SOLUTION: Mr. Martin traveled to a city 150 miles from his home to attend a meeting. Due to car trouble, his average speed returning was 10 mph less than his speed going. If the total time

Question 1008675: Mr. Martin traveled to a city 150 miles from his home to attend a meeting. Due to car trouble, his average speed returning was 10 mph less than his speed going. If the total time for the round trip was 5 hours, at what rate of speed did he travel to the city? (Round your answer to the nearest tenth.)
Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
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R=rate going to other city
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Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=3700 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 65.4138126514911, 4.5861873485089. Here's your graph:

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Since 4.5 does not work (returning was 10 mph less would
result in negative speed), his rate going to the city was
65.4 mph.
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ANSWER: His rate traveling to the city was 65.4 mph
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CHECK:
150 mi/65.4mph+150mi/55.4mph = 5 hours
2.29 hours + 2.71 hours = 5 hours
5 hours = 5 hours
.

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