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

Solution for 1/k=(2x-x^2) equation:

x in (-oo:+oo)

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

1/k+x^2-(2*x) = 0

1/k+x^2-2*x = 0

k^-1+x^2-2*x = 0

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

DELTA = 4-4*k^-1

4-4*k^-1 = 0

-4*k^-1 = -4 // : -4

k^-1 = 1

-1 < 0

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

1 = 1*k^1 // : 1

1 = k^1

k = 1

DELTA = 0  <=>  t_1 = 1

x = 2/(1*2) i k = 1

x = 1 i k = 1

( x = ((4-4*k^-1)^(1/2)+2)/(1*2) or x = (2-(4-4*k^-1)^(1/2))/(1*2) ) i k > 1

( x = ((4-4*k^-1)^(1/2)+2)/2 or x = (2-(4-4*k^-1)^(1/2))/2 ) i k > 1

k-1 > 0

k-1 > 0 // + 1

k > 1

x in { 1, ((4-4*k^-1)^(1/2)+2)/2, (2-(4-4*k^-1)^(1/2))/2 }