5-4/k-1/k^2=0

Solution for 5-4/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)

5-(4/k)-(1/(k^2)) = 0

5-4*k^-1-k^-2 = 0

t_1 = k^-1

5-1*t_1^2-4*t_1^1 = 0

5-t_1^2-4*t_1 = 0

DELTA = (-4)^2-(-1*4*5)

DELTA = 36

DELTA > 0

t_1 = (36^(1/2)+4)/(-1*2) or t_1 = (4-36^(1/2))/(-1*2)

t_1 = -5 or t_1 = 1

t_1 = -5

k^-1+5 = 0

1*k^-1 = -5 // : 1

k^-1 = -5

-1 < 0

1/(k^1) = -5 // * k^1

1 = -5*k^1 // : -5

-1/5 = k^1

k = -1/5

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 { -1/5, 1 }