The minimum number of non real roots of the equation $x^4-2x^3+2x^2-x=k$ is? 
The minimum number of non real roots of the equation
  $x^4-2x^3+2x^2-x=k$ is?

k is any real number.
I plotted this on https://www.desmos.com/calculator/vpfpjwyxz8.It seems that the answer will be 2.But how to solve it manually?

Method $1$
Let $a_i$ be the roots of the polynomial. We then have
$$\sum_{i=1}^4 a_i = 2$$
$$\sum_{\overset{i,j=1}{i \neq j}}^4 a_ia_j = 2$$
This means we have
$$\sum_{i=1}^4 a_i^2 = \left(\sum_{i=1}^4 a_i\right)^2 - 2\cdot\left(\sum_{\overset{i,j=1}{i \neq j}}^4 a_ia_j\right)= 2^2 -2 \cdot 2 = 0$$
Further all $a_i$'s are not zeros. Hence, there is at-least a pair of complex roots.
Method 2
We have
$$x^4-2x^3+2x^2-x = \left(x^2-x+1/2\right)^2-1/4 = k$$
This means we have
$$x^2-x+1/2 = \pm \sqrt{k+1/4} \implies \left(x-1/2\right)^2 = -\dfrac14 \pm \dfrac{\sqrt{4k+1}}2$$
which clearly shows that there are at-least two complex roots.
In fact the second argument shows that if $k < -3/16$, there are $4$ complex roots and if $k \geq -3/16$, there are $2$ complex roots.
