Question

alpha,bita,gamma r the zeroes of polnomia xcube+3xsqare+10x-24then find 1/alpha+1/bita+1/gamma

Answer 1

Answer---->

$\dfrac{1}{ \alpha } + \dfrac{1}{ \beta } + \: \dfrac{1}{ \gamma } = - \dfrac{5}{12}$

Step-by-step explanation:

Given---->

$\alpha \: \beta \: and \: \gamma \: are \: zeroes \: of \: given \: polynomial \\ p( \: x \: ) = {x}^{3 } + 3x ^{2} + 10x - 24$

To find ----->

$value \: of$

$\dfrac{1}{ \alpha } + \dfrac{1}{ \beta } + \dfrac{1}{ \gamma }$

Concept used----->

$if \: \alpha \: \beta \: and \: \gamma \: are \: zeroes \: of \: cubic \: polynomial \: then$

$\alpha + \beta + \gamma = - \dfrac{coefficient \: of \: {x}^{2} }{coefficient \: of \: {x}^{3} }$

$\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{coefficient \: of \: x}{coefficient \: of \: x ^{3} }$

$\alpha \beta \gamma = - \dfrac{constant \: term}{coefficient \: of \: {x}^{3} }$

Solution----> ATQ,

${x}^{3 } + 3 {x}^{2} + 10x - 24$

$now \: zeroes \: of \: above \: polynomial \: are \: \alpha \: \beta \: and \: \gamma so$

$\alpha + \beta + \gamma = - \dfrac{coefficient \: of \: {x}^{2} }{coefficient \: of \: {x}^{3} }$

$\alpha + \beta + \gamma = - \frac{3}{1}$

$\alpha + \beta + \gamma = - 3$

$\alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{coefficient \: of \: x}{coefficient \: of \: {x}^{3} }$

$\alpha \beta + \beta \gamma + \gamma \alpha = \frac{10}{1}$

$\alpha \beta \gamma = - \dfrac{constant \: term}{coefficient \: of \: {x}^{3} }$

$\alpha \beta \gamma = - \dfrac{ - 24}{1}$

$\alpha \beta \gamma = 24$

$Now \: we \: have \: to \: find \: value \: of$

$\dfrac{1}{ \alpha } + \dfrac{1}{ \beta } + \dfrac{1}{ \gamma } = \dfrac{ \alpha + \beta + \gamma }{ \alpha \beta \gamma }$

$\: \: \: \: \: = \dfrac{ - 10}{24}$

$\: \: \: \: = \dfrac{ - 5}{12}$