Question

If alpha and beta be zeros of x²-6x-2=0 With alpha > beta if an= alpha^n-Beta^n N>=1 Find value of[a 10 - 2(a8)]/2a9

Answer 1

Correct Question:

If $\sf{\alpha}$ and $\sf{\beta}$ be the zeros of polynomial $\sf{x^2 - 6x - 2 = 0}$ with $\sf{\alpha > \beta}$$\sf{a_n = {\alpha}^{n} - {\beta}^{n}}$ for all n $\geq 1$Find the value of $\sf{ \dfrac{a_{10}- 2.a_8}{2a_9} }$

$\huge\mathfrak\blue{Answer:}$

Given:

• We have been given an equation $\sf{x^2-6x-2 = 0}$
• $\sf{\alpha}$ and $\sf{\beta}$ are zeros of given equation
• $\sf{a_n = {\alpha}^{n} - {\beta}^{n}}$

To Find:

• We have to find the value of
• $\\ \sf{\dfrac{a_{10} - 2.a_8}{2a_9}}$

Solution:

We have been given that $\sf{\alpha}$ and $\sf{\beta}$ are the zeros of polynomial

$\hookrightarrow \sf{x^2 - 6x - 2 = 0}$

________________________________

$\underline{\large\mathfrak\orange{According \: to \: the \: Question:}}$

$\sf{\alpha}$ and $\sf{\beta}$ are the roots of given polynomial

Hence they will satisfy the given polynomial

$\blue{\bigstar}\: \: \underline{\large\sf{\pink{Putting \: x = \alpha}}}$

$\hookrightarrow \sf{ {\alpha}^{2} - 6 \alpha - 2 = 0}$

$\hookrightarrow \sf{{\alpha}^{2} - 2 = 6 \alpha}$

$\hookrightarrow \boxed{\sf{{\alpha}^{2} - 2 = 6 \alpha}}$ -------------- ( 1 )

$\blue{\bigstar} \: \: \underline{\large\sf{\pink{Putting \: x = \beta}}}$

$\hookrightarrow \sf{{\beta}^{2} - 6 \beta - 2 = 0}$

$\hookrightarrow \sf{{\beta}^{2} - 2 = 6 \beta}$

$\hookrightarrow \boxed{\sf{{\beta}^{2} - 2 = 6 \beta}}$ ---------------- ( 2 )

_________________________________

$\underline{\large\mathfrak\orange{We \: have \: to \: find:}}$

$\implies \sf{\dfrac{a_{10}- 2a_8}{2a_9} }$

Using the given condition

$\boxed{\sf{\green{a_n = {\alpha}^{n} - {\beta}^{n}}}}$

$\implies \sf{\dfrac{({\alpha}^{10} - {\beta}^{10}) - 2({\alpha}^8-{\beta}^8)}{2({\alpha}^9-{\beta}^9)}}$

$\implies \sf{\dfrac{{\alpha}^{10} - {\beta}^{10} - 2{\alpha}^8 - 2{\beta}^8}{2({\alpha}^9-{\beta}^9)}}$

$\implies \sf{\dfrac{({\alpha}^{10} - 2{\alpha}^{8}) - ({\beta}^{10}- 2{\beta}^{8} )} {2({\alpha}^9-{\beta}^9)}}$

$\implies \sf{\dfrac{{\alpha}^8({\alpha}^2 - 2 ) - {\beta}^8({\beta}^2 - 2 )} {2({\alpha} ^9-{\beta}^9)}}$

Putting values from eqn ( 1 ) and ( 2 )

$\\ \\ \implies \sf{\dfrac{{\alpha}^8(6 \alpha) - {\beta}^8(6 \beta)} {2({\alpha}^9-{\beta}^9)}}$

$\implies \sf{\dfrac{6{\alpha}^9- 6{\beta}^9} {2({\alpha}^9-{\beta}^9)}}$

$\implies \sf{\dfrac{6 ( {\alpha}^9 - {\beta}^9 ) } {2({\alpha}^9-{\beta}^9)}}$

$\implies \sf{\dfrac{6 \times \cancel{( {\alpha}^9 - {\beta}^9 )} } {2 \times \cancel{({\alpha}^9-{\beta}^9)}}}$

$\implies \sf{\dfrac{6}{2} = 3 \: ( Answer ) }$

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$\LARGE\sf{\red{A}\green{n}\orange{s}\pink{w}\blue{e}\purple{r}} : \boxed{\sf{\large \red{\dfrac{a_{10} - 2.a_8}{2a_9} = 3 }}}$

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Answer 2

Answer:

ur answer is a10-2.a8/2a9=3

Step-by-step explanation:

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