If alpha, beta, gamma are the roots of the eq | vedclass

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

Question

(C) Given that $\alpha, \beta, \gamma$ are the roots of the equation $4x^3-3x^2+2x-1=0$. Since they are roots,they satisfy the equation: $4\alpha^3-3\

Vedclass · Accepted Answer

$\frac{3}{64}$

Vedclass · Answer

(C) Given that $\alpha, \beta, \gamma$ are the roots of the equation $4x^3-3x^2+2x-1=0$. Since they are roots,they satisfy the equation: $4\alpha^3-3\alpha^2+2\alpha-1=0 \Rightarrow 4\alpha^3=3\alpha^2-2\alpha+1$ $4\beta^3-3\beta^2+2\beta-1=0 \Rightarrow 4\beta^3=3\beta^2-2\beta+1$ $4\gamma^3-3\gamma^2+2\gamma-1=0 \Rightarrow 4\gamma^3=3\gamma^2-2\gamma+1$ Summing these equations: $4(\alpha^3+\beta^3+\gamma^3) = 3(\alpha^2+\beta^2+\gamma^2) - 2(\alpha+\beta+\gamma) + 3$ From Vieta's formulas: $\alpha+\beta+\gamma = \frac{3}{4}$,$\alpha\beta+\beta\gamma+\gamma\alpha = \frac{2}{4} = \frac{1}{2}$,$\alpha\beta\gamma = \frac{1}{4}$ We know $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = (\frac{3}{4})^2 - 2(\frac{1}{2}) = \frac{9}{16} - 1 = -\frac{7}{16}$ Substituting these values: $4(\alpha^3+\beta^3+\gamma^3) = 3(-\frac{7}{16}) - 2(\frac{3}{4}) + 3 = -\frac{21}{16} - \frac{3}{2} + 3 = -\frac{21}{16} + \frac{3}{2} = \frac{-21+24}{16} = \frac{3}{16}$ Therefore,$\alpha^3+\beta^3+\gamma^3 = \frac{3}{64}$.

If $\alpha, \beta, \gamma$ are the roots of the equation $4x^3-3x^2+2x-1=0$,then $\alpha^3+\beta^3+\gamma^3=$

Similar Questions

For the equation $\frac{1}{x + a} - \frac{1}{x + b} = \frac{1}{x + c}$,if the product of the roots is zero,what is the sum of the roots?

Let $\alpha, \beta$ be the roots of the equation $x^2+2 \sqrt{2} x-1=0$. The quadratic equation,whose roots are $\alpha^4+\beta^4$ and $\frac{1}{10}(\alpha^6+\beta^6)$,is :

The quadratic equation in $t$,such that the $A.M.$ of its roots is $A$ and the $G.M.$ of its roots is $G$,is

If the roots of the equation $2x^2 - 3x + 5 = 0$ are the reciprocals of the roots of the equation $ax^2 + bx + 2 = 0$,then:

In a triangle $PQR$,$\angle R = \frac{\pi}{2}$. If $\tan(\frac{P}{2})$ and $\tan(\frac{Q}{2})$ are the roots of the equation $ax^2 + bx + c = 0$ $(a \neq 0)$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module