Key Concepts and Formulas

• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$:
  • Sum of roots: $\alpha + \beta = -\frac{b}{a}$
  • Product of roots: $\alpha \beta = \frac{c}{a}$
• Recurrence Relation for Sums of Powers of Roots: For a quadratic equation $x^2 - Px + Q = 0$ with roots $\alpha$ and $\beta$, where $P = \alpha + \beta$ and $Q = \alpha \beta$, the sum $a_n = \alpha^n + \beta^n$ satisfies the recurrence relation: $a_n = P a_{n-1} - Q a_{n-2} \quad \text{for } n \ge 2$

Step-by-Step Solution

Step 1: Identify Coefficients and Apply Vieta's Formulas We are given the quadratic equation $x^2 - (t^2 - 5t + 6)x + 1 = 0$. Comparing this to the general form $x^2 - Px + Q = 0$, where $\alpha$ and $\beta$ are the roots, we have:

• $P = \alpha + \beta = t^2 - 5t + 6$
• $Q = \alpha \beta = 1$

Step 2: State the Recurrence Relation Since $\alpha$ and $\beta$ are the roots of the given quadratic equation, we can write the recurrence relation for $a_n = \alpha^n + \beta^n$ as: $a_n = (t^2 - 5t + 6) a_{n-1} - a_{n-2}$

Step 3: Apply the Recurrence Relation for n = 2025 We want to find the value of $\frac{a_{2023}+a_{2025}}{a_{2024}}$. Let's use the recurrence relation with $n = 2025$: $a_{2025} = (t^2 - 5t + 6) a_{2024} - a_{2023}$ Rearranging the equation, we get: $a_{2025} + a_{2023} = (t^2 - 5t + 6) a_{2024}$

Step 4: Calculate the Target Expression Now, divide both sides of the equation by $a_{2024}$. We assume $a_{2024} \neq 0$, which is valid since $\alpha$ and $\beta$ are distinct roots and $\alpha\beta = 1$. Thus neither root can be zero, and $a_{2024} = \alpha^{2024} + \beta^{2024} \ne 0$. $\frac{a_{2025} + a_{2023}}{a_{2024}} = t^2 - 5t + 6$

Step 5: Minimize the Quadratic Expression Let $f(t) = t^2 - 5t + 6$. To find the minimum value of this quadratic, we complete the square: $f(t) = t^2 - 5t + \left(\frac{5}{2}\right)^2 - \left(\frac{5}{2}\right)^2 + 6$ $f(t) = \left(t - \frac{5}{2}\right)^2 - \frac{25}{4} + \frac{24}{4}$ $f(t) = \left(t - \frac{5}{2}\right)^2 - \frac{1}{4}$ The minimum value of $f(t)$ occurs when $t = \frac{5}{2}$, and the minimum value is $-\frac{1}{4}$.

Step 6: Check the Distinct Roots Condition We need to ensure the roots are distinct when $t = \frac{5}{2}$. The discriminant of the original quadratic equation is $D = (t^2 - 5t + 6)^2 - 4$. When $t = \frac{5}{2}$, $t^2 - 5t + 6 = -\frac{1}{4}$, so $D = \left(-\frac{1}{4}\right)^2 - 4 = \frac{1}{16} - 4 = -\frac{63}{16}$. Since $D < 0$, the roots are distinct complex conjugates. Also, we need to check $t \ne 1$ and $t \ne 4$. Since $t=5/2$, the condition is satisfied.

Therefore, the minimum value of $\frac{a_{2023}+a_{2025}}{a_{2024}}$ is $-\frac{1}{4}$.

Common Mistakes & Tips

• Forgetting the Recurrence Relation: Remembering the recurrence relation for sums of powers of roots is key to solving this problem efficiently.
• Assuming Real Roots: The problem only states distinct roots, which could be distinct real or distinct complex conjugate roots.
• Not checking distinct root condition: Always make sure that the value of $t$ that minimizes the expression satisfies the distinct root condition.

Summary By applying Vieta's formulas, establishing the recurrence relation for $a_n$, and minimizing the resulting quadratic expression in $t$, we found the minimum value of the given expression to be $-\frac{1}{4}$. We also checked that the value of $t$ which gives the minimum value does not violate the distinct roots condition.

Final Answer The final answer is $\boxed{-1/4}$, which corresponds to option (B).