math home work help?

Question

the length of a rectangle is 4 more than twice the width. the area of the rectangle is 52.08 square feet what are the dimensions

Answer 1

let w = width of rectangle
2w+4 = length

w(2w+4) = 52.08

2w^2 + 4w - 52.08 = 0

w^2 + 2w - 26.04 = 0

From here, you can use the quadratic formula to find w.

:)

Answer 2

Marley K. is correct. When one uses the quadratic formula on these figures, one obtains this:

w = (-2 ± √[(-2)² - (4)(1)(-26.04)]) / 2
w = (-2 ± √(4 + 104.16)) / 2
w = (-2 ± √108.16) / 2
w = (-2 ± √(4)(27.04)) / 2
w = (-2 ± 2 √27.04) / 2

Now we can divide both the top and bottom by 2:
w = -1 ± √27.04

Since a physical length can only be a positive number, then we can discard the negative value of the square root of 27.04.

Then w = -1 + √27.04 = -1 + 5.2 = 4.2.

Since l (length) = 2w + 4, then l = 2(4.2) + 4 = 8.4 + 4 = 12.4.

Then the area of the rectangle, A, is (12.4)(4.2) = 52.08, which is the area you were given.

So, the length of the rectangle is 12.4 ft and the width is 4.2 ft.

Answer 3

ANSWERS:
width = x =16.02
length = 2x+4 =36.05

WORK:
x+2x+4=52.08
3x+4=52.08
3x=48.08
x~16.02

Answer 4

I am assuming it is a right triangle
then
L=2w+4
area= 0.5*(w)(2w+4)=26.04
its a quadratic equation
w^2+2w-26.04=0
w=4.2 & -6.2 (negative width doesn't make sense)
w=4.2
L=2*4.2+4=12.4
Hypoteneus=(12.4^2+4.2^2)^0.5=13.09