(1/x^2)+(4/x)-5=0

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Solution for (1/x^2)+(4/x)-5=0 equation: 


D( x )

x = 0

x^2 = 0

x = 0

x = 0

x^2 = 0

x^2 = 0

1*x^2 = 0 // : 1

x^2 = 0

x = 0

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

4/x+1/(x^2)-5 = 0

4*x^-1+x^-2-5 = 0

t_1 = x^-1

1*t_1^2+4*t_1^1-5 = 0

t_1^2+4*t_1-5 = 0

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

DELTA = 36

DELTA > 0

t_1 = (36^(1/2)-4)/(1*2) or t_1 = (-36^(1/2)-4)/(1*2)

t_1 = 1 or t_1 = -5

t_1 = -5

x^-1+5 = 0

1*x^-1 = -5 // : 1

x^-1 = -5

-1 < 0

1/(x^1) = -5 // * x^1

1 = -5*x^1 // : -5

-1/5 = x^1

x = -1/5

t_1 = 1

x^-1-1 = 0

1*x^-1 = 1 // : 1

x^-1 = 1

-1 < 0

1/(x^1) = 1 // * x^1

1 = 1*x^1 // : 1

1 = x^1

x = 1

x in { -1/5, 1 }

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