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

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

Vedclass · Accepted Answer

$ab-3c$

Vedclass · Answer

(A) Given the cubic equation $x^3-ax^2+bx-c=0$ 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 $\Sigma \alpha^2(\beta+\gamma)$. Since $\alpha+\beta+\gamma = a$,we have $\beta+\gamma = a-\alpha$. Substituting this into the expression: $\Sigma \alpha^2(a-\alpha) = a\Sigma \alpha^2 - \Sigma \alpha^3$ We know that $\Sigma \alpha^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = a^2-2b$. Also,for a cubic equation $x^3-ax^2+bx-c=0$,the sum of cubes $\Sigma \alpha^3$ is given by the identity: $\Sigma \alpha^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\Sigma \alpha^2 - \Sigma \alpha\beta)$ $\Sigma \alpha^3 = a(a^2-2b-b) + 3c = a(a^2-3b) + 3c = a^3-3ab+3c$. Now,substituting these into the expression: $a(a^2-2b) - (a^3-3ab+3c) = a^3-2ab-a^3+3ab-3c = ab-3c$. Thus,the correct option is $A$.

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

Similar Questions

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

Let $f(x) = x^6 - 2x^5 + x^3 + x^2 - x - 1$ and $g(x) = x^4 - x^3 - x^2 - 1$ be two polynomials. Let $a, b, c,$ and $d$ be the roots of $g(x) = 0$. Then,the value of $f(a) + f(b) + f(c) + f(d)$ is

Two roots of the cubic equation $x^3 + 3x^2 + kx + 12 = 0$ are real and unequal but have the same absolute value. Find the value of $k$.

If the roots of the given equation $(2k + 1)x^2 - (7k + 3)x + k + 2 = 0$ are reciprocal to each other,then the value of $k$ will be

Given that $\tan \alpha$ and $\tan \beta$ are the roots of $x^2 - px + q = 0$,then the value of $\sin^2(\alpha + \beta)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module