math Geometry Assessment

Question

A rectangular pool is being created with an area of 578 sq ft. If the width is (2x - 3) and the length is twice the width of the pool

What simplified equation could be used to find the area of the pool?

Answer 1

The area is given, the length is twice the width and width is (2x -3), so it is a matter of substituting in the formula for area of a rectangle A= l w

Answer 2

We are already given the area; we want to find an equation that represents this area.

Expression for width: w = 2x - 3

Expression for length: l = 2w = 2(2x-3) = 4x - 6

Area = width * length = (2x - 3)(4x - 6)

Using FOIL to multiply out, we get Area = 8x² - 12x - 12x + 18 = 8x² - 24x + 18.

So, our equation involving x and the Area is 8x² - 24x + 18 = 578.

Finally, simplifying gives us 4x² - 12x + 9 = 289.

Answer 3

Area = 578 = Lw = 2w(w) = 2w^2

w^2 = 578/2 = 289

w = sqr289 = 17 ft

L = 34 ft

34 x 17 = 578 ft^2

not sure why they through in w=2x-3, except to route you into a more complicated method to find the same solution

w=2x-3

x =w/2 +3/2 = 17/2+3/2 = 10 ft

578 = Lw = 2w^2 = 2(2x-3) = 2(4x^2 -12x +9)

4x^2 -12x +9 = 578/2 = 289

x^2 -3x = (289-9)/4 = 280/4 = 70

complete the square

x^2 -3x +9/4 = 70 + 9/4

x-3/2 = sqr(289/4) = 17/2

x = 3/2 + 17/2 = 20/2 = 10

w= 2x-3 = 2(10)-3 = 20-3 = 17