polynomials - Sum of Non Real Roots of Quartic?

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

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.