algebra precalculus - sum of all non real roots of the equation in a bi-quadratic equation

Consider the equation $8x^4-16x^3+16x^2-8x+a=0\;\left(a\in \mathbb{R}\right)\;,$ Then the sum of

all non real roots of the equation can be

$\bf{OPTIONS::}\;\; (a)\;\; 1\;\;\;\;\;\; (b)\;\; 2\;\;\;\;\;\; (c)\; \displaystyle \frac{1}{2}\;\; \;\;\;\; (d)\;\; None$

$\bf{My\; Try::}$ Let $f(x)=8x^4-16x^3+16x^2-8x+a\;,$ Then $f'(x)=32x^3-48x^2+32x-8$

And $\displaystyle f''(x) = 96x^2-96x+32 = 96\left[x^2-x+\frac{1}{3}\right]=96\left[\left(x-\frac{1}{2}\right)^2+\frac{1}{12}\right]>0\;\forall x \in \mathbb{R}$

So Using $\bf{LMVT\;,}$ We get $f'(x)=0$ has at most $1$ real roots and

$f(x)=0$ has at most $2$ real roots

Now How can i solve it after that, Help me

Thanks

Answer

Now How can i solve it after that

Noting that $f'(1/2)=0$ is a key.

We have
$$f'(x)=8(4x^3-6x^2+4x-1)=8(2x-1)\left(2\left(x-\frac 12\right)^2+\frac 12\right)$$
So, $f(x)$ is decreasing for $x\lt 1/2$ and is increasing for $x\gt 1/2$.

By the way,

$$f\left(\frac 12+s\right)=8s^4+4s^2-\frac 32+a$$
and so $f(1/2-\alpha)=0$ is equivalent to $f(1/2+\alpha)=0$.

Also, by Vieta's formula, the sum of roots is $-(-16)/8=2$.

If $a\le 3/2$, then $f(1/2)\le 0$, and the answer is
$$2-(1/2-\alpha)-(1/2+\alpha)=\color{red}{1}$$

If $a\gt 3/2$, then $f(1/2)\gt 0$, and the answer is $\color{red}{2}$.