Let alpha and beta be two real numbers such t | vedclass

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

Question

(B) Given that $\alpha$ and $\beta$ are roots of the quadratic equation $x^{2} - (\alpha+\beta)x + \alpha\beta = 0$.Substituting the values,we get $x^

Vedclass · Accepted Answer

$324$

Vedclass · Answer

(B) Given that $\alpha$ and $\beta$ are roots of the quadratic equation $x^{2} - (\alpha+\beta)x + \alpha\beta = 0$.Substituting the values,we get $x^{2} - x - 1 = 0$.Since $\alpha$ and $\beta$ are roots,they satisfy the equation:$\alpha^{2} - \alpha - 1 = 0 \Rightarrow \alpha^{n+1} = \alpha^{n} + \alpha^{n-1}$$\beta^{2} - \beta - 1 = 0 \Rightarrow \beta^{n+1} = \beta^{n} + \beta^{n-1}$Adding these two equations,we get:$(\alpha^{n+1} + \beta^{n+1}) = (\alpha^{n} + \beta^{n}) + (\alpha^{n-1} + \beta^{n-1})$This implies the recurrence relation $p_{n+1} = p_{n} + p_{n-1}$.Given $p_{n+1} = 29$ and $p_{n-1} = 11$,we have:$29 = p_{n} + 11$$p_{n} = 29 - 11 = 18$.Therefore,$p_{n}^{2} = 18^{2} = 324$.

Let $\alpha$ and $\beta$ be two real numbers such that $\alpha+\beta=1$ and $\alpha \beta=-1$. Let $p_{n}=\alpha^{n}+\beta^{n}$,$p_{n-1}=11$ and $p_{n+1}=29$ for some integer $n \geq 1$. Then,the value of $p_{n}^{2}$ is .... .

Similar Questions

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $2x^3 - 3x^2 + 6x + 1 = 0$,then $\alpha^2 + \beta^2 + \gamma^2$ is equal to

If the roots of the equation $(x - a)(x - b) = c$ (where $c \neq 0$) are $\alpha$ and $\beta$,then what are the roots of the equation $(x - \alpha)(x - \beta) + c = 0$?

If $\operatorname{cosec} \theta$ and $\cot \theta$ are the roots of $cx^2+bx+a=0$ $(bc \neq 0)$,then $b^2(b^2-4ac)=$

The $H.M.$ between the roots of the equation $x^2 - 10x + 11 = 0$ is

If $\alpha, \beta, \gamma$ are the roots of the equation $2x^3+x^2-13x+6=0$,then $\alpha^3+\beta^3+\gamma^3=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module