Ex 9.5, 4 - Simplify (i) (a^2 - b^2)^2 (ii) (2x + 5)^2 - (2x - 5)^2

Ex 9.5, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 2
Ex 9.5, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 3 Ex 9.5, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 4 Ex 9.5, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 5 Ex 9.5, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 6 Ex 9.5, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 7 Ex 9.5, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 8 Ex 9.5, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 9 Ex 9.5, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 10 Ex 9.5, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 11

  1. Chapter 9 Class 8 Algebraic Expressions and Identities
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Transcript

Ex 9.5, 4 Simplify. (i) (๐‘Ž^2โˆ’๐‘^2 )^2 (๐‘Ž^2โˆ’๐‘^2 )^2 (๐‘ฅโˆ’๐‘ฆ)^2=๐‘ฅ^2+๐‘ฆ^2โˆ’2๐‘ฅ๐‘ฆ Putting ๐‘ฅ = ๐‘Ž^2 & ๐‘ฆ = ๐‘^2 = (๐‘Ž^2 )^2+(๐‘^2 )^2โˆ’2(๐‘Ž^2 )(๐‘^2 ) = ๐‘Ž^(2 ร— 2)+๐‘^(2 ร— 2)โˆ’2๐‘Ž^2 ๐‘^2 = ๐’‚^๐Ÿ’+๐’ƒ^๐Ÿ’โˆ’๐Ÿ๐’‚^๐Ÿ ๐’ƒ^๐Ÿ Ex 9.5, 4 Simplify. (ii) (2๐‘ฅ+5)^2โˆ’(2๐‘ฅโˆ’5)^2 Solving (๐Ÿ๐’™+๐Ÿ“)^๐Ÿ (๐‘Ž+๐‘)^2=๐‘Ž^2+๐‘^2+2๐‘Ž๐‘ Putting ๐‘Ž = 2๐‘ฅ & ๐‘ = 5 = (2๐‘ฅ)^2+(5)^2+2(2๐‘ฅ)(5) = 4๐‘ฅ^2+25+20๐‘ฅ Solving (๐Ÿ๐’™โˆ’๐Ÿ“)^๐Ÿ (๐‘Žโˆ’๐‘)^2=๐‘Ž^2+๐‘^2โˆ’2๐‘Ž๐‘ Putting ๐‘Ž = 2๐‘ฅ & ๐‘ = 5 = (2๐‘ฅ)^2+(5)^2โˆ’2(2๐‘ฅ)(5) = 4๐‘ฅ^2+25โˆ’20๐‘ฅ = 4๐‘ฅ^2+25+20๐‘ฅโˆ’4๐‘ฅ^2โˆ’25+20๐‘ฅ = (4๐‘ฅ^2โˆ’4๐‘ฅ^2 )+(25โˆ’25)+(20๐‘ฅ+20๐‘ฅ) = 0+0+40๐‘ฅ = ๐Ÿ’๐ŸŽ๐’™ Ex 9.5, 4 Simplify. (iii) (7๐‘šโˆ’8๐‘›)^2+(7๐‘š+8๐‘›)^2 Solving (๐Ÿ•๐’Žโˆ’๐Ÿ–๐’)^๐Ÿ (๐‘Žโˆ’๐‘)^2=๐‘Ž^2+๐‘^2โˆ’2๐‘Ž๐‘ Putting ๐‘Ž = 7๐‘š & ๐‘ = 8๐‘› = (7๐‘š)^2+(8๐‘›)^2โˆ’2(7๐‘š)(8๐‘›) = 49๐‘š^2+64๐‘›^2โˆ’(14ร—8)๐‘š๐‘› = 49๐‘š^2+64๐‘›^2โˆ’112๐‘š๐‘› Solving (๐Ÿ•๐’Ž+๐Ÿ–๐’)^๐Ÿ (๐‘Ž+๐‘)^2=๐‘Ž^2+๐‘^2+2๐‘Ž๐‘ Putting ๐‘Ž = 7๐‘š & ๐‘ = 8๐‘› = (7๐‘š)^2+(8๐‘›)^2+2(7๐‘š)(8๐‘›) = 49๐‘š^2+64๐‘›^2+(14ร—8)๐‘š๐‘› = 49๐‘š^2+64๐‘›^2+112๐‘š๐‘› โˆด (7๐‘šโˆ’8๐‘›)^2+(7๐‘š+8๐‘›)^2 = (49๐‘š^2+64๐‘›^2โˆ’112๐‘š๐‘›)+(49๐‘š^2+64๐‘›^2+112๐‘š๐‘›) = (49๐‘š^2+49๐‘š^2 )+(64๐‘›^2+64๐‘›^2 )+(โˆ’112๐‘š๐‘›+112๐‘š๐‘›) = ๐Ÿ—๐Ÿ–๐’Ž^๐Ÿ+๐Ÿ๐Ÿ๐Ÿ–๐’^๐Ÿ Ex 9.5, 4 Simplify. (iv) (4๐‘š+5๐‘›)^2+(5๐‘š+4๐‘›)^2 Solving (๐Ÿ’๐’Ž+๐Ÿ“๐’)^๐Ÿ (๐‘Ž+๐‘)^2=๐‘Ž^2+๐‘^2+2๐‘Ž๐‘ Putting ๐‘Ž = 4๐‘š & ๐‘ = 5๐‘› = (4๐‘š)^2+(5๐‘›)^2+2(4๐‘š)(5๐‘›) = 16๐‘š^2+25๐‘›^2+40๐‘š๐‘› Solving (๐Ÿ“๐’Ž+๐Ÿ’๐’)^๐Ÿ (๐‘Ž+๐‘)^2=๐‘Ž^2+๐‘^2+2๐‘Ž๐‘ Putting ๐‘Ž = 5๐‘š & ๐‘ = 4๐‘› = (5๐‘š)^2+(4๐‘›)^2+2(5๐‘š)(4๐‘›) = 25๐‘š^2+16๐‘›^2+40๐‘š๐‘› โˆด (4๐‘š+5๐‘›)^2+(5๐‘š+4๐‘›)^2 = (16๐‘š^2+25๐‘›^2+40๐‘š๐‘›)+(25๐‘š^2+16๐‘›^2+40๐‘š๐‘›) = (16๐‘š^2+25๐‘š^2 )+(25๐‘›^2+16๐‘›^2 )+(40๐‘š๐‘›+40๐‘š๐‘›) = ๐Ÿ’๐Ÿ๐’Ž^๐Ÿ+๐Ÿ’๐Ÿ๐’^๐Ÿ+๐Ÿ–๐ŸŽ๐’Ž๐’ Ex 9.5, 4 Simplify. (v) (2.5๐‘โˆ’1.5๐‘ž)^2 โˆ’(1.5๐‘โˆ’2.5๐‘ž)^2 Solving (๐Ÿ.๐Ÿ“๐’‘โˆ’๐Ÿ.๐Ÿ“๐’’)^๐Ÿ (๐‘Žโˆ’๐‘)^2=๐‘Ž^2+๐‘^2โˆ’2๐‘Ž๐‘ Putting ๐‘Ž = 2.5๐‘ & ๐‘ = 1.5๐‘ž = (2.5๐‘)^2+(1.5๐‘ž)^2โˆ’ 2(2.5๐‘)(1.5๐‘ž) = (ใ€–2.5ใ€—^2ร—๐‘^2 )+(ใ€–1.5ใ€—^2ร—๐‘ž)^2 โˆ’ (2ร—2.5ร—1.5)ร—(๐‘ร—๐‘ž) = (25/10)^2 ๐‘^2+(15/10)^2 ๐‘ž^2 โˆ’ (2ร—25/10ร—15/10)๐‘๐‘ž = 625/100 ๐‘^2+225/100 ๐‘ž^2โˆ’750/100 ๐‘๐‘ž Solving (๐Ÿ.