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Chapter 5 Class 11 Complex Numbers (Term 1)

Serial order wise

Last updated at May 29, 2018 by Teachoo

Misc 6 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 ππ₯2 + ππ₯ + π = 0 Where a = 9, b = β12, and c = 20 x = (βπΒ±β( π^2 β4ππ ))/2π = (β(β12) Β± β((β12)^2 β 4 Γ 9 Γ 20))/(2 Γ 9) = (12 Β± β(144 β720))/18 = (12 Β± β(β576))/18 = (12 Β± β(β1 Γ 576))/18 = (12 Β± β(β1 ) 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± π4))/3 Thus, π₯ = ((2 Β± π4))/3