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Question Number 155384 by aliyn last updated on 29/Sep/21

$$\mathit{S}\mathit{o}\mathit{l}\mathit{v}\mathit{e}:{\mathit{x}}^2-(5-\mathit{i})\mathit{x}+(12-5\mathit{i})=0$$

Commented by aliyn last updated on 29/Sep/21

$$????$$

Answered by puissant last updated on 30/Sep/21

$$\mathrm{\Delta }={(-5+i)}^2-4(12-5i)=-24+10i$$$$letz=x+iy,$$$$\to x^2+y^2=\mid z\mid =26$$$$\to x^2-y^2=-24$$$$\to xy=\frac{10}{2}=5$$$$\Rightarrow 2x^2=2\Rightarrow x=\pm 1;2y^2=50\Rightarrow y=\pm 5$$$$xy=5>0\Rightarrow (x=1andy=5)or(x=-1andy=-5)$$$$take\sqrt{\mathrm{\Delta }}=1+5i..$$$$x_1=\frac{5-i-1-5i}{2}=\frac{4-6i}{2}=2-3i..$$$$x_2=\frac{5-i+1+5i}{2}=\frac{6+4i}{2}=3+2i..$$$$$$$$S_{\mathbb{C}}=\{2-3i;3+2i\}...$$

Answered by MJS_new last updated on 30/Sep/21

$$\mathrm{just}\mathrm{solve}\mathrm{it}.$$$$x^2+px+q=0\Rightarrow x=-\frac{p}{2}\pm \sqrt{\frac{p^2}{4}-q}$$$$p=-5+\mathrm{i}\wedge q=12-5\mathrm{i}$$$$x=\frac{5-\mathrm{i}}{2}\pm \sqrt{\frac{{(-5+\mathrm{i})}^2}{4}-12+5\mathrm{i}}$$$$x=\frac{5-\mathrm{i}}{2}\pm \frac{\sqrt{-2(12-5\mathrm{i})}}{2}$$$$\sqrt{-2(12-5\mathrm{i})}=1+5\mathrm{i}$$$$x=\frac{5-\mathrm{i}}{2}\pm \frac{1+5\mathrm{i}}{2}$$$$x=3+2\mathrm{i}\vee x=2-3\mathrm{i}$$

Commented by puissant last updated on 30/Sep/21

$$x^2+px+q=0\Rightarrow x^2+px=-q$$$$\Rightarrow x^2+px+\frac{p^2}{4}=\frac{p^2}{4}-q\Rightarrow {(x+\frac{p}{2})}^2=\frac{p^2}{4}-q$$$$\Rightarrow x+\frac{p}{2}=\pm \sqrt{\frac{p^2}{4}-q}$$$$\Rightarrow x=-\frac{p}{2}\pm \sqrt{\frac{p^2}{4}-q}...$$

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