1. Chapter 9 Class 8 Algebraic Expressions and Identities
2. Concept wise
3. Algebra Identities - Identity II

Transcript

Ex 9.5, 3 Find the following squares by using the identities. (i) (πβ7)^2 (πβ7)^2 (π₯βπ¦)^2=π₯^2+π¦^2β2π₯π¦ Putting π₯ = π & π¦ = 7 = (π)^2+(7)^2β2(π)(7) = π^2+49β(2Γ7)Γπ = π^π+ππβπππ Ex 9.5, 3 Find the following squares by using the identities. (ii) (π₯π¦+3π§)^2 (π₯π¦+3π§)^2 (π+π)^2=π^2+π^2+2ππ Putting π = π₯π¦ & π = 3π§ = (π₯π¦)^2+(3π§)^2+2(π₯π¦)(3π§) = (π₯^2Γπ¦^2 )+(3^2Γπ§^2 )+(2Γ3)Γ(π₯Γπ¦Γπ§) = π^π π^π+ππ^π+ππππ Ex 9.5, 3 Find the following squares by using the identities. (iii) (6π₯^2β5π¦)^2 (6π₯^2β5π¦)^2 (πβπ)^2=π^2+π^2β2ππ Putting π = 6π₯^2 & π = 5π¦ = (6π₯^2 )^2+(5π¦)^2β2(6π₯^2 )(5π¦) = (6^2Γπ₯^(2 Γ 2) )+(5^2Γπ¦^2 )β(2Γ6Γ5)Γ(π₯^2Γπ¦) = (36Γπ₯^4 )+(25Γπ¦^2 )β(60Γπ₯^2 π¦) = πππ^π+πππ^πβπππ^π π Ex 9.5, 3 Find the following squares by using the identities. (iv) (2/3 π+3/2 π)^2 (2/3 π+3/2 π)^2 (π+π)^2=π^2+π^2+2ππ Putting π = 2/3 π & π = 3/2 π = (2/3 π)^2+(3/2 π)^2+2(2/3 π)(3/2 π) = (2/3)^2Γπ^2+(3/2)^2Γπ^2+((2 Γ 2 Γ 3)/(3 Γ 2))Γ(πΓπ) = π/π π^π+ π/π π^π+πππ Ex 9.5, 3 Find the following squares by using the identities. (v) (0.4πβ0.5π)^2 (0.4πβ0.5π)^2 (πβπ)^2=π^2+π^2β2ππ Putting π = 0.4π & π = 0.5π = (0.4π)^2+(0.5π)^2β2(0.4π)(0.5π) = (4/10)^2Γπ^2+(5/10)^2Γπ^2β(2Γ4/10Γ5/10)Γ(πΓπ) = 16/100 π^2+ 25/100 π^2β4/10 ππ = π.πππ^π+π.πππ^πβπ.πππ Ex 9.5, 3 Find the following squares by using the identities. (vi) (2π₯π¦+5π¦)^2 (2π₯π¦+5π¦)^2 (π+π)^2=π^2+π^2+2ππ Putting π = 2π₯π¦ & π = 5π¦ = (2π₯π¦)^2+(5π¦)^2+2(2π₯π¦)(5π¦) = (2^2Γπ₯^2Γπ¦^2 )+(5^2Γπ¦^2 )+(2Γ2Γ5)Γπ₯Γ(π¦Γπ¦) = ππ^π π^π+πππ^π+ππππ^π

Algebra Identities - Identity II

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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.