# A garden area is 30 ft long and 20 ft wide. A path of uniform width

is set around the edge. If the remaining garden area is 400 ft2, what is the width of
the path?

I did the following:
Let x = distance between the inner rectangle and the outer boundry; i.e., x is the width of the border. Then
length is 20-x-x and the width is 20-x-x so (30-x-x)(20-x-x)=400
Solve for x.
Post your work if you get stuck.

## To find the width of the path, we will use the quadratic formula to solve the equation (30 - 2x)(20 - 2x) = 400.

1. Start by rewriting the equation as a quadratic equation:
(30 - 2x)(20 - 2x) - 400 = 0

2. Expand the equation:
(30 - 2x)(20 - 2x) = 400
600 - 100x - 40x + 4x^2 = 400
4x^2 - 140x + 200 = 0

x = (-b ± √(b^2 - 4ac)) / 2a

In this equation, a = 4, b = -140, and c = 200.

4. Plug in the values into the quadratic formula and solve:
x = (-(-140) ± √((-140)^2 - 4 * 4 * 200)) / (2 * 4)
x = (140 ± √(19600 - 3200)) / 8
x = (140 ± √16400) / 8
x = (140 ± 128.06) / 8

5. Split the equation into two solutions:
x₁ = (140 + 128.06) / 8 = 268.06 / 8 = 33.51
x₂ = (140 - 128.06) / 8 = 11.94 / 8 = 1.49375

6. Since the width of the path cannot be negative, we discard x₂ as a solution.
Therefore, the width of the path is approximately 33.51 ft.

So, using the quadratic formula, we find that the width of the path is approximately 33.51 ft.