Question

\(\alpha,\beta\) are the roots of \[ x^2-10x-8=0 \] with \(\alpha\gt \beta\). If \[ a_n=\alpha^n-\beta^n \] for \(n\in \mathbb{N}\), then the value of \[ \frac{a_{10}-8a_8}{5a_9} \] is:

• A. \(-3\)
• B. \(3\)
• C. \(-2\)
• D. \(2\)

Hint:

If \(\alpha,\beta\) are roots of a quadratic equation, powers of the roots usually satisfy the same recurrence relation derived from the original quadratic equation.

Answer 1

The Correct Option is D

Step 1: Use the relation satisfied by the roots.
Since \(\alpha,\beta\) are roots of \[ x^2-10x-8=0, \] each root satisfies \[ r^2=10r+8 \] Multiplying by \(r^{n-2}\), \[ r^n=10r^{n-1}+8r^{n-2} \] This holds for both \(\alpha\) and \(\beta\).

Step 2: Obtain the recurrence relation for \(a_n\).
Given, \[ a_n=\alpha^n-\beta^n \] Using \[ \alpha^n=10\alpha^{n-1}+8\alpha^{n-2} \] and \[ \beta^n=10\beta^{n-1}+8\beta^{n-2}, \] we get \[ a_n =(10\alpha^{n-1}+8\alpha^{n-2}) -(10\beta^{n-1}+8\beta^{n-2}) \] \[ a_n=10a_{n-1}+8a_{n-2} \]

Step 3: Apply the recurrence for \(n=10\).
Putting \(n=10\), \[ a_{10}=10a_9+8a_8 \] Thus, \[ a_{10}-8a_8=10a_9 \] Now, \[ \frac{a_{10}-8a_8}{5a_9} = \frac{10a_9}{5a_9} \] \[ =2 \]

Step 4: Final conclusion.
Therefore, \[ \boxed{2} \]