(9/2w)+(5/w)=1

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Solution for (9/2w)+(5/w)=1 equation: 


D( w )

w = 0

w = 0

w = 0

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

(9/2)*w+5/w = 1 // - 1

(9/2)*w+5/w-1 = 0

9/2*w^1+5*w^-1-1*w^0 = 0

(9/2*w^2-1*w^1+5*w^0)/(w^1) = 0 // * w^2

w^1*(9/2*w^2-1*w^1+5*w^0) = 0

w^1

(9/2)*w^2-w+5 = 0

(9/2)*w^2-w+5 = 0

DELTA = (-1)^2-(4*5*(9/2))

DELTA = -89

DELTA < 0

w in { } 

w belongs to the empty set

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