Question

$if \alpha \: \beta \: and \: \gamma \: are \: zeros \: of \: polynomial \: 6{x}^{2} + 3 {x}^{2} - 5x + 1 \: then \: find \: the \: value \: of \: {\alpha}^{ - 1} + {\beta}^{ - 1} + {\gamma}^{ - 1}$
​

Answer 1

Answer:

$\frac{5}{3}$

Step-by-step explanation:

• $A^-^1=\frac{1}{A}$

α, β, γ are the zeroes of polynomial $6x^3+3x^2-5x+1$.

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

$\alpha \beta \gamma =\frac{1}{2}$

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

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

∴$\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }=\frac{5}{6} *2=\frac{5}{3}$

Answer 2

Answer:

Step-by-step explanation:

α, β, γ are the zeroes of polynomial 6x^3+3x^2-5x+16x

3

+3x

2

−5x+1 .

\alpha \beta +\beta \gamma +\gamma \alpha =\frac{5}{6}αβ+βγ+γα=

6

5

\alpha \beta \gamma =\frac{1}{2}αβγ=

2

1

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

α

1

+

β

1

+

γ

1

=

αβγ

αβ+βγ+γα

∴\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }=\frac{5}{6} *2=\frac{5}{3}

αβγ

αβ+βγ+γα

=

6

5

∗2=

3

5