Question

x4-x2-12=0 equaçao biquadrada

Answer 1

Faça $\boxed{x^2= y}$ .
Logo,

$x^4-x^2-12=0\\\\(x^2)^2-x^2-12=0\\\\y^2-y-12=0\\\\a=1\\b=-1\\c=-12\\\\\Delta=b^2-4.a.c\\\Delta=(-1)^2-4.1.(-12)\\\Delta=49$

$y= \frac{-b\pm \sqrt{\Delta} }{2.a} \\\\y=\frac{-(-1)\pm \sqrt{49} }{2.1} \\\\y= \frac{1\pm7}{2}\\\\y'= \frac{8}{2}\\\\\boxed{y'=4} \\\\y''= \frac{-6}{2}\\\\\boxed{y''=-3}$

Soluções:

$x^2=y\\x^2=4\\x=\pm \sqrt{4} \\\boxed{x=\pm2}$

$x^{2} =-3\\x=\pm \sqrt{-3} \\\boxed{x=\pm~i \sqrt{3}}$

$\boxed{S=\{-2,~2,~-i \sqrt{3},~i \sqrt{3} \} }$


$\boldsymbol{\mathfrak{Bons~estudos~~~~:)}}$

Answer 2

x4-x2-12=0 equaçao biquadrada
      
  
            x² = b

b² - b - 12 = 0

Δ= (-1)² - 4.1.(-12) = 1+48=49               √49 = 7
 
 
 
 b = 1 +/-7 ==>  b = 1 +/-7
          2.1                   2
  
 
 b = 1 + 7 ===> b1 = 4
          2

 b = 1 - 7 ==> b2= - 3 
          2



 Comparando      x² = b temos: 

     x² = b1 ==> x² = 4 ==> b =+/- 2

      x² = b1 ==> x² = - 3   não serve pq é negativo