1=(2/x)+(8/x^2)

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Solution for 1=(2/x)+(8/x^2) 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 = 2/x+8/(x^2) // - 2/x+8/(x^2)

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

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

t_1 = x^-1

1-8*t_1^2-2*t_1^1 = 0

1-8*t_1^2-2*t_1 = 0

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

DELTA = 36

DELTA > 0

t_1 = (36^(1/2)+2)/(-8*2) or t_1 = (2-36^(1/2))/(-8*2)

t_1 = -1/2 or t_1 = 1/4

t_1 = -1/2

x^-1+1/2 = 0

1*x^-1 = -1/2 // : 1

x^-1 = -1/2

-1 < 0

1/(x^1) = -1/2 // * x^1

1 = -1/2*x^1 // : -1/2

-2 = x^1

x = -2

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 { -2, 4 }

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