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

Transcript

Ex 9.5, 2 Use the identity (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ to find the following products. (i) (π‘₯+3) (π‘₯+7) (π‘₯+3) (π‘₯+7) (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ Putting π‘Ž = 3 & 𝑏 = 7 = π‘₯^2+(3+7)π‘₯+(3Γ—7) = 𝒙^𝟐+πŸπŸŽπ’™+𝟐𝟏 Ex 9.5, 2 Use the identity (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ to find the following products. (ii) (4π‘₯+5)(4π‘₯+1) (4π‘₯+5)(4π‘₯+1) (𝑦+π‘Ž)(𝑦+𝑏)=𝑦^2+(π‘Ž+𝑏)𝑦+π‘Žπ‘ Putting 𝑦 = 4π‘₯ , π‘Ž = 5 & 𝑏 = 1 = (4π‘₯)^2+(5+1)(4π‘₯)+(5)(1) = (4^2Γ—π‘₯^2 )+(6Γ—4)π‘₯+5 = πŸπŸ”π’™^𝟐+πŸπŸ’π’™+πŸ“ Ex 9.5, 2 Use the identity (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ to find the following products. (iii) (4π‘₯βˆ’5)(4π‘₯βˆ’1) (4π‘₯βˆ’5)(4π‘₯βˆ’1) = [4π‘₯+(βˆ’5)][4π‘₯+(βˆ’1)] (𝑦+π‘Ž)(𝑦+𝑏)=𝑦^2+(π‘Ž+𝑏)𝑦+π‘Žπ‘ Putting 𝑦 = 4π‘₯ , π‘Ž = βˆ’5 & 𝑏 = βˆ’1 = (4π‘₯)^2+(βˆ’5βˆ’1)(4π‘₯)+(βˆ’5)(βˆ’1) = (4^2Γ—π‘₯^2 )+(βˆ’6)(4π‘₯)+5 = πŸπŸ”π’™^πŸβˆ’πŸπŸ’π’™+πŸ“ Ex 9.5, 2 Use the identity (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ to find the following products. (iv) (4π‘₯+5)(4π‘₯βˆ’1) (4π‘₯+5)(4π‘₯βˆ’1) = (4π‘₯+5) (4π‘₯+(βˆ’1)) (𝑦+π‘Ž)(𝑦+𝑏)=𝑦^2+(π‘Ž+𝑏)𝑦+π‘Žπ‘ Putting 𝑦 = 4π‘₯ , π‘Ž = 5 & 𝑏 = βˆ’1 (𝑦+π‘Ž)(𝑦+𝑏)=𝑦^2+(π‘Ž+𝑏)𝑦+π‘Žπ‘ Putting 𝑦 = 4π‘₯ , π‘Ž = 5 & 𝑏 = βˆ’1 Ex 9.5, 2 Use the identity (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ to find the following products. (v) (2π‘₯+5𝑦)(2π‘₯+3𝑦) (2π‘₯+5𝑦)(2π‘₯+3𝑦) (𝑑+π‘Ž)(𝑑+𝑏)=𝑑^2+(π‘Ž+𝑏)𝑑+π‘Žπ‘ Putting 𝑑 = 2π‘₯ , π‘Ž = 5𝑦 & 𝑏 = 3𝑦 (𝑑+π‘Ž)(𝑑+𝑏)=𝑑^2+(π‘Ž+𝑏)𝑑+π‘Žπ‘ Putting 𝑑 = 2π‘₯ , π‘Ž = 5𝑦 & 𝑏 = 3𝑦 (As 𝑦π‘₯=π‘₯𝑦) Ex 9.5, 2 Use the identity (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ to find the following products. (vi) (2π‘Ž^2+9) (2π‘Ž^2+5) (2π‘Ž^2+9) (2π‘Ž^2+5) (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ Putting π‘₯ = 2π‘Ž^2 , π‘Ž = 9 & 𝑏 = 5 = (2π‘Ž^2 )^2+(9+5)(2π‘Ž^2 )+(9)(5) = (2^2Γ—π‘Ž^(2 Γ— 2) )+(14Γ—2) π‘Ž^2+45 = πŸ’π’‚^πŸ’+πŸπŸ–π’‚^𝟐+πŸ’πŸ“ Ex 9.5, 2 Use the identity (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ to find the following products. (vii) (π‘₯π‘¦π‘§βˆ’4) (π‘₯π‘¦π‘§βˆ’2) (π‘₯π‘¦π‘§βˆ’4) (π‘₯π‘¦π‘§βˆ’2) = (π‘₯𝑦𝑧+(βˆ’4)) (π‘₯𝑦𝑧+(βˆ’2)) (𝑑+π‘Ž)(𝑑+𝑏)=𝑑^2+(π‘Ž+𝑏)𝑑+π‘Žπ‘ Putting 𝑑 = π‘₯𝑦𝑧 , π‘Ž = βˆ’4 & 𝑏 = βˆ’2 = (𝑏)^2+(7)^2βˆ’2(𝑏)(7) = 𝑏^2+49βˆ’(2Γ—7)×𝑏 = 𝒃^𝟐+πŸ’πŸ—βˆ’πŸπŸ’π’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 9 years. He provides courses for Maths and Science at Teachoo.