SOLUTION: On this word problem, i guess i need to turn this into a quadratic expression and solve, but i only thought that you solved word problems from linear equations. i have no idea how

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: On this word problem, i guess i need to turn this into a quadratic expression and solve, but i only thought that you solved word problems from linear equations. i have no idea how       Log On


   

Question 144570: On this word problem, i guess i need to turn this into a quadratic expression and solve, but i only thought that you solved word problems from linear equations. i have no idea how this is even done. i would try it on my own if i knew where to begin. If someone can get me started i will glady finish it so i can learn. Thank you.
Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
d=rt Start with the distance-rate-time formula


200=rt Plug in d=200


200%2Fr=t Solve for t



Now if "he gone 10mph faster, the trip would have taken 1 hour less", this would make the equation: 200=%28r%2B10%29%28t-1%29


200=%28r%2B10%29%28200%2Fr-1%29 Plug in t=200%2Fr


200=190-r%2B2000%2Fr Foil


200r=190r-r%5E2%2B2000 Multiply every term by the LCD "r"


0=190r-r%5E2%2B2000-200r Subtract 200r from both sides


0=-r%5E2-10r%2B2000 Combine like terms



0=-%28r%2B50%29%28r-40%29 Factor the right side


Now set each factor equal to zero:
r%2B50=0 or r-40=0

r=-50 or r=40 Now solve for r in each case


Since you cannot travel at -50 mph, this means that our only answer is r=40