Question

O conjunto solução, no campo real, da equação x4 -13x2+36=0

Answer 1

$x^4-13x^2+36=0~~~~fazendo ~x^2=y,temos:\\(x^2)^2-13x^2+36=0~\\y^2-13y+36=0\\\Delta=(-13)^2-4.1.36\\\Delta=169-144\\\Delta=25\\y= \dfrac{13\pm \sqrt{25} }{2.1} \\\\y= \dfrac{13\pm5}{2} \Rightarrow y^{'} = \dfrac{13+5}{2}=9~~e~~y^{"} = \dfrac{13-5}{2} =4\\\\Como~x^2=y,temos:\\x^2=9~~~~~~~~~~ou~~~~~~~x^2=4\\x=\pm \sqrt{9} ~~~~~~~~~~~~~~~~x=\pm \sqrt{4} \\x=\pm3~~~~~~~~~~~~~~~~~~x=\pm2\\S=\{-2,2,-3,3\}$