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

The number of real solutions of the equation $3\left(x^2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+5=0$, is

The equation${e^x} - x - 1 = 0$ has

Let $f(x)={{x}^{2}}-x+k-2,k\in R$ then the complete set of values of $k$ for which $y=\left| f\left( \left| x \right| \right) \right|$ is non-derivable at $5$ distinict points is

The number of real solutions of the equation $e ^{4 x }+4 e ^{3 x }-58 e ^{2 x }+4 e ^{ x }+1=0$ is..........

Let $r$ be a real number and $n \in N$ be such that the polynomial $2 x^2+2 x+1$ divides the polynomial $(x+1)^n-r$. Then, $(n, r)$ can be