If alpha, beta, gamma are the roots of x^3 - | vedclass

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

Question

(D) Given the equation $x^3 - x - 2 = 0$,since $\alpha, \beta, \gamma$ are roots,we have $\alpha^3 = \alpha + 2$,$\beta^3 = \beta + 2$,and $\gamma^3 =

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) Given the equation $x^3 - x - 2 = 0$,since $\alpha, \beta, \gamma$ are roots,we have $\alpha^3 = \alpha + 2$,$\beta^3 = \beta + 2$,and $\gamma^3 = \gamma + 2$.Multiplying by $\alpha^2$,we get $\alpha^5 = \alpha^3 + 2\alpha^2 = (\alpha + 2) + 2\alpha^2 = 2\alpha^2 + \alpha + 2$.Similarly,$\beta^5 = 2\beta^2 + \beta + 2$ and $\gamma^5 = 2\gamma^2 + \gamma + 2$.Summing these,$\sum \alpha^5 = 2\sum \alpha^2 + \sum \alpha + 6$.From the equation $x^3 + 0x^2 - x - 2 = 0$,we have $\sum \alpha = 0$ and $\sum \alpha\beta = -1$.Using the identity $\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta$,we get $\sum \alpha^2 = (0)^2 - 2(-1) = 2$.Substituting these values: $\sum \alpha^5 = 2(2) + 0 + 6 = 4 + 6 = 10$.

If $\alpha, \beta, \gamma$ are the roots of $x^3 - x - 2 = 0$,then the value of $\alpha^5 + \beta^5 + \gamma^5$ is-

Similar Questions

The value of $a$ for which the equations $x^2 - 3x + a = 0$ and $x^2 + ax - 3 = 0$ have a common root is

The value of $2 + \frac{1}{2 + \frac{1}{2 + \dots \infty}}$ is

The roots of the equation $\sqrt{3x + 1} + 1 = \sqrt{x}$ are

Let $x = \frac{\sqrt{13} + \sqrt{11}}{\sqrt{13} - \sqrt{11}}$ and $y = \frac{1}{x}$. Then the value of $3x^2 - 5xy + 3y^2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module