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Last updated at May 29, 2018 by Teachoo

Transcript

Ex 4.3 ,8 A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train. Let the speed of train be x km/hr From (1) (x + 5) (360/ " 1" ) = 360 (x + 5) ((360 )/ ) = 360 (x + 5) (360 x ) = 360x x(360 x) + 5(360 x ) = 360 x 360x x2 +5(360) 5x = 360 x 360x x2 + 1800 5x = 360 x 360x x2 + 1800 5x 360 x = 0 x2 5x 360 x + 360x + 1800 = 0 x2 5x + 1800 = 0 0 = x2 + 5x 1800 x2 + 5x 1800 = 0 We factorize by splitting the middle term method x2 + 45x 40x 1800 = 0 x (x + 45) 40 (x + 45) = 0 (x + 45) (x 40) = 0 Hence x = 45, x = 40 are the roots of the equation We know that Speed of train = x So, x cannot be negative x = 40 is the solution So, Speed of train = x = 40 km/hr

Ex 4.3

Ex 4.3, 1 (i)
Deleted for CBSE Board 2022 Exams

Ex 4.3, 1 (ii) Important Deleted for CBSE Board 2022 Exams

Ex 4.3, 1 (iii) Deleted for CBSE Board 2022 Exams

Ex 4.3, 1 (iv) Important Deleted for CBSE Board 2022 Exams

Ex 4.3, 2 (i)

Ex 4.3, 2 (ii)

Ex 4.3, 2 (iii)

Ex 4.3, 2 (iv) Important

Ex 4.3, 3 (i) Important

Ex 4.3, 3 (ii)

Ex 4.3, 4

Ex 4.3, 5

Ex 4.3, 6

Ex 4.3, 7 Important

Ex 4.3, 8 Important You are here

Ex 4.3, 9 Important

Ex 4.3, 10 Important

Ex 4.3, 11

Chapter 4 Class 10 Quadratic Equations (Term 2)

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.