Question

X⁴+4x²-21=0 como resolver ??

Answer 1

$x^{4}+4x^{2}-21=0 \rightarrow t=x^{2}\textup{ entao }t^{2}+4t-21 =0 \newline \Delta= b^{2}-4ac =4^{2}-4.1(-21) =16+84=100 \Rightarrow \sqrt{\Delta} =10\newline t=\frac{-b+\sqrt{\Delta}}{2a} =\frac{-4+10}{2}=\frac{6}{2}=3 \newline t=\frac{-b-\sqrt{\Delta}}{2a} =\frac{-4-10}{2}=\frac{-14}{2}=-7 \newline\textup{ solucao em x }:\newline i)x^{2}=3 \Rightarrow x=\sqrt{3} \textup{ ou }x=- \sqrt{3}\newline\ ii) x^{2}=-7 \Rightarrow x=\sqrt{-7}=\sqrt{7}.\sqrt{-1}=\sqrt{7}i\textup{ ou }x=-\sqrt{7}i\newline$