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

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

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

1/x-x^-2-2+8 = 0

x^-1-x^-2+6 = 0

t_1 = x^-1

1*t_1^1-1*t_1^2+6 = 0

t_1-t_1^2+6 = 0

DELTA = 1^2-(-1*4*6)

DELTA = 25

DELTA > 0

t_1 = (25^(1/2)-1)/(-1*2) or t_1 = (-25^(1/2)-1)/(-1*2)

t_1 = -2 or t_1 = 3

t_1 = -2

x^-1+2 = 0

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

x^-1 = -2

-1 < 0

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

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

-1/2 = x^1

x = -1/2

t_1 = 3

x^-1-3 = 0

1*x^-1 = 3 // : 1

x^-1 = 3

-1 < 0

1/(x^1) = 3 // * x^1

1 = 3*x^1 // : 3

1/3 = x^1

x = 1/3

x in { -1/2, 1/3 }

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