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

Let $\lambda \neq 0$ be in $\mathbb{R}$. If $\alpha$ and $\beta$ are the roots of the equation $x^{2}-x+2\lambda=0$ and $\alpha$ and $\gamma$ are the roots of the equation $3x^{2}-10x+27\lambda=0$,then $\frac{\beta\gamma}{\lambda}$ is equal to

If the product of roots of the equation $mx^2 + 6x + (2m - 1) = 0$ is $-1$,then the value of $m$ will be

If the roots of the equation $x^2 + bx + c = 0$ are reciprocals of each other,then...

Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up with roots $(4, 3)$. Rahul made a mistake in writing down the coefficient of $x$ and ended up with roots $(3, 2)$. The correct roots of the equation are:

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3+ax^2+bx+c=0$,then the roots of the equation $x^3+(2b-a^2)x^2+(b^2-2ac)x-c^2=0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module