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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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Misc 6 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 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.