x-2/4x+3=x+8/x

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Solution for x-2/4x+3=x+8/x equation: 


D( x )

x = 0

x = 0

x = 0

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

x-((2/4)*x)+3 = x+8/x // - x+8/x

x-((2/4)*x)-x-(8/x)+3 = 0

x+(-1/2)*x-x-8*x^-1+3 = 0

3*x^0-1/2*x^1-8*x^-1 = 0

(3*x^1-1/2*x^2-8*x^0)/(x^1) = 0 // * x^2

x^1*(3*x^1-1/2*x^2-8*x^0) = 0

x^1

(-1/2)*x^2+3*x-8 = 0

(-1/2)*x^2+3*x-8 = 0

DELTA = 3^2-(-8*4*(-1/2))

DELTA = -7

DELTA < 0

x in { } 

x belongs to the empty set

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