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Question Number 37568 by math khazana by abdo last updated on 15/Jun/18

$$let\alpha and\beta therootsoftheequation$$$$x^2-2mx-1=0findintermsoftherealm$$$$A={\alpha }^2+{\beta }^2$$$$B={\alpha }^3+{\beta }^3$$$$c={\alpha }^4+{\beta }^4$$$$D={\alpha }^6+{\beta }^6$$

Answered by Rasheed.Sindhi last updated on 15/Jun/18

$$\alpha \mathrm{\&}\beta \mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}{x}^2-2mx-1=0$$$$\therefore \alpha +\beta =2m$$$$\alpha \beta =-1$$$$A={\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta$$$$={\left(2m\right)}^2-2(-1)=4m^2+2$$$$B={\alpha }^3+{\beta }^3={(\alpha +\beta )}^3-3\alpha \beta (\alpha +\beta )$$$$={\left(2m\right)}^3-3(-1)\left(2m\right)=8m^3+6m$$$$C={\alpha }^4+{\beta }^4={({\alpha }^2+{\beta }^2)}^2-2{\left(\alpha \beta \right)}^2$$$$={(4m^2+2)}^2-2{(-1)}^2$$$$=16m^4+16m^2+2$$$$D={\alpha }^6+{\beta }^6={({\alpha }^3+{\beta }^3)}^2-2{\left(\alpha \beta \right)}^3$$$$={(8m^3+6m)}^2-2{(-1)}^3$$$$=64m^6+96m^3+36m^2+2$$

Answered by Rio Mike last updated on 16/Jun/18

$$\mathrm{a}){\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta$$$${(\alpha +\beta )}^2=(\alpha +\beta )(\alpha +\beta )$$$$={\alpha }^2+\alpha \beta +\alpha \beta +{\beta }^2$$$$={\alpha }^2+{\beta }^2+2\alpha \beta$$$$=(\alpha +\beta )-2\alpha \beta$$$${\mathrm{x}}^2-2\mathrm{mx}-1=0$$$${\mathrm{S}}_{\mathrm{r}}={}^{}\alpha +\beta =\frac{-\mathrm{b}}{\mathrm{a}}$$$$=2\mathrm{m}$$$${\mathrm{P}}_{\mathrm{r}}=\alpha \beta =\frac{\mathrm{c}}{\mathrm{a}}$$$$=-1$$$$\Rightarrow {\alpha }^2+{\beta }^2={\left(2\mathrm{m}\right)}^2-2(-1)$$$$={4\mathrm{m}}^2+2$$$$$$$$$$$$$$$$$$

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