If the roots of the equation x^3 - ax^2 + bx | vedclass

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

Question

(D) Let $\alpha, \beta, \gamma$ be the roots of the equation $x^3 - ax^2 + bx - c = 0$. From Vieta's formulas: $\alpha + \beta + \gamma = a$ $\alpha\b

Vedclass · Accepted Answer

$\frac{3c}{b}$

Vedclass · Answer

(D) Let $\alpha, \beta, \gamma$ be the roots of the equation $x^3 - ax^2 + bx - c = 0$. From Vieta's formulas: $\alpha + \beta + \gamma = a$ $\alpha\beta + \beta\gamma + \gamma\alpha = b$ $\alpha\beta\gamma = c$ Given that $\alpha, \beta, \gamma$ are in $HP$,their reciprocals $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ are in $AP$. The harmonic mean $(HM)$ of three numbers $\alpha, \beta, \gamma$ is defined as $HM = \frac{3}{\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}}$. Substituting the values: $HM = \frac{3}{\frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma}} = \frac{3(\alpha\beta\gamma)}{\alpha\beta + \beta\gamma + \gamma\alpha} = \frac{3c}{b}$.

If the roots of the equation $x^3 - ax^2 + bx - c = 0$ are in harmonic progression $(HP)$,then the harmonic mean of the roots is

Similar Questions

If $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$ are in $H.P.$,then

If $x, y, z$ are in harmonic progression,then $\log(x + z) + \log(x - 2y + z) = \dots$

If $a, b, c$ are in harmonic progression,then $\frac{a-b}{b-c} = \dots$

Let $a_1, a_2, a_3, \ldots$ be in harmonic progression with $a_1 = 5$ and $a_{20} = 25$. The least positive integer $n$ for which $a_n < 0$ is

If $a, b, c$ are in $H.P.$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module