Question Number 123315 by aurpeyz last updated on 24/Nov/20
$${express}\:\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{5}\:{as}\:{a}\:{sum}\:{or}\: \\ $$$${diffrence}\:{of}\:{two}\:{squares} \\ $$
Commented by mr W last updated on 24/Nov/20
$${answer}\:{in}\:{Q}\mathrm{123143}\:{not}\:{good}\:{for}\:{you}? \\ $$
Answered by mathmax by abdo last updated on 24/Nov/20
$$\mathrm{2x}^{\mathrm{2}} \:+\mathrm{3x}+\mathrm{5}\:=\mathrm{2}\left\{\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}+\frac{\mathrm{5}}{\mathrm{2}}\right\} \\ $$$$=\mathrm{2}\left\{\mathrm{x}^{\mathrm{2}} \:+\mathrm{2}.\frac{\mathrm{3}}{\mathrm{4}}\mathrm{x}\:+\frac{\mathrm{9}}{\mathrm{16}}+\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{9}}{\mathrm{16}}\right\}\:=\mathrm{2}\left\{\left(\mathrm{x}+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{31}}{\mathrm{16}}\right\} \\ $$$$=\mathrm{2}\left(\mathrm{x}+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} \:+\frac{\mathrm{31}}{\mathrm{8}}\:=\left(\sqrt{\mathrm{2}}\left(\mathrm{x}+\frac{\mathrm{3}}{\mathrm{4}}\right)\right)^{\mathrm{2}} \:+\left(\sqrt{\frac{\mathrm{31}}{\mathrm{8}}}\right)^{\mathrm{2}} \\ $$
Commented by aurpeyz last updated on 25/Nov/20
$${thank}\:{you}\:{everyone} \\ $$