Question

Resolva essa equação biquadrada
a) 4 2
5 x + 26 x + 5=0

Answer 1

$5x^4+26x^2+5=0 \\ \\ 5x^4+26x^2=-5$

4 e 2 são números pares.
Cuja extremidade esquerda da igual deve ser positivo.
- 5 < 0

Portanto, 
$\forall \ x \\ S= \O$

Answer 2

$5x^4+26x^2+5=0\\ (5x^2)^2+26x^2+5=0\\\\ x^2=y\\\\ 5y^2+26y+5=0\\\\ \Delta=26^2-4\cdot5\cdot5\\ \Delta=676-100\\ \Delta=576\\\\ y= \dfrac{-26\pm \sqrt{576} }{2\cdot5}= \dfrac{-26\pm24}{10}\begin{cases}y_1= \dfrac{2}{10}= \dfrac{1}{5}\\\\ y_2= \dfrac{-50}{10}=-5 \end{cases}\\\\\\ x^2=y\\\\ (x_1)^2= \dfrac{1}{5}\\\\ (x_1)= \pm\sqrt{ \dfrac{1}{5} }= \dfrac{ \sqrt{1} }{ \sqrt{5} }= \dfrac{ \sqrt{1}\cdot \sqrt{5} }{ \sqrt{5}\cdot \sqrt{5} }= \pm\dfrac{ \sqrt{5} }{5}\\\\\\ (x_2)^2=5$

$(x_2)=\pm \sqrt{-5}~~(\notin \mathbb{R})\\\\\\ \text{S}=\left\{ \dfrac{ \sqrt{5} }{5},- \dfrac{ \sqrt{5} }{5}\right\}$