If the range of the function \(f\left(x\right)=\frac{5-x}{{x}^{2}-3x+2}\), \(x\neq 1,2\), is \((-\infty ,\alpha ]\cup [\…

\(y=\frac{5-x}{x^2-3 x+2}\)

\(y x^2+(1-3 y) x+(2 y-5)=0\)

For some real \(x\), this quadratic in \(x\) must have real roots, so

\(D \geq 0\)

\((1-3 y)^2-4 y(2 y-5) \geq 0\)

\(y^2+14 y+1 \geq 0\)

\(y^2+14 y+1=(y+7)^2-48 \geq 0\)

\(y \leq-7-4 \sqrt{3} \) or \(y \geq-7+4 \sqrt{3}\)

Hence, \(\alpha=-7-4 \sqrt{3}, \beta=-7+4 \sqrt{3}\)

Now, \(\alpha+\beta=-14, \alpha \beta=1\)

\(\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta\)

\(\alpha^2+\beta^2=196-2=194\)