Question

determinar o x tal

  x+3  5
  1     x-1  = 0 

Answer 1

Diagonal principal:
(x + 3)(x - 1) = x² - x + 3x - 3 = x² + 2x - 3

Diagonal secundária:
5.1 = 5

Det = x² + 2x - 3 - 5
Det = x² + 2x - 8

Δ = 2² - 4.1.(-8)
Δ = 4 + 32
Δ = 36

x = -2 +/- √36 / 2.1 = -2 +/- 6 / 2

x' = -2 + 6 / 2 = 4 / 2 = 2
x''= -2 - 6 / 2 = -8 / 2 = -4

Det = x² + 2x - 8

Det' = 2² + 2.2 - 8
Det' = 4 + 4 - 8
Det' = 0

Det'' = (-8)² + 2.(-8) - 8
Det'' = 64 - 16 - 8
Det'' = 40.

Answer 2

  x+3  5
  1     x-1  = 0 

(x+3)(x-1)-5=0
x¨2-x+3x-3-5=0
 x¨2 +2x - 8 = 0

delta=2¨2-4.1.(-8)=4+32=36

x= -2+/-V36 = -2+/-6 ==> x1=-2+6 => x1=2  ;x2=-2-6  ==> x2=- 4
          2.1             2                       2                            2