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

Last updated at April 16, 2024 by Teachoo

Algebra Identities and Formulas

Question 1 (i)

Question 1 (ii) Important

Question 1 (iii)

Question 1 (iv)

Question 1 (v) Important

Question 1 (vi) Important

Question 1 (vii)

Question 1 (viii)

Question 1 (ix) Important

Question 1 (x)

Question 2 (i)

Question 2 (ii)

Question 2 (iii)

Question 2 (iv) Important

Question 2 (v)

Question 2 (vi) Important

Question 2 (vii) Important

Question 3 (i)

Question 3 (ii) Important

Question 3 (iii)

Question 3 (iv) Important

Question 3 (v) Important

Question 3 (vi)

Question 4 (i)

Question 4 (ii)

Question 4 (iii) Important

Question 4 (iv)

Question 4 (v) Important

Question 4 (vi)

Question 4 (vii) Important

Question 5 (i)

Question 5 (ii)

Question 5 (iii) Important

Question 5 (iv) You are here

Question 5 (v) Important

Question 6 (i)

Question 6 (ii) Important

Question 6 (iii)

Question 6 (iv)

Question 6 (v) Important

Question 6 (vi)

Question 6 (vii) Important

Question 6 (viii)

Question 6 (ix) Important

Question 7 (i)

Question 7 (ii) Important

Question 7 (iii)

Question 7 (iv) Important

Question 8 (i)

Question 8 (ii)

Question 8 (iii) Important

Question 8 (iv) Important

Last updated at April 16, 2024 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