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

Transcript

Example 12 Using Identity (II), find (i) (4π‘βˆ’3π‘ž)^2 (4π‘βˆ’3π‘ž)^2 (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 4𝑝 & 𝑏 = 3π‘ž = (4𝑝)^2+(3π‘ž)^2βˆ’2(4𝑝)(3π‘ž) = (4^2×𝑝^2 )+(3^2Γ—π‘ž^2 )βˆ’(2Γ—4Γ—3)Γ—(π‘Γ—π‘ž) = πŸπŸ”π’‘^𝟐+πŸ—π’’^πŸβˆ’πŸπŸ’π’‘π’’ Example 12 Using Identity (II), find (ii) (4.9)^2 (4.9)^2 = (5βˆ’0.1)^2 (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 5 & 𝑏 = 0.1 = (5)^2+(0.1)^2βˆ’2(5)(0.1) = 25+(1/10)^2βˆ’(2Γ—5Γ—1/10) = 25+1^2/γ€–10γ€—^2 βˆ’(10/10) = 25+1/100βˆ’1 = 24+1/100 = (24 Γ— 100 + 1)/100 = (2400 + 1)/100 = 2401/100 = πŸπŸ’.𝟎𝟏

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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.