Another QUADRATIC PROBLEM

i'm having some difficulties with this one

to save fuel on the 240km trip to their cottage,the nakamura family reduce their usual average speed by 20km/hr,this lengthens the journey time by 1 hr. what is the slower average speed?

Average usual speed of nakamura family = x

(x - 20)*(240/(x) + 1) = 240

240 + x - 4800/x - 20 = 240

x - 4800/x = 20

x² - 20x - 4800 = 0

x = (20 +/- √(400 - 4(-4800))/2 = (20 +/- 140)/2 = 10 +/- 70

x = 80

Slower average speed = 60 km/hr

Note that I rejected the negative value of x because it does not make sense to have a negative average speed.

Normal speed ---- x km/h
Time at normal speed = 240/x hours
Reduced speed --- (x - 20) km/h
Time at reduced speed = 240/(x - 20)
Solve:
240/(x - 20) - 240/x = 1

Report Abuse

Distance = Velocity x Time

240 = V * t
240 = (V - 20) * (t + 1)

V * t = V * t - 20t + V - 20
0 = -20t + V - 20
20t + 20 = V

240 = (20t + 20) * t
240 = 20 * (t + 1) * t
12 = t^2 + t
t^2 + t - 12 = 0
(t + 4) * (t - 3) = 0
t = -4 , 3

time cannot be negative, so, t = 3

240 = V * 3
80 = V

The slower speed is therefore 60 km/h

To save fuel on the 240 km trip to their cottage, the Nakamura family reduce their usual average speed by 20 km/hr, this lengthens the journey time by 1 hr.
what is the slower average speed?

Let the slower average be x. Then the higher average is x+20.
Then, 240/(x+20) = (240/x) - 1
Solve and find x.

solve system of equations
T=240*v1
T+1=240*v2
v2=v1-20
where v1 is faster speed and v2 is slower and T is trip duration at v1