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Ex 9.5, 5 (iv) - Chapter 9 Class 8 Algebraic Expressions and Identities

Last updated at March 22, 2023 by

Ex 9.5, 5 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 5

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.