Question

If alpha beta and gamma are the zeros of 6 x cube + 3 x square - 5 x + 1 then find the value of one by alpha plus one by beta + 1 by gamma

Answer 1

Answer:

$\text{The value of }\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\text{ is 5}$

Step-by-step explanation:

Given that if $\alpha,\beta\text{ and }\gamma$ are the zeros of

$6x^3 + 3x^2 - 5x + 1$

$\text{we have to find the value of }\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$

$6x^3 + 3x^2 - 5x + 1$

$\text{sum of product of zeroes=}\alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a}=\frac{-5}{6}$

$\text{Product of zeroes=}\alpha\beta\gamma=\frac{-d}{a}=\frac{-1}{6}$

Now,

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

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

$\frac{\frac{-5}{6}}{\frac{-1}{6}}=5$

$\text{The value of }\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\text{ is 5}$

Answer 2

Answer:

$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=5$

Step-by-step explanation:

$Given \: \alpha , \beta \: and \: \gamma \\are \: zeroes\: of \: 6x^{3}+3x^{2}-5x+1$

Compare the polyomial with ax³+bx²+cx+d , we get

a = 6, b = 3, c = -5, d = 1

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

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

$Now,\\\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\\=\frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}\\=\frac{\frac{-5}{6}}{\frac{-1}{6}}\\=5$

Therefore,

$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=5$

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