Example 14 - Use the Identity (x + a) (x + b) = x^2 + (a + b) x + ab

Example 14 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 2
Example 14 - 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 14 Use the Identity (๐‘ฅ+๐‘Ž)(๐‘ฅ+๐‘)=๐‘ฅ^2+(๐‘Ž+๐‘)๐‘ฅ+๐‘Ž๐‘ to find the following: (i) 501 ร— 502 501 ร— 502 = (500+1)ร—(500+2) (๐‘ฅ+๐‘Ž)(๐‘ฅ+๐‘)=๐‘ฅ^2+(๐‘Ž+๐‘)๐‘ฅ+๐‘Ž๐‘ Putting ๐‘ฅ = 500 , ๐‘Ž = 1 & ๐‘ = 2 = (500)^2+(1+2)(500)+(1)(2) = 250000+(3ร—500)+2 = 250000+1500+2 = ๐Ÿ๐Ÿ“๐Ÿ๐Ÿ“๐ŸŽ๐Ÿ Example 14 Use the Identity (๐‘ฅ+๐‘Ž)(๐‘ฅ+๐‘)=๐‘ฅ^2+(๐‘Ž+๐‘)๐‘ฅ+๐‘Ž๐‘ to find the following: (ii) 95 ร— 103 95 ร— 103 = (100โˆ’5)ร—(100+3) = [100+(โˆ’5)]ร—[100+3] (๐‘ฅ+๐‘Ž)(๐‘ฅ+๐‘)=๐‘ฅ^2+(๐‘Ž+๐‘)๐‘ฅ+๐‘Ž๐‘ Putting ๐‘ฅ = 100 , ๐‘Ž = โˆ’5 & ๐‘ = 3 = (100)^2+(โˆ’5+2)(100)+(โˆ’5)(3) = 10000+(โˆ’2ร—100)+(โˆ’15) = 10000โˆ’200โˆ’15 = 1000โˆ’215 = ๐Ÿ—๐Ÿ•๐Ÿ–๐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.