The length l of a 200-square-foot rectangular vegetable garden is 4 feet less than twice the width w. Find the length and width of the garden.

Question

The length l of a 200-square-foot rectangular vegetable garden is 4 feet less than twice the width w. Find the length and width of the garden.

Answer 1

w=w

l=2w-4

A=lw

200=(2w-4)w

200=2w^2-4w

2w^2-4w-200=0

w^2-2w-11=0

(2±√(4+400))/2

(2±√404)/2

(2±20.1)/2

disregard the "-" answer

22.1/2=11.05

w=11.05 feet

l=2(11.05)-4

l=22.1-4

l=18.1 feet

check:

A=18.1*11.05

A≈200 (200.005 actually due to rounding)

Answer 2

Area = length * width

length = 2*width - 4 <--- length is 4 less than 2 times the width

So

A = Lw

L = 2w-4

200 = (2w-4)w

200 = 2w^2 - 4w

100 = w^2 - 2w <--- divides by 2

w^2 - 2w + 1 = 101 <--- completes the square

(w-1)(w-1) = 101 <--- factors

(w-1)^2 = 101

w-1 = +or- sqrt(101)

w = 1 +or- sqrt(101)

however the negative branch is disqualified because it is negative measure

w = 1 + sqrt(101)

L = 2w - 4 = 2 + 2*sqrt(101) - 4 = 2*sqrt(101) - 2

CHECK:

The area is:

length * width =

[2*sqrt(101) - 2 ][ sqrt(101) + 1] =

FOIL:

2*101 + 2*sqrt(101) - 2*sqrt(101) - 2 =

cross terms cancel:

202 - 2 = 200

Yes it checks:

So the width is sqrt(101) + 1

and the length is twice the conjugate