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Ex 5.3
Ex 5.3, 2 Important Deleted for CBSE Board 2023 Exams You are here
Ex 5.3, 3 Deleted for CBSE Board 2023 Exams
Ex 5.3, 4 Deleted for CBSE Board 2023 Exams
Ex 5.3, 5 Deleted for CBSE Board 2023 Exams
Ex 5.3, 6 Important Deleted for CBSE Board 2023 Exams
Ex 5.3, 7 Deleted for CBSE Board 2023 Exams
Ex 5.3, 8 Deleted for CBSE Board 2023 Exams
Ex 5.3, 9 Important Deleted for CBSE Board 2023 Exams
Ex 5.3, 10 Deleted for CBSE Board 2023 Exams
Last updated at March 22, 2023 by Teachoo
Ex 5.3, 2 Solve the equation 2π₯2 + π₯ + 1 = 0 2x2 + x + 1 = 0 The above equation is of the form ππ₯2 + ππ₯ + π = 0 Where a = 2 , b = 1 , c = 1 π₯ = (βπ Β± β( π^2 β 4ππ ))/2π = (β1 Β± β((1)^2 β 4 Γ 2 Γ 1))/(2 Γ 2) = (β1Β± β(1 β 8))/4 = (β1Β± β( β 7))/4 = (β1 Β± β(β1 Γ 7))/4 = (β1 Β± β(β1 Γ 7))/4 = (β1 Β± β(β1 )Γ β7)/4 = (β1 Β± β7 π)/4 Thus, π₯ = (β1 Β± β7 π)/2