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) Since $\alpha, \beta, \gamma$ are roots of $x^3 - x - 2 = 0$,we have $\alpha^3 = \alpha + 2$,$\beta^3 = \beta + 2$,and $\gamma^3 = \gamma + 2$.Mul

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) Since $\alpha, \beta, \gamma$ are roots of $x^3 - x - 2 = 0$,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,$\alpha^5 + \beta^5 + \gamma^5 = 2(\alpha^2 + \beta^2 + \gamma^2) + (\alpha + \beta + \gamma) + 6$.From the equation $x^3 + 0x^2 - x - 2 = 0$,we have $\sum \alpha = 0$ and $\sum \alpha\beta = -1$.Using the identity $\alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2(\sum \alpha\beta) = 0^2 - 2(-1) = 2$.Substituting these values: $\alpha^5 + \beta^5 + \gamma^5 = 2(2) + 0 + 6 = 10$.

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

If $\alpha, \beta$ and $\gamma$ are roots of the equation $x^3+4x-19=0$,then the value of $\frac{\alpha^3}{19-4\alpha}+\frac{\beta^3}{19-4\beta}+\frac{\gamma^3}{19-4\gamma}$ is equal to

If the product of roots of the equation $mx^2 + 6x + (2m - 1) = 0$ is $-1$,then the value of $m$ will be

If $\alpha, \beta$ are the roots of the quadratic equation $x^2+bx+c=0$ such that $\alpha^2+\beta^2=5$ and $\alpha^3+\beta^3=9$,then $b+c=$

If $\alpha$ and $\beta$ are the roots of the equation ${x^2} - (1 + {n^2})x + \frac{1}{2}(1 + {n^2} + {n^4}) = 0$,then the value of ${\alpha ^2} + {\beta ^2}$ is

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 + \frac{a}{2} x + b = 0$ and $(\alpha-\beta)(\alpha-\gamma)$,$(\beta-\alpha)(\beta-\gamma)$,$(\gamma-\alpha)(\gamma-\beta)$ are the roots of the equation $(y+a)^3 + K(y+a)^2 + L = 0$,then $\frac{L}{K} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module