Math question (quadratic equation)

Pfft SG

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 SG

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.