Example 15 - A motor boat whose speed is 18 km/h in still - Solving by quadratic formula - Equation to be formed

Example 15 - Chapter 4 Class 10 Quadratic Equations - Part 2
Example 15 - Chapter 4 Class 10 Quadratic Equations - Part 3 Example 15 - Chapter 4 Class 10 Quadratic Equations - Part 4 Example 15 - Chapter 4 Class 10 Quadratic Equations - Part 5

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Question 8 A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. Given that speed of the boat = 18 km/ hr. Let the speed of the stream = x km / hr. Given that Time taken upstream is 1 hour more than time taken downstream Time upstream = Time downstream + 1 24/((18 โˆ’ ๐‘ฅ)) = 24/((18 + ๐‘ฅ)) + 1 24/((18 โˆ’ ๐‘ฅ)) โ€“ 24/((18 + ๐‘ฅ)) = 1 (24(18 + ๐‘ฅ) โˆ’ 24(18 โˆ’ ๐‘ฅ))/((18 โˆ’ ๐‘ฅ)(18 + ๐‘ฅ)) = 1 24((18 + ๐‘ฅ) โˆ’ (18 โˆ’ ๐‘ฅ))/((18 โˆ’ ๐‘ฅ)(18 + ๐‘ฅ)) = 1 24(18 + ๐‘ฅ โˆ’ 18 + ๐‘ฅ)/((18 โˆ’ ๐‘ฅ)(18 + ๐‘ฅ)) = 1 24(2๐‘ฅ)/((18 โˆ’ ๐‘ฅ)(18 + ๐‘ฅ)) = 1 48๐‘ฅ/((18 โˆ’ ๐‘ฅ)(18 + ๐‘ฅ)) = 1 48x = (18 โ€“ x) (18 + x) 48x = 182 โ€“ x2 48x = 324 โ€“ x2 x2 + 48x โ€“ 324 = 0 Comparing equation with ax2 + bx + c = 0, Here a = 1, b = 48, c = โ€“324 We know that D = b2 โ€“ 4ac D = (48)2 โ€“ 4 ร— 1 ร— (โ€“324) D = 2304 + 4 ร— 324 D = 2304 + 1296 D = 3600 So, the roots to equation are x = (โˆ’๐‘ ยฑ โˆš๐ท)/2๐‘Ž Putting values x = (โˆ’(48) ยฑ โˆš3600)/(2 ร— 1) x = (โˆ’ 48 ยฑ โˆš(60 ร— 60))/(2 ร— 1) x = (โˆ’ 48 ยฑ 60)/2 Solving So, x = 6 & x = โ€“ 54 Since, x is the speed , so it cannot be negative So, x = 6 is the solution of the equation Therefore, speed of the stream (x) = 6 km /hr.

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