1/z+1/2*z=1/2

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Solution for 1/z+1/2*z=1/2 equation:


D( z )

z = 0

z = 0

z = 0

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

(1/2)*z+1/z = 1/2 // - 1/2

(1/2)*z+1/z-(1/2) = 0

(1/2)*z+1/z-1/2 = 0

1/2*z^1+1*z^-1-1/2*z^0 = 0

(1/2*z^2-1/2*z^1+1*z^0)/(z^1) = 0 // * z^2

z^1*(1/2*z^2-1/2*z^1+1*z^0) = 0

z^1

(1/2)*z^2+(-1/2)*z+1 = 0

(1/2)*z^2+(-1/2)*z+1 = 0

DELTA = (-1/2)^2-(1*4*(1/2))

DELTA = -7/4

DELTA < 0

z in { }

z belongs to the empty set

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