Question

If alpha and beta are the zeros of the polynomial 3x square + 2x - 6 then find the value of Alpha minus beta ;Alpha square plus beta Square; One Upon alpha plus one upon beta;
Alpha Cube + beta cube

Answer 1

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Answer 2

Answer:

Given Quadratic Polynomial is 3x² + 2x -6 and α , β are zeroes.

Using relationship between coefficeints and zeroes we get,

$\alpha+\beta=\frac{-b}{a}=\frac{-2}{3}$

$\alpha\,\beta=\frac{c}{a}=\frac{-6}{3}=-2$

1).

$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$

$\alpha^2+\beta^2=(\frac{-2}{3})^2-2(-2)$

$\alpha^2+\beta^2=\frac{4}{9}+4$

$\alpha^2+\beta^2=\frac{40}{9}$

2).

$(\alpha-\beta)^2=\alpha^2+\beta^2-2\alpha\beta$

$(\alpha-\beta)^2=\frac{40}{9}-2(-2)$

$(\alpha-\beta)^2=\frac{40}{9}+4$

$(\alpha-\beta)^2=\frac{76}{9}$

$\alpha-\beta=\pm\frac{\sqrt{76}}{3}$

3).

$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\,\beta}=\frac{\frac{-2}{3}}{-2}=\frac{1}{3}$

4).

$\alpha^3+\beta^3=(\alpha+\beta)(\alpha^2+\beta^2-\alpha\,\beta)$

$\alpha^3+\beta^3=(\frac{-2}{3})(\frac{40}{9}-(-2))$

$\alpha^3+\beta^3=(\frac{-2}{3})(\frac{40}{9}+2)$

$\alpha^3+\beta^3=\frac{-2}{3}\frac{58}{9}$

$\alpha^3+\beta^3=\frac{-116}{27}$