Question

Equações biquadradas, questão de concurso, achar as raízes de x^4-11x^2+16=0 ?

Answer 1

$x^{2} = y\\\\y^{2}-11y+16=0\\delta=121-4.1.16\\delta=121-64\\delta=57\\\\y1=\frac{11+\sqrt{57}}{2}\\y2=\frac{11-\sqrt{57}}{2}\\x^{2}=y(somentes para valores positivos de y)\\x1=\sqrt{\frac{11+\sqrt{57}}{2}}\\x2=-\sqrt{\frac{11+\sqrt{57}}{2}}$

Answer 2

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 $\pm\dfrac{11\sqrt{2}}{2}-6\}$