Question

Quais são as raizes de 2 x³- 6 x²+ 28 x – 24

Answer 1

• de acordo com o enunciado vem:

 2x³- 6x²+ 28x - 24 = 0

 x³- 3x² + 14x - 12 = 0

•  raiz trivial

  x = 1

• divisão por x - 1

 (x³- 3x² + 14x - 12)/(x - 1) = x² - 2x + 12

  x² - 2x + 12 = 0

  delta

  d = 4 - 48 = -44

  delta negativo nao ha raiz real

  S = (1)

Answer 2

Resposta:

2x³- 6x²+ 28 x -24=0

2x³- 4x²-2x²+ 24 x+4x -24=0

2x³- 4x²+24x- 2x²+ 4x -24=0

2x*(x²-2x+12) -2*(x²-2x+12)=0

(x²-2x+12)*(2x-2)=0

(x²-2x+12)*2*(x-1)=0

(x²-2x+12)*(x-1)=0

x-1=0   ==> x'=1

x²-2x+12=0

x''=[2+√(4-48)]/2= [2+√(-44)]/2 =(2+2i√11)/2=1+i√11

x'''=[2-√(4-48)]/2= [2-√(-44)]/2 =(2-2i√11)/2=1-i√11