Math question (quadratic equation)

• Pfft

  11 May 09, 20:45

  A girl and a boy started together at the same constant speed of 2x km/h for a 14km walk. After 1 km,  the boy increased his speed by 1 km/h and walked the remaining distance at this new speed while the girl walked at a constant speed of 2x km/h all the way.

  Given that the boy completed the journey 30 mins earlier than the girl, Form an equation in terms of x and show that it reduces to 2x^2 + x - 13 = 0

• nightzip

  11 May 09, 21:21

  Distance = Speed x Time So, Let time taken by girl as Y. 14-1=(2x)Y => 13=2xy For boy, 14-1=(2x+1)(Y-0.5) => 13=2xy-x+y-0.5 (by expanding above terms) Substituting 2xy=13 as given above first equation. => 13=13-x+y-0.5 =>x-y=-0.5 Now, from 2xy=13 => y=13/(2x) Therefore, x-13/(2x)=-0.5 => (2x^2-13)=-0.5(2x) => 2x^2+x-13=0 Done.