We are given the function $f(x) = \frac{5-x}{x^2-3x + 2}$ where $x \neq 1, 2$. The range is given in the form $(-\infty, \alpha] \cup [\beta, \infty)$. We need to find the value of $\alpha^2 + \beta^2$.

Finding the Range of the Function

Let $y = f(x)$. We set up the equation:

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

Rearranging the equation to form a quadratic in terms of $x$:

$y(x^2 - 3x + 2) = 5 - x$

$yx^2 - 3yx + 2y = 5 - x$

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

For $x$ to be a real number, the discriminant ($\Delta$) of this quadratic equation must be non-negative ($\Delta \ge 0$).

The discriminant is calculated as $\Delta = b^2 - 4ac$, where $a=y$, $b=(1-3y)$, and $c=(2y-5)$.

$\Delta = (1 - 3y)^2 - 4(y)(2y - 5)$

$\Delta = (1 - 6y + 9y^2) - (8y^2 - 20y)$

$\Delta = 1 - 6y + 9y^2 - 8y^2 + 20y$

$\Delta = y^2 + 14y + 1$

Since $\Delta \ge 0$, we have the inequality:

$y^2 + 14y + 1 \ge 0$

To solve this inequality, we first find the roots of the quadratic equation $y^2 + 14y + 1 = 0$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$y = \frac{-14 \pm \sqrt{14^2 - 4(1)(1)}}{2(1)}$

$y = \frac{-14 \pm \sqrt{196 - 4}}{2}$

$y = \frac{-14 \pm \sqrt{192}}{2}$

Simplifying $\sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3}$:

$y = \frac{-14 \pm 8\sqrt{3}}{2}$

$y = -7 \pm 4\sqrt{3}$

The roots are $y_1 = -7 - 4\sqrt{3}$ and $y_2 = -7 + 4\sqrt{3}$. Since the quadratic $y^2 + 14y + 1$ opens upwards, the inequality $y^2 + 14y + 1 \ge 0$ holds when $y$ is less than or equal to the smaller root, or greater than or equal to the larger root.

Thus, the range of the function is:

$(-\infty, -7 - 4\sqrt{3}] \cup [-7 + 4\sqrt{3}, \infty)$

Identifying $\alpha$ and $\beta$

Comparing the obtained range with the given form $(-\infty, \alpha] \cup [\beta, \infty)$, we identify:

• $\alpha = -7 - 4\sqrt{3}$
• $\beta = -7 + 4\sqrt{3}$

Calculating $\alpha^2 + \beta^2$

Now, we calculate the squares of $\alpha$ and $\beta$:

$\alpha^2 = (-7 - 4\sqrt{3})^2 = (-(7 + 4\sqrt{3}))^2 = (7 + 4\sqrt{3})^2$

$\alpha^2 = 7^2 + 2(7)(4\sqrt{3}) + (4\sqrt{3})^2$

$\alpha^2 = 49 + 56\sqrt{3} + (16 \times 3)$

$\alpha^2 = 49 + 56\sqrt{3} + 48 = 97 + 56\sqrt{3}$

And for $\beta$:

$\beta^2 = (-7 + 4\sqrt{3})^2 = (-7)^2 + 2(-7)(4\sqrt{3}) + (4\sqrt{3})^2$

$\beta^2 = 49 - 56\sqrt{3} + (16 \times 3)$

$\beta^2 = 49 - 56\sqrt{3} + 48 = 97 - 56\sqrt{3}$

Finally, we compute $\alpha^2 + \beta^2$:

$\alpha^2 + \beta^2 = (97 + 56\sqrt{3}) + (97 - 56\sqrt{3})$

$\alpha^2 + \beta^2 = 97 + 97 = 194$

The value of $\alpha^2 + \beta^2$ is 194.