Question

Hey Frnd ✨✨✨✨
Pls solve this✔️✔️
if
$\alpha \: and \: \beta$
are zeros of the polynomial f(x)=
$6 {x}^{2} + x - 2$
Find the value of
$\frac{1}{ \alpha } + \frac{1}{ \beta } + \alpha \beta$

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Answer 1

Answer:

1/6

Step-by-step explanation:

1/α + 1/β = α+β / αβ

α+β = -1/6

αβ = -2/6 = -1/3

1/α + 1/β + αβ = (-1/6)/(-2/6) + (-1/3)

1/2 - 1/3 = 1/6

Answer 2

Step-by-step explanation:

Given Equation is 6x² + x - 2.

Here a = 6, b = 1, c = -2.

α,β are the zeroes of the polynomial.

(i) Sum of zeroes:

α + β = -b/a

α + β = -1/6

(ii) Product of zeroes:

αβ = c/a

αβ = -2/6

αβ = -1/3

Now,

(1/α) + (1/β) + αβ

= [(α + β)/αβ] + αβ

= [(-1/6)/-1/3] + (-1/3)

= -5/18

Hope it helps!