Find the real roots of $\sqrt{x-2}(x^2-4x+3)=0 $ I thought the roots of this equation are $1,2,3$.

But the teacher says 1 is not a real root because if we take $x=1$ we get
$$\sqrt{-1}(0)$$

and $\sqrt{-1}$ is not real.
Is this right?
While it is true $x=1$ is a root, the question is: what is the domain in which the equation "exists"?
The answer is that the square root only exists if you consider the inequality  $x-2\ge 0$.  Otherwise your solution does not compile, so to speak.
So the roots coexisting within your domain are $2$ and $3$.
We have indeed 3 roots: 1,2 and 3. But now we are looking for the real roots. And yes 1 is not a real root, because like you said $\sqrt -1$ is not real, it is imaginary.