Last updated at Dec. 16, 2024 by Teachoo
Ex 2.4, 3 Factorise the following using appropriate identities: (ii) 4y2 4y + 1 4y2 4y + 1 = 22 y2 4y + 12 = (2y)2 4y + 12 = (2y)2 2 (2y) (1) + (1)2 Using Identity (a b)2 = a2 + b2 2ab Where a = 2y , b = 1 = (2y 1)2 = (2y 1) (2y 1)
Ex 2.4
Ex 2.4, 1 (ii)
Ex 2.4, 1 (iii) Important
Ex 2.4, 1 (iv)
Ex 2.4, 1 (v)
Ex 2.4, 2 (i)
Ex 2.4, 2 (ii) Important
Ex 2.4, 2 (iii)
Ex 2.4, 3 (i)
Ex 2.4, 3 (ii) Important You are here
Ex 2.4, 3 (iii)
Ex 2.4, 4 (i)
Ex 2.4, 4 (ii)
Ex 2.4, 4 (iii) Important
Ex 2.4, 4 (iv)
Ex 2.4, 4 (v)
Ex 2.4, 4 (vi)
Ex 2.4, 5 (i)
Ex 2.4, 5 (ii) Important
Ex 2.4, 6 (i)
Ex 2.4, 6 (ii)
Ex 2.4, 6 (iii)
Ex 2.4, 6 (iv) Important
Ex 2.4, 7 (i)
Ex 2.4, 7 (ii)
Ex 2.4, 7 (iii) Important
Ex 2.4, 8 (i)
Ex 2.4, 8 (ii)
Ex 2.4, 8 (iii) Important
Ex 2.4, 8 (iv) Important
Ex 2.4, 8 (v)
Ex 2.4, 9 (i)
Ex 2.4, 9 (ii)
Ex 2.4, 10 (i) Important
Ex 2.4, 10 (ii)
Ex 2.4, 11
Ex 2.4,12 Important
Ex 2.4,13
Ex 2.4, 14 (i)
Ex 2.4, 14 (ii) Important
Ex 2.4, 15 (i)
Ex 2.4, 15 (ii) Important
Ex 2.4, 16 (i)
Ex 2.4, 16 (ii) Important
About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo