Algebra?

First of all take $f(x)=x^4+4 x^3-8 x^2$

Differentiate w.r.t. $x$

$f'(x)=4 x^3+12 x^2-16 x \\ =4 x(x^2+3 x-4) \\ =4 x (x+4)(x-1)$

Taking $f'(x)=0$ .There are solutions $x=0,1,-4$

Now $f''(x)=12 x^2+24 x-16$ .

$f(0)=-16>0, f(1)=30>0,f(-4)=80>0$ .So, $x=0$ has a maxima and $x=1,-4$ have mimima.

The function is decreasing in $[0,1]$ and increasing in $[1,\infty)$ .Again $f(x)$ increasing in $[-4,0]$ and decreasing in $(-\infty,-4]$ .

$f(1)=-3$ and $f(-4)=-128$

Now if a integral constant $k$ is added with this function.the value of the function will get increase.Geometrically the graph of the function will be shifted upward according to the value of $k$ .At $k=0$ f(x) has four real zeroes.If we shift the function such a way there will be only be two roots then automatically other two roots will be complex as there are real co-efficients.

Here $f(1)=-3$ .If $k=4$ then the minima part will be above X-axis.The other minima part is still below X-axis.They are contributing two real roots. So minimum value of $k=4$ .Now increase the value of $k$ untill the second minima part should not touch the X-axis.The value of second minima is $128$ .So ,it can be increased up to $k=127$ so that it will still below X-axis.

So the values of $k$ are $4,5,6, \cdots 126,127$ .

The sum is $\boxed{8122}$