If alpha and beta are the roots of x^2 - ax + | 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 $x^2 - ax + b = 0$,they satisfy the equation:$\alpha^2 - a\alpha + b = 0 \implies \alpha^2 = a\alpha - b$$

Vedclass · Accepted Answer

$V_{n+1} = aV_n - bV_{n-1}$

Vedclass · Answer

(C) Since $\alpha$ and $\beta$ are roots of $x^2 - ax + b = 0$,they satisfy the equation:$\alpha^2 - a\alpha + b = 0 \implies \alpha^2 = a\alpha - b$$\beta^2 - a\beta + b = 0 \implies \beta^2 = a\beta - b$Multiplying the first equation by $\alpha^{n-1}$ and the second by $\beta^{n-1}$:$\alpha^{n+1} = a\alpha^n - b\alpha^{n-1}$$\beta^{n+1} = a\beta^n - b\beta^{n-1}$Adding these two equations:$\alpha^{n+1} + \beta^{n+1} = a(\alpha^n + \beta^n) - b(\alpha^{n-1} + \beta^{n-1})$Given $V_n = \alpha^n + \beta^n$,we substitute this into the equation:$V_{n+1} = aV_n - bV_{n-1}$

If $\alpha$ and $\beta$ are the roots of $x^2 - ax + b = 0$ and if $\alpha^n + \beta^n = V_n$,then

Similar Questions

Roots of $2 k x^{2}+5 k x+2=0$ are equal if $k$ is equal to

Let two numbers have an arithmetic mean of $9$ and a geometric mean of $4$. Then these numbers are the roots of the quadratic equation:

The existence of a unique solution for the system of equations $x+y+z=\beta$,$5x-y+\alpha z=10$,and $2x+3y-z=6$ depends on:

If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^{2} + bx + c = 0$,then the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$ is

The number of values of $k$ for which the equation $x^2 - 3x + k = 0$ has two real and distinct roots lying in the interval $(0, 1)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module