Question:

**Step 1: Determine the condition for real and distinct roots** The given quadratic equation is $x^2 + ax - 3x - (a+2) = 0$. Rearranging the terms, we obtain the standard quadratic form $Ax^2 + Bx + C = 0$: $$ x^2 + (a-3)x - (a+2) = 0 $$ For a quadratic equation to have real and distinct roots, its discriminant $\Delta$ must be greater than zero. The discriminant is given by $\Delta = B^2 - 4AC$. In this equation, $A=1$, $B=(a-3)$, and $C=-(a+2)$. Substituting these values into the discriminant formula: $$ \Delta = (a-3)^2 - 4(1)(-(a+2)) $$ $$ \Delta = (a-3)^2 + 4(a+2) $$ Expanding the terms: $$ \Delta = (a^2 - 6a + 9) + (4a + 8) $$ $$ \Delta = a^2 - 2a + 17 $$ To determine if $\Delta > 0$, we complete the square for the expression $a^2 - 2a + 17$: $$ \Delta = (a^2 - 2a + 1) + 16 $$ $$ \Delta = (a-1)^2 + 16 $$ Since $(a-1)^2 \ge 0$ for all real values of $a$, it follows that $(a-1)^2 + 16 \ge 16$. Therefore, $\Delta = (a-1)^2 + 16 > 0$ for all real values of $a$. This condition implies that the quadratic equation always has real and distinct roots for any real number $a$. Consequently, there are no restrictions on the value of $a$ other than it being a real number. **Step 2: Minimize the given expression** We need to find the minimum value of the expression $\frac{a^2+1}{a^2+2}$. Let $y = a^2$. Since $a$ is a real number, $a^2 \ge 0$, which means $y \ge 0$. The expression can be rewritten as a function of $y$: $$ f(y) = \frac{y+1}{y+2} \quad \text{for } y \ge 0 $$ To find the minimum value of $f(y)$, we analyze its derivative with respect to $y$: $$ f'(y) = \frac{d}{dy} \left( \frac{y+1}{y+2} \right) $$ Using the quotient rule, $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$: $$ f'(y) = \frac{(1)(y+2) - (y+1)(1)}{(y+2)^2} $$ $$ f'(y) = \frac{y+2 - y-1}{(y+2)^2} $$ $$ f'(y) = \frac{1}{(y+2)^2} $$ For $y \ge 0$, the denominator $(y+2)^2$ is always positive. Thus, $f'(y) = \frac{1}{(y+2)^2} > 0$ for all $y \ge 0$. Since the derivative $f'(y)$ is always positive, the function $f(y)$ is strictly increasing for $y \ge 0$. Therefore, the minimum value of $f(y)$ occurs at the smallest possible value of $y$, which is $y=0$. Substituting $y=0$ into the function $f(y)$: $$ f(0) = \frac{0+1}{0+2} = \frac{1}{2} $$ The minimum value of the expression $\frac{a^2+1}{a^2+2}$ is $\frac{1}{2}$.