Respostas

  • Usuário do Brainly
2013-06-03T15:07:49-03:00

Temos que:

 

\text{x}^2-\text{x}-1=0

 

Vamos determinar as raízes:

 

\text{x}=\dfrac{-(-1)\pm\sqrt{(-1)^2-4\cdot1\cdot(-1)}}{2\cdot1}=\dfrac{1\pm\sqrt{5}}{2}

 

Logo, as raízes são:

 

\text{x}'=\dfrac{1+\sqrt{5}}{2}

 

\text{x}"=\dfrac{1-\sqrt{5}}{2}

 

Desta maneira, podemos afirmar que \text{p}=\dfrac{1+\sqrt{5}}{2} e, portanto:

 

\text{p}^2=\left(\dfrac{1+\sqrt{5}}{2}\right)^2=\dfrac{1+2\sqrt{5}+5}{4}=\dfrac{6+2\sqrt{5}}{4}=\dfrac{3+\sqrt{5}}{2}

3 1 3