Miscellaneous

Chapter 4 Class 11 Complex Numbers
Serial order wise

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### Transcript

Question 2 Solve the equation 3π₯2 β 4π₯ + 20/3 = 0 3π₯2 β 4π₯ + 20/3 = 0 Multiplying both sides by 3 3 Γ (3π₯2 β 4π₯ "+ " 20/3) = 3 Γ 0 3 Γ 3x2 β 3 Γ 4x + 3 Γ 20/3 = 0 9π₯2 β 12π₯ + 20 = 0 The above equation is of the form ππ₯2 + ππ₯ + π = 0 Where a = 9, b = β12, and c = 20 x = (βπΒ±β( π^2 β4ππ ))/2π = (β(β12) Β± β((β12)^2 β 4 Γ 9 Γ 20))/(2 Γ 9) = (12 Β± β(144 β720))/18 = (12 Β± β(β576))/18 = (12 Β± β(β1 Γ 576))/18 = (12 Β± β(β1 ) Γ β( 576))/18 = (12 Β± π Γ β( 576))/18 = (12 Β± π Γ β( 242))/18 = (12 Β± π Γ 24)/18 = (6(2 Β± π Γ 4))/18 = ((2 Β± π Γ 4))/3 = ((2 Β± π4))/3 Thus, π₯ = ((2 Β± π4))/3

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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.