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  1. Chapter 9 Class 8 Algebraic Expressions and Identities
  2. Serial order wise

Transcript

Ex 9.3, 4 (a) (a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) x = 1/2 3π‘₯ (4π‘₯βˆ’5)+3 = (3π‘₯Γ—4π‘₯)βˆ’(3π‘₯Γ—5)+3 = 12π‘₯^2βˆ’15π‘₯+3 (i) For 𝒙=πŸ‘ Putting π‘₯=3 in expression 12π‘₯^2βˆ’15π‘₯+3 = 12(3)^2βˆ’15(3)+3 = (12Γ—9)βˆ’(15Γ—3) +3 = 108βˆ’45 +3 = 108βˆ’42 = 66 (ii) For 𝒙=𝟏/𝟐 Putting π‘₯=1/2 in expression 12π‘₯^2βˆ’15π‘₯+3 = 12(1/2)^2βˆ’15(1/2)+3 = 12(1/2Γ—1/2)βˆ’ 15/2 +3 = 12Γ—(1/4)βˆ’15/2+3 = 3βˆ’15/2+3 = 6βˆ’15/2 = (6 Γ— 2 βˆ’ 15)/2 = (12 βˆ’ 15)/2 = (βˆ’πŸ‘)/𝟐 Ex 9.3, 4 (b) (b) Simplify a (a2 + a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1 (iii) a = – 1. π‘Ž (π‘Ž^2+π‘Ž+1)+5 = (π‘ŽΓ—π‘Ž^2 )+(π‘ŽΓ—π‘Ž)+(π‘ŽΓ—1)+5 = π‘Ž^3+π‘Ž^2+π‘Ž+5 (i) For 𝒂=𝟎 Putting π‘Ž=0 in expression π‘Ž^3+π‘Ž^2+π‘Ž+5 = (0)^3+(0)^2+0+5 = 0+0+0+5 = πŸ“ (i) For 𝒂=𝟏 Putting π‘Ž=1 in expression π‘Ž^3+π‘Ž^2+π‘Ž+5 = γ€–(1)γ€—^3+ (1)^2+1+5 = 1+1+1+5 = πŸ– (iii) For 𝒂=βˆ’πŸ Putting π‘Ž=βˆ’1 in expression π‘Ž^3+π‘Ž^2+π‘Ž+5 = γ€–(βˆ’1)γ€—^3+ (βˆ’1)^2+(βˆ’1)+5 = (βˆ’1)^2Γ—(βˆ’1)+(1)+(βˆ’1)+5 = 1Γ—(βˆ’1)+1βˆ’1+5 = βˆ’1+1βˆ’1+5 = βˆ’2+6 = πŸ’

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.