<s>[INST] 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?
 [/INST] 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 $2$). 
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