Ex 9.5, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 7

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Ex 9.5, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 8

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Ex 9.5, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 9

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

Transcript

Ex 9.5, 4 Simplify. (v) (2.5π‘βˆ’1.5π‘ž)^2 βˆ’(1.5π‘βˆ’2.5π‘ž)^2 Solving (𝟐.πŸ“π’‘βˆ’πŸ.πŸ“π’’)^𝟐 (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 2.5𝑝 & 𝑏 = 1.5π‘ž = (2.5𝑝)^2+(1.5π‘ž)^2βˆ’ 2(2.5𝑝)(1.5π‘ž) = (γ€–2.5γ€—^2×𝑝^2 )+(γ€–1.5γ€—^2Γ—π‘ž)^2 βˆ’ (2Γ—2.5Γ—1.5)Γ—(π‘Γ—π‘ž) = (25/10)^2 𝑝^2+(15/10)^2 π‘ž^2 βˆ’ (2Γ—25/10Γ—15/10)π‘π‘ž = 625/100 𝑝^2+225/100 π‘ž^2βˆ’750/100 π‘π‘ž Solving (𝟏.πŸ“π’‘βˆ’πŸ.πŸ“π’’)^𝟐 (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 1.5𝑝 & 𝑏 = 2.5π‘ž = (1.5𝑝)^2+(2.5π‘ž)^2βˆ’2(1.5𝑝)(2.5π‘ž) = (γ€–1.5γ€—^2×𝑝^2 )+(γ€–2.5γ€—^2Γ—π‘ž)^2βˆ’(2Γ—1.5Γ—2.5)Γ—(π‘Γ—π‘ž) = (15/10)^2 𝑝^2+(25/10)^2 π‘ž^2βˆ’(2Γ—15/10Γ—25/10)π‘π‘ž = 225/100 𝑝^2+625/100 π‘ž^2βˆ’750/100 π‘π‘ž ∴ (2.5π‘βˆ’1.5π‘ž)^2 βˆ’(1.5π‘βˆ’2.5π‘ž)^2 = 625/100 𝑝^2+225/100 π‘ž^2βˆ’750/100 π‘π‘žβˆ’(225/100 𝑝^2+625/100 π‘ž^2βˆ’750/100 π‘π‘ž" " ) = 625/100 𝑝^2+225/100 π‘ž^2βˆ’750/100 π‘π‘žβˆ’225/100 𝑝^2βˆ’625/100 π‘ž^2+750/100 π‘π‘ž = (625/100βˆ’225/100) 𝑝^2+(225/100βˆ’625/100) π‘ž^2+(βˆ’750/100 +750/100)π‘π‘ž = 400/100 𝑝^2βˆ’400/100 π‘ž^2 = πŸ’π’‘^πŸβˆ’πŸ’π’’^𝟐

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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.