Question

If alpha, Beta,gama are the zeroes of the polynomial
6xsquare+ 3xsquare- 5x+1, find the value of a alpha di power-1+beta di power-1+gama di power-1​

Answer 1

Answer:

Step-by-step explanation:

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

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Answer 2

Answer:

α⁻¹ + β⁻¹ + γ⁻¹ = 5

Step-by-step explanation:

The zeroes of the given cubic polynomial are α, β and γ respectively.

Given polynomial ;

F(x) = 6x³ + 3x² - 5x + 1

On comparing the given equation with ax³ + bx² + cx + d, we get -

• a = 6
• b = 3
• c = -5
• d = 1

Sum of zeroes = -b/a

⇒ α + β + γ = -3/6

⇒ α + β + γ = -1/2

Sum of product of zeroes = c/a

⇒ αβ + βγ + αγ = -5/6

Product of zeroes = -d/a

⇒ αβγ = -1/6

We need to find;

$\alpha {}^{ - 1} + { \beta }^{ - 1} + \gamma {}^{ - 1} \\ \\ \implies \frac{1}{ \alpha } + \frac{1}{ \beta } + \frac{1}{ \gamma } \\ \\ \implies \frac{ \alpha \beta + \beta \gamma + \alpha \gamma }{ \alpha \beta \gamma }$

Now, put the above values here;

$\implies \frac{ - ( \frac{5}{6}) }{ - ( \frac{1}{6} )} \\ \\ \implies \frac{5}{6} \times \frac{6}{1} \\ \\ \implies 5$

Hence, the answer is 5.