1/x+1/x^2=5/16

Solution for 1/x+1/x^2=5/16 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)

1/x+1/(x^2) = 5/16 // - 5/16

1/x+1/(x^2)-(5/16) = 0

1/x+1/(x^2)-5/16 = 0

x^-1+x^-2-5/16 = 0

t_1 = x^-1

1*t_1^2+1*t_1^1-5/16 = 0

t_1^2+t_1-5/16 = 0

DELTA = 1^2-(-5/16*1*4)

DELTA = 9/4

DELTA > 0

t_1 = ((9/4)^(1/2)-1)/(1*2) or t_1 = (-(9/4)^(1/2)-1)/(1*2)

t_1 = 1/4 or t_1 = -5/4

t_1 = -5/4

x^-1+5/4 = 0

1*x^-1 = -5/4 // : 1

x^-1 = -5/4

-1 < 0

1/(x^1) = -5/4 // * x^1

1 = -5/4*x^1 // : -5/4

-4/5 = x^1

x = -4/5

t_1 = 1/4

x^-1-1/4 = 0

1*x^-1 = 1/4 // : 1

x^-1 = 1/4

-1 < 0

1/(x^1) = 1/4 // * x^1

1 = 1/4*x^1 // : 1/4

4 = x^1

x = 4

x in { -4/5, 4 }