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Professor de Matemática no Colégio Estadual Dinah Gonçalves
 E Biologia na rede privada de Salvador-Bahia
Professor Antonio Carlos carneiro Barroso
email accbarroso@hotmail.com
Blog HTTP://ensinodematemtica.blogspot.com


Equações de 2º grau Completa:




                       −b+ ∆                           −b
                xi =         e         x i + x ii =
     − b± ∆              2a                            a
x=                     −b− ∆                       c
        2a      x ii =
                         2a
                                       x .x ii =
                                                   a


Resolvendo:
x 2 − +=
     5x 6 0
a = 1
b = − 5
c = 6
∆b 2 −
 =   4 ac
∆( 5 )
                 2
 =−                  −1.6 = − =
                      4.   25 24 1
      − b ±    ∆
x =
          2a
      − 5)
        (−±       1
x =
            2.1
     5 ±  1
x =
        2
    5 + 1     6
xi         = =    3
      2       2
       5 −  1   4
x ii =        = =   2
          2     2
S = ,3)
     ( 2
x     2
           − 15 =
            8x + 0
 a         =
           1
 b         =
           −8
 c         =
           15
 ∆ −
 = 4 ac
  b2
 ∆
 =( 8) −
   −   4.1.15
                                2
                                                                =
 ∆ − 4
 = 60 =
  64
      − b ±     ∆
 x  =
          2a

 x =
        ( 8)
      − ± −         4
            2. 1
      8 ±  2
 x =
        2
       8 +  2    10
 xi   =        =      =
                      5
          2       2
       8 −6  2
 x ii =        =3    =
          2       2
 S   =( 5)
       3,



X2-4x+4=0
▲=b2-4ac
▲=42-4.1.4=16-16=0
   − ( − 4) ± 0
                          4±0              4+0 4            4−0 4
X=       2.1         x=             x1 =      = =2   x2 =      = =2
                           2                2  2             2  2


S = ( 2)


x 2 + 2 x +1 = 0
                                           −b  ∆         − ( + 2) ± 0
∆ = b 2 − 4ac                       x=               x=
                                             2a                 2.1
∆ = ( 2 ) − 4.1.1 = 4 − 4 = 0
           2


      −2 +0 −2
x1 =          =      = −1
          2
      −2 −0 −2
                  2
                                S = ( − 1)
x2 =          =      = −1
          2       2
x 2 −4 x + =0
          5
a = , b =− , c =5
    1      4
∆=b 2 −4ac
∆=(−4 )
          2
              − .1.5 =
               4      16 −20 =−4
∆0
 〈
S =[ ]


x 2 −4 x − =0
          5
a = , b =− , c =−
    1      4     5
∆=b 2 −4ac
∆=(−4 ) −4.1.(− ) =
               5   16 +20 =36
      − ± ∆
       b
x =
         2a
      −(−4 ) ± 36     4 ±6
x   =               =
           2 .1         2
       4 +6     10
x    =        =    =5
         2       2
       4 −6     −2
x    =        =    =− 1
         2       2
S   =(− ,5)
        1

4x    −4 x + =0
            1
a =4, b =− , c =1
          4
∆=b 2 −4ac =( −4 ) −4.4.1 =16 −
                    2
                               16 =0
   −b ± ∆   −( −4 ) ± 0   4 ±0  4  1
x =       =             =      = =
      2a        2.4         8   8  2
   1 
S = 
   2 
x 2 − x + = → = 2,3)
     5   6 0 S (
x 2 − x + = → = 1,5 )
     6   5 0 S (
x 2 − x − = → = − ,3)
     2   3 0 S ( 1
x 2 − x − = → = − ,5 )
     4   5 0 S ( 1
x 2 − x + = → = 1,6 )
     7   6 0 S (
x2 − x +
    7   10 = → = 2,5 )
            0 S (
x 2 + x + = → = − ,− )
     7   6 0 S ( 1 6
x2 − x + = → =
    4   6 0 S [                             ]
x 2 − x + = → = 1,2 )
     3   2 0 S (
x 2 + x + = → (− ,− )
     3   2 0 S  1 2
x 2 − x + = → = 2)
     4   4 0 S (
x2 − x +
    8   15 = → = 3,5)
            0 S (
x2 − x +
    9   14 = → = 2,7 )
            0 S (
x2 −10 +25 = → = 5 )
            0 S (
x2 + x +
    11  30 = → = − ,− )
            0 S ( 5 6
− 2 + x+
 x   3  10 = → = 2,5)
            0 S (
                   4
3 x 2 − x + = → = 1, 
       7   4  0 S 
                   3
x2 + x −
      4   21 = → = 3,− )
              0 S (   7
x2 + x +
    8   16 = → = − )
            0 S ( 4
3x 2 − x +
      2   24 = → =
              0 S [                             ]
x2 − x +
    10  24 = → = 4,6 )
            0 S (
x 2 − x + = → = 1,3)
     4   3 0 S (
x2 − x −
    4   12 = → = − ,6 )
            0 S ( 2


