Question

If
$\alpha \beta \gamma \: are \: zeroes \: of \: 6 {x}^{2} + 3 {x }^{2} - 5x \: + 1 \: then \:$
then find value of
${ \alpha }^{ - 1} + { \beta }^{ - 1} + { \gamma }^{ - 1}$
​

Answer 1

Answer:

$\bold{\alpha ^{ - 1} + \beta ^{ - 1} + \gamma ^{ -1} = 5}$

Step-by-step explanation:

$\bold{Given} = > \\ \alpha \: \beta \: and \: \gamma \: are \: zeroes \: of \: polynomial \\ 6 {x}^{3} \: + \: 3 {x}^{2} \: - 5x \: + 1$

$\bold{To \: find \:} = > \\ value \: of \: ( { \alpha }^{ - 1} + \beta ^{ - 1} \: + { \gamma }^{ - 1} \: )$

$\bold{Concept \: used }= > \\ if \: \alpha \: \beta \: and \: \gamma \: are \: zeroes \: of \: any \: cubic \: polynomial \: then$

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

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

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

$\bold{Solution} = > \\ 6 {x}^{3} \: + 3 {x}^{2} - 5x + 1$

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

$= > \alpha + \beta + \gamma = - \dfrac{3}{6}$

$= > \alpha + \beta + \gamma = - \dfrac{1}{2}$

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

$= > \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{ - 5}{6}$

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

$= > \alpha \: \beta \: \gamma = - \dfrac{1}{6}$

$now$

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

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

$= \dfrac{ - \dfrac{ 5}{6} }{ - \dfrac{1}{6} }$

$= \: 5$

Answer 2

Correct Question:

If α, β, γ are zeroes of the polynomial 6x³ + 3x² - 5x + 1 then find the value of α⁻¹ + β⁻¹ + γ⁻¹.

Solution:

6x³ + 3x² - 5x + 1

Comparing the given polynomial with ax³ + bx² + cx + d, we get

• a = 6
• b = 3
• c = - 5
• d = 1

Product of zeroes = αβγ = - d / a = - 1 / 6

Sum of product of two zeroes taken at a time = αβ + βγ + γα = c / a = - 5 / 6

Now, α⁻¹ + β⁻¹ + γ⁻¹

α⁻¹ + β⁻¹ + γ⁻¹ = ( 1 / α ) + ( 1 / β ) + ( 1 / γ )

= ( βγ + γα + αβ ) / ( αβγ )

= ( αβ + βγ + γα ) / ( αβγ )

= ( - 5 / 6 ) ÷ ( - 1 / 6 )

= ( - 5 / 6 ) × ( - 6)

= 5

Therefore the value of α⁻¹ + β⁻¹ + γ⁻¹ is 5.