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

Last updated at Dec. 16, 2024 by

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

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

Share on WhatsApp

Transcript

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

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