SOLUTION: Steve traveled 150 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 2 hours less. Find the speed of his vehicle

Click here to see ALL problems on Quadratic Equations

Question 180502: Steve traveled 150 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 2 hours less. Find the speed of his vehicle
Found 2 solutions by stanbon, ankor@dixie-net.com:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Steve traveled 150 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 2 hours less. Find the speed of his vehicle
---------------------------
Steve DATA:
distance = 150 miles ; rate = x mph ; time = d/r = 150/x hrs
-----------------------------------------------
Improved Steve DATA:
distance = 150 miles : rate = x+20 ph ; time = d/r = 150/(x+2) hrs
---------------------------------
Equations:
Old Steve time - Improved Steve time = 2 hrs
150/x - 150/(x+2) = 2
150(x+2) - 150x = 2x(x+2)
300 = 2(x^2+2x)
x^2 + 2x - 150 = 0
x = [-2 +- sqrt(4 - 4*1*-150)]/2
x = [-2 +- sqrt(604)]/2
Positive solution:
x = 11.29 mph (Old rate)
================
Cheers,
Stan H.

Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
Steve traveled 150 miles at a certain speed. Had he gone 20mph faster, the trip
would have taken 2 hours less. Find the speed of his vehicle
:
Let s = the speed of his vehicle
then
(s+20) = the faster speed
:
Write a time equation: Time =
:
Actual time = faster speed time + two hrs
= + 2
Multiply equation by s(s+20) to eliminate the denominators, results
150(s+20) = 150s + 2s(s+20)
:
150s + 3000 = 150s + 2s^2 + 40s
:
0 = 150s - 150s + 2s^2 + 40s - 3000
:
A quadratic equation
2s^2 + 40s - 3000 = 0
:
Simplify, divide by 2:
s^2 + 20s - 1500
:
Factors to:
(s + 50)(s - 30) = 0
Positive solution
s = 30 mph is his actual speed
;
:
Check solution by finding the times
150/30 = 5 hrs
150/50 = 3 hrs
--------------
difference 2hr