Question

x4-16x²=0

 

x4-8x²+16=0

Answer 1

Olá, Giulia.

$x^4-16x^2=0\Rightarrow(x^2+4x)(x^2-4x)=0\Rightarrow\\\\x(x+4)\cdot x(x-4)=0\Rightarrow x^2 (x+4) (x-4)=0\Rightarrow \\\\\boxed{x=0,x=-4\text{ ou }x=4}$


$x^4-8x^2+16=0\Rightarrow(x^2-4)^2=0\Rightarrow\\\\\ [(x+2)(x-2)]^2=0\Rightarrow\boxed{x=-2\text{ ou }x=2}$