Question

Qual a solução S da equação biquadrada abaixo: (Aplicar / Efetuar)
(1 Ponto)


a) S= {1,9}

b) S= {1,-1,3,-3}

c) S= {3,-3}

d) S= {-1, 1}

e) Não existe raiz real​

Answer 1

$4\text x^4-17\text x^2+4=0$

Façamos :

$\text x^2 = \text a \ \to \ \text x^4=\text a^2$

Substituindo :

$4\text a^2-17\text a+4=0$

$\displaystyle \text a= \frac{-(-17)\pm\sqrt{(-17)^2-4.4.4 }}{2.4}\\\\\\ \text a = \frac{17\pm\sqrt{289-64}}{8 } \to \text a = \frac{17\pm\sqrt{225}}{8} \\\\ \text a=\frac{17\pm15}{8} \\\\\\ \underline{\text a' = 4 \ ; \ \text a'' = \frac{1}{4}}$

Desfazendo a troca de variável :

$\displaystyle \text x^2 = \text a' \ \ \ \ \ \ \ \ \ \ \text{ou} \ \ \ \ \ \ \ \ \ \ \ \text x^2=\text a'' \\\\\\\text x^2 = 4 \ \ \ \ \ \ \ \ \ \ \ \text{ou} \ \ \ \ \ \ \ \ \ \ \ \text x^2=\frac{1}{4} \\\\\\\text x = \pm2 \ \ \ \ \ \ \ \ \ \ \text{ou} \ \ \ \ \ \ \ \ \ \ \ \text x = \pm\frac{1}{2}$

Solução :

$\huge\boxed{\text S =\{2,-2,\frac{1}{2}, \frac{-1}{2}\}\ }\checkmark$