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

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

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

Transcript

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
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.