For every real number a the equation $8x^4-16x^3+16x^2-8x+a=0$ ,

Substituting $x=y+\left(\frac{1}{2}\right)$ in the equation, we obtain the equa- tion in y:
 $8y^4+4y^2+a-\frac{3}{2}=0$  …(2)
Using the transformation z = y², we get a quadratic equation in z:
$8x^2+4z+a-\frac{3}{2}=0$   …(3)
The discriminant of this equation is $\text{32}(\text{2}-\text{a})$ which is nonnegative if and only if  $\text{a}\le \text{2}$. For  $\text{a}\le \text{2}$ , we obtain the roots$z_1=\frac{-1+\sqrt{2(2-a)}}{4},z_2=\frac{-1-\sqrt{2(2-a)}}{4}$
  
For getting real y we need  $\text{z }\ge 0$. Obviously $Z_{\text{2}}<0$ and hence it gives only non-real  values of y. But $Z_{\text{1}}>0$ if and only if $a\le \frac{3}{2}$ . In this case we obtain two real values for y and  hence two real roots for the original equation (1). Thus we conclude that there are two real  roots  and two non-real roots for  $a\le \frac{3}{2}$ and four non-real roots for $a>\frac{3}{2}$ . Obviously the  sum of all the roots of the equation is 2. For $a\le \frac{3}{2}$ , two real roots of (2) are given by  $y_1=+\sqrt{z_1}$   and  $y_2=-\sqrt{z_1}\arcsin \theta$. Hence the sum of real roots of (1) is given by $y_1+\frac{1}{2}+y_2+\frac{1}{2}$   which reduces to 1. It follows the sum of the non-real roots of (1) for $a\le \frac{3}{2}$   is also 1. Thus
The sum of nonreal roots =  $\{\begin{array}{cc}1& fora\le \frac{3}{2}\\ 2& fora>\frac{3}{2}\end{array}$