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Question Number 62611 by hovea cw last updated on 23/Jun/19

$$\mathrm{If}\alpha \mathrm{and}\beta \mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}$$$$x^2-(a+1)x+\frac{1}{2}(a^2+a+1)=0\mathrm{then}$$$${\alpha }^2+{\beta }^2=\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}.$$

Commented by Prithwish sen last updated on 23/Jun/19

$$\alpha +\beta =(\mathrm{a}+1)$$$$2\alpha \beta =({\mathrm{a}}^2+\mathrm{a}+1)$$$$\therefore {\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta ={(\mathrm{a}+1)}^2-({\mathrm{a}}^2+\mathrm{a}+1)$$$$=\mathrm{a}$$$$$$

Commented by mathmax by abdo last updated on 24/Jun/19

$${\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta ={(a+1)}^2-2\frac{1}{2}(a^2+a+1)$$$$=a^2+2a+1-a^2-a-1=a.$$

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