   Question Answers

# Let $\alpha$ and $\beta$ be the roots of equation ${{x}^{2}}-6x-2=0$ . If ${{a}_{n}}={{\alpha }^{n}}-{{\beta }^{n}}$ , for $n\ge 1$ , then the value of $\dfrac{{{a}_{10}}-2{{a}_{8}}}{2{{a}_{9}}}$ is equals to( a ) 3( b ) -3( c ) 6( d ) -6

Hint: To solve this question, what we will do is first we will write the given quadratic equation ${{x}^{2}}-6x-2=0$ in terms of $\alpha$ and $\beta$. Then, we will multiply both the equations with factor ${{\alpha }^{8}}$ and ${{\beta }^{8}}$ respectively. Then we will arrange the equations together in such a way that we get factors of the same power separately and hence, we will try to get the condition asked in question.

As in question it is clearly given that, $\alpha$ and $\beta$ be the roots of equation ${{x}^{2}}-6x-2=0$ .
then, we can put x = $\alpha$ and x = $\beta$,
so, putting x = $\alpha$ in ${{x}^{2}}-6x-2=0$, we get
${{\alpha }^{2}}-6\alpha -2=0$…..( i )
so, putting x = $\beta$ in ${{x}^{2}}-6x-2=0$, we get
${{\beta }^{2}}-6\beta -2=0$…. ( ii )
Multiplying equation ( i ) by ${{\alpha }^{8}}$, we get
${{\alpha }^{8}}({{\alpha }^{2}}-6\alpha -2)=0({{\alpha }^{8}})$
$({{\alpha }^{10}}-6{{\alpha }^{9}}-2{{\alpha }^{8}})=0$
${{\alpha }^{10}}=6{{\alpha }^{9}}+2{{\alpha }^{8}}$……( iii )
Multiplying equation ( i ) by ${{\beta }^{8}}$, we get
${{\beta }^{8}}({{\beta }^{2}}-6\beta -2)=0({{\beta }^{8}})$
$({{\beta }^{10}}-6{{\beta }^{9}}-2{{\beta }^{8}})=0$
${{\beta }^{10}}=6{{\beta }^{9}}+2{{\beta }^{8}}$…..( iv )
Subtracting ( iv ) from ( iii ), we get
${{\alpha }^{10}}-{{\beta }^{10}}=6{{\alpha }^{9}}+2{{\alpha }^{8}}-(6{{\beta }^{9}}+2{{\beta }^{8}})$
On simplifying we get
${{\alpha }^{10}}-{{\beta }^{10}}=6({{\alpha }^{9}}-{{\beta }^{9}})+2({{\alpha }^{8}}-{{\beta }^{8}})$
As, in question it is given that ${{a}_{n}}={{\alpha }^{n}}-{{\beta }^{n}}$, so ${{a}_{9}}={{\alpha }^{9}}-{{\beta }^{9}}$and ${{a}_{10}}={{\alpha }^{10}}-{{\beta }^{10}}$ and ${{a}_{8}}={{\alpha }^{8}}-{{\beta }^{8}}$
${{a}_{10}}=6({{a}_{9}})+2({{a}_{8}})$
Moving, $2({{a}_{8}})$ from right hand side to left hand side, we get
${{a}_{10}}-{{a}_{8}}=6{{a}_{9}}$
Re – writing above equation, we get
${{a}_{10}}-{{a}_{8}}=3\cdot 2{{a}_{9}}$
Using cross-multiplication, we get
$\dfrac{{{a}_{10}}-{{a}_{8}}}{2{{a}_{9}}}=3$

So, the correct answer is “Option A”.

Note: Always remember if it is given that a and b are roots of quadratic equation, then we can replace x in quadratic equation with a and b. This question is a bit tricky so what you can do in this type of question is just try to re – arrange the quadratic equation according to the given equation in question to prove or any condition provided this will make the question a bit easier. Try to avoid silly mistakes as it may make the question incorrect.