Question 1042685: Can somebody walk me through a step on this quadratic word problem?
A small motorboat travels 13 mph in still water. It takes 2 hours longer to travel 60 miles going upstream than it does going downstream. Find the rate of the current.
I am at a point where I have:
60/13-x - 60/13+x =2
I know that I need to multiply both sides of the equation by the LCD and that it will turn into:
2x^2+20x-338=0
But I just cant seem to figure out HOW to actually do that. I think the denominators of 13-x/13+x are really throwing me.
Or if somebody has an easier way to get to the quadratic equation in this word problem I am all ears.
Thanks so much in advance!
Found 4 solutions by ikleyn, Boreal, ankor@dixie-net.com, stanbon: Answer by ikleyn(52756) (Show Source): Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! The speed upstream is 13-c, where c is the current.
The time it takes is 60/(13-c), as you have with units mi/mi/hr, with units hr.
The time it takes to go downstream is 60/(13+c), and that is 2 hours less than upstream
Therefore, 60/(13-c)-2=60/(13+c)
Your LCD is (13-c)(13+c)
60(13+c)-2(169-c^2)=60(13-c)
780 +60c-338+2c^2=780-60c
780 cancels,
2c^2+120c-338=0; Notice, this is 120c, not 20c.
divide by 2
c^2+60c-169=0
c=(1/2)(-60+/-sqrt (3600+676) sqrt (4276)=65.39
use positive root (1/2)5.39=2.695 mph
Upstream speed is 10.305 mph
Downstream speed is 15.695 mph
60/10.305=5.822 hours
60/15.695=3.822 hours
Answer by ankor@dixie-net.com(22740) (Show Source): Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! A small motorboat travels 13 mph in still water. It takes 2 hours longer to travel 60 miles going upstream than it does going downstream. Find the rate of the current.
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I am at a point where I have:
60/(13-x) - 60/(13+x) =2
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60(13+x) - 60(13-x) = 2(169-x^2)
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120x = 2*169 - 2x^2
2x^2+120x-2*169 = 0
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x^2 + 60x - 169 = 0
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x = [-60 +- sqrt(3600-4*-169)]/(2)
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Positive solution::
x = 2.32 mph
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Cheers,
Stan H.
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