Ex 9.5, 3 - Find the squares by using identities (i) (b - 7)^2

Ex 9.5, 3 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 2
Ex 9.5, 3 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 3 Ex 9.5, 3 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 4 Ex 9.5, 3 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 5 Ex 9.5, 3 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 6

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

Transcript

Ex 9.5, 3 Find the following squares by using the identities. (i) (๐‘โˆ’7)^2 (๐‘โˆ’7)^2 (๐‘ฅโˆ’๐‘ฆ)^2=๐‘ฅ^2+๐‘ฆ^2โˆ’2๐‘ฅ๐‘ฆ Putting ๐‘ฅ = ๐‘ & ๐‘ฆ = 7 = (๐‘)^2+(7)^2โˆ’2(๐‘)(7) = ๐‘^2+49โˆ’(2ร—7)ร—๐‘ = ๐’ƒ^๐Ÿ+๐Ÿ’๐Ÿ—โˆ’๐Ÿ๐Ÿ’๐’ƒ Ex 9.5, 3 Find the following squares by using the identities. (ii) (๐‘ฅ๐‘ฆ+3๐‘ง)^2 (๐‘ฅ๐‘ฆ+3๐‘ง)^2 (๐‘Ž+๐‘)^2=๐‘Ž^2+๐‘^2+2๐‘Ž๐‘ Putting ๐‘Ž = ๐‘ฅ๐‘ฆ & ๐‘ = 3๐‘ง = (๐‘ฅ๐‘ฆ)^2+(3๐‘ง)^2+2(๐‘ฅ๐‘ฆ)(3๐‘ง) = (๐‘ฅ^2ร—๐‘ฆ^2 )+(3^2ร—๐‘ง^2 )+(2ร—3)ร—(๐‘ฅร—๐‘ฆร—๐‘ง) = ๐’™^๐Ÿ ๐’š^๐Ÿ+๐Ÿ—๐’›^๐Ÿ+๐Ÿ”๐’™๐’š๐’› Ex 9.5, 3 Find the following squares by using the identities. (iii) (6๐‘ฅ^2โˆ’5๐‘ฆ)^2 (6๐‘ฅ^2โˆ’5๐‘ฆ)^2 (๐‘Žโˆ’๐‘)^2=๐‘Ž^2+๐‘^2โˆ’2๐‘Ž๐‘ Putting ๐‘Ž = 6๐‘ฅ^2 & ๐‘ = 5๐‘ฆ = (6๐‘ฅ^2 )^2+(5๐‘ฆ)^2โˆ’2(6๐‘ฅ^2 )(5๐‘ฆ) = (6^2ร—๐‘ฅ^(2 ร— 2) )+(5^2ร—๐‘ฆ^2 )โˆ’(2ร—6ร—5)ร—(๐‘ฅ^2ร—๐‘ฆ) = (36ร—๐‘ฅ^4 )+(25ร—๐‘ฆ^2 )โˆ’(60ร—๐‘ฅ^2 ๐‘ฆ) = ๐Ÿ‘๐Ÿ”๐’™^๐Ÿ’+๐Ÿ๐Ÿ“๐’š^๐Ÿโˆ’๐Ÿ”๐ŸŽ๐’™^๐Ÿ ๐’š Ex 9.5, 3 Find the following squares by using the identities. (iv) (2/3 ๐‘š+3/2 ๐‘›)^2 (2/3 ๐‘š+3/2 ๐‘›)^2 (๐‘Ž+๐‘)^2=๐‘Ž^2+๐‘^2+2๐‘Ž๐‘ Putting ๐‘Ž = 2/3 ๐‘š & ๐‘ = 3/2 ๐‘› = (2/3 ๐‘š)^2+(3/2 ๐‘›)^2+2(2/3 ๐‘š)(3/2 ๐‘›) = (2/3)^2ร—๐‘š^2+(3/2)^2ร—๐‘›^2+((2 ร— 2 ร— 3)/(3 ร— 2))ร—(๐‘šร—๐‘›) = ๐Ÿ’/๐Ÿ— ๐’Ž^๐Ÿ+ ๐Ÿ—/๐Ÿ’ ๐’^๐Ÿ+๐Ÿ๐’Ž๐’ Ex 9.5, 3 Find the following squares by using the identities. (v) (0.4๐‘โˆ’0.5๐‘ž)^2 (0.4๐‘โˆ’0.5๐‘ž)^2 (๐‘Žโˆ’๐‘)^2=๐‘Ž^2+๐‘^2โˆ’2๐‘Ž๐‘ Putting ๐‘Ž = 0.4๐‘ & ๐‘ = 0.5๐‘ž = (0.4๐‘)^2+(0.5๐‘ž)^2โˆ’2(0.4๐‘)(0.5๐‘ž) = (4/10)^2ร—๐‘^2+(5/10)^2ร—๐‘ž^2โˆ’(2ร—4/10ร—5/10)ร—(๐‘ร—๐‘ž) = 16/100 ๐‘^2+ 25/100 ๐‘ž^2โˆ’4/10 ๐‘๐‘ž = ๐ŸŽ.๐Ÿ๐Ÿ”๐’‘^๐Ÿ+๐ŸŽ.๐Ÿ๐Ÿ“๐’’^๐Ÿโˆ’๐ŸŽ.๐Ÿ’๐’‘๐’’ Ex 9.5, 3 Find the following squares by using the identities. (vi) (2๐‘ฅ๐‘ฆ+5๐‘ฆ)^2 (2๐‘ฅ๐‘ฆ+5๐‘ฆ)^2 (๐‘Ž+๐‘)^2=๐‘Ž^2+๐‘^2+2๐‘Ž๐‘ Putting ๐‘Ž = 2๐‘ฅ๐‘ฆ & ๐‘ = 5๐‘ฆ = (2๐‘ฅ๐‘ฆ)^2+(5๐‘ฆ)^2+2(2๐‘ฅ๐‘ฆ)(5๐‘ฆ) = (2^2ร—๐‘ฅ^2ร—๐‘ฆ^2 )+(5^2ร—๐‘ฆ^2 )+(2ร—2ร—5)ร—๐‘ฅร—(๐‘ฆร—๐‘ฆ) = ๐Ÿ’๐’™^๐Ÿ ๐’š^๐Ÿ+๐Ÿ๐Ÿ“๐’š^๐Ÿ+๐Ÿ๐ŸŽ๐’™๐’š^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.