(1/x^2)+(13/x)+42=0

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Solution for (1/x^2)+(13/x)+42=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)

13/x+1/(x^2)+42 = 0

13*x^-1+x^-2+42 = 0

t_1 = x^-1

1*t_1^2+13*t_1^1+42 = 0

t_1^2+13*t_1+42 = 0

DELTA = 13^2-(1*4*42)

DELTA = 1

DELTA > 0

t_1 = (1^(1/2)-13)/(1*2) or t_1 = (-1^(1/2)-13)/(1*2)

t_1 = -6 or t_1 = -7

t_1 = -7

x^-1+7 = 0

1*x^-1 = -7 // : 1

x^-1 = -7

-1 < 0

1/(x^1) = -7 // * x^1

1 = -7*x^1 // : -7

-1/7 = x^1

x = -1/7

t_1 = -6

x^-1+6 = 0

1*x^-1 = -6 // : 1

x^-1 = -6

-1 < 0

1/(x^1) = -6 // * x^1

1 = -6*x^1 // : -6

-1/6 = x^1

x = -1/6

x in { -1/7, -1/6 }

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