quadratic word problem?

englsinger_J

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Jan 13, 2006
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A speedboat takes 1 hour longer to go 24 miles up a river than to return. If the boat cruises at 10 miles per hour in still water, what is the rate of the current?

I think I am supposed to use the quadratic equation to solve this, but I am not positve.
 
It takes t hours to go up stream a distance d.
The rivers velocity is v.
Up stream:
d = (10-v)*t
24 = 10t-vt
t = 24/(10-v)

Down stream:
d = (10+v)*(t-1)
24 = 10t+vt-10-v
t = (34+v)/(10+v)

t=t
24/(10-v)=(34+v)/(10+v)
24(10+v)=(34+v)(10-v)
240+24v=340-34v+10v-v²
v²+48v-100 = 0
You could use the quadratic equation but it factors
(v+50)(v-2) = 0
v=-50 or v=2
It isn't flowing backwards so
v=2
 
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