If alpha, beta, gamma are the roots of x^3+4x | vedclass

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(C) Given,$\alpha, \beta, \gamma$ are the roots of $x^3+4x+1=0$. From Vieta's formulas: $\alpha+\beta+\gamma=0$,$\alpha\beta+\beta\gamma+\gamma\alpha=

Vedclass · Accepted Answer

$x^3+4x-1=0$

Vedclass · Answer

(C) Given,$\alpha, \beta, \gamma$ are the roots of $x^3+4x+1=0$. From Vieta's formulas: $\alpha+\beta+\gamma=0$,$\alpha\beta+\beta\gamma+\gamma\alpha=4$,and $\alpha\beta\gamma=-1$. Since $\alpha+\beta+\gamma=0$,we have $\beta+\gamma=-\alpha$,$\gamma+\alpha=-\beta$,and $\alpha+\beta=-\gamma$. The roots of the new equation are $y_1 = \frac{\alpha^2}{-\alpha} = -\alpha$,$y_2 = \frac{\beta^2}{-\beta} = -\beta$,and $y_3 = \frac{\gamma^2}{-\gamma} = -\gamma$. Let $y = -x$,then $x = -y$. Substituting $x = -y$ into the original equation $x^3+4x+1=0$: $(-y)^3+4(-y)+1=0 \implies -y^3-4y+1=0 \implies y^3+4y-1=0$. Thus,the required equation is $x^3+4x-1=0$.

$X = x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X = x_i)$	$\frac{1}{6}$	$k$	$\frac{1}{4}$	$k$	$\frac{1}{6}$

If $\alpha, \beta, \gamma$ are the roots of $x^3+4x+1=0$,then the equation whose roots are $\frac{\alpha^2}{\beta+\gamma}, \frac{\beta^2}{\gamma+\alpha}, \frac{\gamma^2}{\alpha+\beta}$ is

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