SOLUTION: how do you solve a problem like this: One pipe can fill a reservoir 1 hr faster than another pipe. Together they can fill the reservoir in 4 hrs. How long to the nearest tenth of

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Question 311523: how do you solve a problem like this:
One pipe can fill a reservoir 1 hr faster than another pipe. Together they can fill the reservoir in 4 hrs. How long to the nearest tenth of an hour does it take the faster pipe to fill the reservoir by itself?

Answer by nerdybill(7384)   (Show Source): You can put this solution on YOUR website!
One pipe can fill a reservoir 1 hr faster than another pipe. Together they can fill the reservoir in 4 hrs. How long to the nearest tenth of an hour does it take the faster pipe to fill the reservoir by itself?
.
Let x = rate of slower pipe
then
x-1 = rate of faster pipe
.
4(1/x + 1/(x-1)) = 1
4((x-1) + x) = x(x-1)
4(2x-1) = x(x-1)
8x-4 = x^2-x
-4 = x^2-9x
0 = x^2-9x+4
Because we can't factor, we must resort to the quadratic formula. Doing so yields:
x = {0.5, 8.5}
Tossing out the 0.5 (because calculating the faster pipe will produce a negative answer) leaves:
x = 8.5 hrs
.
Time for faster pipe to fill: 8.5-1 = 7.5 hours
.
Details of quadratic follows:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=65 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 8.53112887414927, 0.468871125850725. Here's your graph:


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