Respuesta :
Using given facts, the length of the parking lot to be increased is given by: Option B: 20 feet.
How does area of a rectangle, and its length and width are related?
Area of a rectangle is the product of its length and width.
If a rectangle has length L units and width of W units, then
Area = L × W squared units.
How to find the solution to a standard quadratic equation?
Suppose the given quadratic equation is
[tex]ax^2 + bx + c = 0[/tex]
Then its solutions are given as
[tex]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
For the given case, we can use variable in place of unknown increment in length and width of the parking lot.
- Current length of the parking lot = 120 ft
- Current width of the parking lot = 80 ft.
Thus, current area of the parking lot = [tex]120 \times 80 = 9600 \:\rm ft^2[/tex]
Let the increment in length and width of the parking lot be of 'x' feet.
Then, we get:
- New length of the parking lot = 120 + x ft.
- New width of the parking lot = 80 + x ft.
Thus, new area of the parking lot = [tex](120 + x)(80 +x )[/tex]
It is given that the new area = current area + 4400 ft squared.
Thus, we get an equation as:
[tex](120 + x)(80 +x ) = 9600 + 4400 = 14000\\9600 + 200x + x^2 = 14000\\x^2 + 200x -4400 = 0\\[/tex]
Comparing it with [tex]ax^2 + bx + c = 0[/tex], we get a = 1, b = 200, c = -4400
Thus, its roots are:
[tex]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\\x = \dfrac{-200 \pm \sqrt{(200)^2 - 4(-4400)(1)}}{2} = \dfrac{-200 \pm \sqrt{57600}}{2}\\\\x = \dfrac{-200 \pm 240}{2} \\\\x = -220, x = 20[/tex]
x is denoting the length increased, so it cannot be negative, as increment is in positive sign always.
Thus, x = 20 (feet).
Learn more about solution of quadratic equations here:
https://brainly.com/question/3358603
