Ex 9.5, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 4

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Ex 9.5, 4 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 5

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

Transcript

Ex 9.5, 4 Simplify. (iii) (7π‘šβˆ’8𝑛)^2+(7π‘š+8𝑛)^2 Solving (πŸ•π’Žβˆ’πŸ–π’)^𝟐 (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 7π‘š & 𝑏 = 8𝑛 = (7π‘š)^2+(8𝑛)^2βˆ’2(7π‘š)(8𝑛) = 49π‘š^2+64𝑛^2βˆ’(14Γ—8)π‘šπ‘› = 49π‘š^2+64𝑛^2βˆ’112π‘šπ‘› Solving (πŸ•π’Ž+πŸ–π’)^𝟐 (π‘Ž+𝑏)^2=π‘Ž^2+𝑏^2+2π‘Žπ‘ Putting π‘Ž = 7π‘š & 𝑏 = 8𝑛 = (7π‘š)^2+(8𝑛)^2+2(7π‘š)(8𝑛) = 49π‘š^2+64𝑛^2+(14Γ—8)π‘šπ‘› = 49π‘š^2+64𝑛^2+112π‘šπ‘› ∴ (7π‘šβˆ’8𝑛)^2+(7π‘š+8𝑛)^2 = (49π‘š^2+64𝑛^2βˆ’112π‘šπ‘›)+(49π‘š^2+64𝑛^2+112π‘šπ‘›) = (49π‘š^2+49π‘š^2 )+(64𝑛^2+64𝑛^2 )+(βˆ’112π‘šπ‘›+112π‘šπ‘›) = πŸ—πŸ–π’Ž^𝟐+πŸπŸπŸ–π’^𝟐

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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.