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Ex 5.3, 2 - Chapter 5 Class 11 Complex Numbers - Part 2


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Ex 5.3, 2 Solve the equation 2π‘₯2 + π‘₯ + 1 = 0 2x2 + x + 1 = 0 The above equation is of the form π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 = 0 Where a = 2 , b = 1 , c = 1 π‘₯ = (βˆ’π‘ Β± √( 𝑏^2 βˆ’ 4π‘Žπ‘ ))/2π‘Ž = (βˆ’1 Β± √((1)^2 βˆ’ 4 Γ— 2 Γ— 1))/(2 Γ— 2) = (βˆ’1Β± √(1 βˆ’ 8))/4 = (βˆ’1Β± √( βˆ’ 7))/4 = (βˆ’1 Β± √(βˆ’1 Γ— 7))/4 = (βˆ’1 Β± √(βˆ’1 Γ— 7))/4 = (βˆ’1 Β± √(βˆ’1 )Γ— √7)/4 = (βˆ’1 Β± √7 𝑖)/4 Thus, π‘₯ = (βˆ’1 Β± √7 𝑖)/2

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.