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Algebra Identities and Formulas

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Last updated at May 29, 2023 by Teachoo

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