CHAPITRE

1

Second degré
Entraînement (page 32)

EXERCICES

Ά

DE TÊTE

(2x – 3) (8 + 2x) = 0 ; ᏿ = – 4 ;

50 2[(x – 3)2 – 4] = 0.

39 1. A(1) ≠ 0.
Donc 1 n’est pas solution de l’équation A(x) = 0.
2. A(x) > 0.

c) A(12) > 0.

ÉQUATIONS DU SECOND DEGRÉ
PARTICULIÈRES

Ά

·

1
(x + 3) (2x + 1) = 0 ; ᏿ = – 3 ; –
.
2

42 [(x – 1) + (2 – 3x)] [(x – 1) – (2 – 3x)] = 0.
(– 2x + 1) (4x – 3) = 0 ; ᏿ =

2[(x – 3 – 2) (x – 3 + 2)] = 0.
2(x – 5) (x – 1) = 0 ; ᏿ = {1 ; 5}.

51 a) [3 – 2(x – 2)] [3 + 2(x – 2)] = 0.

40 1. (x – 1) (x + 2) = A(x).
2. ᏿ = {– 2 ; 1}.
3. a) A(0) < 0.
b) A(– 3) > 0.

41

·

3
.
2

Ά 1 ; 3 ·.
2 4

43 (x – 4) [(x – 4) + (x + 2)] = 0.

(x – 4) (2x – 2) = 0 ; ᏿ = {1 ; 4}.

(– 2x + 7) (2x – 1) = 0 ; ᏿ =
b) ᏿ = Ø.

Ά 1 ; 7 ·.
2 2

52 a) [512 – (2x + 3)] [512 + (2x + 3)] = 0.
(– 2x + 512 – 3) (2x + 512 + 3) = 0 ;
512 – 3 – 512 – 3
᏿=
;
.
2
2
b) ᏿ = Ø.

Ά

·

DEUX MÉTHODES DE RÉSOLUTION
53 • [2x + (x – 2)] [2x – (x – 2)] = 0.

Ά

(3x – 2) (x + 2) = 0 ; ᏿ = – 2 ; –
2

2

·

2
.
3

(x + 2) [(x – 2) – 3(x + 2)] = 0.
(x + 2) (– 2x – 8) = 0 ; ᏿ = {– 4 ; – 2}.

• 4x – (x – 4x + 4) = 0.
3x2 + 4x – 4 = 0.
Δ = 64, deux solutions :
–4 – 8
–4 + 8 2
= – 2 et x2 =
= . x1 =
6
6
3
2
᏿ = – 2; –
.
3
54 • (x + 3) (1 – x) = 0 ; ᏿ = {– 3 ; 1}.
• x + 3 – x2 – 3x = 0.
– x2 – 2x + 3 = 0.
Δ = 16, deux solutions :
2–4
2+4 x1 =
= 1 et x2 =
= – 3.
–2
–2
᏿ = {– 3 ; 1}.

48 3(x2 – 4x + 4) = 0.

55 • 2(5 + 2 – 3x) (5 – 2 + 3x) = 0.

44 2(9x2 + 6x + 1) = 0.

Ά 1 ·.
3

2(3x + 1)2 = 0 ; ᏿ = –

45 (2 + 5x) [(3x) – (x + 1)] = 0.

Ά

·

2 1
;
.
5 2
46 2x(x + 3) – (x + 3) (x – 3) = 0.
(x + 3) [2x – (x – 3)] = 0.
(x + 3) (x + 3) = 0 ; ᏿ = {– 3}.
(2 + 5x) (2x – 1) = 0 ; ᏿ = –

47 (x + 2) (x – 2) – 3(x + 2)2 = 0.

3(x – 2)2 = 0 ; ᏿ = {2}.

49 5(2x – 3) – (3 + 2x) (3 – 2x) = 0.

Ά

·

Ά

2(7 – 3x) (3 + 3x) = 0 ; ᏿ = – 1 ;
• 50 – 2(4