SOLUTION: This question I come up with the wrong answer: Pete drove from Buffalo to Boston, a distance of 720km. On return trip he increased his speed by 10km/h. If the total trip took 17h

Question 214899: This question I come up with the wrong answer:
Pete drove from Buffalo to Boston, a distance of 720km. On return trip he increased his speed by 10km/h. If the total trip took 17h, what was his speed from Boston to Buffalo?
I'm having trouble working it out, I begin with this:
I start with (720/x)+(720/x+10)
Expanded it to: 17x^2 - 1270x + 7200
Speed I got next using the quadratic equation: 62.5
But thats apparently wrong, what am I doing wrong?
Found 3 solutions by Alan3354, stanbon, josmiceli:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Pete drove from Buffalo to Boston, a distance of 720km. On return trip he increased his speed by 10km/h. If the total trip took 17h, what was his speed from Boston to Buffalo?
----------------
(720/r) + 720/(r+10) = 17
720r + 720(r+10) = 17r*(r+10)
1440r + 7200 = 17r^2 + 170r
17r^2 - 1270r - 7200 = 0

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

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=2102500 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 80, -5.29411764705882. Here's your graph:

r = 80 km/hr Boston to Buffalo
r = 90 km/hr return

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
Pete drove from Buffalo to Boston, a distance of 720km. On return trip he increased his speed by 10km/h. If the total trip took 17h, what was his speed from Boston to Buffalo?
I'm having trouble working it out, I begin with this:
I start with (720/x)+(720/x+10)
Expanded it to: 17x^2 - 1270x + 7200
Speed I got next using the quadratic equation: 62.5
But thats apparently wrong, what am I doing wrong?
---------------------------------------------------
Buff to Bost DATA:
distance = 720 km ; rate = x km/h ; time = d/r = 720/x hrs
----------------------------------------------------------
Bost to Buff DATA:
distance = 720 km ; rate = x+10 km/h ; time = 720/(x+10) hrs
--------------------------------------------
Equation:
time + time = 17 hrs.
(720/x) + (720/(x+10)) = 17
Multiply thru by x(x+10) to get:
720(x+10) + 720(x) = 17x(x+10)
720x + 7200 + 720x = 17x^2 + 170x
17x^2 - 1270x - 7200 = 0
(x-80)(17x + 90) = 0
Positive solution:
x+ 10 = 90 km/h (rate from Boston to Buffalo)
================================================
Cheers,
Stan H.
Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
I write 2 equations, 1 for each
leg of the trip.
Buffalo to Boston:

given:
km
km
km/hr
Hr
-----------------------
I can rewrite the equations

and

By substitution:

Multiply both sides by

Using quadratic equation:

km/hr
His speed from Buffalo to Boston was 80 km/hr
check:
km/hr
km/hr

hr

and

OK
It looks like you had a sign wrong

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