Consider $$f(x)=8x^4-16x^3+16x^2-8x+k=0$$ where $k \in \mathbb{R}$,then find sum of non Real roots of f(x).
My approach:
we have $$f'(x)=32x^3-48x^2+32x-8=0$$ Also
$$f''(x)=96x^2-96x+32=96(x-\frac{1}{2})^2+8 \gt 0$$ so $f'(x)$ is strictly increasing and hence has only one real root $x_0$ in $(0 \:\:1)$, So $f(x)$ has Local Minima at $x=x_0$ and no Local Maxima. So $f(x)$ must have eactly two Real roots and two Complex Roots. Also $f(0)=k=f(1)$ and since $f(x)$ has Local Minima at $x_0$ where $0<x_0<1$ ,$k \gt 0$
Now let the roots of $f(x)$ be $a$,$b$, $p+iq$ and $p-iq$ where $$a,b,p,q \in \mathbb{R}$$ We need to find value of $2p$...Now using relationships between coefficients and roots of $f(x)$ we have
$$a+b+2p=2 \tag{1}$$
$$ab+(p^2+q^2)=(2p-1)^2+1 \tag{2}$$
$$ab(2p)+(p^2+q^2)(2-2p)=1 \tag{3}$$
$$(ab)(p^2+q^2)=\frac{k}{8} \tag{4}$$
Since $a$ and $b$ lie in the interval $(0\:\: 1)$ we have
$0\le a+b \le 1$ $\implies$
$0 \le 2-2p \le 1$ $\implies$
$1 \le 2p \le 2$ but unable to find exact value of $2p$...any help will be greatly appreciated.
Note that the function is symmetric with respect to $\dfrac{1}{2}$. That is $$8x^4-16x^3+16x^2-8x+k=\frac{1}{2}(2x-1)^4+(2x-1)^2-\frac{3}{2}+k$$
This function - as you have noted - is concave up everywhere which means when counted with multiplicity, we have two real roots when $k\le\dfrac{3}{2}$ and zero real roots when $k>\dfrac{3}{2}$.
We now make use Vieta's formulas to get that the sum of all the roots is $-\dfrac{-16}{8}=2$.
When we have all imaginary roots, we can directly apply this to get the final sum as $2$. When we have two real roots, their average is $\dfrac{1}{2}$, so their sum is $1$, and hence the sum of the imaginary roots is also $1$.