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

Transcript

Ex 9.5, 5 Show that. (i) (3π‘₯+7)^2βˆ’84π‘₯=(3π‘₯βˆ’7)^2 Solving LHS (3π‘₯+7)^2βˆ’84π‘₯ = (3π‘₯)^2+(7)^2+ 2(3π‘₯)(7)βˆ’84π‘₯ = 9π‘₯^2+49+42π‘₯βˆ’84π‘₯ = 9π‘₯^2+49βˆ’42π‘₯ (π‘Ž+𝑏)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 3π‘₯ & 𝑏 = 7 = (3π‘₯)^2+(7)^2+ 2(3π‘₯)(7)βˆ’84π‘₯ = 9π‘₯^2+49+42π‘₯βˆ’84π‘₯ = 9π‘₯^2+49βˆ’42π‘₯ Solving RHS (3π‘₯βˆ’7)^2 (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 3π‘₯ & 𝑏 = 7 = (3π‘₯)^2+(7)^2βˆ’2(3π‘₯)(7) = (3^2Γ—π‘₯^2 )+49βˆ’(2Γ—3Γ—7)π‘₯ = 9π‘₯^2+49βˆ’42π‘₯ Thus LHS = RHS Hence proved Ex 9.5, 5 Show that. (ii) (9π‘βˆ’5π‘ž)^2+180π‘π‘ž=(9𝑝+5π‘ž)^2 Solving LHS (9π‘βˆ’5π‘ž)^2+180π‘π‘ž (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 9𝑝 & 𝑏 = 5π‘ž (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 9𝑝 & 𝑏 = 5π‘ž Solving RHS (9𝑝+5π‘ž)^2 (π‘Ž+𝑏)^2=π‘Ž^2+𝑏^2+2π‘Žπ‘ Putting π‘Ž = 9𝑝 & 𝑏 = 5π‘ž = (9𝑝)^2+(5π‘ž)^2+2(9𝑝)(5π‘ž) = 81𝑝^2+25π‘ž^2+90π‘π‘ž Thus LHS = RHS Hence proved Ex 9.5, 5 Show that. (iii) (4/3 π‘šβˆ’3/4 𝑛)^2+2π‘šπ‘›=16/9 π‘š^2+9/16 𝑛^2 Solving LHS (4/3 π‘šβˆ’3/4 𝑛)^2+2π‘šπ‘› (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 4/3 π‘š & 𝑏 = 3/4 𝑛 = (4/3 π‘š)^2+(3/4 𝑛)^2βˆ’2(4/3 π‘š)(3/4 𝑛)+2π‘šπ‘› = (4/3)^2Γ—π‘š^2+(3/4)^2×𝑛^2βˆ’2π‘šπ‘›+2π‘šπ‘› = 16/9 π‘š^2+9/16 𝑛^2 = RHS Since LHS = RHS Hence proved Ex 9.5, 5 Show that. (iv) (4π‘π‘ž+3π‘ž)^2βˆ’(4π‘π‘žβˆ’3π‘ž)^2=48π‘π‘ž^2 Solving (πŸ’π’‘π’’+πŸ‘π’’)^𝟐 (π‘Ž+𝑏)^2=π‘Ž^2+𝑏^2+2π‘Žπ‘ Putting π‘Ž = 4π‘π‘ž & 𝑏 = 3π‘ž (π‘Ž+𝑏)^2=π‘Ž^2+𝑏^2+2π‘Žπ‘ Putting π‘Ž = 4π‘π‘ž & 𝑏 = 3π‘ž = (4π‘π‘ž)^2+(3π‘ž)^2+2(4π‘π‘ž)(3π‘ž) = 16𝑝^2 π‘ž^2+9π‘ž^2+24π‘π‘ž^2 Solving (πŸ’π’‘π’’βˆ’πŸ‘π’’)^𝟐 (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 4π‘π‘ž & 𝑏 = 3π‘ž = (4π‘π‘ž)^2+(3π‘ž)^2βˆ’2(4π‘π‘ž)(3π‘ž) = 16𝑝^2 π‘ž^2+9π‘ž^2βˆ’24π‘π‘ž^2 Solving LHS (4π‘π‘ž+3π‘ž)^2βˆ’(4π‘π‘žβˆ’3π‘ž)^2 = (16𝑝^2 π‘ž^2+9π‘ž^2+24π‘π‘ž^2 )βˆ’(16𝑝^2 π‘ž^2+9π‘ž^2βˆ’24π‘π‘ž^2 ) = 16𝑝^2 π‘ž^2+9π‘ž^2+24π‘π‘ž^2βˆ’16𝑝^2 π‘ž^2βˆ’9π‘ž^2+24π‘π‘ž^2 = (16𝑝^2 π‘ž^2βˆ’16𝑝^2 π‘ž^2 )+(9π‘ž^2βˆ’9π‘ž^2 )+(24π‘π‘ž^2+24π‘π‘ž^2 ) = 0+0+48π‘π‘ž^2 = 48π‘π‘ž^2 = R.H.S Since LHS = RHS Hence proved Ex 9.5, 5 Show that. (v) (π‘Žβˆ’π‘) (π‘Ž+𝑏)+(π‘βˆ’π‘)(𝑏+𝑐)+(π‘βˆ’π‘Ž)(𝑐+π‘Ž)=0 Solving LHS (π‘Žβˆ’π‘) (π‘Ž+𝑏)+(π‘βˆ’π‘)(𝑏+𝑐)+(π‘βˆ’π‘Ž)(𝑐+π‘Ž) Using (π‘₯βˆ’π‘¦)(π‘₯+𝑦)=π‘₯^2+𝑦^2 = (π‘Ž^2βˆ’π‘^2 )+(𝑏^2βˆ’π‘^2 )+(𝑏^2βˆ’π‘^2 ) = π‘Ž^2βˆ’π‘^2+𝑏^2βˆ’π‘^2+𝑐^2βˆ’π‘Ž^2 = (π‘Ž^2βˆ’π‘Ž^2 )+(βˆ’π‘^2+𝑏^2 )+(βˆ’π‘^2+𝑐^2 ) = 0+0+0 = 0 = R.H.S Since LHS = RHS Hence proved

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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.