Question 1 - End-of-Chapter Exercises - Chapter 4 Class 9 - Exploring Algebraic Identities (Ganita Manjari I)

Transcript

Question 1 (i) Use suitable identities to find the following products: (i) (−3𝑥+4)^2 Now, (−3𝑥+4)^2=(𝟒−𝟑𝒙)^𝟐 =4^2+(3𝑥)^2−2 × 4 × 3𝑥 =16+9𝑥^2−24𝑥 =𝟗𝒙^𝟐−𝟐𝟒𝒙+𝟏𝟔 (𝑎−𝑏)^2=𝑎^2+𝑏^2−2𝑎𝑏 Putting 𝑎 = 4 & 𝑏 = 3𝑥 Question 1 (ii) Use suitable identities to find the following products: (ii) (2𝑠+7)(2𝑠−7) (2𝑠+7)(2𝑠−7) =(2s)^2−7^2 =𝟒𝐬^𝟐−𝟒𝟗 Using (𝑎+𝑏)(𝑎−𝑏)=𝑎^2−𝑏^2 Putting 𝒂 = 𝟐𝐬, 𝒃 = 𝟕 Question 1 (iii) Use suitable identities to find the following products: (iii) (𝑝^2+1/2)(𝑝^2−1/2) (𝑝^2+1/2)(𝑝^2−1/2) = (𝑝^2 )^2−(1/2)^2 = 𝒑^𝟒−𝟏/𝟒 Using (𝑎+𝑏)(𝑎−𝑏)=𝑎^2−𝑏^2 Putting 𝒂 = 𝒑^𝟐, 𝒃 = 𝟏/𝟐 Question 1 (iv) Use suitable identities to find the following products: (iv) (2𝑛+7)(2𝑛−7) (2𝑛+7)(2𝑛−7) =(2n)^2−7^2 =𝟒𝐧^𝟐−𝟒𝟗 Using (𝑎+𝑏)(𝑎−𝑏)=𝑎^2−𝑏^2 Putting 𝒂 = 𝟐𝒏, 𝒃 = 𝟕 Question 1 (v) Use suitable identities to find the following products: (v) (𝑠−2𝑡)(𝑠^2+2𝑠𝑡+4𝑡^2 ) (𝑠−2𝑡)(𝑠^2+2𝑠𝑡+4𝑡^2 ) =(𝑠−2𝑡)(𝑠^2+𝑠 × 2𝑡+(2𝑡)^2 ) =𝑠^3−〖(2𝑡)〗^3 =𝒔^𝟑−𝟖𝒕^𝟑 Using 𝑎^3−𝑏^3=(𝑎 −𝑏)(𝑎^2+𝑎𝑏+𝑏^2 ) Putting 𝒂 = 𝐬, 𝒃 = 𝟐𝐭 Question 1 (vi) Use suitable identities to find the following products: (vi) (1/2𝑟−4𝑟)^2 (1/2𝑟−4𝑟)^2 = (1/2𝑟)^2+(4𝑟)^2−2 ×1/2𝑟 × 4𝑟 = 1/(2^2 × 𝑟^2 )+16𝑟^2−4 = 𝟏/(𝟒𝒓^𝟐 )+𝟏𝟔𝒓^𝟐−𝟒 (𝑎−𝑏)^2=𝑎^2+𝑏^2−2𝑎𝑏 Putting 𝑎 = 4 & 𝑏 = 3𝑥 Question 1 (vii) Use suitable identities to find the following products: (vii) (−3𝑚+4𝑘−𝑙)^2 (−3𝑚+4𝑘−𝑙)^2 = 〖(−3𝑚)〗^2 + 〖(4𝑘)〗^2 + 〖(−𝑙)〗^2 + 2 × (−3𝑚) × (4𝑘) +2 × (−3𝑚) ×(−𝑙)+2 × (4𝑘) × (−𝑙) =𝟗𝒎^𝟐+𝟏𝟔𝒌^𝟐+𝒍^𝟐−𝟐𝟒𝒎𝒌+𝟔𝒎𝒍−𝟖𝒌𝒍 Using (𝑎+𝑏+𝑐)^2=𝑎^2+𝑏^2+𝑐^2+2𝑎𝑏+2𝑏𝑐+2𝑎𝑐 Putting 𝑎 = −3𝑚, 𝑏 = 4𝑘 & 𝑐 = −𝑙 Question 1 (viii) Use suitable identities to find the following products: (viii)(𝑥−1/3 𝑦)^3 (𝑥−1/3 𝑦)^3 = 𝑥^3−3 × 𝑥^2 × (1/3 𝑦)+3 × 𝑥 ×(1/3 𝑦)^2−(1/3 𝑦)^3 = 𝑥^3−𝑥^2 𝑦+3 × 𝑥 ×1/9 × 𝑦^2−1/27 × 𝑦^3 = 𝒙^𝟑−𝒙^𝟐 𝒚+𝟏/𝟑 𝒙𝒚^𝟐−𝟏/𝟐𝟕 𝒚^𝟑 Using (𝑎−𝑏)^3=𝑎^3−3𝑎^2 𝑏+3𝑎𝑏^2−𝑏^3 Putting 𝑎 = 𝑥, 𝑏 = 1/3 𝑦 Question 1 (ix) Use suitable identities to find the following products: (ix) (7/2 𝑘−2/3 𝑚)^3 (7/2 𝑘−2/3 𝑚)^3 = (7/2 𝑘)^3−3 × (7/2 𝑘)^2 ×(2/3 𝑚)+3 ×(7/2 𝑘)×(2/3 𝑚)^2−(2/3 𝑚)^3 = 7^3/2^3 𝑥^3−3 ×7^2/2^2 𝑘^2 ×2/3 𝑚+3 ×7/2 𝑘 ×2^2/3^2 𝑚^2−2^3/3^3 𝑚^3 = 343/8 𝑥^3−3 ×49/4 𝑘^2 ×2/3 𝑚+3 ×7/2 𝑘 ×4/9 𝑚^2−8/27 𝑚^3 = 𝟑𝟒𝟑/𝟖 𝒙^𝟑−𝟒𝟗/𝟐 𝒌^𝟐 𝒎+𝟏𝟒/𝟑 𝒌𝒎^𝟐−𝟖/𝟐𝟕 𝒎^𝟑 Using (𝑎−𝑏)^3=𝑎^3−3𝑎^2 𝑏+3𝑎𝑏^2−𝑏^3 Putting 𝑎 = 7/2 𝑘, 𝑏 = 2/3 𝑚