Let alpha and beta be the roots of x^2-6 x-2= | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Since $\alpha$ and $\beta$ are the roots(solutions) of the equation

$x^2-6 x-2=0$

So, it will satisfy the equation

$\alpha^2-6 \alpha-2=0$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

Since $\alpha$ and $\beta$ are the roots(solutions) of the equation

$x^2-6 x-2=0$

So, it will satisfy the equation

$\alpha^2-6 \alpha-2=0$

$\beta^2-6 \beta-2=0$

Now, we have to find the value of $\frac{a_{10}-2 a_8}{2 a_9}$

Given, $a_n=\alpha^n-\beta^n$

So, $a_{10}=\alpha^{10}-\beta^{10}$

So, to get the values of $\alpha^{10}$ and $\beta^{10}$, multiplying equation $(1)$ and $(2)$ by $\alpha^8$ and $\beta^8$ respectively.

$\alpha^{10}-6 \alpha^9-2 \alpha^8=0$

$	ext { or, } \alpha^{10}=6 \alpha^9+2 \alpha^8 \ldots . . 	ext { (4) }$

$\beta^{10}-6 \beta^9-2 \beta^8=0$

$	ext { or, } \beta^{10}=6 \beta^9+2 \beta^8 \ldots . . 	ext { (5) }$

So, using equation$(4)$ and $(5)$ in $(3)$, we get

$a_{10}=6 \alpha^9+2 \alpha^8-\left(6 \beta^9+2 \beta^8ight)$

$a_{10}=6 \alpha^9+2 \alpha^8-6 \beta^9-2 \beta^8$

$=6\left(\alpha^9-\beta^9ight)+2\left(\alpha^8-\beta^8ight)$

$a_{10}=6 a_9+2 a_8$

$a_{10}-2 a_8=6 a_9$

$\frac{a_{10}-2 a_8}{2 a_9}=3$

Let $\alpha$ and $\beta$ be the roots of $x^2-6 x-2=0$, with $\alpha>\beta$. If $a_n=\alpha^n-\beta^n$ for $n \geq 1$, then the value of $\frac{a_{10}-2 a_8}{2 a_9}$ is

Similar Questions

If $|x - 2| + |x - 3| = 7$, then $x =$

If $x,\;y,\;z$ are real and distinct, then $u = {x^2} + 4{y^2} + 9{z^2} - 6yz - 3zx - zxy$ is always

$\{ x \in R:|x - 2|\,\, = {x^2}\} = $

The sum of all non-integer roots of the equation $x^5-6 x^4+11 x^3-5 x^2-3 x+2=0$ is

If $\alpha , \beta , \gamma$ are roots of equation $x^3 + qx -r = 0$ then the equation, whose roots are

$\left( {\beta \gamma + \frac{1}{\alpha }} \right),\,\left( {\gamma \alpha + \frac{1}{\beta }} \right),\,\left( {\alpha \beta + \frac{1}{\gamma }} \right)$