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Ex 5.3
Ex 5.3, 2 Important Deleted for CBSE Board 2023 Exams
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Ex 5.3, 5 Deleted for CBSE Board 2023 Exams
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Ex 5.3, 10 Deleted for CBSE Board 2023 Exams
Last updated at March 22, 2023 by Teachoo
Ex5.3, 4 Solve the equation βπ₯2 + π₯ β 2 = 0 βπ₯2 + π₯ β 2 = 0 The above equation is of the form ππ₯2 + ππ₯ + π = 0 Where a = β1, b = 1, and c = β2 π₯ = (βπ Β± β( π^2 β 4ππ ))/2π π₯ = (β1 Β± β((1)^2 β 4 Γ (β1) (β2)))/(2 Γ "(β" 1)) = (β1Β± β(1 β 4 Γ (+2)))/("β" 2) = (1 Β± β(1 β 8))/(β2) = (β 1 Β± β(β 7))/(β2) = (β 1 Β± β( 7 Γβ1))/(β2) = (β1 Β± β7 π)/(β 2) = (β(β1 Β± β7 π))/( 2) = (1 β β7 π)/2 Thus, π₯ = (1 β β7 π)/2