Theory of Equations 1 Question 17
18. Let $\alpha$ and $\beta$ be the roots of equation $x^{2}-6 x-2=0$. If $a _n=\alpha^{n}-\beta^{n}$, for $n \geq 1$, then the value of $\frac{a _{10}-2 a _8}{2 a _9}$ is
(a) 6
(b) -6
(c) 3
(d) -3
(2015 Main)
Show Answer
Answer:
Correct Answer: 18. (c)
Solution:
- Given, $\alpha$ and $\beta$ are the roots of the equation $x^{2}-6 x-2=0$.
$ \therefore \quad \begin{aligned} a _n & =\alpha^{n}-\beta^{n} \text { for } n \geq 1 \\ a _{10} & =\alpha^{10}-\beta^{10} \\ a _8 & =\alpha^{8}-\beta^{8} \\ a _9 & =\alpha^{9}-\beta^{9} \end{aligned} $
Now, consider
$ \begin{aligned} \frac{a _{10}-2 a _8}{2 a _9}= & \frac{\alpha^{10}-\beta^{10}-2\left(\alpha^{8}-\beta^{8}\right)}{2\left(\alpha^{9}-\alpha^{9}\right)} \\ = & \frac{\alpha^{8}\left(\alpha^{2}-2\right)-\beta^{8}\left(\beta^{2}-2\right)}{2\left(\alpha^{9}-\beta^{9}\right)} \\ = & \frac{\alpha^{8} \cdot 6 \alpha-\beta^{8} 6 \beta}{2\left(\alpha^{9}-\beta^{9}\right)}=\frac{6 \alpha^{9}-6 \beta^{9}}{2\left(\alpha^{9}-6 \beta^{9}\right)}=\frac{6}{2}=3 \\ & \because \alpha \text { and } \beta \text { are the roots of } \\ & \Rightarrow \quad x^{2}-6 x-2=0 \text { or } x^{2}=6 x+2 \\ & \text { and } \alpha^{2}=6 \alpha+2 \Rightarrow \alpha^{2}-2=6 \alpha+2 \Rightarrow \beta^{2}-2=6 \beta \end{aligned} $
Alternate Solution
Since, $\alpha$ and $\beta$ are the roots of the equation
$ \begin{aligned} & x^{2}-6 x-2=0 . \\ & \text { or } \quad x^{2}=6 x+2 \\ & \therefore \quad \alpha^{2}=6 \alpha+2 \\ & \Rightarrow \quad \alpha^{10}=6 \alpha^{9}+2 \alpha^{8} \\ & \text { Similarly, } \quad \beta^{10}=6 \beta^{9}+2 \beta^{8} \end{aligned} $
On subtracting Eq. (ii) from Eq. (i), we get
$ \begin{array}{rlrl} & \alpha^{10}-\beta^{10} =6\left(\alpha^{9}-\beta^{9}\right)+2\left(\alpha^{8}-\beta^{8}\right) \quad\left(\because a _n=\alpha^{n}-\beta^{n}\right) \\ \Rightarrow \quad a _{10} & =6 a _9+2 a _8 \\ \Rightarrow \quad a _{10}-2 a _8 & =6 a _9 \Rightarrow \frac{a _{10}-2 a _8}{2 a _9}=3 \end{array} $
