Resposta :
[tex]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}}[/tex]
Temos a equação:
[tex]\text{x}^4-11\text{x}^2+16=0[/tex]
[tex](\text{x}^2)^2-11\text{x}^2+16=0[/tex]
Façamos [tex]\text{x}^2=\text{y}[/tex]
Desta maneira, obtemos:
[tex]\text{y}^2-11\text{y}+16=0[/tex]
[tex]\text{y}=\dfrac{-(-11)\pm\sqrt{(-11)^2-4\cdot1\cdot16}}{2\cdot1}=\dfrac{11\pm\sqrt{57}}{2}[/tex]
[tex]\text{y}'=\dfrac{11+\sqrt{19\cdot3}}{2}[/tex]
[tex]\text{y}"=\dfrac{11-\sqrt{19\cdot3}}{2}[/tex]
Como [tex]\text{x}^2=\text{y}[/tex], podemos afirmar que:
[tex]\text{x}'=\pm\sqrt{\dfrac{11+\sqrt{19\cdot3}}{2}}[/tex]
[tex]\text{x}'=\pm\dfrac{\sqrt{11+\sqrt{19\cdot3}}}{\sqrt{2}}[/tex]
[tex]\text{x}'=\pm\dfrac{11\sqrt{2}+\sqrt{144}}{2}[/tex]
[tex]\text{x}'=\pm\dfrac{11\sqrt{2}+12}{2}[/tex]
[tex]\text{x}'=\pm\dfrac{11\sqrt{2}}{2}+6[/tex]
Analogamente, temos:
[tex]\text{x}''=\pm\sqrt{\dfrac{11-\sqrt{19\cdot3}}{2}}[/tex]
[tex]\text{x}''=\pm\dfrac{\sqrt{11-\sqrt{19\cdot3}}}{\sqrt{2}}[/tex]
[tex]\text{x}''=\pm\dfrac{11\sqrt{2}-\sqrt{144}}{2}[/tex]
[tex]\text{x}''=\pm\dfrac{11\sqrt{2}-12}{2}[/tex]
[tex]\text{x}''=\pm\dfrac{11\sqrt{2}}{2}-6[/tex]
Logo, as raízes da equação supracitada são:
[tex]\text{S}=\{\pm\dfrac{11\sqrt{2}}{2}+6[/tex] e [tex]\pm\dfrac{11\sqrt{2}}{2}-6\}[/tex]