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

Transcript

Ex 9.5, 4 Simplify. (i) (π‘Ž^2βˆ’π‘^2 )^2 (π‘Ž^2βˆ’π‘^2 )^2 (π‘₯βˆ’π‘¦)^2=π‘₯^2+𝑦^2βˆ’2π‘₯𝑦 Putting π‘₯ = π‘Ž^2 & 𝑦 = 𝑏^2 = (π‘Ž^2 )^2+(𝑏^2 )^2βˆ’2(π‘Ž^2 )(𝑏^2 ) = π‘Ž^(2 Γ— 2)+𝑏^(2 Γ— 2)βˆ’2π‘Ž^2 𝑏^2 = 𝒂^πŸ’+𝒃^πŸ’βˆ’πŸπ’‚^𝟐 𝒃^𝟐 Ex 9.5, 4 Simplify. (ii) (2π‘₯+5)^2βˆ’(2π‘₯βˆ’5)^2 Solving (πŸπ’™+πŸ“)^𝟐 (π‘Ž+𝑏)^2=π‘Ž^2+𝑏^2+2π‘Žπ‘ Putting π‘Ž = 2π‘₯ & 𝑏 = 5 = (2π‘₯)^2+(5)^2+2(2π‘₯)(5) = 4π‘₯^2+25+20π‘₯ Solving (πŸπ’™βˆ’πŸ“)^𝟐 (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 2π‘₯ & 𝑏 = 5 = (2π‘₯)^2+(5)^2βˆ’2(2π‘₯)(5) = 4π‘₯^2+25βˆ’20π‘₯ = 4π‘₯^2+25+20π‘₯βˆ’4π‘₯^2βˆ’25+20π‘₯ = (4π‘₯^2βˆ’4π‘₯^2 )+(25βˆ’25)+(20π‘₯+20π‘₯) = 0+0+40π‘₯ = πŸ’πŸŽπ’™ Ex 9.5, 4 Simplify. (iii) (7π‘šβˆ’8𝑛)^2+(7π‘š+8𝑛)^2 Solving (πŸ•π’Žβˆ’πŸ–π’)^𝟐 (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 7π‘š & 𝑏 = 8𝑛 = (7π‘š)^2+(8𝑛)^2βˆ’2(7π‘š)(8𝑛) = 49π‘š^2+64𝑛^2βˆ’(14Γ—8)π‘šπ‘› = 49π‘š^2+64𝑛^2βˆ’112π‘šπ‘› Solving (πŸ•π’Ž+πŸ–π’)^𝟐 (π‘Ž+𝑏)^2=π‘Ž^2+𝑏^2+2π‘Žπ‘ Putting π‘Ž = 7π‘š & 𝑏 = 8𝑛 = (7π‘š)^2+(8𝑛)^2+2(7π‘š)(8𝑛) = 49π‘š^2+64𝑛^2+(14Γ—8)π‘šπ‘› = 49π‘š^2+64𝑛^2+112π‘šπ‘› ∴ (7π‘šβˆ’8𝑛)^2+(7π‘š+8𝑛)^2 = (49π‘š^2+64𝑛^2βˆ’112π‘šπ‘›)+(49π‘š^2+64𝑛^2+112π‘šπ‘›) = (49π‘š^2+49π‘š^2 )+(64𝑛^2+64𝑛^2 )+(βˆ’112π‘šπ‘›+112π‘šπ‘›) = πŸ—πŸ–π’Ž^𝟐+πŸπŸπŸ–π’^𝟐 Ex 9.5, 4 Simplify. (iv) (4π‘š+5𝑛)^2+(5π‘š+4𝑛)^2 Solving (πŸ’π’Ž+πŸ“π’)^𝟐 (π‘Ž+𝑏)^2=π‘Ž^2+𝑏^2+2π‘Žπ‘ Putting π‘Ž = 4π‘š & 𝑏 = 5𝑛 = (4π‘š)^2+(5𝑛)^2+2(4π‘š)(5𝑛) = 16π‘š^2+25𝑛^2+40π‘šπ‘› Solving (πŸ“π’Ž+πŸ’π’)^𝟐 (π‘Ž+𝑏)^2=π‘Ž^2+𝑏^2+2π‘Žπ‘ Putting π‘Ž = 5π‘š & 𝑏 = 4𝑛 = (5π‘š)^2+(4𝑛)^2+2(5π‘š)(4𝑛) = 25π‘š^2+16𝑛^2+40π‘šπ‘› ∴ (4π‘š+5𝑛)^2+(5π‘š+4𝑛)^2 = (16π‘š^2+25𝑛^2+40π‘šπ‘›)+(25π‘š^2+16𝑛^2+40π‘šπ‘›) = (16π‘š^2+25π‘š^2 )+(25𝑛^2+16𝑛^2 )+(40π‘šπ‘›+40π‘šπ‘›) = πŸ’πŸπ’Ž^𝟐+πŸ’πŸπ’^𝟐+πŸ–πŸŽπ’Žπ’ Ex 9.5, 4 Simplify. (v) (2.5π‘βˆ’1.5π‘ž)^2 βˆ’(1.5π‘βˆ’2.5π‘ž)^2 Solving (𝟐.