If α, β are roots of the equation x² + 5(√2)x + 10 = 0, α β and Pn = αⁿ - βⁿ for each positive int

Q: Wait, why can't I just solve for $\alpha$ and $\beta$ using the quadratic formula and plug them into $P_n$?

You absolutely could, but that is the path of infinite pain! You would be dealing with powers of irrational numbers like $(\frac{-5\sqrt{2} + i\sqrt{10}}{2})^{20}$. Newton's Sums allow us to bypass the calculation of the roots entirely and work directly with the structure of the equation. In JEE Advanced, the goal is often to find the elegant path, not the brute-force one.

Q: Is it always true that $P_n$ follows the recurrence of the quadratic equation?

Yes! Because both $\alpha^n$ and $\beta^n$ satisfy the recurrence $x^n + 5\sqrt{2}x^{n-1} + 10x^{n-2} = 0$, their difference $P_n = \alpha^n - \beta^n$ must also satisfy it by the principle of linearity. It is a beautiful consequence of the algebraic structure of polynomials.

Q: How do I know when to use Newton's Sums in an exam?

Whenever you see a problem involving powers of roots (like $\alpha^n + \beta^n$ or $\alpha^n - \beta^n$) and a quadratic or higher-degree polynomial, your brain should immediately switch to Newton's Sums. It is the standard tool for handling sequences generated by roots.

Question

If $\alpha, \beta$ are roots of the equation $x^2 + 5(\sqrt{2})x + 10 = 0, \alpha > \beta$ and $P_n = \alpha^n - \beta^n$ for each positive integer $n$, then the value of $\frac{P_{17}P_{20}+5\sqrt{2}P_{17}P_{19}}{P_{18}P_{19}+5\sqrt{2}P_{18}^2}$ is equal to

Sigma Percentile · Accepted Answer

### The Beauty of Hidden Patterns

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of calculation.

We are given the quadratic equation $x^2 + 5\sqrt{2}x + 10 = 0$ and asked to evaluate a complex fraction involving $P_{17}, P_{18}, P_{19},$ and $P_{20}$, where $P_n = \alpha^n - \beta^n$.

If you try to calculate these powers directly, you will be lost in a sea of surds and exponents. But here is the secret: in mathematics, when you see a problem that looks like it requires brute force, it is almost always a trap. There is an elegant path waiting for you.

### Phase 1: The Root's Identity

Let us start with the foundation. We know that $\alpha$ and $\beta$ are roots of $x^2 + 5\sqrt{2}x + 10 = 0$.

This means that $\alpha$ satisfies the equation perfectly:
$$\alpha^2 + 5\sqrt{2}\alpha + 10 = 0$$

This is not just an equation; it is a statement of identity. Now, imagine we want to reach higher powers. If we multiply this entire equation by $\alpha^{n-2}$, we get:
$$\alpha^n + 5\sqrt{2}\alpha^{n-1} + 10\alpha^{n-2} = 0$$

This is the birth of our recurrence. By the exact same logic, $\beta$ also satisfies:
$$\beta^n + 5\sqrt{2}\beta^{n-1} + 10\beta^{n-2} = 0$$

We now have two parallel universes of equations, one for $\alpha$ and one for $\beta$.

### Phase 2: The Birth of the Recurrence

We define $P_n = \alpha^n - \beta^n$. To create this, we simply subtract the $\beta$ equation from the $\alpha$ equation.

When we group the terms, we get:
$$(\alpha^n - \beta^n) + 5\sqrt{2}(\alpha^{n-1} - \beta^{n-1}) + 10(\alpha^{n-2} - \beta^{n-2}) = 0$$

Look at that! The terms inside the parentheses are exactly our sequence $P_n$. This gives us the golden recurrence relation:
$$P_n + 5\sqrt{2}P_{n-1} + 10P_{n-2} = 0$$

If we rearrange this to isolate the first two terms, we get:
$$P_n + 5\sqrt{2}P_{n-1} = -10P_{n-2}$$

This is the key that unlocks the entire problem.

### Phase 3: The Art of Simplification

Now, let us look at the numerator of our expression: $P_{17}P_{20} + 5\sqrt{2}P_{17}P_{19}$. Factoring out $P_{17}$, we get $P_{17}(P_{20} + 5\sqrt{2}P_{19})$.

Does the term in the bracket look familiar? It is exactly our recurrence relation for $n=20$!