๐Ÿ“๐’‘โˆ’๐Ÿ.๐Ÿ“๐’’)^๐Ÿ (๐‘Žโˆ’๐‘)^2=๐‘Ž^2+๐‘^2โˆ’2๐‘Ž๐‘ Putting ๐‘Ž = 1.5๐‘ & ๐‘ = 2.5๐‘ž = (1.5๐‘)^2+(2.5๐‘ž)^2โˆ’2(1.5๐‘)(2.5๐‘ž) = (ใ€–1.5ใ€—^2ร—๐‘^2 )+(ใ€–2.5ใ€—^2ร—๐‘ž)^2โˆ’(2ร—1.5ร—2.5)ร—(๐‘ร—๐‘ž) = (15/10)^2 ๐‘^2+(25/10)^2 ๐‘ž^2โˆ’(2ร—15/10ร—25/10)๐‘๐‘ž = 225/100 ๐‘^2+625/100 ๐‘ž^2โˆ’750/100 ๐‘๐‘ž โˆด (2.5๐‘โˆ’1.5๐‘ž)^2 โˆ’(1.5๐‘โˆ’2.5๐‘ž)^2 = 625/100 ๐‘^2+225/100 ๐‘ž^2โˆ’750/100 ๐‘๐‘žโˆ’(225/100 ๐‘^2+625/100 ๐‘ž^2โˆ’750/100 ๐‘๐‘ž" " ) = 625/100 ๐‘^2+225/100 ๐‘ž^2โˆ’750/100 ๐‘๐‘žโˆ’225/100 ๐‘^2โˆ’625/100 ๐‘ž^2+750/100 ๐‘๐‘ž = (625/100โˆ’225/100) ๐‘^2+(225/100โˆ’625/100) ๐‘ž^2+(โˆ’750/100 +750/100)๐‘๐‘ž = 400/100 ๐‘^2โˆ’400/100 ๐‘ž^2 = ๐Ÿ’๐’‘^๐Ÿโˆ’๐Ÿ’๐’’^๐Ÿ Ex 9.5, 4 Simplify. (vi) (๐‘Ž๐‘+๐‘๐‘)^2โˆ’2๐‘Ž๐‘^2 ๐‘ (๐‘Ž๐‘+๐‘๐‘)^2โˆ’2๐‘Ž๐‘^2 ๐‘ (๐‘ฅ+๐‘ฆ)^2=๐‘ฅ^2+๐‘ฆ^2+2๐‘ฅ๐‘ฆ Putting ๐‘ฅ = ๐‘Ž๐‘ & ๐‘ฆ = ๐‘๐‘ = (๐‘Ž๐‘)^2+(๐‘๐‘)^2+2(๐‘Ž๐‘)(๐‘๐‘)โˆ’2๐‘Ž๐‘^2 ๐‘ = (๐‘Ž^2ร—๐‘^2 )+(๐‘^2ร—๐‘^2 )+(2ร—๐‘Žร—(๐‘ร—๐‘)ร—๐‘)โˆ’2๐‘Ž๐‘^2 ๐‘ = ๐‘Ž^2 ๐‘^2+๐‘^2 ๐‘^2+2๐‘Ž๐‘^2 ๐‘โˆ’2๐‘Ž๐‘^2 ๐‘ = ๐’‚^๐Ÿ ๐’ƒ^๐Ÿ+๐’ƒ^๐Ÿ ๐’„^๐Ÿ Ex 9.5, 4 Simplify. (vii) (๐‘š^2โˆ’๐‘›^2 ๐‘š)^2+2๐‘š^3 ๐‘›^2 (๐‘š^2โˆ’๐‘›^2 ๐‘š)^2+2๐‘š^3 ๐‘›^2 (๐‘Žโˆ’๐‘)^2=๐‘Ž^2+๐‘^2โˆ’2๐‘Ž๐‘ Putting ๐‘Ž = ๐‘š^2 & ๐‘ = ๐‘›^2 ๐‘š = (๐‘š^2 )^2+(๐‘›^2 ๐‘š)^2โˆ’2(๐‘š^2 )(๐‘›^2 ๐‘š)+2๐‘š^3 ๐‘›^2 = ๐‘š^(2 ร— 2)+(๐‘›^2 )^2ร—๐‘š^2โˆ’2ร—(๐‘š^2ร—๐‘š)ร—๐‘›^2+2๐‘š^3 ๐‘›^2 = ๐‘š^4+๐‘›^4 ๐‘š^2โˆ’2๐‘š^3 ๐‘›^2+2๐‘š^3 ๐‘›^2 = ๐’Ž^๐Ÿ’+๐’^๐Ÿ’ ๐’Ž^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.