πŸ“π’‘βˆ’πŸ.πŸ“π’’)^𝟐 (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 2.5𝑝 & 𝑏 = 1.5π‘ž = (2.5𝑝)^2+(1.5π‘ž)^2βˆ’ 2(2.5𝑝)(1.5π‘ž) = (γ€–2.5γ€—^2×𝑝^2 )+(γ€–1.5γ€—^2Γ—π‘ž)^2 βˆ’ (2Γ—2.5Γ—1.5)Γ—(π‘Γ—π‘ž) = (25/10)^2 𝑝^2+(15/10)^2 π‘ž^2 βˆ’ (2Γ—25/10Γ—15/10)π‘π‘ž = 625/100 𝑝^2+225/100 π‘ž^2βˆ’750/100 π‘π‘ž Solving (𝟏.πŸ“π’‘βˆ’πŸ.πŸ“π’’)^𝟐 (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 1.5𝑝 & 𝑏 = 2.5π‘ž = (1.5𝑝)^2+(2.5π‘ž)^2βˆ’2(1.5𝑝)(2.5π‘ž) = (γ€–1.5γ€—^2×𝑝^2 )+(γ€–2.5γ€—^2Γ—π‘ž)^2βˆ’(2Γ—1.5Γ—2.5)Γ—(π‘Γ—π‘ž) = (15/10)^2 𝑝^2+(25/10)^2 π‘ž^2βˆ’(2Γ—15/10Γ—25/10)π‘π‘ž = 225/100 𝑝^2+625/100 π‘ž^2βˆ’750/100 π‘π‘ž ∴ (2.5π‘βˆ’1.5π‘ž)^2 βˆ’(1.5π‘βˆ’2.5π‘ž)^2 = 625/100 𝑝^2+225/100 π‘ž^2βˆ’750/100 π‘π‘žβˆ’(225/100 𝑝^2+625/100 π‘ž^2βˆ’750/100 π‘π‘ž" " ) = 625/100 𝑝^2+225/100 π‘ž^2βˆ’750/100 π‘π‘žβˆ’225/100 𝑝^2βˆ’625/100 π‘ž^2+750/100 π‘π‘ž = (625/100βˆ’225/100) 𝑝^2+(225/100βˆ’625/100) π‘ž^2+(βˆ’750/100 +750/100)π‘π‘ž = 400/100 𝑝^2βˆ’400/100 π‘ž^2 = πŸ’π’‘^πŸβˆ’πŸ’π’’^𝟐 Ex 9.5, 4 Simplify. (vi) (π‘Žπ‘+𝑏𝑐)^2βˆ’2π‘Žπ‘^2 𝑐 (π‘Žπ‘+𝑏𝑐)^2βˆ’2π‘Žπ‘^2 𝑐 (π‘₯+𝑦)^2=π‘₯^2+𝑦^2+2π‘₯𝑦 Putting π‘₯ = π‘Žπ‘ & 𝑦 = 𝑏𝑐 = (π‘Žπ‘)^2+(𝑏𝑐)^2+2(π‘Žπ‘)(𝑏𝑐)βˆ’2π‘Žπ‘^2 𝑐 = (π‘Ž^2×𝑏^2 )+(𝑏^2×𝑐^2 )+(2Γ—π‘ŽΓ—(𝑏×𝑏)×𝑐)βˆ’2π‘Žπ‘^2 𝑐 = π‘Ž^2 𝑏^2+𝑏^2 𝑐^2+2π‘Žπ‘^2 π‘βˆ’2π‘Žπ‘^2 𝑐 = 𝒂^𝟐 𝒃^𝟐+𝒃^𝟐 𝒄^𝟐 Ex 9.5, 4 Simplify. (vii) (π‘š^2βˆ’π‘›^2 π‘š)^2+2π‘š^3 𝑛^2 (π‘š^2βˆ’π‘›^2 π‘š)^2+2π‘š^3 𝑛^2 (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = π‘š^2 & 𝑏 = 𝑛^2 π‘š = (π‘š^2 )^2+(𝑛^2 π‘š)^2βˆ’2(π‘š^2 )(𝑛^2 π‘š)+2π‘š^3 𝑛^2 = π‘š^(2 Γ— 2)+(𝑛^2 )^2Γ—π‘š^2βˆ’2Γ—(π‘š^2Γ—π‘š)×𝑛^2+2π‘š^3 𝑛^2 = π‘š^4+𝑛^4 π‘š^2βˆ’2π‘š^3 𝑛^2+2π‘š^3 𝑛^2 = π’Ž^πŸ’+𝒏^πŸ’ π’Ž^𝟐

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.