So, $P_{20} + 5\sqrt{2}P_{19} = -10P_{18}$. The numerator becomes:
$$P_{17}(-10P_{18}) = -10P_{17}P_{18}$$

We apply the same logic to the denominator: $P_{18}P_{19} + 5\sqrt{2}P_{18}^2$. Factoring out $P_{18}$, we get $P_{18}(P_{19} + 5\sqrt{2}P_{18})$.

Again, using our recurrence for $n=19$, this bracket becomes $-10P_{17}$. The denominator becomes:
$$P_{18}(-10P_{17}) = -10P_{17}P_{18}$$

### The Final Victory

When we place our simplified numerator over our simplified denominator, we get:
$$\frac{-10P_{17}P_{18}}{-10P_{17}P_{18}} = 1$$

Everything cancels out perfectly to leave us with **1**. This is the thrill of mathematics.

We didn't need to know the values of $\alpha$ or $\beta$. We didn't need to calculate $P_{20}$. We only needed to understand the structure of the equation. Keep this mindset, and you will conquer any problem the JEE throws at you.

Enter Numerical Value:

Visualized Solution

The Quadratic Equation

Defining the Sequence $P_{n}$

Newton's Sums Concept

Substituting $α$

Scaling for Higher Powers

Symmetry for $β$

Subtracting to Form $P_{n}$

The Final Recurrence Relation

Factoring the Numerator

Applying Recurrence to Numerator

Factoring the Denominator

Applying Recurrence to Denominator

The Final Result

The Sigma Insight: Relation Between Roots and Coefficients

The Beauty of Hidden Patterns

Phase 1

Phase 2

Phase 3

The Final Victory

Similar Questions

Let $α, β; α > β$ , be the roots of the equation $x^{2} - 2 x - 3 = 0$ . Let $P_{n} = α^{n} - β^{n}, n \in N$ . Then $(113 - 102) P_{10} + (112 + 10) P_{11} - 11 P_{12}$ is equal to

Let $α$ and $β$ be the roots of $x^{2} + 3 x - 16 = 0$ , and $γ$ and $δ$ be the roots of $x^{2} + 3 x - 1 = 0$ . If $P_{n} = α^{n} + β^{n}$ and $Q_{n} = γ^{n} + δ^{n}$ , then $\frac{P _{25} + 3 P _{24}}{2 P _{23}} + \frac{Q _{25} - Q _{23}}{Q _{24}}$ is equal to

Let $α, β$ ( $α > β$ ) be the roots of the quadratic equation $x^{2} - x - 4 = 0$ . If $P_{n} = α^{n} - β^{n}, n \in N$ , then $\frac{P _{15} P _{16} - P _{14} P _{16} - P _{15}^{2} + P _{14} P _{15}}{P _{13} P _{14}}$ is equal to ______.

Let $α$ and $β$ be the roots of the equation $x^{2} - x - 1 = 0$ . If $p_{k} = (α)^{k} + (β)^{k}, k \geq 1$ , then which of the following statements is not true?

Let $α, β$ be the roots of the equation $x^{2} - a x - b = 0$ with $I m (α) < I m (β)$ . Let $P_{n} = α^{n} - β^{n}$ . If $P_{3} = - 57 i, P_{4} = - 37 i, P_{5} = 117 i$ and $P_{6} = 457 i,$ then $∣ α^{4} + β^{4} ∣$ is equal to

Let $α$ and $β$ be the roots of $x^{2} - 6 x - 2 = 0$ , with $α > β$ . If $a_{n} = α^{n} - β^{n}$ for $n \geq 1$ , then the value of $\frac{a _{10} - 2 a _{8}}{2 a _{9}}$ is

Let $α, β$ be roots of $x^{2} + 2 x - 8 = 0$ . If $U_{n} = α^{n} + β^{n}$ , then $\frac{U _{10} + 2 U _{9}}{2 U _{8}}$ is equal to______

Let $α$ and $β$ are the roots of the equation $x^{2} - x - 1 = 0$ . If $p_{k} = (α)^{k} + (β)^{k}, k \geq 1$ then which one of the following statements is not true?

Let $α$ and $β$ be the roots of $x^{2} - 6 x - 2 = 0$ . If $a_{n} = α^{n} - β^{n}$ for $n \geq 1$ , then the value of $\frac{a _{10} - 2 a _{8}}{3 a _{9}}$ is:

Let $a, b \in R, a \neq = 0$ , such that the equation, $a x^{2} - 2 b x + 5 = 0$ has a repeated root $α$ , which is also a root of the equation $x^{2} - 2 b x - 10 = 0$ . If $β$ is the root of this equation, then $α^{2} + β^{2}$ is equal to: