Question

realizar esta ecuacion U^4+8=0 ,por el método de completar cuadrados...con procedimiento gracias

Answer 1

$u^4+8=0\\\\ u^4+2\cdot2\sqrt2u^2+8-2\cdot2\sqrt2u^2=0\\\\ (u^2+2\sqrt2)^2-4\sqrt2u^2=0\\\\ (u^2-2\sqrt[4]2u+2\sqrt2)(u^2+2\sqrt[4]2u+2\sqrt2)=0\\\\(i)~u^2-2\sqrt[4]2u+2\sqrt2=0~~o~~(ii)~u^2+2\sqrt[4]2u+2\sqrt2=0$

$(i)~u^2-2\sqrt[4]2u+2\sqrt2=0\\\\ \Delta=b^2-4ac\\\\ \Delta=(-2\sqrt[4]{2})^2-4\cdot1\cdot2\sqrt2\\\\ \Delta=4\sqrt2-8\sqrt2\\\\ \Delta=-4\sqrt2\\\\\\ u=\dfrac{-b\pm\sqrt\Delta}{2a}=\dfrac{-(-2\sqrt[4]2)\pm\sqrt{-4\sqrt2}}{2\cdot1}=\dfrac{2\sqrt[4]2\pm2\sqrt[4]2i}{2}\\\\ \boxed{u=\sqrt[4]2(1\pm i)}$

$(ii)~u^2+2\sqrt[4]2u+2\sqrt2=0\\\\ \Delta=b^2-4ac\\\\ \Delta=(2\sqrt[4]{2})^2-4\cdot1\cdot2\sqrt2\\\\ \Delta=4\sqrt2-8\sqrt2\\\\ \Delta=-4\sqrt2\\\\\\ u=\dfrac{-b\pm\sqrt\Delta}{2a}=\dfrac{-(2\sqrt[4]2)\pm\sqrt{-4\sqrt2}}{2\cdot1}=\dfrac{-2\sqrt[4]2\pm2\sqrt[4]2i}{2}\\\\ \boxed{u=\sqrt[4]2(-1\pm i)}$

Por lo tanto, las soluciones son:

$S=\{\sqrt[4]2(-1- i),\sqrt[4]2(-1+i),\sqrt[4]2(1-i),\sqrt[4]2(1+i)\}$