SOLUTION: A 150 mile trip is driven at R miles per hour. The same trip would take 2 hours less if you increase the speed by 20 mph. What is R? I have tried R=150/t R=150/t-2 (150/t+2=150/

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Question 28932: A 150 mile trip is driven at R miles per hour. The same trip would take 2 hours less if you increase the speed by 20 mph. What is R? I have tried R=150/t
R=150/t-2 (150/t+2=150/t-2).t(t-1)= 150t-150t+2t2-2t= t2-t-150=0 = (t+2)(t+75)and that is as far as I get because I know that that is not right.
Thank you for your help. These always put me in a tail spin. Cheryl Vogel
Found 2 solutions by Paul, stanbon:
Answer by Paul(988) About Me  (Show Source):
You can put this solution on YOUR website!
let the normal speed be
If increased by 20 the speed will be x+20
Subtract the increase speed by the normal speed respect to the distance = time.

Equation:
150%2Fx-150%2F%28x%2B20%29=2
150%28%28x%29-%28x%2B20%29%29=2%28%28x%29%28x%2B20%29%29
150%2820%29=2%28x%5E2%2B20x%29
3000=2x%5E2%2B40x
2x%5E2%2B40x-3000
x%5E2%2B20x-1500=0
a=1, b=20, c=-1500

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x=%28-20%2B-sqrt%2820%5E2-4%2A1%2A-1500%29%29%2F%282%2A1%29
Simplfy that and you get 2 solutions:

x=-50 and x=30

Remove the negative

Hence, the normal speed (R) is 30mph.
Paul.


Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
original data: 150=Rt. So, t=150/R
Modified time and rate result in the following:
150=(R+20)(t-2)
Substitue for "t" to get:
150=(R+20)[(150/R)-2]
150=150-2R+3000/R-40
0=-2R^2+3000-40R
0=2R^2+40R-3000
0=R^2+20R-1500
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aR%5E2%2BbR%2Bc=0 (in our case 1R%5E2%2B20R%2B-1500+=+0) has the following solutons:

R%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2820%29%5E2-4%2A1%2A-1500=6400.

Discriminant d=6400 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-20%2B-sqrt%28+6400+%29%29%2F2%5Ca.

R%5B1%5D+=+%28-%2820%29%2Bsqrt%28+6400+%29%29%2F2%5C1+=+30
R%5B2%5D+=+%28-%2820%29-sqrt%28+6400+%29%29%2F2%5C1+=+-50

Quadratic expression 1R%5E2%2B20R%2B-1500 can be factored:
1R%5E2%2B20R%2B-1500+=+1%28R-30%29%2A%28R--50%29
Again, the answer is: 30, -50. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B20%2Ax%2B-1500+%29

Hope this helps.
Cheers,
Stan H.