Question 1175418: The length of a rectangular is 2 ft more than twice the width. The area is 60 ft^2. Find the length and width if the rectangle Found 3 solutions by ewatrrr, VFBundy, ikleyn:Answer by ewatrrr(24785) (Show Source):
Hi
.
length is 2 ft more than twice the width 'w'
Lw = A
(2w+2)w = 60Ft^2
2w^2 + 2w -60 = 0
w^2 + w - 30 = 0
(w + 6)(w-5) = 0
w = 5ft |tossing out a negative value for length
and
L = 12ft
12ft(5ft) = 60ft^2 checks.
Wish You the Best in your Studies.
You can put this solution on YOUR website! Width = w
Length = 2w + 2
w(2w + 2) = 60
2w² + 2w = 60
2w² + 2w - 60 = 0
2(w² + w - 30) = 0
w² + w - 30 = 0
(w - 5)(w + 6) = 0
w = 5 and w = -6
Since the width (w) cannot be a negative number, we can eliminate w = -6. Therefore, were are left with w = 5.
Width = w = 5
Length = 2w + 2 = 2(5) + 2 = 10 + 2 = 12
You can put this solution on YOUR website! .
The length of a RECTANGLE is 2 ft more than twice the width.
The area is 60 ft^2. Find the length and width if the rectangle
~~~~~~~~~~~~~
Mental solution
w*(2w+2) = 60
w*(w+1) = 30
the product of two consecutive integers is 30 -------------> hence, the numbers are 5 and 6.
ANSWER. The width is 5 units; the length is 2*5+2 = 12 units.
.
length is 2 ft more than twice the width 'w'
Lw = A
(2w+2)w = 60Ft^2
2w^2 + 2w -60 = 0
w^2 + w - 30 = 0
(w + 6)(w-5) = 0
w = 5ft |tossing out a negative value for length
and
L = 12ft 
