Last updated at Dec. 8, 2020 by Teachoo

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Ex 4.3 ,1 Find the roots of the following quadratic equations, if they exist, by the method of completing the square: (ii) 2x2 + x โ 4 = 0 2x2 + x โ 4 = 0 Dividing whole equation by 2 (2๐ฅ2 + ๐ฅ โ 4)/2=0/2 2๐ฅ2/2+๐ฅ/2โ4/2=0 x2 + ๐ฅ/2โ2=0 We know that (a + b)2 = a2 + 2ab + b2 Here, a = x & 2ab = ๐ฅ/2 2xb = ๐ฅ/2 2b = 1/2 b = 1/2 ร 1/2 b = 1/4 Now, in our equation x2 + ๐ฅ/2โ2=0 Adding and subtracting (1/4)^2 x2 + ๐ฅ/2โ2+(1/4)^2โ(1/4)^2= 0 x2 + ๐ฅ/2+(1/4)^2โ 2 โ (1/4)^2=0 (๐ฅ+1/4)^2โ2โ(1/4)^2= 0 (๐ฅ+1/4)^2โ2 โ1/16=0 (๐ฅ+1/4)^2=2+1/16 (๐ฅ+1/4)^2=(2(16) + 1)/16 (๐ฅ+1/4)^2=(32 + 1)/16 (๐ฅ+1/4)^2=33/16 (๐ฅ+1/4)^2=(โ33/4)^2 Cancelling square both sides ๐ฅ+1/4 = ยฑ โ33/4 Solving So, the root of the equation are x = (โ33 โ 1)/4 & x = (โ(โ33 + 1))/4

Ex 4.3

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

Ex 4.3, 1 (ii) Deleted for CBSE Board 2022 Exams You are here

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

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

Ex 4.3, 2

Ex 4.3, 3

Ex 4.3, 4

Ex 4.3, 5

Ex 4.3, 6

Ex 4.3, 7

Ex 4.3, 8 Important

Ex 4.3, 9 Important

Ex 4.3, 10

Ex 4.3, 11

Chapter 4 Class 10 Quadratic Equations (Term 2)

Serial order wise

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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.