The length of a rectangular garden is 5 feet greater than the width. The area of the rectangle is 300 square feet. Find the length and the width.
2006-10-17 16:13:03 · 7 answers · asked by dawn h 1 in Science & Mathematics ➔ Mathematics
L = W + 5
A = L*W
A = 300
300 = (W+5)*W
300 = W^2 + 5W
Subtract 300 from each side
0 = W^2 + 5W - 300
0 = (W-15)(W+20)
W-15 = 0 or W+20 = 0
W = 15 or W = -20
Since we cannot have a negative length, we exclude W = -20
So W = 15; therefore, L = 15 + 5 = 20
2006-10-17 16:19:27 · answer #1 · answered by MsMath 7 · 1⤊ 0⤋
Call the width w and the length w + 5.
w(w+5) = 300
w^2+5w-300 = 0
(w - 15 )(w + 20) = 0
w = {15, -20}
Reject the -20 because width can't be negative.
So w = 15 and w+5 = 20
2006-10-17 16:21:26 · answer #2 · answered by PatsyBee 4 · 0⤊ 0⤋
L = 5 + W
L*W = 300
(5 + W) W = 300 => W^2 + 5W - 300 = 0
it will be the form of (ax + c) ( bx + d) = 0
abx^2 +(ad +cd) x + cd = 0
ab =1
cd= -300 and c+d = 5
c=20 and d=-15
so you'll have (W + 20) ( W - 15) = 0
W should be positive so W=15 ft is the acceptable solution
and L = 20 ft
2006-10-17 16:29:39 · answer #3 · answered by Anonymous · 0⤊ 0⤋
L = 5 + W
L x W = 300
(5+W)W = 300
W^2 + 5W = 300
W^2 + 5W -300 = 0
(W -15)(W+20) = 0
Since the width must be a positive number, W = 15
L = W + 5 = 20
2006-10-17 16:20:34 · answer #4 · answered by jg 2 · 1⤊ 0⤋
the equation is w(w+5) = 300
w^2 +5w - 300 = 0, (w-15)(w+20)=0, width = 15, length = 20
2006-10-17 16:20:33 · answer #5 · answered by shamu 2 · 1⤊ 0⤋
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2006-10-17 16:21:42 · answer #6 · answered by thewordofgodisjesus 5 · 0⤊ 0⤋
w=7.74 and L=38.729
2006-10-17 16:32:38 · answer #7 · answered by Gina456 1 · 0⤊ 0⤋