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Question Number 123315 by aurpeyz last updated on 24/Nov/20
express 2x^2 +3x+5 as a sum or   diffrence of two squares
$${express}\:\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{5}\:{as}\:{a}\:{sum}\:{or}\: \\ $$$${diffrence}\:{of}\:{two}\:{squares} \\ $$
Commented by mr W last updated on 24/Nov/20
answer in Q123143 not good for you?
$${answer}\:{in}\:{Q}\mathrm{123143}\:{not}\:{good}\:{for}\:{you}? \\ $$
Answered by mathmax by abdo last updated on 24/Nov/20
2x^2  +3x+5 =2{x^2  +(3/2)x+(5/2)}  =2{x^2  +2.(3/4)x +(9/(16))+(5/2)−(9/(16))} =2{(x+(3/4))^2 +((31)/(16))}  =2(x+(3/4))^2  +((31)/8) =((√2)(x+(3/4)))^2  +((√((31)/8)))^2
$$\mathrm{2x}^{\mathrm{2}} \:+\mathrm{3x}+\mathrm{5}\:=\mathrm{2}\left\{\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}+\frac{\mathrm{5}}{\mathrm{2}}\right\} \\ $$$$=\mathrm{2}\left\{\mathrm{x}^{\mathrm{2}} \:+\mathrm{2}.\frac{\mathrm{3}}{\mathrm{4}}\mathrm{x}\:+\frac{\mathrm{9}}{\mathrm{16}}+\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{9}}{\mathrm{16}}\right\}\:=\mathrm{2}\left\{\left(\mathrm{x}+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{31}}{\mathrm{16}}\right\} \\ $$$$=\mathrm{2}\left(\mathrm{x}+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} \:+\frac{\mathrm{31}}{\mathrm{8}}\:=\left(\sqrt{\mathrm{2}}\left(\mathrm{x}+\frac{\mathrm{3}}{\mathrm{4}}\right)\right)^{\mathrm{2}} \:+\left(\sqrt{\frac{\mathrm{31}}{\mathrm{8}}}\right)^{\mathrm{2}} \\ $$
Commented by aurpeyz last updated on 25/Nov/20
thank you everyone
$${thank}\:{you}\:{everyone} \\ $$

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