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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

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

Similar Questions

The harmonic mean of the roots of the equation $(5 + \sqrt{2})x^2 - (4 + \sqrt{5})x + 8 + 2\sqrt{5} = 0$ is

If the sum of the roots of the equation $ax^2 + bx + c = 0$ is equal to the sum of the reciprocals of their squares,then $bc^2, ca^2, ab^2$ will be in

If $\alpha, \beta$ are the roots of the equation $x^2+bx+c=0$ and $\alpha+h, \beta+h$ are the roots of the equation $x^2+qx+r=0$,then $h$ is equal to

If one root of the equation $i x^2 - 2(i + 1) x + (2 - i) = 0$ is $(2 - i)$,then the other root is

If $a, -a, b$ are the roots of $x^{3}-5x^{2}-x+5=0$,then $b$ is a root of

Vedclass Test Series

Exam Paper Generator

Online Exam Module