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Question-210142




Question Number 210142 by Abdullahrussell last updated on 01/Aug/24
Commented by Frix last updated on 02/Aug/24
I think there is only one “nice” solution:  x=(3/2)−((√(11))/2)i     y=(3/2)+((√(11))/2)i     z=3  ⇒ x+y+z=6  But there should be more solutions with  x+y+z∈R
$$\mathrm{I}\:\mathrm{think}\:\mathrm{there}\:\mathrm{is}\:\mathrm{only}\:\mathrm{one}\:“\mathrm{nice}''\:\mathrm{solution}: \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\mathrm{i}\:\:\:\:\:{y}=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\mathrm{i}\:\:\:\:\:{z}=\mathrm{3} \\ $$$$\Rightarrow\:{x}+{y}+{z}=\mathrm{6} \\ $$$$\mathrm{But}\:\mathrm{there}\:\mathrm{should}\:\mathrm{be}\:\mathrm{more}\:\mathrm{solutions}\:\mathrm{with} \\ $$$${x}+{y}+{z}\in\mathbb{R} \\ $$

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