Question 27170
{{{x^2 + kx + k = 0}}} has no real roots
that means the roots are imaginary- they involve {{{sqrt(-1)}}}, which is i.
The quadratic formula is:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
where the equation is 
{{{a*x^2 +b*x + c = 0}}}
but we have
{{{x^2 + k*x + k = 0}}}
so,
a = 1
b = k
c = k
substitute these in the formula
{{{x = (-k +- sqrt( k^2-4*1*k ))/(2*1) }}}
{{{x = (-k +- sqrt( k^2-4*k ))/2 }}}
The discriminant, which is everything under the square root sign,
needs to be negative if the roots are imaginary
that means 
{{{ k^2 - 4*k  < 0}}}
factoring
{{{ k(k - 4) < 0}}}
k can't be greater than 4 or equal to 4.
that would make the inequality false
{{{ 4(4 - 4) < 0}}}
{{{4 * 0 < 0}}}
not true
or
{{{ 5(5 - 4) < 0}}}
{{{5 * 1 < 0}}}
not true either
k can't be zero or less than zero either
I would get
{{{ 0(0 - 4) < 0}}}
not true
or
{{{ -1(-1 - 4) < 0}}}
{{{-1 * -5 < 0}}}
not true either
so, that means
{{{4 > k > 0}}}
meaning k is between 0 and 4 but doesn't include 0 or 4.