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Question Number 132594 by Study last updated on 15/Feb/21

Commented by mr W last updated on 15/Feb/21

$t h e s u m o f t w o c o m p l e x n u m b e r s$ $c a n b e a r e a l n u m b e r, e . g .$ $(1 + i) + (1 - i) = 2$ $t h e p r o d u c t o f t w o c o m p l e x n u m b e r s$ $c a n a l s o b e a r e a l n u m b e r, e . g .$ $(1 + i) (1 - i) = 2$
	Answered by liberty last updated on 15/Feb/21
	$2x^2 +6x+5=0 , x∈C x^2 +3x+(5/2)=0 (x+(3/2))^2 −(9/4)+((10)/4)=0 (x+(3/2))^2 =−(1/4) x+(3/2)= ±(1/2)i → { ((x_1 =−(3/2)+(1/2)i)),((x_2 =−(3/2)−(1/2)i)) :} → { ((x_1 +x_2 = −3)),((x_1 .x_2 =(−(3/2)+(1/2)i)(−(3/2)−(1/2)i)=(5/2))) :}$
	${2 x}^{2} + 6 x + 5 = 0, x \in C$ $x^{2} + 3 x + \frac{5}{2} = 0$ ${(x + \frac{3}{2})}^{2} - \frac{9}{4} + \frac{10}{4} = 0$ ${(x + \frac{3}{2})}^{2} = - \frac{1}{4}$ $x + \frac{3}{2} = \pm \frac{1}{2} i$ $\to {\begin{cases} x_{1} = - \frac{3}{2} + \frac{1}{2} i \\ x_{2} = - \frac{3}{2} - \frac{1}{2} i \end{cases}$ $\to {\begin{cases} x_{1} + x_{2} = - 3 \\ x_{1} . x_{2} = (- \frac{3}{2} + \frac{1}{2} i) (- \frac{3}{2} - \frac{1}{2} i) = \frac{5}{2} \end{cases}$
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