SOLUTION: Paul drove to the city, 200 miles away, and returned home at the speed which was 10 mph faster than on his trip to the city. If the trip to the city was one hour longer than his re

Question 871931: Paul drove to the city, 200 miles away, and returned home at the speed which was 10 mph faster than on his trip to the city. If the trip to the city was one hour longer than his return trip, what was Paul's speed from home to the city?
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
going speed x
returning speed x + 10

Distance = same 200 miles 1
original time � time with increased speed = 1
t=d/t
200 / x - 200 / ( x + 10 ) = 1
LCD= x ( x + 10 )
multiply by LCD
200 ( x + 10 ) - 200 x = 1 x ( x + 10 )
200 x + 2000 - 200 x = 1 X^2 + 10 x
2000 = 1 X^2 + 10 x
1 X^2 + 10 x - -2000 = 0
Find roots of the quadratic equation
a= 1 b= 10 c= -2000

x1= ( -10 + sqrt( 100 - 8000 )) / 2
x1=( -10 + 90 )/ 2
x1= 40

x2= ( -7 - sqrt( 49 - 20 ) / 2
x2=( -10 - 90 )/ 2
x2= -50
ignore negative
going speed = 40 mph
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