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

x^4 +x^2 +16=0

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

Answered by MJS last updated on 10/May/19

x=±(√t) t^2 +t+16=0 t=−(1/2)±((3(√7))/2)i x=±(√(−(1/2)±((3(√7))/2)i))=±((√7)/2)±(3/2)i

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

Answered by malwaan last updated on 10/May/19

x^2 =((−1±(√(1^2 −4×16)))/(2×1))=((−1±(√(1−64)))/2) =((−1±(√(−63)))/2)=((−1±3(√7)i)/2) ⇒x=±((√((((−1)/2)+(√((((−1)/2))^2 +(((3(√7))/2))^2 )))/2))±i(√((((1/2))^2 +(((3(√7))/2))^2 )/2))) =±(((√7)/2)±(3/2)i)

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

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