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
$$\alpha+\beta\:={s}\:\:;\:\alpha\beta={p} \\ $$$$\alpha−\beta\:=\:\sqrt{\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{4}\alpha\beta} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\mathrm{2}\alpha\:=\:{s}+\sqrt{{s}^{\mathrm{2}} −\mathrm{4}{p}} \\ $$$$\alpha\:=\:\frac{{s}}{\mathrm{2}}+\frac{\sqrt{{s}^{\mathrm{2}} −\mathrm{4}{p}}}{\mathrm{2}} \\ $$$$\beta\:=\:\frac{{s}}{\mathrm{2}}−\frac{\sqrt{{s}^{\mathrm{2}} −\mathrm{4}{p}}}{\mathrm{2}}\:\:\:. \\ $$
Commented by hassentimol last updated on 28/Oct/18
$$ \\ $$$$ \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$