Question

resolva as  equações  biquadradas em r  
x elevado a 4 + 6x² + 8 =0
9x elevado a 4 - 13x² + 4 =0
x elevado a  4\2 -  x²-1/3  =7
(x²-3)² = (x+1) (x-1)

Answer 1

$x^{4}+6x^{2}+8=0\\(x^{2})^{2}+6x^{2}+8=0$

Chamando x² de y:

$y^{2}+6y+8=0$

$S=-b/a=-6/1=-6\\P=c/a=8/1=8$

Raízes: 2 números que quando somados dão -6 e quando multiplicados dão 8

$y'=-2\\y''=-4$

Como y = x²:

$y=-2\\x^{2}=-2\\x=\pm\sqrt{-2}$

Raiz quadrada de número negativo não pertence aos reais

$y=-4\\x^{2}=-4\\x=\pm\sqrt{-4}$

O mesmo. A solução da equação, portanto, será vazia.

$S=\{\ ~\}$
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$9x^{4}-13x^{2}+4=0\\9(x^{2})^{2}-13x^{2}+4=0\\9y^{2}-13y+4=0$

$\Delta=b^{2}-4ac\\\Delta=(-13)^{2}-4*9*4\\\Delta=169-144\\\Delta=25$

$y=(-b\pm\sqrt{\Delta})/2a\\y=(-[13]\pm\sqrt{25})/(2*9)\\y=(13\pm5)/18$

$y'=(13+5)/18\\y'=18/18\\y'=1\\\\y''=(13-5)/18\\y''=8/18\\y''=4/9$

$y=1\\x^{2}=1\\x=\pm\sqrt{1}\\x=\pm1\\\\y=4/9\\x^{2}=4/9\\x=\pm\sqrt{4/9}\\x=\pm2/3$

$S=\{-1,~1,-2/3,~2/3\}$
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$(x^{4}/2)-[(x^{2}-1)/3]=7$

Multiplicando todos os membros pelo mmc entre 2 e 3 (6):

$6(x^{4}/2)-6[(x^{2}-1)/3]=6.7\\3x^{4}-2(x^{2}-1)=42\\3x^{4}-2x^{2}+2=42\\3x^{4}-2x^{2}+2-42=0\\3x^{4}-2x^{2}-40=0\\3(x^{2})^{2}-2x^{2}-40=0\\3y^{2}-2y-40=0$

$\Delta=b^{2}-4ac\\\Delta=(-2)^{2}-4*3*(-40)\\\Delta=4+480\\\Delta=484$

$y=(-b\pm\sqrt{\Delta})/2a\\y=(-[-2]\pm\sqrt{484})/(2*3)\\y=(2\pm22)/(2*3)\\y=2*(1\pm11)/(2*3)\\y=(1\pm11)/3$

$y'=(1+11)/3\\y'=12/3\\y'=4\\\\y''=(1-11)/3\\y''=-10/3$

$y=4\\x^{2}=4\\x=\pm\sqrt{4}\\x=\pm2\\\\y=-10/3\\x^{2}=-10/3\\x=\pm\sqrt{-10/3}$

Só consideraremos -2 e 2 como raízes:

$S=\{-2,2\}$
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$(x^{2}-3)^{2}=(x+1)(x-1)\\(x^{2})^{2}-2*x^{2}*3+3^{2}=x^{2}-1^{2}\\(x^{2})^{2}-6x^{2}+9=x^{2}-1\\(x^{2})^{2}-6x^{2}-x^{2}+9+1=10\\(x^{2})^{2}-7x^{2}+10=0\\y^{2}-7y+10=0$

$\Delta=b^{2}-4ac\\\Delta=(-7)^{2}-4*1*10\\\Delta=49-40\\\Delta=9$

$y=(-b\pm\sqrt{\Delta})/2a\\y=(-[-7]\pm\sqrt{9})/(2*1)\\y=(7\pm3)/2\\\\y'=(7+3)/2\\y'=10/2\\y'=5\\\\y''=(7-3)/2\\y''=4/2\\y''=2$

$y=5\\x^{2}=5\\x=\pm\sqrt{5}\\\\y=2\\x^{2}=2\\x=\pm\sqrt{2}$

$S=\{-\sqrt{5},~\sqrt{5},-\sqrt{2},~\sqrt{2}\}$