Resposta:
[tex]\textsf{Leia abaixo}[/tex]
Explicação passo a passo:
[tex]\sf{x^4 + 5x^2 + 6 = 0}[/tex]
[tex]\sf{y = x^2}[/tex]
[tex]\sf{y^2 + 5y + 6 = 0}[/tex]
[tex]\sf{\Delta = b^2 - 4.a.c}[/tex]
[tex]\sf{\Delta = (5)^2 - 4.1.6}[/tex]
[tex]\sf{\Delta = 25 - 24}[/tex]
[tex]\sf{\Delta = 1}[/tex]
[tex]\mathsf{y = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{-5 \pm \sqrt{1}}{2} \rightarrow \begin{cases}\mathsf{y' = \dfrac{-5 + 1}{2} = -\dfrac{4}{2} = -2}\\\\\mathsf{y'' = \dfrac{-5 - 1}{2} = -\dfrac{6}{2} = -3}\end{cases}}[/tex]
[tex]\sf{x^2 = -2}[/tex]
[tex]\sf{x = \pm\:\sqrt{2}\:.\:\sqrt{-1}}[/tex]
[tex]\sf{x = \pm\:\sqrt{2}\:.\:i}[/tex]
[tex]\sf{x^2 = -3}[/tex]
[tex]\sf{x = \pm\:\sqrt{3}\:.\:\sqrt{-1}}[/tex]
[tex]\sf{x = \pm\:\sqrt{3}\:.\:i}[/tex]
[tex]\boxed{\boxed{\mathsf{S = \{\sqrt{2}i\:;\:-\sqrt{2}i\:;\:\sqrt{3}i\:;\:-\sqrt{3}i\}}}}[/tex]