Professor Antonio Carlos Carneiro Barroso
Equção de 2º grau completa  autor professor antonio carneiro

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Equção de 2º grau completa autor professor antonio carneiro

  • 1. Professor de Matemática no Colégio Estadual Dinah Gonçalves E Biologia na rede privada de Salvador-Bahia Professor Antonio Carlos carneiro Barroso email accbarroso@hotmail.com Blog HTTP://ensinodematemtica.blogspot.com Equações de 2º grau Completa: −b+ ∆ −b xi = e x i + x ii = − b± ∆ 2a a x= −b− ∆ c 2a x ii = 2a x .x ii = a Resolvendo: x 2 − += 5x 6 0 a = 1 b = − 5 c = 6 ∆b 2 − = 4 ac ∆( 5 ) 2 =− −1.6 = − = 4. 25 24 1 − b ± ∆ x = 2a − 5) (−± 1 x = 2.1 5 ± 1 x = 2 5 + 1 6 xi = = 3 2 2 5 − 1 4 x ii = = = 2 2 2 S = ,3) ( 2
  • 2. x 2 − 15 = 8x + 0 a = 1 b = −8 c = 15 ∆ − = 4 ac b2 ∆ =( 8) − − 4.1.15 2 = ∆ − 4 = 60 = 64 − b ± ∆ x = 2a x = ( 8) − ± − 4 2. 1 8 ± 2 x = 2 8 + 2 10 xi = = = 5 2 2 8 −6 2 x ii = =3 = 2 2 S =( 5) 3, X2-4x+4=0 ▲=b2-4ac ▲=42-4.1.4=16-16=0 − ( − 4) ± 0 4±0 4+0 4 4−0 4 X= 2.1 x= x1 = = =2 x2 = = =2 2 2 2 2 2 S = ( 2) x 2 + 2 x +1 = 0 −b  ∆ − ( + 2) ± 0 ∆ = b 2 − 4ac x= x= 2a 2.1 ∆ = ( 2 ) − 4.1.1 = 4 − 4 = 0 2 −2 +0 −2 x1 = = = −1 2 −2 −0 −2 2 S = ( − 1) x2 = = = −1 2 2
  • 3. x 2 −4 x + =0 5 a = , b =− , c =5 1 4 ∆=b 2 −4ac ∆=(−4 ) 2 − .1.5 = 4 16 −20 =−4 ∆0 〈 S =[ ] x 2 −4 x − =0 5 a = , b =− , c =− 1 4 5 ∆=b 2 −4ac ∆=(−4 ) −4.1.(− ) = 5 16 +20 =36 − ± ∆ b x = 2a −(−4 ) ± 36 4 ±6 x = = 2 .1 2 4 +6 10 x = = =5 2 2 4 −6 −2 x = = =− 1 2 2 S =(− ,5) 1 4x −4 x + =0 1 a =4, b =− , c =1 4 ∆=b 2 −4ac =( −4 ) −4.4.1 =16 − 2 16 =0 −b ± ∆ −( −4 ) ± 0 4 ±0 4 1 x = = = = = 2a 2.4 8 8 2 1  S =  2 
  • 4. x 2 − x + = → = 2,3) 5 6 0 S ( x 2 − x + = → = 1,5 ) 6 5 0 S ( x 2 − x − = → = − ,3) 2 3 0 S ( 1 x 2 − x − = → = − ,5 ) 4 5 0 S ( 1 x 2 − x + = → = 1,6 ) 7 6 0 S ( x2 − x + 7 10 = → = 2,5 ) 0 S ( x 2 + x + = → = − ,− ) 7 6 0 S ( 1 6 x2 − x + = → = 4 6 0 S [ ] x 2 − x + = → = 1,2 ) 3 2 0 S ( x 2 + x + = → (− ,− ) 3 2 0 S 1 2 x 2 − x + = → = 2) 4 4 0 S ( x2 − x + 8 15 = → = 3,5) 0 S ( x2 − x + 9 14 = → = 2,7 ) 0 S ( x2 −10 +25 = → = 5 ) 0 S ( x2 + x + 11 30 = → = − ,− ) 0 S ( 5 6 − 2 + x+ x 3 10 = → = 2,5) 0 S (  4 3 x 2 − x + = → = 1,  7 4 0 S   3 x2 + x − 4 21 = → = 3,− ) 0 S ( 7 x2 + x + 8 16 = → = − ) 0 S ( 4 3x 2 − x + 2 24 = → = 0 S [ ] x2 − x + 10 24 = → = 4,6 ) 0 S ( x 2 − x + = → = 1,3) 4 3 0 S ( x2 − x − 4 12 = → = − ,6 ) 0 S ( 2 Professor Antonio Carlos Carneiro Barroso