Algebra Identities and Formulas
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 You are here
Question 5 (iv)
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. (iii) (4/3 𝑚−3/4 𝑛)^2+2𝑚𝑛=16/9 𝑚^2+9/16 𝑛^2 Solving LHS (4/3 𝑚−3/4 𝑛)^2+2𝑚𝑛 (𝑎−𝑏)^2=𝑎^2+𝑏^2−2𝑎𝑏 Putting 𝑎 = 4/3 𝑚 & 𝑏 = 3/4 𝑛 = (4/3 𝑚)^2+(3/4 𝑛)^2−2(4/3 𝑚)(3/4 𝑛)+2𝑚𝑛 = (4/3)^2×𝑚^2+(3/4)^2×𝑛^2−2𝑚𝑛+2𝑚𝑛 = 16/9 𝑚^2+9/16 𝑛^2 = RHS Since LHS = RHS Hence proved