Example 12 - Using Identity (II), find (i) (4p - 3q)^2 (ii) (4.9)^2

Example 12 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 2
Example 12 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 3

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

Transcript

Example 12 Using Identity (II), find (i) (4๐‘โˆ’3๐‘ž)^2 (4๐‘โˆ’3๐‘ž)^2 (๐‘Žโˆ’๐‘)^2=๐‘Ž^2+๐‘^2โˆ’2๐‘Ž๐‘ Putting ๐‘Ž = 4๐‘ & ๐‘ = 3๐‘ž = (4๐‘)^2+(3๐‘ž)^2โˆ’2(4๐‘)(3๐‘ž) = (4^2ร—๐‘^2 )+(3^2ร—๐‘ž^2 )โˆ’(2ร—4ร—3)ร—(๐‘ร—๐‘ž) = ๐Ÿ๐Ÿ”๐’‘^๐Ÿ+๐Ÿ—๐’’^๐Ÿโˆ’๐Ÿ๐Ÿ’๐’‘๐’’ Example 12 Using Identity (II), find (ii) (4.9)^2 (4.9)^2 = (5โˆ’0.1)^2 (๐‘Žโˆ’๐‘)^2=๐‘Ž^2+๐‘^2โˆ’2๐‘Ž๐‘ Putting ๐‘Ž = 5 & ๐‘ = 0.1 = (5)^2+(0.1)^2โˆ’2(5)(0.1) = 25+(1/10)^2โˆ’(2ร—5ร—1/10) = 25+1^2/ใ€–10ใ€—^2 โˆ’(10/10) = 25+1/100โˆ’1 = 24+1/100 = (24 ร— 100 + 1)/100 = (2400 + 1)/100 = 2401/100 = ๐Ÿ๐Ÿ’.๐ŸŽ๐Ÿ

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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.