<s>[INST] Given that $m$ is a real number not less than $-1$ The question is:
Given that $m$ is a real number not less than $-1$, such that the equation in $x$ is $x^2+2(m-2)x+m^2-3m+3=0$ has two distinct roots $r$ and $s$. If $r^2+s^2=6$, find the value of $m$.  
Here's what I've tried:. 
Using Vieta's formulas: 
$rs = m^2-3m+3$
$r+s = -2m+4$   
Then I set the equation as such; $(r+s)^2 - 2rs = 6$. I got $2m^2-10m+4=0$. Using the quadratic equation, I got $\frac {5+\sqrt {17}}2$ and $\frac {5-\sqrt {17}}2$. Is there anything wrong with my solution?
 [/INST] The posted solution is correct up to that point, though it remains to be verified that $\,r \ne s\,$ for those values of $\,m\,$.
Alt. hint:   $\color{blue}{r+s}=-2m+4$, and each root satisfies $\,x^2= -\big(2(m-2)x+m^2-3m+3\big)\,$, so:
$$
\begin{align}
6 = r^2+s^2 &= -2(m-2)(\color{blue}{r+s})- 2(m^2-3m+3) \\
 &= 2(m-2)(2m-4)- 2(m^2-3m+3) \\
 &= 2 (m^2 - 5 m + 5)
\end{align}
$$
Then what remains to solve is $\,m^2-5m+5=3\,$ for solutions $\,\ge -1\,$ (and, again, verify that they yield distinct roots $\,r,s\,$).
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