Question

If α,β,γ be zeros of polynomial 6x3 + 3x2− 5x + 1, then find the value of α−1+β−1+γ−1.​

Answer 1

Answer:

Given α,β & ɣ are the roots of the polynomials 6x³ + 3x² - 5x + 1 = 0 then

α+β+ɣ  = -3/6 = -1/2

αβ+βɣ+αɣ = -5/6

αβɣ = -1/6

α⁻¹ + β⁻¹ + ɣ⁻¹  = 1/α + 1/β + 1/ɣ

                      = (αβ+βɣ+αɣ) / αβɣ

                       = -5/6 / -1/6

                      = 5

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Answer 2

$\underline\mathfrak{Given:-}$

• 6x³ + 3x² - 5x + 1

$\underline\mathfrak{To \: \: Find:-}$

• find the value of α−1+β−1+γ−1 ...?

$\underline\mathfrak{Solutions:-}$

• The equation is in the form of:-

6x³ + 3x² - 5x + 1 = 0

• a = 6
• b = 3
• c = -5
• d = 1

$\: \: \: \: \: \therefore {Sum \: \: of \: \: zeroes}:-$

$\: \: \: \: \: \leadsto \: \: \alpha\: + \: \beta \: + \: \gamma \: \: = \: \: - \: \frac{-b}{a}$

$\: \: \: \: \: \leadsto \: \: \alpha\: + \: \beta \: + \: \gamma \: \: = \: \: - \: \frac{-3}{6}$

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

$\: \: \: \: \: \therefore {Sum \: \: of \: \: the \: \: product \: \: of \: \: zeroes}:-$

$\: \: \: \: \: \leadsto \: \: \alpha\beta \: + \: \beta\gamma \: + \: \alpha\gamma \: \: = \: \: \frac{c}{a}$

$\: \: \: \: \: \leadsto \: \: \alpha\beta \: + \: \beta\gamma \: + \: \alpha\gamma \: \: = \: \: \frac{-5}{6}$

$\: \: \: \: \: \therefore {Product \: \: of \: \: zeroes}:-$

$\: \: \: \: \: \leadsto \: \: \alpha\beta\gamma \: \: = \: \: \frac{-d}{a}$

$\: \: \: \: \: \leadsto \: \: \alpha\beta\gamma \: \: = \: \: \frac{-1}{6}$

Now,

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

$\: \: \: \: \: \leadsto \: \: \frac{\alpha\beta \: + \: \beta\gamma \: + \: \alpha\gamma}{\alpha\beta\gamma}$

$\: \: \: \: \: \leadsto \: \: \frac{\frac{-5}{6}}{\frac{-1}{6}}$

$\: \: \: \: \: \leadsto \: \: \frac{-5}{\cancel{6}} \: \times \: \frac{\cancel{6}}{-1}$

$\: \: \: \: \: \leadsto \: \: {5}$

Hence, the value of α−1+β−1+γ−1 is 5.