If alpha, beta, gamma are the roots of x^3+ax | vedclass

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Question

(C) Given the cubic equation $x^3+ax^2+bx+c=0$. Let the roots be $\alpha, \beta, \gamma$. From Vieta's formulas,we have: $\alpha + \beta + \gamma = -a

Vedclass · Accepted Answer

$\frac{-b}{c}$

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(C) Given the cubic equation $x^3+ax^2+bx+c=0$. Let the roots be $\alpha, \beta, \gamma$. From Vieta's formulas,we have: $\alpha + \beta + \gamma = -a$ $\alpha\beta + \beta\gamma + \gamma\alpha = b$ $\alpha\beta\gamma = -c$ We need to find $\sum \frac{1}{\alpha} = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}$. Taking the common denominator,we get: $\frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma}$. Substituting the values from Vieta's formulas: $\frac{b}{-c} = -\frac{b}{c}$. Thus,the correct option is $C$.

If $\alpha, \beta, \gamma$ are the roots of $x^3+ax^2+bx+c=0$,then find the value of $\sum \frac{1}{\alpha}$,given that $\alpha, \beta, \gamma$ are non-zero.

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