SOLUTION: 3. 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. i think it is 30mph but I canno

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: 3. 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. i think it is 30mph but I canno      Log On


   

Question 199563: 3. 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.
i think it is 30mph but I cannot figure out the equation
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Remember, the distance-rate-time formula is

d=rt

Since he "traveled 150 miles at a certain speed", this means that the first equation is 150=rt (simply plug in d=150)

The statement "Had gone 20mph faster, the trip would have taken 2 hour less", tells us that the next equation is 150=%28r%2B20%29%28t-2%29 (the distance is the same, but the speed is now 20 mph faster and the time is 2 hours shorter)


The goal is to use this system to solve for "r" (and if we want, "t" also)


150=rt Start with the first equation.


150%2Fr=t Divide both sides by r.


t=150%2Fr Rearrange the equation


150=%28r%2B20%29%28t-2%29 Move onto the second equation.


150=%28r%2B20%29%28150%2Fr-2%29 Plug in t=150%2Fr


150=150-2r%2B3000%2Fr-40 FOIL


150r=150r-2r%5E2%2B3000-40r Multiply EVERY term by the LCD r to clear out the fraction.


0=150r-2r%5E2%2B3000-40r-150r Subtract 600r from both sides.


0=-2r%5E2-40r%2B3000 Combine and rearrange the terms.


Notice we have a quadratic in the form of Ar%5E2%2BBr%2BC where A=-2, B=-40, and C=3000


Let's use the quadratic formula to solve for "r":


r+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


r+=+%28-%28-40%29+%2B-+sqrt%28+%28-40%29%5E2-4%28-2%29%283000%29+%29%29%2F%282%28-2%29%29 Plug in A=-2, B=-40, and C=3000


r+=+%2840+%2B-+sqrt%28+%28-40%29%5E2-4%28-2%29%283000%29+%29%29%2F%282%28-2%29%29 Negate -40 to get 40.


r+=+%2840+%2B-+sqrt%28+1600-4%28-2%29%283000%29+%29%29%2F%282%28-2%29%29 Square -40 to get 1600.


r+=+%2840+%2B-+sqrt%28+1600--24000+%29%29%2F%282%28-2%29%29 Multiply 4%28-2%29%283000%29 to get -24000


r+=+%2840+%2B-+sqrt%28+1600%2B24000+%29%29%2F%282%28-2%29%29 Rewrite sqrt%281600--24000%29 as sqrt%281600%2B24000%29


r+=+%2840+%2B-+sqrt%28+25600+%29%29%2F%282%28-2%29%29 Add 1600 to 24000 to get 25600


r+=+%2840+%2B-+sqrt%28+25600+%29%29%2F%28-4%29 Multiply 2 and -2 to get -4.


r+=+%2840+%2B-+160%29%2F%28-4%29 Take the square root of 25600 to get 160.


r+=+%2840+%2B+160%29%2F%28-4%29 or r+=+%2840+-+160%29%2F%28-4%29 Break up the expression.


r+=+%28200%29%2F%28-4%29 or r+=++%28-120%29%2F%28-4%29 Combine like terms.


r+=+-50 or r+=+30 Simplify.


So the possible solutions are r+=+-50 or r+=+30


Since a negative speed doesn't make any sense, this means that we must ignore r+=+-50


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Answer:

So the only solution is r+=+30 which means that he was traveling 30 mph. So you are correct, good job.