Question

Let $\alpha$ and $\beta$ be the roots of $x^2–6x–2=0$. If$a_n={\alpha }^n–{\beta }^n$ for $n\ge 1$, then the value of $\frac{(a_{10}–2a_8)}{(3a_9)}$ is

• A. $4$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C

$2$

Explanation for the correct option:

Step 1: Equating the quadratic equation by substituting $\alpha$and $\beta$ in the equation

The given quadratic equation is

$x^2–6x–2=0...\left(i\right)$

And $\alpha$ and $\beta$ be the roots of this equation

Substituting $x=\alpha$ in $\left(i\right)$

$$\begin{gathered} \Rightarrow {\alpha }^2–6\alpha –2=0 \\ \Rightarrow {\alpha }^2–2=6\alpha ....\left(ii\right) \end{gathered}$$

And substituting $x=\beta$ in $\left(i\right)$

$$\begin{gathered} \Rightarrow {\beta }^2–6\beta –2=0 \\ \Rightarrow {\beta }^2–2=6\beta ...\left(iii\right) \end{gathered}$$

Step 2: Equating for $\frac{(a_{10}–2a_8)}{(3a_9)}$

Since given that $a_n={\alpha }^n–{\beta }^n$

Therefore,

$$\begin{gathered} \Rightarrow \frac{(a_{10}–2a_8)}{(3a_9)}=\frac{\left({\alpha }^{10}-{\beta }^{10}\right)-2\left({\alpha }^8-{\beta }^8\right)}{3\left({\alpha }^9-{\beta }^9\right)} \\ =\frac{\left({\alpha }^{10}-2{\alpha }^8\right)-\left({\beta }^{10}-2{\beta }^8\right)}{3\left({\alpha }^9-{\beta }^9\right)} \\ =\frac{{\alpha }^8\left({\alpha }^2-2\right)-{\beta }^8\left({\beta }^2-2\right)}{3\left({\alpha }^9-{\beta }^9\right)} \\ =\frac{{\alpha }^8\left(6\alpha \right)-{\beta }^8\left(6\beta \right)}{3\left({\alpha }^9-{\beta }^9\right)}[\text{from equation}(ii)\text{and}(iii\left)\right] \\ =\frac{6\left({\alpha }^9-{\beta }^9\right)}{3\left({\alpha }^9-{\beta }^9\right)} \\ =\frac{6}{3} \\ =2 \end{gathered}$$

Thus, $\frac{(a_{10}–2a_8)}{(3a_9)}=2$

Therefore, option (C) is the correct answer.