Respostas

  • Usuário do Brainly
2014-08-15T02:21:43-03:00
Temos que:

\Delta=(3+\sqrt{3})^2-4\cdot1\cdot(3\sqrt{3})

\Delta=12+6\sqrt{3}-12\sqrt{3}

\Delta=12-6\sqrt{3}.

Portanto, x=\dfrac{-(3+\sqrt{3})\pm\sqrt{12-6\sqrt{3}}}{2}.

Ou seja, x'=\dfrac{-3-\sqrt{3}+\sqrt{12-6\sqrt{3}}}{2} e x''=\dfrac{-3-\sqrt{3}-\sqrt{12-6\sqrt{3}}}{2}.