Question

Quadratic equation

Answer 1

Do you only need Qualified Helpers to be here? :P

Answer 2

let \(\alpha\) and \(\beta\) be the roots of the equation.

\(\large \color{black }{\begin{align}
x^2-6x-2=0\hspace{.33em}\\~\\
\end{align}}\)

.If

\(\large \color{black }{\begin{align}a_n=\alpha^n-\beta^n \end{align}}\)


for \(n\geq 1\)


then the value of


\(\large \color{black }{\begin{align}\dfrac{a_{10}-2a_{8}}{2a_9} \end{align}}\) is ?

Answer 3

\(\large \color{black }{\begin{align}
&a.)\quad 3\hspace{.33em}\\~\\
&b.)-3 \hspace{.33em}\\~\\
&c.)\quad6 \hspace{.33em}\\~\\
&d.)-6 \hspace{.33em}\\~\\
\end{align}}\)

Answer 4

Can you find the roots?

Answer 5

yes

Answer 6

What are the roots?

Answer 7

\(3+\sqrt{11},3-\sqrt{11}\)

Answer 8

What is \(a_1\)?

Answer 9

\(2\sqrt {11}\)

Answer 10

What is \(a_2\) ? I am just trying to find out some pattern..

Answer 11

\(12\sqrt{11}\)

Answer 12

we may use this identity \[\large x^n-y^n = (x+y)(x^{n-1}-y^{n-1}) -xy(x^{n-2}-y^{n-2})\]

Answer 13

plugin \((x,y,n) = (\alpha, \beta, 10)\)

Answer 14

this question reminds me of another similar q
http://openstudy.com/users/eliassaab#/updates/5499f402e4b0b8a54faea29e

Answer 15

is this the only way i m bad at remembering the aukward formula's

Answer 16

in that link there are at least 3 different ways which we can use to solve this problem

i have used the least painful one :)

Answer 17

@rational

m stuck at this

\(\large \color{black }{\begin{align}
&\dfrac{(\alpha+\beta)(\alpha^{9}-\beta^{9}) -\alpha\beta(\alpha^{8}-\beta^{8})-2(\alpha^{8}-\beta^{8})}{2(\alpha^{9}-\beta^{9}) }\hspace{.33em}\\~\\
&=\dfrac{(\alpha+\beta)}{2}-\dfrac{\alpha\beta(\alpha^{8}-\beta^{8})-2(\alpha^{8}-\beta^{8})}{2(\alpha^{9}-\beta^{9}) }\hspace{.33em}\\~\\
& =\dfrac{(\alpha+\beta)}{2}-\dfrac{(\alpha\beta -2)(\alpha^{8}-\beta^{8})}{2(\alpha^{9}-\beta^{9}) }\hspace{.33em}\\~\\
\end{align}}\)

Answer 18

I got another way, here are the steps:
\[\begin{gathered}
\frac{{{a_{10}} - 2{a_8}}}{{2{a_9}}} = \frac{{{a_{10}} - 2{a_8} + {a_6} - {a_6}}}{{2{a_9}}} = \frac{{{{\left( {{\alpha ^5} - {\alpha ^3}} \right)}^2} - {{\left( {{\beta ^5} - {\beta ^3}} \right)}^2} - {\alpha ^6} + {\beta ^6}}}{{2\left( {{\alpha ^9} - {\beta ^9}} \right)}} = \hfill \\
= \frac{{{\alpha ^6}\left[ {{{\left( {{\alpha ^2} - 1} \right)}^2} - 1} \right] - {\beta ^6}\left[ {{{\left( {{\beta ^2} - 1} \right)}^2} - 1} \right]}}{{2\left( {{\alpha ^9} - {\beta ^9}} \right)}} = \hfill \\
= \frac{{{\alpha ^8}\left( {{\alpha ^2} - 2} \right) - {\beta ^8}\left( {{\beta ^2} - 2} \right)}}{{2\left( {{\alpha ^9} - {\beta ^9}} \right)}} = \hfill \\
= \frac{{{\alpha ^8} \cdot 6\alpha - {\beta ^8} \cdot 6\beta }}{{2\left( {{\alpha ^9} - {\beta ^9}} \right)}} = 3 \hfill \\
\end{gathered} \]

Answer 19

Nice :)

Answer 20

@mathmath333 the solution it is a one liner using that identity, let me work it..

Answer 21

thanks! @rational

Answer 22

\( \begin{gathered}
\frac{{{a_{10}} - 2{a_8}}}{{2{a_9}}} = \frac{{{a_{10}} - 2{a_8} + {a_6} - {a_6}}}{{2{a_9}}} = \color{red }{\frac{{{{\left( {{\alpha ^5} - {\alpha ^3}} \right)}^2} - {{\left( {{\beta ^5} - {\beta ^3}} \right)}^2} - {\alpha ^6} + {\beta ^6}}}{{2\left( {{\alpha ^9} - {\beta ^9}} \right)}}} = \hfill \\
= \frac{{{\alpha ^6}\left[ {{{\left( {{\alpha ^2} - 1} \right)}^2} - 1} \right] - {\beta ^6}\left[ {{{\left( {{\beta ^2} - 1} \right)}^2} - 1} \right]}}{{2\left( {{\alpha ^9} - {\beta ^9}} \right)}} = \hfill \\
= \frac{{{\alpha ^8}\left( {{\alpha ^2} - 2} \right) - {\beta ^8}\left( {{\beta ^2} - 2} \right)}}{{2\left( {{\alpha ^9} - {\beta ^9}} \right)}} = \hfill \\
= \frac{{{\alpha ^8} \cdot 6\alpha - {\beta ^8} \cdot 6\beta }}{{2\left( {{\alpha ^9} - {\beta ^9}} \right)}} = 3 \hfill \\
\end{gathered}\)

i didnt understand the red part

Answer 23

\[\large x^n-y^n = (x+y)(x^{n-1}-y^{n-1}) -xy(x^{n-2}-y^{n-2})\]

plugin \((x,y,n) = (\alpha, \beta, 10)\) and get

\[\large \begin{align} \alpha^{10}-\beta^{10} &= (\alpha+\beta)(\alpha^{9}-\beta^{9}) -\alpha\beta(\alpha^{8}-\beta^{8})\\~\\
a_{10}&= 6a_9 +2a_8\\~\\
a_{10}-2a_{8} &= 6a_9
\end{align}\]

divide \(a_9\) both sides and you're done!

Answer 24

ok that makes sense now ,like a rabit out of a hat

Answer 25

please compute the squares of binomial in order to get the origibal expression @mathmath333

Answer 26

oops..original*

Answer 27

thnks