Ex 9.3, 4 - (a) Simplify 3x (4x - 5) + 3 and find its values for

Ex 9.3, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 2
Ex 9.3, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 3 Ex 9.3, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 4 Ex 9.3, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 5

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

Transcript

Ex 9.3, 4 (a) (a) Simplify 3x (4x โ€“ 5) + 3 and find its values for (i) x = 3 (ii) x = 1/2 3๐‘ฅ (4๐‘ฅโˆ’5)+3 = (3๐‘ฅร—4๐‘ฅ)โˆ’(3๐‘ฅร—5)+3 = 12๐‘ฅ^2โˆ’15๐‘ฅ+3 (i) For ๐’™=๐Ÿ‘ Putting ๐‘ฅ=3 in expression 12๐‘ฅ^2โˆ’15๐‘ฅ+3 = 12(3)^2โˆ’15(3)+3 = (12ร—9)โˆ’(15ร—3) +3 = 108โˆ’45 +3 = 108โˆ’42 = 66 (ii) For ๐’™=๐Ÿ/๐Ÿ Putting ๐‘ฅ=1/2 in expression 12๐‘ฅ^2โˆ’15๐‘ฅ+3 = 12(1/2)^2โˆ’15(1/2)+3 = 12(1/2ร—1/2)โˆ’ 15/2 +3 = 12ร—(1/4)โˆ’15/2+3 = 3โˆ’15/2+3 = 6โˆ’15/2 = (6 ร— 2 โˆ’ 15)/2 = (12 โˆ’ 15)/2 = (โˆ’๐Ÿ‘)/๐Ÿ 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 = โ€“ 1. ๐‘Ž (๐‘Ž^2+๐‘Ž+1)+5 = (๐‘Žร—๐‘Ž^2 )+(๐‘Žร—๐‘Ž)+(๐‘Žร—1)+5 = ๐‘Ž^3+๐‘Ž^2+๐‘Ž+5 (i) For ๐’‚=๐ŸŽ Putting ๐‘Ž=0 in expression ๐‘Ž^3+๐‘Ž^2+๐‘Ž+5 = (0)^3+(0)^2+0+5 = 0+0+0+5 = ๐Ÿ“ (i) For ๐’‚=๐Ÿ Putting ๐‘Ž=1 in expression ๐‘Ž^3+๐‘Ž^2+๐‘Ž+5 = ใ€–(1)ใ€—^3+ (1)^2+1+5 = 1+1+1+5 = ๐Ÿ– (iii) For ๐’‚=โˆ’๐Ÿ Putting ๐‘Ž=โˆ’1 in expression ๐‘Ž^3+๐‘Ž^2+๐‘Ž+5 = ใ€–(โˆ’1)ใ€—^3+ (โˆ’1)^2+(โˆ’1)+5 = (โˆ’1)^2ร—(โˆ’1)+(1)+(โˆ’1)+5 = 1ร—(โˆ’1)+1โˆ’1+5 = โˆ’1+1โˆ’1+5 = โˆ’2+6 = ๐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 10 years. He provides courses for Maths and Science at Teachoo.