Example 13 - Using Identity (III), find (i) (3/2 m + 2/3n) (ii) 983^2

Example 13 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 2
Example 13 - 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 13 Using Identity (III), find (i) (3/2 ๐‘š+2/3 ๐‘›)(3/2 ๐‘šโˆ’2/3 ๐‘›) (3/2 ๐‘š+2/3 ๐‘›)(3/2 ๐‘šโˆ’2/3 ๐‘›) (๐‘Ž+๐‘)(๐‘Žโˆ’๐‘)=๐‘Ž^2โˆ’๐‘^2 Putting ๐‘Ž = 3/2 ๐‘š & ๐‘ = 2/3 ๐‘› = (3/2 ๐‘š)^2โˆ’(2/3 ๐‘›)^2 = (3/2)^2ร—๐‘š^2โˆ’(2/3)^2ร—๐‘›^2 = ๐Ÿ—/๐Ÿ’ ๐’Ž^๐Ÿโˆ’๐Ÿ’/๐Ÿ— ๐’^๐Ÿ Example 13 Using Identity (III), find (ii) ใ€–983ใ€—^2โˆ’ ใ€–17ใ€—^2 ใ€–983ใ€—^2โˆ’ ใ€–17ใ€—^2 ๐‘Ž^2โˆ’๐‘^2=(๐‘Ž+๐‘)(๐‘Žโˆ’๐‘) Putting ๐‘Ž = 983 & ๐‘ = 17 = (983+17)ร—(983โˆ’17) = 1000ร—966 = 9,66,000 Example 13 Using Identity (III), find (iii) 194 ร— 206 194 ร— 206 = (200โˆ’6)ร—(200+6) 194 ร— 206 = (200โˆ’6)ร—(200+6) = (200)^2โˆ’(6)^2 = 40000โˆ’36 = 39964

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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.