1/x^4+10/x^2+22=0

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Solution for 1/x^4+10/x^2+22=0 equation: 


D( x )

x^4 = 0

x^2 = 0

x^4 = 0

x^4 = 0

1*x^4 = 0 // : 1

x^4 = 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)

10/(x^2)+1/(x^4)+22 = 0

10*x^-2+x^-4+22 = 0

t_1 = x^-2

1*t_1^2+10*t_1^1+22 = 0

t_1^2+10*t_1+22 = 0

DELTA = 10^2-(1*4*22)

DELTA = 12

DELTA > 0

t_1 = (12^(1/2)-10)/(1*2) or t_1 = (-12^(1/2)-10)/(1*2)

t_1 = (2*3^(1/2)-10)/2 or t_1 = (-2*3^(1/2)-10)/2

t_1 = (-2*3^(1/2)-10)/2

x^-2-((-2*3^(1/2)-10)/2) = 0

1*x^-2 = (-2*3^(1/2)-10)/2 // : 1

x^-2 = (-2*3^(1/2)-10)/2

-2 < 0

1/(x^2) = (-2*3^(1/2)-10)/2 // * x^2

1 = ((-2*3^(1/2)-10)/2)*x^2 // : (-2*3^(1/2)-10)/2

2*(-2*3^(1/2)-10)^-1 = x^2

t_1 = (2*3^(1/2)-10)/2

x^-2-((2*3^(1/2)-10)/2) = 0

1*x^-2 = (2*3^(1/2)-10)/2 // : 1

x^-2 = (2*3^(1/2)-10)/2

-2 < 0

1/(x^2) = (2*3^(1/2)-10)/2 // * x^2

1 = ((2*3^(1/2)-10)/2)*x^2 // : (2*3^(1/2)-10)/2

2*(2*3^(1/2)-10)^-1 = x^2

x belongs to the empty set

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