If alpha, beta are the roots of x^2 - 2x + 4 | vedclass

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

Question

(B) Given the quadratic equation $x^2 - 2x + 4 = 0$.For this equation,the sum of roots $\alpha + \beta = 2$ and the product of roots $\alpha \beta = 4

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) Given the quadratic equation $x^2 - 2x + 4 = 0$.For this equation,the sum of roots $\alpha + \beta = 2$ and the product of roots $\alpha \beta = 4$.We know that $\alpha$ and $\beta$ satisfy the equation,so $\alpha^2 = 2\alpha - 4$ and $\beta^2 = 2\beta - 4$.Using the recurrence relation $S_n = \alpha^n + \beta^n$,we have $S_n - 2S_{n-1} + 4S_{n-2} = 0$.For $n=1$,$S_1 = \alpha + \beta = 2$.For $n=2$,$S_2 = (\alpha + \beta)^2 - 2\alpha \beta = (2)^2 - 2(4) = 4 - 8 = -4$.For $n=3$,$S_3 = 2S_2 - 4S_1 = 2(-4) - 4(2) = -8 - 8 = -16$.For $n=4$,$S_4 = 2S_3 - 4S_2 = 2(-16) - 4(-4) = -32 + 16 = -16$.For $n=5$,$S_5 = 2S_4 - 4S_3 = 2(-16) - 4(-16) = -32 + 64 = 32$.Thus,$\alpha^5 + \beta^5 = 32$.

If $\alpha, \beta$ are the roots of $x^2 - 2x + 4 = 0$,then $\alpha^5 + \beta^5$ is equal to

Similar Questions

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}+5x+6=0$
$II.$ $y^{2}+3y+2=0$

The solution of $2-x=\frac{x-2}{x}$ would include

If $a, b, c, d$ and $p$ are distinct real numbers such that $(a^2 + b^2 + c^2)p^2 - 2p(ab + bc + cd) + (b^2 + c^2 + d^2) \le 0$,then

If $\lambda$ is the ratio of the roots of the quadratic equation in $x$,$3m^2x^2 + m(m - 4)x + 2 = 0$,then the least value of $m$ for which $\lambda + \frac{1}{\lambda} = 1$ is:

If $a, b, c$ are in $G.P.$,then the equations $ax^2 + 2bx + c = 0$ and $dx^2 + 2ex + f = 0$ have a common root if $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in

Vedclass Test Series

Exam Paper Generator

Online Exam Module