your own homework , you'll benifit more from doing it your self. No one can help you in the future when your testing so learn the material.
2007-01-30 06:06:26 · answer #6 · answered by Ramona T 2 · 0⤊ 3⤋
CHEATER
How about you listen to your math teacher.
Haven't you thought of that ?
2007-01-30 06:30:25 · answer #7 · answered by Anonymous · 0⤊ 3⤋