Question

Sol. $x^{4} - 2x^{3} + 2x^{2} - x = - \frac{a}{8}$

Let

$$\begin{matrix} f(x) & \ = x^{4} - 2x^{3} + 2x^{2} - x = x(x - 1)\left( x^{2} - x + 1 \right) \\ f'(x) & \ = 4x^{3} - 6x^{2} + 4x - 1 \\ f^{''}(x) & \ = 12x^{2} - 12x + 4 > 0\ \forall x \in R \end{matrix}$$

$$f'\left( \frac{1}{2} \right) = 0$$

Also $f(1 - x) = f(x)$

If

$$- \frac{a}{8} \geq - \frac{3}{16}\ \Rightarrow \ a \leq \frac{3}{2}$$

then equation has two real roots $\alpha,1 - \alpha$

⇒ Sum of all non real roots $= 2 - 1 = 1$

If $a > \frac{3}{2}$, then equation has 4 non real roots

⇒ Sum of all non real roots $= 2$.