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Question 2 - Miscellaneous - Chapter 4 Class 11 Complex Numbers

Last updated at May 29, 2023 by

Misc 6 - Solve 3x2 - 4x + 20/3 = 0 - Chapter 5 NCERT - Quadaratic equation

Misc 6 - Chapter 5 Class 11 Complex Numbers - Part 2

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Question 2 Solve the equation 3π‘₯2 – 4π‘₯ + 20/3 = 0 3π‘₯2 – 4π‘₯ + 20/3 = 0 Multiplying both sides by 3 3 Γ— (3π‘₯2 – 4π‘₯ "+ " 20/3) = 3 Γ— 0 3 Γ— 3x2 – 3 Γ— 4x + 3 Γ— 20/3 = 0 9π‘₯2 – 12π‘₯ + 20 = 0 The above equation is of the form π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 = 0 Where a = 9, b = –12, and c = 20 x = (βˆ’π‘Β±βˆš( 𝑏^2 βˆ’4π‘Žπ‘ ))/2π‘Ž = (βˆ’(βˆ’12) Β± √((βˆ’12)^2 βˆ’ 4 Γ— 9 Γ— 20))/(2 Γ— 9) = (12 Β± √(144 βˆ’720))/18 = (12 Β± √(βˆ’576))/18 = (12 Β± √(βˆ’1 Γ— 576))/18 = (12 Β± √(βˆ’1 ) Γ— √( 576))/18 = (12 Β± 𝑖 Γ— √( 576))/18 = (12 Β± 𝑖 Γ— √( 242))/18 = (12 Β± 𝑖 Γ— 24)/18 = (6(2 Β± 𝑖 Γ— 4))/18 = ((2 Β± 𝑖 Γ— 4))/3 = ((2 Β± 𝑖4))/3 Thus, π‘₯ = ((2 Β± 𝑖4))/3

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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.