Question

Solve the following equation;
$x^4-3x^2-4=0$

Answer 1

$x^4-3x^2-4=0$

$x^4-4x^2+x^2-4=0$

$x^2(x^2-4)+1(x^2-4)=0$

$(x^2-4)(x^2+1)=0$

Now, $x^2-4=0x=\pm 2$ or $x^2+1=0x^2=-1x=\pm \sqrt{-1}x=\pm i\{i^2=-1\}$

So, $x=-2,+2,i,-i$