Question 814153: Andres paddled a canoe 11 miles upstream in a river that has a current of 4 mph. He then turned around and paddled downstream until he reached his original starting place. If the entire trip took him 2 hours, how fast would Andres paddle in still water?
answer in MPH
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! s = d / t
t = d / s
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w = 4
11/(c + w) + 11/(c - w) = 2
11/(c + 4) + 11/(c - 4) = 2
11(c - 4)/(c + 4)(c - 4) + 11(c + 4)/(c - 4)(c + 4) = 2
11(c - 4)/(cc - 16) + 11(c + 4)/(cc - 16) = 2
11(c - 4) + 11(c + 4) = 2(cc - 16)
11c - 44 + 11c + 44 = 2cc - 32
22c = 2cc - 32
2cc - 22c - 32
1cc - 11c - 16
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the above quadratic equation is in standard form, with a=2, b=-22, and c=-32
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to solve the quadratic equation, plug this:
2 -22 -32
into this: https://sooeet.com/math/quadratic-equation-solver.php
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the roots of the quadratic are:
12.3007353
-1.30073525
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the negative root doesn't make sense as the canoe speed, so use the positive root:
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c = 12.3007353 ~= 12.3 mph
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