2x² – 2x – 4 = 0

7
2x² – 2x – 4 = 0 a = b = c = 2 – 2 – 4 Calcul du discriminant : ac b 4 2 ) 4 ( 2 4 ) 2 ( 2 = 36 > 0, donc l’équation admet deux solutions 4 4 4 6 2 2 2 36 2 2 1 1 x a b x 4 8 4 6 2 2 2 36 2 2 2 2 x a b x x 1 = 1 x 2 = 2

description

2x² – 2x – 4 = 0. Calcul du discriminant :. a = b = c =. 2 – 2 – 4. D = 36. D > 0, donc l’équation admet deux solutions. x 2 = 2. x 1 = – 1. x² + 4x + 3 = 0. Calcul du discriminant :. a = b = c =. 1 4 3. D = 4. - PowerPoint PPT Presentation

Transcript of 2x² – 2x – 4 = 0

Page 1: 2x²  –  2x  –  4 = 0

2x² – 2x – 4 = 0

a =

b =

c =

2

– 2

– 4

Calcul du discriminant :

acb 42 )4(24)2( 2

= 36

> 0, donc l’équation admet deux solutions

4

4

4

62

22

362

2

1

1

x

a

bx

4

8

4

62

22

362

2

2

2

x

a

bx

x1 = – 1 x2 = 2

Page 2: 2x²  –  2x  –  4 = 0

x² + 4x + 3 = 0

a =

b =

c =

1

4

3

Calcul du discriminant :

acb 42 31442

= 4

> 0, donc l’équation admet deux solutions

2

6

2

24

12

44

2

1

1

x

a

bx

2

2

2

24

12

44

2

2

2

x

a

bx

x1 = – 3 x2 = – 1

Page 3: 2x²  –  2x  –  4 = 0

6x² + 6x – 12 = 0

a =

b =

c =

6

6

–12

Calcul du discriminant :

acb 42 31442

= 324

> 0, donc l’équation admet deux solutions

12

24

12

186

62

3246

2

1

1

x

a

bx

12

12

12

186

62

3246

2

2

2

x

a

bx

x1 = – 2 x2 = 1

Page 4: 2x²  –  2x  –  4 = 0

x² + 8x + 16 = 0

a =

b =

c =

1

8

16

Calcul du discriminant :

acb 42 161482

= 0

= 0, donc l’équation admet une solution

2

8

12

82

x

a

bx

x = – 4

Page 5: 2x²  –  2x  –  4 = 0

x² – 6x + 7 = –2

a =

b =

c =

1

– 6

9

Calcul du discriminant :

acb 42 914)6( 2

= 0

= 0, donc l’équation admet une solution

2

6

12

62

x

a

bx

x = 3

x² – 6x + 9 = 0

différent de zéro !!!

Page 6: 2x²  –  2x  –  4 = 0

– 2x² +3x – 4 = 0

a =

b =

c =

– 2

3

– 4

Calcul du discriminant :

acb 42 )4()2(432

= – 23

< 0, donc l’équation n’admet pas de solution

Page 7: 2x²  –  2x  –  4 = 0

4x² – x + 1 = 0

a =

b =

c =

4

– 1

1

Calcul du discriminant :

acb 42 144)1( 2

= – 15

< 0, donc l’équation n’admet pas de solution