Let alpha, beta be the roots of the equation | vedclass

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

Question

(C) Given the quadratic equation $x^{2}-6x-2=0$. Since $\alpha$ and $\beta$ are roots,they satisfy the equation: $x^{n}-6x^{n-1}-2x^{n-2}=0$ $\Rightar

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Given the quadratic equation $x^{2}-6x-2=0$. Since $\alpha$ and $\beta$ are roots,they satisfy the equation: $x^{n}-6x^{n-1}-2x^{n-2}=0$ $\Rightarrow x^{n}-2x^{n-2}=6x^{n-1}$ For $x = \alpha$ and $x = \beta$,we have: $\alpha^{n}-2\alpha^{n-2}=6\alpha^{n-1}$ $\beta^{n}-2\beta^{n-2}=6\beta^{n-1}$ Subtracting the two equations: $(\alpha^{n}-\beta^{n})-2(\alpha^{n-2}-\beta^{n-2})=6(\alpha^{n-1}-\beta^{n-1})$ Using the definition $a_{n}=\alpha^{n}-\beta^{n}$,we get: $a_{n}-2a_{n-2}=6a_{n-1}$ For $n=10$: $a_{10}-2a_{8}=6a_{9}$ Therefore,the value of $\frac{a_{10}-2a_{8}}{2a_{9}} = \frac{6a_{9}}{2a_{9}} = 3$.

Let $\alpha, \beta$ be the roots of the equation $x^{2}-6x-2=0$ with $\alpha>\beta$. If $a_{n}=\alpha^{n}-\beta^{n}$ for $n \geq 1$,then the value of $\frac{a_{10}-2a_{8}}{2a_{9}}$ is

Similar Questions

If $\frac{x-4}{x^2-5x-2k} = \frac{2}{x-2} - \frac{1}{x+k}$,then $k$ is equal to

For the quadratic equation $2x^2 - 2(p - 2)x - p - 1 = 0$,what should be the value of $p$ such that the sum of the squares of its roots is minimized?

If $\sin \alpha$ and $\cos \alpha$ are the roots of the equation $ax^2 + bx + c = 0$,then:

For the equation $x^4+x^3-4x^2+x-1=0$,the ratio of the sum of the squares of all the roots to the product of the distinct roots is

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+ax^2+bx+c=0$,then $\alpha^{-1}+\beta^{-1}+\gamma^{-1}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module