Suppose alpha, beta, gamma are the roots of x | vedclass

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(A) Given that $\alpha, \beta, \gamma$ are the roots of $x^3+x^2+x+2=0$. From Vieta's formulas,$\alpha+\beta+\gamma = -1$. Let $y = \frac{\alpha+\beta

Vedclass · Accepted Answer

$-\frac{47}{2}$

Vedclass · Answer

(A) Given that $\alpha, \beta, \gamma$ are the roots of $x^3+x^2+x+2=0$. From Vieta's formulas,$\alpha+\beta+\gamma = -1$. Let $y = \frac{\alpha+\beta-2\gamma}{\gamma}$. Since $\alpha+\beta+\gamma = -1$,we have $\alpha+\beta = -1-\gamma$. Substituting this into the expression for $y$: $y = \frac{-1-\gamma-2\gamma}{\gamma} = \frac{-1-3\gamma}{\gamma} = -3 - \frac{1}{\gamma}$. Thus,$\frac{1}{\gamma} = -y-3$,which implies $\gamma = -\frac{1}{y+3}$. Since $\gamma$ is a root of $x^3+x^2+x+2=0$,we substitute $x = -\frac{1}{y+3}$: $(-\frac{1}{y+3})^3 + (-\frac{1}{y+3})^2 + (-\frac{1}{y+3}) + 2 = 0$. Multiplying by $-(y+3)^3$: $1 - (y+3) + (y+3)^2 - 2(y+3)^3 = 0$. $1 - y - 3 + y^2 + 6y + 9 - 2(y^3 + 9y^2 + 27y + 27) = 0$. $-2y^3 - 17y^2 - 49y - 47 = 0$. $2y^3 + 17y^2 + 49y + 47 = 0$. The product of the roots of this cubic equation in $y$ is given by $-\frac{d}{a} = -\frac{47}{2}$.

Suppose $\alpha, \beta, \gamma$ are the roots of $x^3+x^2+x+2=0$. Then,the value of $\left(\frac{\alpha+\beta-2 \gamma}{\gamma}\right)\left(\frac{\beta+\gamma-2 \alpha}{\alpha}\right)\left(\frac{\gamma+\alpha-2 \beta}{\beta}\right)$ is

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