Arectangle has side lengths of (11−x) m and (x−4) m. the area of the rectangle is 15m2. which quadratic equation in standard form represents the area of this rectangle?

Area of the rectangle = -(x - 15/2)² + 49/4

Step-by-step explanation:

* Lets revise the rule of the area of the rectangle

- Any rectangle has two dimensions,length and width

-The area of the rectangle = length × width

* In our problem the dimensions of the rectangle are

 (11 - x) and (x - 4)

∴ The area of the rectangle = (11 - x)(x - 4)

- Lets multiply the two brackets

∴ (11 - x)(x - 4) = 11 × x + 11 × (-4) + (-x) × (x) + (-x) × (-4)

∴ (11 - x)(x - 4) = 11x - 44 - x² + 4x ⇒ collect the like terms

∴ (11 - x)(x - 4) = -x² + 15x - 44

* The area of the rectangle = -x² + 15x - 44

- To put them in the standard form a(x + h)² + k, lets equate

  the two forms to find a , h and k

∵ -x² + 15x - 44 = a(x + h)² + k

∴ -x² + 15x - 44 = a(x² + 2hx + h²) + k

∴ -x² + 15x - 44 = ax² + 2ahx +ah² + k

* Compare the two sides

∵ -1 = a ⇒ coefficient of x²

∴ a = -1

∵ 2ah = 15 ⇒ coefficient of x

∴ 2(-1)h = 15 ⇒ -2h = 15 ⇒ divide the both sides by -2

∴ h = -15/2 = -7.5

∵ ah² + k = -44

∴ (-1)(-15/2)² + k = -44

∴ -225/4 + k = -44 ⇒ add -225/4 to the both sides

∴ k = 49/4 = 12.25

* The standard form of the equation is -(x - 15/2)² + 49/4

∴ The quadratic equation represented the area is

  A = -(x - 15/2)² + 49/4

part a: 1.08

part b: 27.59