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

(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\

Vedclass · Accepted Answer

$\frac{b^2 - 2ac}{c^2}$

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 find $\alpha^{-2} + \beta^{-2} + \gamma^{-2} = \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2}$. This can be written as: $\frac{\beta^2\gamma^2 + \alpha^2\gamma^2 + \alpha^2\beta^2}{(\alpha\beta\gamma)^2} = \frac{(\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma)}{(\alpha\beta\gamma)^2}$. Substituting the values: $= \frac{b^2 - 2(c)(a)}{c^2} = \frac{b^2 - 2ac}{c^2}$.

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

Similar Questions

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-5x^2-2x+24=0$,then $\frac{\beta\gamma}{\alpha}+\frac{\gamma\alpha}{\beta}+\frac{\alpha\beta}{\gamma}=$

If $\alpha, \beta$ are the roots of $(x - a)(x - b) = c$,where $c \neq 0$,then the roots of $(x - \alpha)(x - \beta) + c = 0$ are

For the equation $ax^2 + bx + c = 0$,the roots are $\alpha, \beta$,and for the equation $Ax^2 + Bx + C = 0$,the roots are $\alpha - k, \beta - k$. Then $\frac{B^2 - 4AC}{b^2 - 4ac} = \dots$

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $2x^3 - 3x^2 + 6x + 1 = 0$,then $\alpha^2 + \beta^2 + \gamma^2$ is equal to

The cubic equation whose roots are thrice to each of the roots of $x^3+2x^2-4x+1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module