In a stream running at 3 km/h, a motorboat goes 12 km upstream and ba…

We know that,

x = [-b ± √(b² – 4ac)]/2a

We have,

In a stream running at 3 km/h, a motorboat goes 12 km upstream and back to the starting point in 60 min

Let the speed of boat in still water be x, then

According to the question,

= 12/(x – 3) + 12/(x + 3) = 1

= 12 [(x + 3 + x – 3)/(x² – 9)] = 1

= 12 × 2x = x2 – 9

= x² – 24x – 9 = 0

Compare on ax² + bx + c = 0

a = 1, b = -24 and c = -9

As we know,

= x = [-b ± √(b² – 4ac)]/2a

= x = [24 ± √(24² – 4 × 1 × (-9)]/2

= x = [24 ± √(576 + 36)]/2

= x = [24 ± √612]/2

= x = (24 ± 6√17)/2

= x = 12 ± 3√17

= x = 3 (4 ± √17)

Speed of boat in still water = 3 (4 + √17)