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

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

Question

(C) Given that $\alpha$ and $\beta$ are the roots of $x^2 - 6x - 2 = 0$.Since $\alpha$ and $\beta$ are roots,they satisfy the equation:$\alpha^2 - 6\a

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Given that $\alpha$ and $\beta$ are the roots of $x^2 - 6x - 2 = 0$.Since $\alpha$ and $\beta$ are roots,they satisfy the equation:$\alpha^2 - 6\alpha - 2 = 0 \implies \alpha^2 - 2 = 6\alpha$$\beta^2 - 6\beta - 2 = 0 \implies \beta^2 - 2 = 6\beta$We need to evaluate $\frac{a_{10} - 2a_8}{2a_9}$.Substituting $a_n = \alpha^n - \beta^n$:$\frac{(\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8)}{2(\alpha^9 - \beta^9)}$$= \frac{\alpha^8(\alpha^2 - 2) - \beta^8(\beta^2 - 2)}{2(\alpha^9 - \beta^9)}$Using the relations $\alpha^2 - 2 = 6\alpha$ and $\beta^2 - 2 = 6\beta$:$= \frac{\alpha^8(6\alpha) - \beta^8(6\beta)}{2(\alpha^9 - \beta^9)}$$= \frac{6\alpha^9 - 6\beta^9}{2(\alpha^9 - \beta^9)}$$= \frac{6(\alpha^9 - \beta^9)}{2(\alpha^9 - \beta^9)} = \frac{6}{2} = 3$.

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

Similar Questions

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

Let $f(x) = x^3 - 2x + 2$. If real numbers $a, b, c$ are such that $|f(a)| + |f(b)| + |f(c)| = 0$,then the value of $f^2(a^2 + \frac{2}{a}) + f^2(b^2 + \frac{2}{b}) - f^2(c^2 + \frac{2}{c})$ is equal to:

If $\alpha, \beta, \gamma$ are the roots of $x^3-6x^2+11x-6=0$,then the equation having the roots $\alpha^2+\beta^2, \beta^2+\gamma^2$ and $\gamma^2+\alpha^2$ is

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-3x^2+3x+1=0$,then $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=$

If the sum of the squares of the roots of the equation $x^2-(a-2)x-(a+1)=0$ is least for an appropriate value of the variable parameter $a$, then the value of $a$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module