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Question 5 (iv) - Algebra Identities and Formulas - Chapter 8 Class 8 Algebraic Expressions and Identities

Last updated at Dec. 16, 2024 by Teachoo

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Question 5 Show that. (iv) (4𝑝𝑞+3𝑞)^2−(4𝑝𝑞−3𝑞)^2=48𝑝𝑞^2 Solving (𝟒𝒑𝒒+𝟑𝒒)^𝟐 (𝑎+𝑏)^2=𝑎^2+𝑏^2+2𝑎𝑏 Putting 𝑎 = 4𝑝𝑞 & 𝑏 = 3𝑞 (𝑎+𝑏)^2=𝑎^2+𝑏^2+2𝑎𝑏 Putting 𝑎 = 4𝑝𝑞 & 𝑏 = 3𝑞 = (4𝑝𝑞)^2+(3𝑞)^2+2(4𝑝𝑞)(3𝑞) = 16𝑝^2 𝑞^2+9𝑞^2+24𝑝𝑞^2 Solving (𝟒𝒑𝒒−𝟑𝒒)^𝟐 (𝑎−𝑏)^2=𝑎^2+𝑏^2−2𝑎𝑏 Putting 𝑎 = 4𝑝𝑞 & 𝑏 = 3𝑞 = (4𝑝𝑞)^2+(3𝑞)^2−2(4𝑝𝑞)(3𝑞) = 16𝑝^2 𝑞^2+9𝑞^2−24𝑝𝑞^2 Solving LHS (4𝑝𝑞+3𝑞)^2−(4𝑝𝑞−3𝑞)^2 = (16𝑝^2 𝑞^2+9𝑞^2+24𝑝𝑞^2 )−(16𝑝^2 𝑞^2+9𝑞^2−24𝑝𝑞^2 ) = 16𝑝^2 𝑞^2+9𝑞^2+24𝑝𝑞^2−16𝑝^2 𝑞^2−9𝑞^2+24𝑝𝑞^2 = (16𝑝^2 𝑞^2−16𝑝^2 𝑞^2 )+(9𝑞^2−9𝑞^2 )+(24𝑝𝑞^2+24𝑝𝑞^2 ) = 0+0+48𝑝𝑞^2 = 48𝑝𝑞^2 = R.H.S Since LHS = RHS Hence proved
  1. Chapter 8 Class 8 Algebraic Expressions and Identities
  2. Serial order wise

    3. Algebra Identities and Formulas

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Chapter 8 Class 8 Algebraic Expressions and Identities
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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo