Question

The length of a rectangle is 3ft more than twice the width, and the area of the rectangle is 54ft^2. Find the dimensions of the rectangle.

Answer 1

Let wedth = x
Length = 2x+3
Area = 54
Area = l X b
54= (2x + 3) X x
54 = 2 x² + 3x
The equation will be:
2 x² + 3x - 54
2 x² + 12x -9x - 54
2x (x+6) -9(x + 6)
(2x-9) (x+6)
We will use (2x-9)
2x-9= 0
2x = 9
x = 9/2
x = 4.5

Width = 4.5
Length = (2x +3)
Length = (2X 4.5 + 3)
Length = 9.0 + 3 = 12

Answer 2

Answer: Let length = 2x + 3

Let width = x

Area = 54 ft2

length × width = Area

x(2x + 3) = 54

2x2 + 3x = 54

2x2 + 3x - 54 = 0

(2x - 9)(x + 6) = 0

x = 9/2 and x = -6

x = 4.5 and x = -6

We accept x = 4.5 because length cannot be a negative value. Substituting this value into the dimensions:

width = 4.5 ft

length = 2(4.5) + 3 = 9 + 3 = 12 ft

Explanation: