The length of a rectangle is 1mm more than twice the width. If the area is 300 square mm, find the dimension of the rectangle?
L=Length, W=Width
Length is 1 more than twice the width: L=1+2W
Area is 300: L*W=300
System of Eq's, plug and solve
L*W=300
(1+2W)*W=300
2W^2+W=300
2W^2+W-300=0
Quadratic Formula
W1= -12.5 W2= 12
reject W1 because you cannot have negative width
W=12
L=1+2W=1+2*12=25
Check:
25*12=300
300=300 checks out
w(2w+1)=300 , 2w^2+w-300=0 , w=12 or -25/8 , length=25 and width=12
L = 1 + 2W
WL = A
W(1 + 2W) = 300
W + 2W^2 = 300
2W^2 + W - 300 = 0
(-1 ± â2401) / 4
W = {-12.5, 12} ignore negative answer
W = 12 mm
L = 25 mm
Comments
L=Length, W=Width
Length is 1 more than twice the width: L=1+2W
Area is 300: L*W=300
System of Eq's, plug and solve
L*W=300
(1+2W)*W=300
2W^2+W=300
2W^2+W-300=0
Quadratic Formula
W1= -12.5 W2= 12
reject W1 because you cannot have negative width
W=12
L=1+2W=1+2*12=25
Check:
L*W=300
25*12=300
300=300 checks out
w(2w+1)=300 , 2w^2+w-300=0 , w=12 or -25/8 , length=25 and width=12
L = 1 + 2W
WL = A
W(1 + 2W) = 300
W + 2W^2 = 300
2W^2 + W - 300 = 0
(-1 ± â2401) / 4
W = {-12.5, 12} ignore negative answer
W = 12 mm
L = 25 mm