Question

Find all complex solutions of the equation x^4 + x^2 - 6 = 0

Answer 1

$x^4+x^2-6=0 \\ (x^2)^2+x^2-6=0 \\ \\ a=1 \\ b=1 \\ c=-6 \\ b^2-4ac=1^2-4 \times 1 \times (-6)=1+24=25 \\ \\ x^2=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-1 \pm \sqrt{25}}{2 \times 1}=\frac{-1 \pm 5}{2} \\ x^2=\frac{-1-5}{2} \ \lor \ x^2=\frac{-1+5}{2} \\ x^2=\frac{-6}{2} \ \lor \ x^2=\frac{4}{2} \\ x^2=-3 \ \lor \ x^2=2 \\ x=\pm \sqrt{-3} \ \lor \ x=\pm \sqrt{2} \\ \boxed{x=-i\sqrt{3} \hbox{ or } x=i\sqrt{3} \hbox{ or } x=-\sqrt{2} \hbox{ or } x=\sqrt{2}}$

Answer 2

$x^4+x^2-6=0\\ x^4+3x^2-2x^2-6=0\\ x^2(x^2+3)-2(x^2+3)=0\\ (x^2-2)(x^2+3)=0\\ x^2-2=0 \vee x^2+3=0\\ x^2=2 \vee x^2=-3\\ x=-\sqrt2 \vee x=\sqrt2 \vee x=-i\sqrt3 \vee x=i\sqrt3$