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Ex 9.5, 4 (v) - Chapter 9 Class 8 Algebraic Expressions and Identities

Last updated at March 22, 2023 by

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

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

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.