Ex 9.5, 5 - Show that (i) (3x + 7)^2 - 84x = (3x - 7)^2 (ii) (9p - 5q)

Ex 9.5, 5 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 2
Ex 9.5, 5 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 3 Ex 9.5, 5 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 4 Ex 9.5, 5 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 5 Ex 9.5, 5 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 6 Ex 9.5, 5 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 7 Ex 9.5, 5 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 8

  1. Chapter 9 Class 8 Algebraic Expressions and Identities
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Transcript

Ex 9.5, 5 Show that. (i) (3๐‘ฅ+7)^2โˆ’84๐‘ฅ=(3๐‘ฅโˆ’7)^2 Solving LHS (3๐‘ฅ+7)^2โˆ’84๐‘ฅ = (3๐‘ฅ)^2+(7)^2+ 2(3๐‘ฅ)(7)โˆ’84๐‘ฅ = 9๐‘ฅ^2+49+42๐‘ฅโˆ’84๐‘ฅ = 9๐‘ฅ^2+49โˆ’42๐‘ฅ (๐‘Ž+๐‘)^2=๐‘Ž^2+๐‘^2โˆ’2๐‘Ž๐‘ Putting ๐‘Ž = 3๐‘ฅ & ๐‘ = 7 = (3๐‘ฅ)^2+(7)^2+ 2(3๐‘ฅ)(7)โˆ’84๐‘ฅ = 9๐‘ฅ^2+49+42๐‘ฅโˆ’84๐‘ฅ = 9๐‘ฅ^2+49โˆ’42๐‘ฅ Solving RHS (3๐‘ฅโˆ’7)^2 (๐‘Žโˆ’๐‘)^2=๐‘Ž^2+๐‘^2โˆ’2๐‘Ž๐‘ Putting ๐‘Ž = 3๐‘ฅ & ๐‘ = 7 = (3๐‘ฅ)^2+(7)^2โˆ’2(3๐‘ฅ)(7) = (3^2ร—๐‘ฅ^2 )+49โˆ’(2ร—3ร—7)๐‘ฅ = 9๐‘ฅ^2+49โˆ’42๐‘ฅ Thus LHS = RHS Hence proved Ex 9.5, 5 Show that. (ii) (9๐‘โˆ’5๐‘ž)^2+180๐‘๐‘ž=(9๐‘+5๐‘ž)^2 Solving LHS (9๐‘โˆ’5๐‘ž)^2+180๐‘๐‘ž (๐‘Žโˆ’๐‘)^2=๐‘Ž^2+๐‘^2โˆ’2๐‘Ž๐‘ Putting ๐‘Ž = 9๐‘ & ๐‘ = 5๐‘ž (๐‘Žโˆ’๐‘)^2=๐‘Ž^2+๐‘^2โˆ’2๐‘Ž๐‘ Putting ๐‘Ž = 9๐‘ & ๐‘ = 5๐‘ž Solving RHS (9๐‘+5๐‘ž)^2 (๐‘Ž+๐‘)^2=๐‘Ž^2+๐‘^2+2๐‘Ž๐‘ Putting ๐‘Ž = 9๐‘ & ๐‘ = 5๐‘ž = (9๐‘)^2+(5๐‘ž)^2+2(9๐‘)(5๐‘ž) = 81๐‘^2+25๐‘ž^2+90๐‘๐‘ž Thus LHS = RHS Hence proved Ex 9.5, 5 Show that. (iii) (4/3 ๐‘šโˆ’3/4 ๐‘›)^2+2๐‘š๐‘›=16/9 ๐‘š^2+9/16 ๐‘›^2 Solving LHS (4/3 ๐‘šโˆ’3/4 ๐‘›)^2+2๐‘š๐‘› (๐‘Žโˆ’๐‘)^2=๐‘Ž^2+๐‘^2โˆ’2๐‘Ž๐‘ Putting ๐‘Ž = 4/3 ๐‘š & ๐‘ = 3/4 ๐‘› = (4/3 ๐‘š)^2+(3/4 ๐‘›)^2โˆ’2(4/3 ๐‘š)(3/4 ๐‘›)+2๐‘š๐‘› = (4/3)^2ร—๐‘š^2+(3/4)^2ร—๐‘›^2โˆ’2๐‘š๐‘›+2๐‘š๐‘› = 16/9 ๐‘š^2+9/16 ๐‘›^2 = RHS Since LHS = RHS Hence proved Ex 9.5, 5 Show that. (iv) (4๐‘๐‘ž+3๐‘ž)^2โˆ’(4๐‘๐‘žโˆ’3๐‘ž)^2=48๐‘๐‘ž^2 Solving (๐Ÿ’๐’‘๐’’+๐Ÿ‘๐’’)^๐Ÿ (๐‘Ž+๐‘)^2=๐‘Ž^2+๐‘^2+2๐‘Ž๐‘ Putting ๐‘Ž = 4๐‘๐‘ž & ๐‘ = 3๐‘ž (๐‘Ž+๐‘)^2=๐‘Ž^2+๐‘^2+2๐‘Ž๐‘ Putting ๐‘Ž = 4๐‘๐‘ž & ๐‘ = 3๐‘ž = (4๐‘๐‘ž)^2+(3๐‘ž)^2+2(4๐‘๐‘ž)(3๐‘ž) = 16๐‘^2 ๐‘ž^2+9๐‘ž^2+24๐‘๐‘ž^2 Solving (๐Ÿ’๐’‘๐’’โˆ’๐Ÿ‘๐’’)^๐Ÿ (๐‘Žโˆ’๐‘)^2=๐‘Ž^2+๐‘^2โˆ’2๐‘Ž๐‘ Putting ๐‘Ž = 4๐‘๐‘ž & ๐‘ = 3๐‘ž = (4๐‘๐‘ž)^2+(3๐‘ž)^2โˆ’2(4๐‘๐‘ž)(3๐‘ž) = 16๐‘^2 ๐‘ž^2+9๐‘ž^2โˆ’24๐‘๐‘ž^2 Solving LHS (4๐‘๐‘ž+3๐‘ž)^2โˆ’(4๐‘๐‘žโˆ’3๐‘ž)^2 = (16๐‘^2 ๐‘ž^2+9๐‘ž^2+24๐‘๐‘ž^2 )โˆ’(16๐‘^2 ๐‘ž^2+9๐‘ž^2โˆ’24๐‘๐‘ž^2 ) = 16๐‘^2 ๐‘ž^2+9๐‘ž^2+24๐‘๐‘ž^2โˆ’16๐‘^2 ๐‘ž^2โˆ’9๐‘ž^2+24๐‘๐‘ž^2 = (16๐‘^2 ๐‘ž^2โˆ’16๐‘^2 ๐‘ž^2 )+(9๐‘ž^2โˆ’9๐‘ž^2 )+(24๐‘๐‘ž^2+24๐‘๐‘ž^2 ) = 0+0+48๐‘๐‘ž^2 = 48๐‘๐‘ž^2 = R.H.S Since LHS = RHS Hence proved Ex 9.5, 5 Show that. (v) (๐‘Žโˆ’๐‘) (๐‘Ž+๐‘)+(๐‘โˆ’๐‘)(๐‘+๐‘)+(๐‘โˆ’๐‘Ž)(๐‘+๐‘Ž)=0 Solving LHS (๐‘Žโˆ’๐‘) (๐‘Ž+๐‘)+(๐‘โˆ’๐‘)(๐‘+๐‘)+(๐‘โˆ’๐‘Ž)(๐‘+๐‘Ž) Using (๐‘ฅโˆ’๐‘ฆ)(๐‘ฅ+๐‘ฆ)=๐‘ฅ^2+๐‘ฆ^2 = (๐‘Ž^2โˆ’๐‘^2 )+(๐‘^2โˆ’๐‘^2 )+(๐‘^2โˆ’๐‘^2 ) = ๐‘Ž^2โˆ’๐‘^2+๐‘^2โˆ’๐‘^2+๐‘^2โˆ’๐‘Ž^2 = (๐‘Ž^2โˆ’๐‘Ž^2 )+(โˆ’๐‘^2+๐‘^2 )+(โˆ’๐‘^2+๐‘^2 ) = 0+0+0 = 0 = R.H.S Since LHS = RHS Hence proved

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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.