Question

If $\alpha ,\beta $ are roots of the equation ${{x}^{2}}-6x-2=0$ and we define ${{a}_{n}}={{\alpha }^{n}}-{{\beta }^{n}}$ then find the value of $\dfrac{{{a}_{10}}-2{{a}_{8}}}{2{{a}_{9}}}$

Answer 1

Hint: Now we know that $\alpha ,\beta $ are roots of the equation ${{x}^{2}}-6x-2=0$. Now it will hence also satisfy the equation ${{x}^{8}}\left( {{x}^{2}}-6x-2 \right)=0$. Hence after substituting x = $\alpha $ and x = $\beta $ in this equation we will get two equations in $\alpha $ and $\beta $ . Now we know that ${{a}_{n}}={{\alpha }^{n}}-{{\beta }^{n}}$ using this we will evaluate $\dfrac{{{a}_{10}}-2{{a}_{8}}}{2{{a}_{9}}}$ and then substitute the equations we have in in $\alpha $ and $\beta $. Hence we will find the value of $\dfrac{{{a}_{10}}-2{{a}_{8}}}{2{{a}_{9}}}$.

Complete step by step answer:
Now let $\alpha ,\beta $ ne the roots of the equation ${{x}^{2}}-6x-2=0$
Hence the roots will satisfy the equation
Hence we have ${{\alpha }^{2}}-6\alpha -2=0$
Multiplying the whole equation by ${{\alpha }^{8}}$ we get.
${{\alpha }^{10}}-6{{\alpha }^{9}}-2{{\alpha }^{8}}=0$
${{\alpha }^{10}}-2{{\alpha }^{8}}=6{{\alpha }^{9}}...................\left( 1 \right)$
Similarly we also have ${{\beta }^{2}}-6\beta -2=0$
Multiplying the whole equation by ${{\alpha }^{8}}$ we get.
${{\beta }^{10}}-6{{\beta }^{9}}-2{{\beta }^{8}}=0$
${{\beta }^{10}}-2{{\beta }^{8}}=6{{\beta }^{9}}.................(2)$
Now we are given that ${{a}_{n}}={{\alpha }^{n}}-{{\beta }^{n}}$
Substituting the equation in $\dfrac{{{a}_{10}}-2{{a}_{8}}}{{{a}_{9}}}$ we get.
$\dfrac{{{\alpha }^{10}}-{{\beta }^{10}}-2\left( {{\alpha }^{8}}-{{\beta }^{8}} \right)}{2\left( {{\alpha }^{9}}-{{\beta }^{9}} \right)}$
Now opening the bracket and rearranging we get
$=\dfrac{{{\alpha }^{10}}-2{{\alpha }^{8}}-{{\beta }^{10}}+2{{\beta }^{8}}}{{{\alpha }^{9}}-{{\beta }^{9}}}$
Taking -1 common we get
$=\dfrac{{{\alpha }^{10}}-2{{\alpha }^{8}}-\left( {{\beta }^{10}}-2{{\beta }^{8}} \right)}{{{\alpha }^{9}}-{{\beta }^{9}}}$
Now substituting the values from equation (1) and equation (2) we get.
$=\dfrac{6{{\alpha }^{9}}-6{{\beta }^{9}}}{{{\alpha }^{9}}-{{\beta }^{9}}}$
Now we can say taking – 6 common from numerator we get
$=\dfrac{6({{\alpha }^{9}}-{{\beta }^{9}})}{{{\alpha }^{9}}-{{\beta }^{9}}}$
= 6.
Hence we have the value of $\dfrac{{{a}_{10}}-2{{a}_{8}}}{2{{a}_{9}}}$ is 6.

Note:
Here we can also find the roots of the equation with the help of formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and then substitute the values to find ${{a}^{10}},{{a}^{8}}{{a}^{9}}$ but this will be a very lengthy and tedious process. Hence we substitute the roots in the equation and solve accordingly. Also note that when we get $=\dfrac{6({{\alpha }^{9}}-{{\beta }^{9}})}{{{\alpha }^{9}}-{{\beta }^{9}}}$ we can cancel terms ${{\alpha }^{9}}-{{\beta }^{9}}$ because ${{\alpha }^{9}}-{{\beta }^{9}}\ne 0$ and this is because the roots are not same as discriminant is not 0. Hence it is necessary to check that first