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

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Question

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

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,we have $\alpha+\beta+\gamma=0$,$\alpha\beta+\beta\gamma+\gamma\alpha=4$,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 $x^3+4x+1=0$: $(-y)^3 + 4(-y) + 1 = 0$ $-y^3 - 4y + 1 = 0$ $y^3 + 4y - 1 = 0$. Thus,the required equation is $x^3+4x-1=0$.

List-$I$	List-$II$
$(i) \alpha = \beta$	$(A) (ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii) \alpha = 2\beta$	$(B) 2b^2 = 9ac$
$(iii) \alpha = 3\beta$	$(C) b^2 = 6ac$
$(iv) \alpha = \beta^2$	$(D) 3b^2 = 16ac$
	$(E) b^2 = 4ac$
	$(F) (ac^2)^{1/3} + (a^2c)^{1/3} = b$

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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