Question

If alpha and beta are the zeros of the polynomial f x is equals to 2 X square + 11 x + 5 then find the value of Alpha cube beta + alpha beta cube

Answer 1

Step-by-step explanation:

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}