Question 269238
A cement walk of uniform width is built around a 20 foot by 40 foot rectangular pool and the area of the walk is 700 square feet. How wide will the walk be?

Let x be the uniform width of the walk. Then the length and width of the rectangle bounded by the outer edge of the walk are 40+2x and 20+2x respectively. The area of the walk is the total area of the outer rectangle minus the area of the pool or:

(20+2x)*(40+2x) - 20*40 = 700

40*20 + 40x + 80x + 4x^2 - 800 = 700
800 +120x + 4x^2 - 800 = 700
4x^2 + 120x - 700 = 0

Divide both sides by 4:

x^2 + 30x - 175 = 0

Factor the above:

(x+35)*(x-5) = 0 

The solution for x must satisfy either x+35 = 0 and/or x-5 = 0

These solutions are x = -35 and x = 5. Since the width must be positive, the correct value for x must be 5.