SOLUTION: Nancy traveled 6 miles upstream to do some fly fishing. It took her 20 minutes longer to get there than to return. If the current in the river is 2 miles per hour, then how fast wi

Question 433314: Nancy traveled 6 miles upstream to do some fly fishing. It took her 20 minutes longer to get there than to return. If the current in the river is 2 miles per hour, then how fast will her boat go in still water.
Found 2 solutions by stanbon, jorel1380:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Nancy traveled 6 miles upstream to do some fly fishing. It took her 20 minutes longer to get there than to return. If the current in the river is 2 miles per hour, then how fast will her boat go in still water.
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Upstream DATA:
distance = 6 miles ; rate = b-2 mph; time = 6/(b-2) hrs
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Downstream DATA:
distance = 6 miles; rate = b+2 mph; time = 6/(b+2) mph
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Equation:
time up - time down = 1/3 hr
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6/(b-2) - 6/(b+2) = 1/3
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Multiply thru by 3(b^2-9) to get:
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18(b+2)-18(b-2) = b^2-9
----
72 = b^2-9
------
b^2= 81
b = 9 mph (speed of the boat in still water)
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Cheers,
Stan H.
Answer by jorel1380(3719) (Show Source): You can put this solution on YOUR website!
To find Nancy's rate of travel, we have to solve for:
6/x-2=6/(x+2)+1/3hr.
Multiply both sides by 3(x-2)(x+2), we get:
18(x+2)=18(x-2)+x2-4
18x+36=18x-36+x2-4
0=x2-76
76=x2
√76=x
√4√19=x
2√19=x
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Nancy's boat goes about 8.72 mph in still water.

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=304 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 8.71779788708135, -8.71779788708135. Here's your graph: