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Misc 9 - Solve 21x2 - 28x + 10 = 0 - Chapter 5 NCERT - Miscellaneous

Misc 9 - Chapter 5 Class 11 Complex Numbers - Part 2

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Misc 9 Solve the equation 21π‘₯2 – 28π‘₯ + 10 = 0 21π‘₯2 – 28π‘₯ + 10 = 0 The Above equation of the form π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 = 0 Where a = 21, b = –28, and c = 10 π‘₯ = (βˆ’π‘ Β± √( 𝑏^2 βˆ’ 4π‘Žπ‘ ))/2π‘Ž π‘₯ = (βˆ’ ( βˆ’ 28) Β± √((βˆ’28)^2 βˆ’4 Γ— 21 Γ—10))/(2 Γ— 21) = (28 Β± √(784 βˆ’ 840))/42 = (28 Β± √(βˆ’56))/42 = (28 Β± √(βˆ’ 14 Γ—4 ))/42 = (28 Β± 2√(βˆ’ 14))/42 = (2 [14 Β± √(βˆ’ 14) ])/42 = (2 [ 14 Β± √(βˆ’ 14) ])/(2 Γ— 21) = ( [ 14 Β± √(βˆ’ 14) ] )/21 = ( 14 Β± √(βˆ’ 1 Γ— 14) )/21 = ( 14 Β± √(βˆ’ 1) Γ—βˆš14)/21 = ( 14 Β± √14 𝑖)/21 Thus, π‘₯ = ( 14 Β± √14 𝑖)/21

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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.