This is the visual
The length of the rectangle is 3 more than twice the width
Let the width of the rectangle = x
The length = 2(x) + 3 = 2x + 3
Its area = length x width
Area = (2x + 3) x x
Area = x(2x + 3)
The area of the rectangle = 44 square meters
Equate the area by 44
x(2x + 3) = 44
The equation is x(2x + 3) = 44
Let us solve the equation
Multiply the bracket by x
[tex]\begin{gathered} x(2x)+x(3)\text{=44} \\ 2x^2+3x=44 \end{gathered}[/tex]
Subtract 44 from both sides
[tex]\begin{gathered} 2x^2+3x-44=44-44 \\ 2x^2+3x-44=0 \end{gathered}[/tex]
Factorize it into two factors
[tex]\begin{gathered} 2x\times x=2x^2 \\ -4\times11=-44 \\ 2x\times-4=-8x \\ x\times11=11x \\ 11x-8x=3x \end{gathered}[/tex]
The factors are (2x + 11)(x - 4)
(2x + 11)(x - 4) = 0
Equate each factor by 0 to find x
x - 4 = 0
Add 4 to both sides
x - 4 + 4 = 0 + 4
x = 4
The width = 4
The length = 2(4) + 3 = 8 + 3 = 11
The dimensions are 4 m and 11 m
The second factor will give a negative value of x, then we will refuse it, no dimension with a negative value
