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Question Number 59478 by subhankar10 last updated on 10/May/19

$${\mathrm{x}}^4+{\mathrm{x}}^2+16=0$$

Answered by MJS last updated on 10/May/19

$$x=\pm \sqrt{t}$$$$t^2+t+16=0$$$$t=-\frac{1}{2}\pm \frac{3\sqrt{7}}{2}\mathrm{i}$$$$x=\pm \sqrt{-\frac{1}{2}\pm \frac{3\sqrt{7}}{2}\mathrm{i}}=\pm \frac{\sqrt{7}}{2}\pm \frac{3}{2}\mathrm{i}$$

Answered by malwaan last updated on 10/May/19

$$x^2=\frac{-1\pm \sqrt{1^2-4\times 16}}{2\times 1}=\frac{-1\pm \sqrt{1-64}}{2}$$$$=\frac{-1\pm \sqrt{-63}}{2}=\frac{-1\pm 3\sqrt{7}i}{2}$$$$\Rightarrow x=\pm (\sqrt{\frac{\frac{-1}{2}+\sqrt{{\left(\frac{-1}{2}\right)}^2+{\left(\frac{3\sqrt{7}}{2}\right)}^2}}{2}}\pm i\sqrt{\frac{{\left(\frac{1}{2}\right)}^2+{\left(\frac{3\sqrt{7}}{2}\right)}^2}{2}})$$$$=\pm (\frac{\sqrt{7}}{2}\pm \frac{3}{2}i)$$

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