Question

1) Resolva as equações biquadradas, transformando-as em equação do 2º grau.



c) 4x4 – 10x2 + 9 = 0
d) x4 + 3x2 – 4 = 0
e) 4x4 -37x2 + 9 = 0
f) 16x4 – 40x2 + 9 = 0
g) x4 -7x2 + 12 = 0

Answer 1

c) $4x^4-10x^2+9=0$

$y=x^2$

$4y^2-10y+9=0$

$\Delta=(-10)^2-4\cdot4\cdot9=100-144=-44$

Não há solução real.


d) $x^4+3x^2-4=0$

$y=x^2$

$y^2+3y-4=0$

$\Delta=3^2-4\cdot1\cdot(-4)=9+16=25$

$y=\dfrac{-3\pm\sqrt{25}}{2}=\dfrac{-3\pm5}{2}$

$y'=\dfrac{-3+5}{2}=1$ e $y"=\dfrac{-3-5}{2}=-4$ (não serve).

$x^2=1$

$x=\pm1$

$S=\{-1,1\}$


e) $4x^4-37x^2+9=0$

$y=x^2$

$4y^2-37y+9=0$

$\Delta=(-37)^2-4\cdot4\cdot9=1~369-144=1~225$

$y=\dfrac{-(-37)\pm\sqrt{1~225}}{2\cdot4}=\dfrac{37\pm35}{8}$

$y'=\dfrac{37+35}{8}=9$ e $y"=\dfrac{37-35}{8}=\dfrac{1}{4}$

$x^2=9$ ou $x^2=\dfrac{1}{4}$

$x=\pm3$ ou $x=\pm\dfrac{1}{2}$

$S=\{-3,\frac{-1}{2},\frac{1}{2},3\}$


f) $16x^4-40x^2+9=0$

$y=x^2$

$16y^2-40y+9=0$

$\Delta=(-40)^2-4\cdot16\cdot9=1~600-576=1~024$

$y=\dfrac{-(-40)\pm\sqrt{1~024}}{2\cdot16}=\dfrac{40\pm32}{32}=\dfrac{5\pm4}{4}$

$y'=\dfrac{5+4}{4}=\dfrac{9}{4}$ e $y"=\dfrac{5-4}{4}=\dfrac{1}{4}$

$x^2=\dfrac{9}{4}$ ou $x^2=\dfrac{1}{4}$

$x=\pm\dfrac{3}{2}$ ou $x=\pm\dfrac{1}{2}$.

$S=\{-\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{3}{2}\}$.


g) $x^4-7x^2+12=0$

$x^2=y$

$y^2-7y+12=0$

$\Delta=(-7)^2-4\cdot1\cdot12=49-48=1$

$y=\dfrac{-(-7)\pm\sqrt{1}}{2}=\dfrac{7\pm1}{2}$

$y'=\dfrac{7+1}{2}=4$ e $y"=\dfrac{7-1}{2}=3$

$x^2=4$ ou $x^2=3$

$x=\pm2$ ou $x=\pm\sqrt{3}$.

$S=\{-\sqrt{3},-2,\sqrt{3},2\}$.