Question

Let $\alpha$ and $\beta$ are the roots of the ${x^2} - 6x - 2 = 0$, with $\alpha$>$\beta$. If ${a_n} = {\alpha ^n} - {\beta ^n}$ for $n \geqslant 1,$ then the value of $\dfrac{{{a_{10}} - 2{a_8}}}{{2{a_9}}}$ is
A) 1
B) 2
C) 3
D) 4

Answer 1

Since it is given that $\alpha$ and $\beta$ are the roots of the ${x^2} - 6x - 2 = 0$, then we will put the value of x as $\alpha$ and $\beta$ in the given equation and find the value.

Complete step-by-step answer:
It is given that $\alpha$ and $\beta$ are the roots of the ${x^2} - 6x - 2 = 0$.
Hence, ${\alpha ^2} - 6\alpha - 2 = 0$
${\alpha ^2} = 6\alpha + 2$
Multiply ${\alpha ^8}$ to both sides of the given equation, we get
${\alpha ^2} \times {\alpha ^8} = {\alpha ^8}(6\alpha + 2)$
${\alpha ^{10}} = 6{\alpha ^9} + 2{\alpha ^8}$ (Since bases are same, we can add the powers) ……………… (1)
Similarly, we will put the value of x as $\beta$, we get
${\beta ^2} - 6\beta - 2 = 0$
${\beta ^2} = 6\beta + 2$
Multiply ${\beta ^8}$ to both sides of the given equation, we get
${\beta ^2} \times {\beta ^8} = {\beta ^8}(6\beta + 2)$
${\beta ^{10}} = 6{\beta ^9} + 2{\beta ^8}$ (Since bases are same, we can add the powers) ……………… (2)
Now, we have to find the value of $\dfrac{{{a_{10}} - 2{a_8}}}{{2{a_9}}}$ and it is given that ${a_n} = {\alpha ^n} - {\beta ^n}$ (where $\alpha$>$\beta$)
Now, we will put n = 10 in ${a_n} = {\alpha ^n} - {\beta ^n}$, we get
${a_{10}} = {\alpha ^{10}} - {\beta ^{10}}$, ${a_8} = {\alpha ^8} - {\beta ^8}$
Now, we have
$\dfrac{{{a_{10}} - 2{a_8}}}{{2{a_9}}}$= $\dfrac{{{\alpha ^{10}} - {\beta ^{10}} - 2({\alpha ^8} - {\beta ^8})}}{{2{a_9}}}$
Put the value of (1) and (2), we get
$\dfrac{{{a_{10}} - 2{a_8}}}{{2{a_9}}}$= $\dfrac{{6{\alpha ^9} + 2{\alpha ^8} - 6{\beta ^9} - 2{\beta ^8} - 2{\alpha ^8} + 2{\beta ^8}}}{{2{a_9}}}$
= $\dfrac{{6({\alpha ^9} - {\beta ^9})}}{{2{a_9}}}$
= $\dfrac{{6{a_9}}}{{2{a_9}}}$= 3

Note: In this question, we are given an equation, of which $\alpha$ and $\beta$ are the two roots, so we have put in the value of x as $\alpha$ and $\beta$ and solve it then we will find the value of a10 (since the value of an is also given as ${a_n} = {\alpha ^n} - {\beta ^n}$ for $n \geqslant 1,$). Put all the values in $\dfrac{{{a_{10}} - 2{a_8}}}{{2{a_9}}}$ to get the desired result.