Question Number 123143 by aurpeyz last updated on 23/Nov/20
$${write}\:{ax}^{\mathrm{2}} +{bx}+{c}\:{as}\:{a}\:{sum}\:{or}\:{difference}\: \\ $$$${of}\:{two}\:{squares} \\ $$
Commented by mr W last updated on 23/Nov/20
$$=\left(\sqrt{{a}}{x}\right)^{\mathrm{2}} +\mathrm{2}\left(\frac{{b}}{\:\mathrm{2}\sqrt{{a}}}×\sqrt{{a}}\right){x}+\left(\frac{{b}}{\mathrm{2}\sqrt{{a}}}\right)^{\mathrm{2}} +{c}−\frac{{b}^{\mathrm{2}} }{\mathrm{4}{a}} \\ $$$$=\left(\sqrt{{a}}{x}+\frac{{b}}{\mathrm{2}\sqrt{{a}}}\right)^{\mathrm{2}} −\left(\sqrt{\frac{{b}^{\mathrm{2}} }{\mathrm{4}{a}}−{c}}\right)^{\mathrm{2}} \:{for}\:{c}<\frac{{b}^{\mathrm{2}} }{\mathrm{4}{a}} \\ $$$$=\left(\sqrt{{a}}{x}+\frac{{b}}{\mathrm{2}\sqrt{{a}}}\right)^{\mathrm{2}} +\left(\sqrt{{c}−\frac{{b}^{\mathrm{2}} }{\mathrm{4}{a}}}\right)^{\mathrm{2}} \:{for}\:{c}\geqslant\frac{{b}^{\mathrm{2}} }{\mathrm{4}{a}} \\ $$