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

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

Question

(B) Given the cubic equation $x^3 - 3x^2 + x + 5 = 0$.By Vieta's formulas,we have:$\alpha + \beta + \gamma = 3$$\alpha \beta + \beta \gamma + \gamma \

Vedclass · Accepted Answer

$y^3 - y^2 - y - 2 = 0$

Vedclass · Answer

(B) Given the cubic equation $x^3 - 3x^2 + x + 5 = 0$.By Vieta's formulas,we have:$\alpha + \beta + \gamma = 3$$\alpha \beta + \beta \gamma + \gamma \alpha = 1$$\alpha \beta \gamma = -5$We need to find the value of $y = \sum \alpha^2 + \alpha \beta \gamma$.Recall that $\sum \alpha^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha \beta + \beta \gamma + \gamma \alpha)$.Substituting the values:$\sum \alpha^2 = (3)^2 - 2(1) = 9 - 2 = 7$.Now,$y = 7 + (\alpha \beta \gamma) = 7 + (-5) = 2$.Since $y = 2$,we check which equation is satisfied by $y = 2$:For option $B$: $y^3 - y^2 - y - 2 = (2)^3 - (2)^2 - 2 - 2 = 8 - 4 - 2 - 2 = 0$.Thus,$y = 2$ satisfies the equation $y^3 - y^2 - y - 2 = 0$.

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3 - 3x^2 + x + 5 = 0$,then $y = \sum \alpha^2 + \alpha \beta \gamma$ satisfies which of the following equations?

Similar Questions

If $2$ and $3$ are two roots of the equation $2x^3 + mx^2 - 13x + n = 0$,then the values of $m$ and $n$ are respectively

If $p$ and $q$ are the roots of $x^2 + px + q = 0$,then

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) = $

If $\alpha, \beta$ are the roots of the equation $ax^2 + bx + c = 0$,then the equation whose roots are $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$ is

If $\alpha, \beta, \gamma$ are the roots of $x^3-x+1=0$,then $\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module