Let alpha and beta be the roots of the equati | vedclass

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

Question

(C) Since $\alpha$ and $\beta$ are roots of $x^2 - 6x - 2 = 0$,we have $\alpha + \beta = 6$ and $\alpha\beta = -2$.Also,$\alpha^2 = 6\alpha + 2$ and $

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Since $\alpha$ and $\beta$ are roots of $x^2 - 6x - 2 = 0$,we have $\alpha + \beta = 6$ and $\alpha\beta = -2$.Also,$\alpha^2 = 6\alpha + 2$ and $\beta^2 = 6\beta + 2$.Given $a_n = \alpha^n - \beta^n$,we evaluate the expression:$\frac{a_{10} - 2a_8}{2a_9} = \frac{(\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8)}{2(\alpha^9 - \beta^9)}$$= \frac{\alpha^8(\alpha^2 - 2) - \beta^8(\beta^2 - 2)}{2(\alpha^9 - \beta^9)}$Since $\alpha^2 - 2 = 6\alpha$ and $\beta^2 - 2 = 6\beta$,we substitute these values:$= \frac{\alpha^8(6\alpha) - \beta^8(6\beta)}{2(\alpha^9 - \beta^9)}$$= \frac{6(\alpha^9 - \beta^9)}{2(\alpha^9 - \beta^9)}$$= \frac{6}{2} = 3$.

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

Similar Questions

The cubic equation whose roots are thrice to each of the roots of $x^3+2x^2-4x+1=0$ is

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

The expression $x^3 - 3x^2 - 9x + c$ can be written in the form $(x - a)^2 (x - b)$ if the value of $c$ is

If the roots of the equation $x^3+3px^2+3qx-8=0$ are in a geometric progression,then $\frac{q^3}{p^3}=$

If $3p^2 = 5p + 2$ and $3q^2 = 5q + 2$,where $p \ne q$,then $pq$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module