If alpha, beta, gamma are the roots of the eq | vedclass

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

Question

(D) Given the cubic equation $x^3+a x^2-b x+c=0$ with roots $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha+\beta+\gamma = -a$ $\alpha \beta+\be

Vedclass · Accepted Answer

$a b+3 c$

Vedclass · Answer

(D) Given the cubic equation $x^3+a x^2-b x+c=0$ with roots $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha+\beta+\gamma = -a$ $\alpha \beta+\beta \gamma+\gamma \alpha = -b$ $\alpha \beta \gamma = -c$ We need to evaluate $\sum \beta^2(\gamma+\alpha) = \beta^2(\gamma+\alpha) + \gamma^2(\alpha+\beta) + \alpha^2(\beta+\gamma)$. Since $\alpha+\beta+\gamma = -a$,we have $(\gamma+\alpha) = -a-\beta$,$(\alpha+\beta) = -a-\gamma$,and $(\beta+\gamma) = -a-\alpha$. Substituting these: $\sum \beta^2(-a-\beta) = -a(\alpha^2+\beta^2+\gamma^2) - (\alpha^3+\beta^3+\gamma^3)$. Alternatively,using the identity $\sum \alpha^2(\beta+\gamma) = (\alpha+\beta+\gamma)(\alpha \beta+\beta \gamma+\gamma \alpha) - 3 \alpha \beta \gamma$: $= (-a)(-b) - 3(-c) = ab + 3c$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+a x^2-b x+c=0$,then $\sum \beta^2(\gamma+\alpha) = $

Similar Questions

If the sum of the squares of the roots of the equation $x^2-(a-2)x-(a+1)=0$ is least for an appropriate value of the variable parameter $a$, then the value of $a$ will be

If the product of the roots of the equation $2x^2 + 6x + \alpha^2 + 1 = 0$ is $-\alpha$,then the value of $\alpha$ will be

The condition that one root of the equation $ax^2 + bx + c = 0$ is three times the other is:

If $\tan \alpha$ and $\tan \beta$ are the roots of the equation $x^2 - px + q = 0$,then find the value of $\sin^2 (\alpha + \beta)$.

Find $\alpha^4+\beta^4$ if $\alpha, \beta$ are the roots of the equation $x^2+x+1=0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module