Respostas

2014-07-22T16:23:19-03:00
Olá Moreninha>

x^4-5x^2+4=0\\
(x^2)^2-5 x^{2} +4=0\\\\
x^2=k\\\\
k^2-5k+4=0\\\\
\Delta=b^2-4ac\\
\Delta=(-5)^2-4*1*4\\
\Delta=25-16\\
\Delta=9\\\\
k= \dfrac{-b\pm \sqrt{\Delta} }{2a}= \dfrac{-(-5)\pm \sqrt{9} }{2*1}= \dfrac{5\pm3}{2}\begin{cases}k'= \dfrac{5-3}{2}= \dfrac{2}{2}=1\\\\
k''= \dfrac{5+3}{2}= \dfrac{8}{2}=4    \end{cases}

Se x² = k, teremos:

 x^{2} =1~~~~~~~~~~~~ x^{2} =4\\
x=\pm \sqrt{1}~~~~~~~~~x=\pm \sqrt{4} \\
x=\pm1~~~~~~~~~~~x=\pm2\\\\\\
\boxed{S=\{1,-1,2-2\}}

Tenha ótimos estudos =))
1 5 1