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Question Number 58963 by Tony Lin last updated on 02/May/19

solve x^2 −2(1+i)x−5+14i=0

$${solve}\:{x}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}+{i}\right){x}−\mathrm{5}+\mathrm{14}{i}=\mathrm{0} \\ $$

Answered by tanmay last updated on 02/May/19

x^2 −2x(1+i)+(1+i)^2 −5+14i−(1+i)^2 =0 {x−(1+i)}^2 −5+14i−1−2i−i^2 =0 {x−(1+i)}^2 =5−12i {x−(1+i)}=±(3−2i) x=(1+i)±(3−2i) so x=(1+3+i−2i)and(1−3+i+2i) x=(4−i) and(−2+3i)

$${x}^{\mathrm{2}} −\mathrm{2}{x}\left(\mathrm{1}+{i}\right)+\left(\mathrm{1}+{i}\right)^{\mathrm{2}} −\mathrm{5}+\mathrm{14}{i}−\left(\mathrm{1}+{i}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left\{{x}−\left(\mathrm{1}+{i}\right)\right\}^{\mathrm{2}} −\mathrm{5}+\mathrm{14}{i}−\mathrm{1}−\mathrm{2}{i}−{i}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left\{{x}−\left(\mathrm{1}+{i}\right)\right\}^{\mathrm{2}} =\mathrm{5}−\mathrm{12}{i} \\ $$$$\left\{{x}−\left(\mathrm{1}+{i}\right)\right\}=\pm\left(\mathrm{3}−\mathrm{2}{i}\right) \\ $$$${x}=\left(\mathrm{1}+{i}\right)\pm\left(\mathrm{3}−\mathrm{2}{i}\right) \\ $$$${so}\:{x}=\left(\mathrm{1}+\mathrm{3}+{i}−\mathrm{2}{i}\right){and}\left(\mathrm{1}−\mathrm{3}+{i}+\mathrm{2}{i}\right) \\ $$$${x}=\left(\mathrm{4}−{i}\right)\:{and}\left(−\mathrm{2}+\mathrm{3}{i}\right) \\ $$

Commented by Tony Lin last updated on 02/May/19

thanks,sir

$${thanks},{sir} \\ $$

Commented by tanmay last updated on 02/May/19

most welcome

$${most}\:{welcome} \\ $$

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