Animated Solution for Mathematics - Quadratic Equations: Let $\alpha ,\beta$ be the distinct roots of the equation $x^2-(t^2-5t+6)x+1=0,t\in \mathbb{R}$ and $a_n={\alpha }^n+{\beta }^n$. Then the minimum value of $\frac{a_{2023}+a_{2025}}{a_{2024}}$ is

Facing the Monster

Imagine you are sitting in the examination hall. The clock is ticking, the pressure is mounting, and you encounter a problem that asks for the minimum value of a fraction involving $a_{2023}$, $a_{2024}$, and $a_{2025}$. Your first instinct might be panic.

You see the indices—2023, 2024, 2025—and your brain immediately screams, "How on earth am I supposed to calculate the 2025th power of a root?"

Stop. Take a deep breath. This is exactly where the JEE examiners want you to stumble. They want you to try and find the roots $\alpha$ and $\beta$ using the quadratic formula, get lost in a sea of irrational numbers, and eventually give up.

But you are not going to do that. You are going to look past the numbers and see the structure. This problem is not about calculation; it is about recognizing a pattern.

Newton's Theorem

Whenever you see a sequence defined as the sum of the $n$-th powers of the roots of a polynomial, your mind should immediately pivot to one of the most elegant tools in algebra: Newton's Sums (or Newton's Theorem).

Consider a general quadratic equation $A{x}^2+Bx+C=0$ with roots $\alpha$ and $\beta$. If we define $a_n={\alpha }^n+{\beta }^n$, Newton's Theorem tells us that these terms are not random; they are bound by a linear recurrence relation:

Why does this work? Think about it. Since $\alpha$ and $\beta$ are roots, they satisfy the equation $A{x}^2+Bx+C=0$. This means $A{\alpha }^2+B\alpha +C=0$.

If you multiply this by ${\alpha }^n$, you get $A{\alpha }^{n+2}+B{\alpha }^{n+1}+C{\alpha }^n=0$. Similarly, for $\beta$, you get $A{\beta }^{n+2}+B{\beta }^{n+1}+C{\beta }^n=0$.

When you add these two equations together, the magic happens: the sum of the powers emerges naturally. It is not magic; it is the inherent symmetry of roots.

From Algebra to Geometry

Let us apply this to our specific equation: $x^2-(t^2-5t+6)x+1=0$. By comparing this with the standard form $A{x}^2+Bx+C=0$, we identify our coefficients:

$A=1$ $B=-(t^2-5t+6)$ $C=1$

Substituting these into our recurrence relation, we get:

Now, look at the expression we need to minimize: $\frac{a_{2023}+a_{2025}}{a_{2024}}$. To get this, we simply set $n=2023$ in our recurrence relation:

Rearranging this is the moment of truth. Move the middle term to the right side:

Divide both sides by $a_{2024}$, and behold! The terrifying fraction collapses into a simple quadratic expression:

Minimization

We have successfully transformed a problem about high-power sequences into a problem about a simple parabola. Let $f(t)=t^2-5t+6$.

Since the coefficient of $t^2$ is positive, this is an upward-opening parabola. We know from our study of quadratic functions that the minimum value occurs at the vertex, where $t=-\frac{b}{2a}$.

Here, $a=1$ and $b=-5$. So, the minimum occurs at:

Now, we just need to evaluate the function at this point to find the minimum value:

Let's calculate this carefully:

To combine these, we use a common denominator of 4:

The Beauty of the Method

Look at what we have achieved. We started with a problem that seemed to require calculating massive powers, and we ended with a simple arithmetic operation. This is the essence of JEE Advanced preparation.

It is not about brute force; it is about finding the elegant path. Whenever you see $a_n={\alpha }^n+{\beta }^n$, let your mind immediately jump to Newton's Theorem.

It is a powerful, reliable, and beautiful tool that will save you time and boost your confidence. Keep practicing, keep visualizing, and remember: every complex problem is just a simple concept in disguise. The final answer is $-\frac{1}{4}$.