10-(9/k)-(1/k^2)=0

Solution for 10-(9/k)-(1/k^2)=0 equation:

D( k )

k^2 = 0

k = 0

k^2 = 0

k^2 = 0

1*k^2 = 0 // : 1

k^2 = 0

k = 0

k = 0

k = 0

k in (-oo:0) U (0:+oo)

10-(9/k)-(1/(k^2)) = 0

10-9*k^-1-k^-2 = 0

t_1 = k^-1

10-1*t_1^2-9*t_1^1 = 0

10-t_1^2-9*t_1 = 0

DELTA = (-9)^2-(-1*4*10)

DELTA = 121

DELTA > 0

t_1 = (121^(1/2)+9)/(-1*2) or t_1 = (9-121^(1/2))/(-1*2)

t_1 = -10 or t_1 = 1

t_1 = -10

k^-1+10 = 0

1*k^-1 = -10 // : 1

k^-1 = -10

-1 < 0

1/(k^1) = -10 // * k^1

1 = -10*k^1 // : -10

-1/10 = k^1

k = -1/10

t_1 = 1

k^-1-1 = 0

1*k^-1 = 1 // : 1

k^-1 = 1

-1 < 0

1/(k^1) = 1 // * k^1

1 = 1*k^1 // : 1

1 = k^1

k = 1

k in { -1/10, 1 }