Question

If α ,β,γbe the zeroes of polynomial
6x^3+3x^2-5x+1
then find the value of
α^-1+β^-1 + γ ^-1​

Answer 1

Answer:

HELLO DEAR,

Given that:-

we have to find,

α^-1 + β^-1 + γ^-1

⇒1/α + 1/β + 1/γ

⇒(βγ + αγ + αβ)/αβγ-------------(1)

Now,

From the GIVEN Equation:-

6x³ + 3x² - 5x + 1 = 0

Where,

a = 6

b = 3

c = (-5)

d = 1

we know that:-

α + β + γ = -b/a = -3/6 = -1/2

αβγ = -d/a = -1/6

αβ + βγ + αγ = c/a = -5/6

Put the values in -----(1)

We get,

⇒(βγ + αγ + αβ)/αβγ

⇒ (-5/6) / (-1/6)

⇒30/6

⇒5

I HOPE ITS HELP YOU DEAR,

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Answer 2

$\huge\mathfrak\blue{Answer:}$

Given:

• We have been given a polynomial
• 6x³ + 3x² - 5x + 1
• α,β,γ are the zeros of given polynomial

To Find:

• We have to find the value of
• ${ \alpha }^{ - 1} + { \beta }^{ - 1} + { \gamma }^{ - 1}$

Solution:

The given polynomial is 6x³ + 3x² - 5x + 1

α,β,γ are the zeros of given polynomial

Sum of zeros ( α+β+γ ) :

= -b/a

= -3/6

= - 1/2

Sum of product of zeros ( αβ+βγ+γα ) :

= c/a

= -5/6

Product of zeros ( αβγ ) :

= - d/a

= - 1/6

We need to find the value of :

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

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

$= > \dfrac{αβ+βγ+γα}{αβγ}$

= > (-5/6) / (-1/6)

= > 5

Hence value of α¯¹+β¯¹+γ¯¹ is 5