Bonjour pouvez vous m'aider a resoudre ces 2 equations : a. (3x+1)(4x-2)=0 et b. (3x+1)(4x-2)=-2. Merci !!

Bonsoir

(3x + 1) (4x - 2) = 0
<=> 3x + 1 = 0      OU      4x - 2 = 0
<=> 3x = -1      OU      4x = 2
<=> x = -1 / 3      OU      x = 2 / 4
<=> x = -1 / 3      OU      x = 1 / 2

Donc S = {-1 / 3 ; 1 / 2}

(3x + 1) (4x - 2) = -2
<=> (3x + 1) (4x - 2) + 2 = 0
<=> (3x * 4x + 3x * (-2) + 1 * 4x + 1 * (-2)) +  2 = 0
<=> (12x² - 6x + 4x - 2) + 2 = 0
<=> 12x² - 2x - 2 + 2 = 0
<=> 12x² - 2x = 0
<=> 2x * (6x - 1) = 0                         => produit nul
<=> 2x = 0      OU      6x - 1 = 0
<=> x = 0      OU      6x = 1
<=> x = 0      OU      x = 1 / 6

Donc S = {0 ; 1 / 6}

Voilà j'espère avoir pu t'aider ^^