Question

if a,b,y be zeros of the polynomial 6x^3+3x^2-5x+1 , then find the value of a^-1+b^-1+y^-1 ?​

Answer 1

Answer:

5

Explanation:

Given polynomial,

⇒ 6x³ + 3x² - 5x + 1

On comapring with ax³ + bx² + cx + d

• a = 6
• b = 3
• c = -5
• d = -1

Then,

• α + β + γ = -b/a = -3/6 = -½
• αβ + βγ + αγ = c/a = -5/6
• αβγ = d/a = -1/6

Solving for :-

⇒ 1/α + 1/β + 1/γ

⇒ αβ + βγ + αγ/αβγ

Substituting the values,

⇒ (-5/6)/(-1/6)

⇒ -5/6 × -6/1

⇒ 5

Required answer: 5

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Answer 2

$*Question:-$

if a,b,y be zeros of the polynomial 6x^3+3x^2-5x+1 , then find the value of a^-1+b^-1+y^-1 ?

$*Answer:-$

$\alpha , \beta , \gamma$

↑↑These are the zeros of polynomial

6x³ + 3x² - 5x + 1

$= > \alpha + \beta + \gamma = \frac{ - b}{a} = \frac{ - 3}{6}$

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

$= > \alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a} = \frac{ - 5}{6}$

$= > \alpha \beta \gamma = \frac{ - d}{a} = \frac{ - 1}{6}$

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

$= > \frac{ -5/6}{ - 1/6} = \frac{ - 5}{6} \times \frac{6}{ - 1}$

Hence,

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

So the required answer is 5

Hope it helps you.

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