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

(D) Given that $\alpha, \beta, \gamma$ are roots of $2x^3-5x^2+4x-3=0$. By Vieta's formulas: $\alpha+\beta+\gamma = \frac{5}{2}$,$\alpha\beta+\beta\ga

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(D) Given that $\alpha, \beta, \gamma$ are roots of $2x^3-5x^2+4x-3=0$. By Vieta's formulas: $\alpha+\beta+\gamma = \frac{5}{2}$,$\alpha\beta+\beta\gamma+\gamma\alpha = 2$,and $\alpha\beta\gamma = \frac{3}{2}$. We know that $\sum \alpha\beta(\alpha+\beta) = (\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha) - 3\alpha\beta\gamma$. Substituting the values: $\sum \alpha\beta(\alpha+\beta) = (\frac{5}{2})(2) - 3(\frac{3}{2})$ $= 5 - \frac{9}{2} = \frac{1}{2}$.

If $\alpha, \beta, \gamma$ are the roots of the equation $2x^3-5x^2+4x-3=0$,then $\sum \alpha \beta(\alpha+\beta)=$

Similar Questions

If the sum of two of the roots of $x^3+p x^2-q x+r=0$ is zero,then $pq$ is equal to

If $x^2 + px + 1$ is a factor of $ax^3 + bx + c$,then:

If the roots of $x^2 - 7x + 6 = 0$ are $\alpha$ and $\beta$,then find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.

If $\alpha, \beta$ are the roots of the equation $x^2 - 6x - 2 = 0$,$\alpha > \beta$ and $a_n = \alpha^n - \beta^n$,$n > 1$,then the value of $\frac{a_{10} - 2a_8}{2a_9}$ is equal to

If $\frac{1}{\sqrt{\alpha}}$ and $\frac{1}{\sqrt{\beta}}$ are the roots of the equation $ax^2 + bx + 1 = 0$ $(a \ne 0, a, b \in R)$,then the equation $x(x + b^3) + (a^3 - 3abx) = 0$ has roots

Vedclass Test Series

Exam Paper Generator

Online Exam Module