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

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