Consider an equation $$x^4-2x^3+2x^2-x+k=0$$ For every real number $k$
Find the minimum number of non real roots.
The sum of non real roots can be___
My work:
For the first part I got the right answer, $2$ by using Rule of Signs. But what about second part$?$ I have no idea how to solve that. Maybe we have to some kind of pairing$?$ And unfortunately I am not familiar with discriminant of a quartic function.
Any help is greatly appreciated.
Let $f(x)=x^4-2x^3+2x^2-x+k\implies f'(x)=4x^3-6x^2+4x-1$ As $f''(x)=12x^2-12x+4>0, \forall x \in \Re$, $f''(x)=0$ does not have real roots, then by reverse logic of Rolle's theorem $f(x)=0$ will have at most 2 real roots.
$f'(x)=0$ has only one real root at $x=1/2$, $f''(1/2)=1>0$. $f(x)$ has only one min at $x=1/2$. $f_{min}=f(1/2)=k-3/16$ if $f_{min}<0 \implies k< \frac{3}{16}$, there will be only two real roots of $f(x)=0.$ Then the sum of all non real roots is 2, when $k>\frac{3}{16}$.
Next, luckily $f(x)=0$ can also be rewritten as $$(2x^2-2x+1)^2=1-4k \implies x=\frac{1}{2}\left(1\pm\sqrt{-1\pm 2\sqrt{1-4k}}\right).$$
One can check that when $k<\frac{3}{16}$ only 2 roots of $f(x)=0$ are real and sum of these real roots is 1. Next, when $k>\frac{3}{16}$, there is no real root.