Animated Solution for Mathematics - Quadratic Equations: Let $\alpha ,\beta$ be roots of $x^2+\sqrt{2}x-8=0$. If $U_n={\alpha }^n+{\beta }^n$, then $\frac{U_{10}+\sqrt{2}{U}_9}{2U_8}$ is equal to______

Imagine you are standing before a daunting mountain. The problem asks you to calculate the value of the expression:

Here, $U_n={\alpha }^n+{\beta }^n$, where $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2+\sqrt{2}x-8=0$. Your first instinct might be to reach for the quadratic formula to find $\alpha$ and $\beta$ explicitly, but that is the path of the novice. In the world of JEE Advanced, we do not calculate; we manipulate to find the underlying structure.

The most fundamental property of a root is that it satisfies its parent equation. If $\alpha$ is a root of $x^2+\sqrt{2}x-8=0$, it must obey:

This is not just an equation; it is a definition of $\alpha$'s existence. We do not need to know the specific values of $\alpha$ or $\beta$; we only need to understand how they behave.

We need to reach $U_{10}$, which involves ${\alpha }^{10}$. To bridge the gap from ${\alpha }^2$ to ${\alpha }^{10}$, we scale the equation by multiplying ${\alpha }^2+\sqrt{2}\alpha -8=0$ by ${\alpha }^8$:

This creates a recurrence relation. Because $\beta$ is also a root, it must obey the exact same law:

Now, we bring these two parallel universes together by adding the equations vertically:

The structure of $U_n={\alpha }^n+{\beta }^n$ emerges naturally. We can rewrite this as:

This recurrence relation connects the powers without ever requiring us to solve for the roots individually.

Look back at our target expression: $\frac{U_{10}+\sqrt{2}{U}_9}{2U_8}$. Our recurrence relation tells us that:

Substitute this into the numerator of the target expression:

The $U_8$ terms cancel out, leaving us with $\frac{8}{2}=4$. The final answer is 4.