Q.Solve: $x^4 + 4 = 0$

Split $x^4+4$ as a difference of squares by completing the square in $x^2$, then solve the two resulting quadratics.

1. $x^4+4=0\Rightarrow x^4=-4$.
2. Add and subtract $4x^2$: $x^4+4x^2+4-4x^2=0\Rightarrow(x^2+2)^2-(2x)^2=0$.
3. Factor as a difference of squares: $\big[(x^2+2)-2x\big]\big[(x^2+2)+2x\big]=0$, i.e. $(x^2-2x+2)(x^2+2x+2)=0$.
4. Solve $x^2-2x+2=0$: $x=\dfrac{2\pm\sqrt{4-8}}{2}=\dfrac{2\pm2i}{2}=1\pm i$.
5. Solve $x^2+2x+2=0$: $x=\dfrac{-2\pm\sqrt{4-8}}{2}=\dfrac{-2\pm2i}{2}=-1\pm i$.
6. So the four roots are $1+i,\,1-i,\,-1+i,\,-1-i$ — note they occur as two conjugate pairs, as expected for a real-coefficient equation. …