Ex 9.5, 2 - Use the identity (x + a) (x + b) = x^2 + (a + b) x + ab to

Ex 9.5, 2 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 2
Ex 9.5, 2 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 3 Ex 9.5, 2 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 4 Ex 9.5, 2 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 5 Ex 9.5, 2 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 6 Ex 9.5, 2 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 7

  1. Chapter 9 Class 8 Algebraic Expressions and Identities
  2. Serial order wise

Transcript

Ex 9.5, 2 Use the identity (๐‘ฅ+๐‘Ž)(๐‘ฅ+๐‘)=๐‘ฅ^2+(๐‘Ž+๐‘)๐‘ฅ+๐‘Ž๐‘ to find the following products. (i) (๐‘ฅ+3) (๐‘ฅ+7) (๐‘ฅ+3) (๐‘ฅ+7) (๐‘ฅ+๐‘Ž)(๐‘ฅ+๐‘)=๐‘ฅ^2+(๐‘Ž+๐‘)๐‘ฅ+๐‘Ž๐‘ Putting ๐‘Ž = 3 & ๐‘ = 7 = ๐‘ฅ^2+(3+7)๐‘ฅ+(3ร—7) = ๐’™^๐Ÿ+๐Ÿ๐ŸŽ๐’™+๐Ÿ๐Ÿ Ex 9.5, 2 Use the identity (๐‘ฅ+๐‘Ž)(๐‘ฅ+๐‘)=๐‘ฅ^2+(๐‘Ž+๐‘)๐‘ฅ+๐‘Ž๐‘ to find the following products. (ii) (4๐‘ฅ+5)(4๐‘ฅ+1) (4๐‘ฅ+5)(4๐‘ฅ+1) (๐‘ฆ+๐‘Ž)(๐‘ฆ+๐‘)=๐‘ฆ^2+(๐‘Ž+๐‘)๐‘ฆ+๐‘Ž๐‘ Putting ๐‘ฆ = 4๐‘ฅ , ๐‘Ž = 5 & ๐‘ = 1 = (4๐‘ฅ)^2+(5+1)(4๐‘ฅ)+(5)(1) = (4^2ร—๐‘ฅ^2 )+(6ร—4)๐‘ฅ+5 = ๐Ÿ๐Ÿ”๐’™^๐Ÿ+๐Ÿ๐Ÿ’๐’™+๐Ÿ“ Ex 9.5, 2 Use the identity (๐‘ฅ+๐‘Ž)(๐‘ฅ+๐‘)=๐‘ฅ^2+(๐‘Ž+๐‘)๐‘ฅ+๐‘Ž๐‘ to find the following products. (iii) (4๐‘ฅโˆ’5)(4๐‘ฅโˆ’1) (4๐‘ฅโˆ’5)(4๐‘ฅโˆ’1) = [4๐‘ฅ+(โˆ’5)][4๐‘ฅ+(โˆ’1)] (๐‘ฆ+๐‘Ž)(๐‘ฆ+๐‘)=๐‘ฆ^2+(๐‘Ž+๐‘)๐‘ฆ+๐‘Ž๐‘ Putting ๐‘ฆ = 4๐‘ฅ , ๐‘Ž = โˆ’5 & ๐‘ = โˆ’1 = (4๐‘ฅ)^2+(โˆ’5โˆ’1)(4๐‘ฅ)+(โˆ’5)(โˆ’1) = (4^2ร—๐‘ฅ^2 )+(โˆ’6)(4๐‘ฅ)+5 = ๐Ÿ๐Ÿ”๐’™^๐Ÿโˆ’๐Ÿ๐Ÿ’๐’™+๐Ÿ“ Ex 9.5, 