Respostas

  • Usuário do Brainly
2014-08-16T16:57:49-03:00
Temos que, \dfrac{x+1}{x}=\dfrac{x+3}{4}.

Assim, x(x+3)=4(x+1).

x^2+3x=4x+4

x^2-x-4=0

\Delta=(-1)^2-4\cdot1\cdot(-4)=17

Logo, x=\dfrac{(-1)\pm\sqrt{17}}{2}=\dfrac{1\pm\sqrt{17}}{2}.

Assim, x'=\dfrac{1+\sqrt{17}}{2} e x''=\dfrac{1-\sqrt{17}}{2},