Let - and - be the roots of x2 -6x-2 - 0- with - - - If an - - n - -n for n - 1- then the value of a10-2a8-2a9is

Αβ are the roots of x^2 6x 2= 0 ∴α^2 6α 2= 0 ⇒α^10 6α^9 2α^8 = 0 ⇒α^10 2α^8 =6α^9 ....(1) Similarly β ^10 2β^8 = 6β ^9 ...

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Byju's Answer

Standard XII

Mathematics

Roots of a Quadratic Equation

Let α and β b...

Question

Let $α$ and $β$ be the roots of $x^{2} - 6 x - 2 = 0$ , with $α > β$ . If $a_{n} = α^{n} - β^{n} f o r n \geq 1$ , then the value of $\frac{a_{10} - 2 a_{8}}{2 a_{9}}$ is

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Solution

The correct option is C 3
$α β$ are the roots of $x^{2} - 6 x - 2 = 0 ∴ α^{2} - 6 α - 2 = 0 \Rightarrow α^{10} - 6 α^{9} - 2 α^{8} = 0 \Rightarrow α^{10} - 2 α^{8} = 6 α^{9} . . . . (1)$
Similarly $β^{10} - 2 β^{8} = 6 β^{9} . . . . . (2)$
From equation (1) and (2) -
$α^{10} - β^{10} - 2 (α^{8} - β^{8}) = 6 (α^{9} - β^{9}) \Rightarrow a_{10} - 2 a_{8} = 6 a_{9} \Rightarrow \frac{a_{10} - 2 a_{8}}{2 q_{9}} = 3$

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Similar questions

Let $α$ and $β$ be the roots of $x^{2} - 6 x - 2 = 0$ with $α > β$ if $a_{n} = α^{n} - β^{n}$ for $n \geq 1$ then the value of $\frac{a_{10} - 2 a_{8}}{3 a_{9}} =$

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:

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}}

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Let α and β be the roots of x2−6x−2=0, with α>β . If an=αn−βnforn≥1, then the value of a10−2a82a9is

The correct option is C 3αβ are the roots of x2−6x−2=0∴α2−6α−2=0⇒α10−6α9−2α8=0⇒α10−2α8=6α9....(1) Similarly β10−2β8=6β9.....(2) From equation (1) and (2) - α10−β10−2(α8−β8)=6(α9−β9)⇒a10−2a8=6a9⇒a10−2a82q9=3

Let $α$ and $β$ be the roots of $x^{2} - 6 x - 2 = 0$ , with $α > β$ . If $a_{n} = α^{n} - β^{n} f o r n \geq 1$ , then the value of $\frac{a_{10} - 2 a_{8}}{2 a_{9}}$ is