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

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Question 5 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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.