2 Use the identity (๐‘ฅ+๐‘Ž)(๐‘ฅ+๐‘)=๐‘ฅ^2+(๐‘Ž+๐‘)๐‘ฅ+๐‘Ž๐‘ to find the following products. (iv) (4๐‘ฅ+5)(4๐‘ฅโˆ’1) (4๐‘ฅ+5)(4๐‘ฅโˆ’1) = (4๐‘ฅ+5) (4๐‘ฅ+(โˆ’1)) (๐‘ฆ+๐‘Ž)(๐‘ฆ+๐‘)=๐‘ฆ^2+(๐‘Ž+๐‘)๐‘ฆ+๐‘Ž๐‘ Putting ๐‘ฆ = 4๐‘ฅ , ๐‘Ž = 5 & ๐‘ = โˆ’1 (๐‘ฆ+๐‘Ž)(๐‘ฆ+๐‘)=๐‘ฆ^2+(๐‘Ž+๐‘)๐‘ฆ+๐‘Ž๐‘ Putting ๐‘ฆ = 4๐‘ฅ , ๐‘Ž = 5 & ๐‘ = โˆ’1 Ex 9.5, 2 Use the identity (๐‘ฅ+๐‘Ž)(๐‘ฅ+๐‘)=๐‘ฅ^2+(๐‘Ž+๐‘)๐‘ฅ+๐‘Ž๐‘ to find the following products. (v) (2๐‘ฅ+5๐‘ฆ)(2๐‘ฅ+3๐‘ฆ) (2๐‘ฅ+5๐‘ฆ)(2๐‘ฅ+3๐‘ฆ) (๐‘ก+๐‘Ž)(๐‘ก+๐‘)=๐‘ก^2+(๐‘Ž+๐‘)๐‘ก+๐‘Ž๐‘ Putting ๐‘ก = 2๐‘ฅ , ๐‘Ž = 5๐‘ฆ & ๐‘ = 3๐‘ฆ (๐‘ก+๐‘Ž)(๐‘ก+๐‘)=๐‘ก^2+(๐‘Ž+๐‘)๐‘ก+๐‘Ž๐‘ Putting ๐‘ก = 2๐‘ฅ , ๐‘Ž = 5๐‘ฆ & ๐‘ = 3๐‘ฆ (As ๐‘ฆ๐‘ฅ=๐‘ฅ๐‘ฆ) Ex 9.5, 2 Use the identity (๐‘ฅ+๐‘Ž)(๐‘ฅ+๐‘)=๐‘ฅ^2+(๐‘Ž+๐‘)๐‘ฅ+๐‘Ž๐‘ to find the following products. (vi) (2๐‘Ž^2+9) (2๐‘Ž^2+5) (2๐‘Ž^2+9) (2๐‘Ž^2+5) (๐‘ฅ+๐‘Ž)(๐‘ฅ+๐‘)=๐‘ฅ^2+(๐‘Ž+๐‘)๐‘ฅ+๐‘Ž๐‘ Putting ๐‘ฅ = 2๐‘Ž^2 , ๐‘Ž = 9 & ๐‘ = 5 = (2๐‘Ž^2 )^2+(9+5)(2๐‘Ž^2 )+(9)(5) = (2^2ร—๐‘Ž^(2 ร— 2) )+(14ร—2) ๐‘Ž^2+45 = ๐Ÿ’๐’‚^๐Ÿ’+๐Ÿ๐Ÿ–๐’‚^๐Ÿ+๐Ÿ’๐Ÿ“ Ex 9.5, 2 Use the identity (๐‘ฅ+๐‘Ž)(๐‘ฅ+๐‘)=๐‘ฅ^2+(๐‘Ž+๐‘)๐‘ฅ+๐‘Ž๐‘ to find the following products. (vii) (๐‘ฅ๐‘ฆ๐‘งโˆ’4) (๐‘ฅ๐‘ฆ๐‘งโˆ’2) (๐‘ฅ๐‘ฆ๐‘งโˆ’4) (๐‘ฅ๐‘ฆ๐‘งโˆ’2) = (๐‘ฅ๐‘ฆ๐‘ง+(โˆ’4)) (๐‘ฅ๐‘ฆ๐‘ง+(โˆ’2)) (๐‘ก+๐‘Ž)(๐‘ก+๐‘)=๐‘ก^2+(๐‘Ž+๐‘)๐‘ก+๐‘Ž๐‘ Putting ๐‘ก = ๐‘ฅ๐‘ฆ๐‘ง , ๐‘Ž = โˆ’4 & ๐‘ = โˆ’2 = (๐‘)^2+(7)^2โˆ’2(๐‘)(7) = ๐‘^2+49โˆ’(2ร—7)ร—๐‘ = ๐’ƒ^๐Ÿ+๐Ÿ’๐Ÿ—โˆ’๐Ÿ๐Ÿ’๐’ƒ

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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.