Respostas

  • Usuário do Brainly
2013-01-11T21:20:36-02:00

Temos a equação:

 

\text{x}^4-11\text{x}^2+16=0

 

(\text{x}^2)^2-11\text{x}^2+16=0

 

Façamos \text{x}^2=\text{y}

 

Desta maneira, obtemos:

 

\text{y}^2-11\text{y}+16=0

 

\text{y}=\dfrac{-(-11)\pm\sqrt{(-11)^2-4\cdot1\cdot16}}{2\cdot1}=\dfrac{11\pm\sqrt{57}}{2}

 

\text{y}'=\dfrac{11+\sqrt{19\cdot3}}{2}

 

\text{y}"=\dfrac{11-\sqrt{19\cdot3}}{2}

 

Como \text{x}^2=\text{y}, podemos afirmar que:

 

\text{x}'=\pm\sqrt{\dfrac{11+\sqrt{19\cdot3}}{2}}

 

\text{x}'=\pm\dfrac{\sqrt{11+\sqrt{19\cdot3}}}{\sqrt{2}}

 

\text{x}'=\pm\dfrac{11\sqrt{2}+\sqrt{144}}{2}

 

\text{x}'=\pm\dfrac{11\sqrt{2}+12}{2}

 

\text{x}'=\pm\dfrac{11\sqrt{2}}{2}+6

 

Analogamente, temos:

 

\text{x}''=\pm\sqrt{\dfrac{11-\sqrt{19\cdot3}}{2}}

 

\text{x}''=\pm\dfrac{\sqrt{11-\sqrt{19\cdot3}}}{\sqrt{2}}

 

\text{x}''=\pm\dfrac{11\sqrt{2}-\sqrt{144}}{2}

 

\text{x}''=\pm\dfrac{11\sqrt{2}-12}{2}

 

\text{x}''=\pm\dfrac{11\sqrt{2}}{2}-6

 

Logo, as raízes da equação supracitada são:

 

\text{S}=\{\pm\dfrac{11\sqrt{2}}{2}+6 e \pm\dfrac{11\sqrt{2}}{2}-6\}