Question

ajuda aqui por favor!!!

Answer 1

$2x^{4}-5x^{2}+3 = 0$

Podemos escreve-la assim:

$2(x^{2})^{2}-5x^{2}+3 = 0$

Perceba que o x² repete. Então podemos chama-lo de y.

$\boxed{x^{2} = y}$

E substituir:

$2(x^{2})^{2}-5x^{2}+3 = 0 \\\\ 2y^{2}-5y+3 = 0 \\\\ \Delta = b^{2}-4 \cdot a \cdot c \\\\ \Delta = (-5)^{2}-4 \cdot (2) \cdot (3) \\\\ \Delta = 25-24 \\\\ \Delta = 1 \\\\\\ y = \frac{-b \pm \sqrt{\Delta}}{2a} \\\\ y = \frac{-(-5) \pm \sqrt{1}}{2 \cdot 2} \\\\ y = \frac{5 \pm 1}{4} \\\\\\ \rightarrow y' = \frac{5 + 1}{4} = \frac{6}{4} = \boxed{\frac{3}{2}} \\\\ \rightarrow y'' = \frac{5 - 1}{4} = \frac{4}{4} = \boxed{1}$


Voltando:

$\Rightarrow y = \frac{3}{2} \\\\ x^{2} = \frac{3}{2} \\\\ x = \pm \sqrt{\frac{3}{2}} \\\\ x = \pm \frac{\sqrt{3}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \boxed{\pm \frac{\sqrt{6}}{2}} \\\\\\ \Rightarrow y = 1 \\\\ x^{2} = 1 \\\\ x = \pm \sqrt{1} \\\\ \boxed{x = \pm 1}$

Então possui quatro raízes reais, são elas:

$\boxed{\boxed{S = \{-\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}, -1, 1\}}}$