View Full Version : Algebra Quadratic wordproblem
sean080
Dec 13, 2007, 12:09 AM
Hi, I'm trying to figure this out but always have trouble with problems like this, here it is:
Jennifer traveled at a constant speed on an old road for 160 miles. She then traveled 5 miles per hour faster on a newer road for 90 miles. If she drove for 6 hours, find the car's speed for the part of the trip that was traveled on the new road.
jiten55
Dec 13, 2007, 03:56 AM
Let speed on new road = x
Original speed = x - 5
Total time =
160/(x - 5) + 90/x = 6
Solve this equation to get value of x.
asterisk_man
Dec 13, 2007, 07:00 AM
jiten55 is essentially correct. One extra thing to be aware of is that since this will result in a quadratic equation you'll end up with two mathematic solutions for x. However, only one of them will make sense with your problem, you should be able to figure out which one that is and discard the other.
Shy But Nice
Dec 21, 2007, 07:40 AM
so u feel x = 5.52631578947368
jiten55
Dec 21, 2007, 08:27 AM
Answer is 45.
Reason:
The quadraticd equation reduces to:
3 x^2 - 140 x + 225 = 0
It has roots: 45 and 10/3
Only 45 is is valid, because (x -5 ) should not be negative.
New Road: 45, Old Road = 40
galactus
Dec 21, 2007, 08:31 AM
so u feel x = 5.52631578947368
No, x is a nice integer solution. I will use r for rate.
\frac{160}{r}+\frac{90}{r+5}=6
Multiply by the GCF, which is r(r+5):
This results in a quadratic 3r^{2}-110r-400=0
Factor: (r-40)(3r+10)=0
r=40 \;\ and \;\ \frac{-10}{3}
Which concurs with Jiten's post. Excellent.
Of course, ignore the negative solution.