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?
 A: Let $p(x)=x^4-2x^3+2x^2-x$. Since $p''(x)=12x^2-12x+4$ is a second-degree polynomial with a negative discriminant, $p''(x)>0$, hence $p(x)$ is a convex function and the maximum number of real roots of $p(x)=k$ is $\color{red}{2}$.
A: 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.
A: How many turning points does $y=x^4-2x^3+2x^2-x-k$ have? Recall that a polynomial has at least one turning point between each pair of roots. Jack's answer reflects a systematic approach to answering that question.
A: The derivative of $p(x) = x^4-2x^3+2x^2-x$ is $4 x^3-6 x^2+4 x-1= (2 x-1) (2 x^2-2 x+1)$, $<0$ on $(-\infty, 1/2)$, $0$ at $x=1/2$, and $>0$ on $(1/2, \infty)$. Moreover, $\lim_{x\ \to -\infty} p(x) = \lim_{x\ \to \infty} p(x)= +\infty$. Now, $f(1/2) =-3/16$.  Therefore, the number of real solutions of the equation $p(x) = k$ is $0$ if $k < -3/16$, $1$ if $k = -3/16$, and $2$ if $k > -3/16$. Thus, for every $k$ the equation $p(x)=k$ has at least $2$ non-real solutions. Well, in fact we'd better check whether those complex roots are not multiple. The zeroes of $p'(x)$ are $1/2 \pm i/2$ and $p(1/2 \pm i/2) = -1/4$ ( $< -3/16$  of course),so yes, the equation $p(x) = -1/4$ has two non-real roots ( conjugate, each multiplicity $2$). 
