A pool with a volume of 160 m3 is filled with two pipes. After opening the pipes, the volume of water

A pool with a volume of 160 m3 is filled with two pipes. After opening the pipes, the volume of water in the pool changes according to the law V (t) = 2t ^ 2-14t + 100, where t is the time in hours. How long will the pool be completely filled with water if one pipe is opened?

First, it is necessary to determine how long it will take 2 pipes to fill the pool.

2t ^ 2 + 14t +100 = 160

2t ^ 2 + 14t – 60 = 0

Divide by all 2, we get

t ^ 2 + 7t – 30 = 0

d = b ^ 2-4ac = (7) ^ 2 -4 * 1 * (- 30) = 49 + 120 = 169

x1 = (- 7 + 13) / 2 * 1 = 3

x2 = (-7 -13) / 2 * 1 = -10 (this root does not fit according to the condition of the problem, since the time t> 0)

We found that two pipes fill the pool in 3 hours.

Then one pipe in 2t = 3 * 2 = 6 hours.

Answer: 6 hours.