3=(k-5/k)-(1/k)

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Solution for 3=(k-5/k)-(1/k) equation: 


D( k )

k = 0

k = 0

k = 0

k in (-oo:0) U (0:+oo)

3 = k-(5/k)-(1/k) // - k-(5/k)-(1/k)

5/k-k+1/k+3 = 0

6*k^-1-1*k^1+3*k^0 = 0

(3*k^1-1*k^2+6*k^0)/(k^1) = 0 // * k^2

k^1*(3*k^1-1*k^2+6*k^0) = 0

k^1

3*k-k^2+6 = 0

3*k-k^2+6 = 0

DELTA = 3^2-(-1*4*6)

DELTA = 33

DELTA > 0

k = (33^(1/2)-3)/(-1*2) or k = (-33^(1/2)-3)/(-1*2)

k = (33^(1/2)-3)/(-2) or k = (33^(1/2)+3)/2

k in { (33^(1/2)-3)/(-2), (33^(1/2)+3)/2} 

k in { (33^(1/2)-3)/(-2), (33^(1/2)+3)/2 }

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