solve x^2 −2(1+i)x−5+14i=0


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

$s o l v e x^{2} - 2 (1 + i) x - 5 + 14 i = 0$
	Answered by tanmay last updated on 02/May/19
	$x^2 −2x(1+i)+(1+i)^2 −5+14i−(1+i)^2 =0 {x−(1+i)}^2 −5+14i−1−2i−i^2 =0 {x−(1+i)}^2 =5−12i {x−(1+i)}=±(3−2i) x=(1+i)±(3−2i) so x=(1+3+i−2i)and(1−3+i+2i) x=(4−i) and(−2+3i)$
	$x^{2} - 2 x (1 + i) + {(1 + i)}^{2} - 5 + 14 i - {(1 + i)}^{2} = 0$ ${x - (1 + i)}^{2} - 5 + 14 i - 1 - 2 i - i^{2} = 0$ ${x - (1 + i)}^{2} = 5 - 12 i$ ${x - (1 + i)} = \pm (3 - 2 i)$ $x = (1 + i) \pm (3 - 2 i)$ $s o x = (1 + 3 + i - 2 i) a n d (1 - 3 + i + 2 i)$ $x = (4 - i) a n d (- 2 + 3 i)$
	Commented by Tony Lin last updated on 02/May/19

	$t h a n k s, s i r$
	Commented by tanmay last updated on 02/May/19

	$m o s t w e l c o m e$
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