x^4-4\sqrt{3}x^2+4=0
wprowadzamy zmienną t=x^2
t^2-4\sqrt{3}t+4=0
\Delta=b^2-4ac=(-4\sqrt{3})^2-4*1*4=48-16=32
\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=4\sqrt{2}
t_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{4\sqrt{3}-4\sqrt{2}}{2}=2(\sqrt{3}-\sqrt{2})
t_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{4\sqrt{3}+4\sqrt{2}}{2}=2(\sqrt{3}+\sqrt{2})
ponieważ
t=x^2
to
x_1=\sqrt{2(\sqrt{3}-\sqrt{2})} lub x_2=-\sqrt{2(\sqrt{3}-\sqrt{2})}
lub
x_3=\sqrt{2(\sqrt{3}+\sqrt{2})} lub x_4=-\sqrt{2(\sqrt{3}+\sqrt{2})}
(\sqrt{2(\sqrt{3}-\sqrt{2})})^4+(-\sqrt{2(\sqrt{3}-\sqrt{2})})^4+(\sqrt{2(\sqrt{3}+\sqrt{2})})^4+(-\sqrt{2(\sqrt{3}+\sqrt{2})})^4=(2(\sqrt{3}-\sqrt{2}))^2+(2(\sqrt{3}-\sqrt{2}))^2+(2(\sqrt{3}+\sqrt{2}))^2+(2(\sqrt{3}+\sqrt{2}))^2=(2\sqrt{3}-2\sqrt{2})^2+(2\sqrt{3}-2\sqrt{2})^2+(2\sqrt{3}+2\sqrt{2})^2+(2\sqrt{3}+2\sqrt{2})^2=4*3-8\sqrt{6}+4*2+4*3+8\sqrt{6}+4*2+4*3-8\sqrt{6}+4*2+4*3+8\sqrt{6}+4*2=12+8+12+8+12+8+12+8=80
nie wiem, czy się gdzieś nie zaplątałam… :-////