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Question Number 46553 by hassentimol last updated on 28/Oct/18

$$\mathrm{Can}\mathrm{someone}\mathrm{please}\mathrm{explain}\mathrm{me}\mathrm{how}\mathrm{to}$$$$\mathrm{solve}\mathrm{a}\mathrm{quadratic}\mathrm{equation}\mathrm{without}\mathrm{the}$$$$\mathrm{common}\mathrm{formulas}\mathrm{but}\mathrm{with}\mathrm{the}\mathrm{sums}\mathrm{and}$$$$\mathrm{products}\mathrm{method}\mathrm{please}?$$$$\mathrm{Thank}$$

Answered by ajfour last updated on 28/Oct/18

$$\alpha +\beta =s;\alpha \beta =p$$$$\alpha -\beta =\sqrt{{(\alpha +\beta )}^2-4\alpha \beta }$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$2\alpha =s+\sqrt{s^2-4p}$$$$\alpha =\frac{s}{2}+\frac{\sqrt{s^2-4p}}{2}$$$$\beta =\frac{s}{2}-\frac{\sqrt{s^2-4p}}{2}.$$

Commented by hassentimol last updated on 28/Oct/18

$$$$$$$$$$\mathrm{Thank}\mathrm{you}\mathrm{sir}$$

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