If alpha ,beta are the roots of {x^2} - ax + | vedclass

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

Question

(c) Multiplying ${x^2} - ax + b = 0$by ${x^{n - 1}}$

${x^{n + 1}} - a{x^n} + b{x^{n - 1}} = 0$.....$(i)$

$\alpha ,\beta $are roots of ${x^2} - a

Vedclass · Accepted Answer

${V_{n + 1}} = a{V_n} - b{V_{n - 1}}$

Vedclass · Answer

(c) Multiplying ${x^2} - ax + b = 0$by ${x^{n - 1}}$

${x^{n + 1}} - a{x^n} + b{x^{n - 1}} = 0$.....$(i)$

$\alpha ,\beta $are roots of ${x^2} - ax + b = 0$, therefore they will satisfy $(i)$ also

${\alpha ^{n + 1}} - a{\alpha ^n} + b{\alpha ^{n - 1}} = 0$.....$(ii)$

and ${\beta ^{n + 1}} - a{\beta ^n} + b{\beta ^{n - 1}} = 0$.....$(iii)$

Adding $(ii)$ and $(iii)$

$({\alpha ^{n + 1}} + {\beta ^{n + 1}}) - a({\alpha ^n} + {\beta ^n}) + b({\alpha ^{n - 1}} + {\beta ^{n - 1}}) = 0$

or ${V_{n + 1}} - a{V_n} + b{V_{n - 1}} = 0$

or ${V_{n + 1}} = a{V_n} - b{V_{n - 1}} = 0$ (Given ${\alpha ^n} + {\beta ^n} = {V_n}$)

Trick : Put $n = 0$, $1,\,\,2$

${V_0} = {\alpha ^0} + {\beta ^0} = 2$, ${V_1} = \alpha + \beta = a$,

${\alpha ^2} + {\beta ^2} = {V_2} = {a^2} - 2b$

Now the option $(c)$ ==> ${V_2} = a{V_1} - b{V_0} = {a^2} - 2b$

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

Similar Questions

For a real number $x$, let $[x]$ denote the largest integer less than or equal to $x$, and let $\{x\}=x-[x]$. The number of solutions $x$ to the equation $[x]\{x\}=5$ with $0 \leq x \leq 2015$ is

The number of real roots of the equation ${e^{\sin x}} - {e^{ - \sin x}} - 4$ $ = 0$ are

The equation $x^2-4 x+[x]+3=x[x]$, where $[x]$ denotes the greatest integer function, has:

The number of real solutions of the equation $\mathrm{x}|\mathrm{x}+5|+2|\mathrm{x}+7|-2=0$ is .....................