Question 560125
During the first part of a trip, a canoeist travels 34 miles at a certain speed.
 The canoeist travels 9 miles on the second part of the trip at a speed 5mph slower.
 The total time for the trip is 5hrs. What was the speed on each part of the trip?
:
Let s = the speed on the 1st 34 mi
then
(s-5) = speed on the last 9 mi
:
Write a time equation, time = dist/speed
:
fast time + slow time = 5 hrs
{{{34/s}}} + {{{9/((s-5))}}} = 5
multiply by s(s-5), results:
34(s-5) + 9s = 5s(s-5)
34s - 170 + 9s = 5s^2 - 25s
43s - 170 = 5s^2 - 25s
0 = 5s^2 - 25s - 43s + 170 
A quadratic equation\
5s^2 - 68s + 170 = 0
use the quadratic formula to find s:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this problem x=s; a=5; b=-68; c=170
{{{s = (-(-68) +- sqrt(-68^2-4*5*170 ))/(2*5) }}}
:
{{{x = (68 +- sqrt(4624-3400 ))/10 }}}
:
I'll let you do the rest of the math here
you will get two positive solutions, but only one will make sense
Check it in the original time equation