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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Since $\alpha$ and $\beta$ are the roots of $x^2 - ax + b = 0$,they satisfy the equation:$\alpha^2 - a\alpha + b = 0 \implies \alpha^2 = a\alpha - b$$\beta^2 - a\beta + b = 0 \implies \beta^2 = a\beta - b$Multiply the first equation by $\alpha^{n-1}$ and the second by $\beta^{n-1}$:$\alpha^{n+1} = a\alpha^n - b\alpha^{n-1}$$\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

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

The condition that the roots of $x^3-b x^2+c x-d=0$ are in geometric progression is

If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^{2}+bx+c=0$ and $3b^{2}=16ac$,then:

If the difference of the roots of $x^2 - px + 8 = 0$ is $2$,then the value of $p$ is

The minimum value of the sum of the squares of the roots of $x^{2}+(3-a)x+1=2a$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module