Example 11
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Example, 11 Solve β5x2 + x + β5 = 0 β5x2 + x + β5 = 0 The above equation is of the form ππ₯2 + ππ₯ + π = 0 where a = β5 , b = 1, and c = β5 x = (βπ Β± β( π^2 β 4ππ ))/2π Putting values x = (β1 Β± β(1^2 β 4 Γ β(5 )Γ β5) )/(2 Γ β5) = (β1 Β± β(1 β 4 Γ 5))/(2β5) = (β1 Β± β(1 β 20))/(2β5) = (β1 Β± β(β19))/(2β5) = (β1 Β± β19 Γ β(β1))/(2β5) = (β1 Β± β19 π)/(2β5) Thus, π₯ = (β1 Β± β19 π)/(2β5)
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Davneet Singh
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