Describe quaternion solutions to $x^2-3x+3=0$

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So I have the complex solutions $\frac{3}{2}+\frac{\sqrt{3}}{2}i$ and $\frac{3}{2}-\frac{\sqrt{3}}{2}i$. And I want to describe the rest of the solutions, which I think are supposed to be rotations of these. But I'm not sure how to do that.

Can I just pick any unit quarternion as an axis and all rotations of these solutions are solutions?

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Hint Ugly solution.

Let $x=a+bi+cj+dk$ be any quaternion.

Then $$x^2=(a+bi+cj+dk)^2=..\\ -3x=-3a-3b-3c-3d \\ x^2-3x+3=...$$

Nicer solution

If you are familiar with the Matrix representation of the quaternions, the problem becomes simpler. You want a matrix $$X= \begin{bmatrix} a+bi & c+di \\ -c+di & a-bi \end{bmatrix}$$ satisfying $$X^2-3X+3I =0 $$

Now, the minimal polynomial of $X$ over $\mathbb C$ is either linear or quadratic, and divides $X^2-3X+2$.

If it is quadratic, you must have $tr(X)=3, \det(X)=2$ and this gives all the solutions in this case.

If it is linear, then it is $X-\lambda$ for $\lambda$ being one of the roots you described. In this case, your matrix having this linear minimal polynomial means $$X= \lambda I$$ giving the two complex solutions.