Animated Solution for Mathematics - Quadratic Equations: Comprehension Passage

My dear student, welcome to a problem that is not merely about algebra, but about the hidden architecture of numbers. When you first look at the equation $x^2-x-1=0$, you might see a simple quadratic.

But I want you to see something more. I want you to see the Fibonacci sequence hiding in plain sight. This problem is a masterclass in how we can use the properties of roots to bypass tedious calculations and arrive at the truth with elegance.

The Magic of Degree Reduction

Let us start with the foundation. We are given the quadratic equation $x^2-x-1=0$, with roots $\alpha$ and $\beta$.

In the heat of an exam, the instinct is often to reach for the quadratic formula immediately. But hold that thought! Before we calculate the messy values of $\alpha$ and $\beta$, let us look at what it means to be a root.

If $\alpha$ is a root, it must satisfy the equation: ${\alpha }^2-\alpha -1=0$. This gives us the beautiful identity: ${\alpha }^2=\alpha +1$.

This is what I call the 'Degree Reduction' trick. It allows us to express higher powers of $\alpha$ in terms of lower powers.

If we multiply this equation by ${\alpha }^n$, we get ${\alpha }^{n+2}={\alpha }^{n+1}+{\alpha }^n$. This is the DNA of our sequence. It tells us that the powers of $\alpha$ follow the Fibonacci recurrence. Since $\beta$ satisfies the same equation, it follows the exact same logic.

The Power of Linearity

Now, consider the sequence $a_n=p{\alpha }^n+q{\beta }^n$. We want to find a relationship between the terms. Let us look at $a_{n+2}$.

By definition, $a_{n+2}=p{\alpha }^{n+2}+q{\beta }^{n+2}$. Using our degree reduction identity, we can substitute ${\alpha }^{n+2}$ with ${\alpha }^{n+1}+{\alpha }^n$ and ${\beta }^{n+2}$ with ${\beta }^{n+1}+{\beta }^n$.

When we substitute these, we get:

If we rearrange this, we get:

Look closely! The first parenthesis is exactly $a_{n+1}$, and the second is $a_n$. Thus, we have derived the recurrence relation:

This is the moment where the problem becomes simple. We have proven that our sequence, regardless of the values of $p$ and $q$, must obey the Fibonacci recurrence. For the first part of the question, finding $a_{12}$ is now trivial: $a_{12}=a_{11}+a_{10}$.

The Irrationality Gatekeeper

Now, let us tackle the second part. We are given $a_4=28$. We need to find $p+2q$. To do this, we must express $a_4$ in terms of $p$ and $q$.

Using our recurrence, we can build up from $a_0$ and $a_1$:

Substituting $a_0=p+q$ and $a_1=p\alpha +q\beta$, we get:

Now, we finally use the quadratic formula to find $\alpha =\frac{1+\sqrt{5}}{2}$ and $\beta =\frac{1-\sqrt{5}}{2}$. Substituting these into our expression for $a_4$, we get a term involving $\sqrt{5}$.

This is where the 'Fact' provided in the question becomes our best friend. We arrive at an equation of the form $A+B\sqrt{5}=28$. By grouping the rational parts and the irrational parts, we get:

Because $\sqrt{5}$ is irrational, the only way this sum can be zero is if the rational part is zero AND the coefficient of $\sqrt{5}$ is zero. This gives us a system of two equations:

1) $3(p-q)=0⟹p=q$

2) $7(p+q)-56=0⟹p+q=8$

Solving this is straightforward: $2p=8$, so $p=4$, and consequently $q=4$.

The Elegance of the Solution

Finally, we calculate $p+2q=4+2(4)=12$.

Do you see what happened here? We didn't just 'solve' a problem; we navigated a structure. We used the recurrence to simplify the sequence, we used the roots to simplify the expression, and we used the property of irrational numbers to lock in our variables.

This is the JEE Advanced mindset: don't fight the math, understand its structure, and let the structure do the heavy lifting for you. Keep practicing this way of thinking, and you will find that even the most intimidating problems become beautiful puzzles waiting to be solved.