Let - - be the roots of x2 - 6x - 2- 0 which - - - if an- -n -n for n-1 - then the value of a10-2a8-2a9 is

Α^2 6α 2=0————(1) β^2 6β 2 =0 ————(2) α_10 2α_8/2α_9 = (α^10 β^10) 2(α^8 –β^8)/ 2(α^9 β^9) nbsp;= α^8(α^2 2) –β^8(β^2 2)/2(α^9 – β^9) = α^36α –β^86 ...

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Byju's Answer

Standard XI

Mathematics

Algebra of Roots of Quadratic Equations

Let α β be ...

Question

Let α & β be the roots of ${x^2}$– 6x – 2= 0 which α >β . if $a_{n}$ = $α^{n}$ – $β^{n}$ for n $\geq$ 1 , then the

value of $\frac{a_{10} - 2 a_{8}}{2 a_{9}}$ is

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Solution

The correct option is B
3

$α^{2}$ - 6α - 2=0------------(1)
$β^{2}$ -6β -2 =0 ------------(2)
$\frac{α_{1} 0 - 2 α_{8}}{2 α_{9}}$ = $\frac{(α^{10} - β^{10}) - 2 (α^{8} - β^{8})}{2 (α^{9} - β^{9})}$ = $\frac{α^{8} (α^{2} - 2) - β^{8} (β^{2} - 2)}{2 (α^{9} - β^{9})}$
= $\frac{α^{3} 6 α - β^{8} 6 β}{2 (α^{9} - β^{9})}$ =3

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Similar questions

Q. Let

α

and

β

be the roots of

x^{2} - 6 x - 2 = 0

, with

α > β

. lf

a_{n} = α^{n} - β^{n}

for

n \geq 1

, then the value of

\frac{a_{10} - 2 a_{8}}{2 a_{9}}

Q. Let

α

and

β

be the roots of equation

x^{2} - 6 x - 2 = 0

. If

a_{n} = α^{n} - β^{n}

, for

n \geq 1

, then the value of

\frac{a_{10} - 2 a_{8}}{2 a_{9}}

Q. Let

α

and

β

be the roots of equation

x^{2} - 6 x - 2 = 0

. If

a_{n} = α^{n} - β^{n}

, for

n ⩾ 1

, then the value of

\frac{a_{10} - 2 a_{8}}{2 a_{9}}

is equal to

Q. Let

α

and

β

be the roots of equation

x^{2} - 6 x - 2 = 0

. If

a_{n} = α^{n} - β^{n}

, for

n \geq 1

, then the value of

\frac{a_{10} - 2 a_{8}}{2 a_{9}}

Q. Let

α

and

β

be the roots of

x^{2} - 6 x - 2 = 0

. If

a_{n} = α^{n} - β^{n}

for

n \geq 1

, then the value of

\frac{a_{10} - 2 a_{8}}{3 a_{9}}

is:

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Let α & β be the roots of \({x^2}\) – 6x – 2= 0 which α >β . if an= αn –βn for n≥1 , then the value of a10−2a82a9 is

The correct option is B 3 α2 - 6α - 2=0------------(1) β2 -6β -2 =0 ------------(2) α10−2α82α9 = (α10−β10)−2(α8–β8)2(α9−β9) = α8(α2−2)–β8(β2−2)2(α9–β9) = α36α–β86β2(α9−β9) =3

Let α & β be the roots of \({x^2}\)– 6x – 2= 0 which α >β . if $a_{n}$ = $α^{n}$ – $β^{n}$ for n $\geq$ 1 , then the

value of $\frac{a_{10} - 2 a_{8}}{2 a_{9}}$ is