Bonjour veuillez m’aider s’il vous plaît. Voici la fonction f : f(x) = 3(x-1) (x+1) + x (x-4) résoudre les équations suivantes : a) f(x) = -4 b) f(x) = 0 c) f(x

bonsoir

f(x) = 3 ( x - 1) ( x + 1) + x ( x - 4 )

f (x) = ( 3 x - 3 ) ( x + 1 ) + ( x² - 4 x )

f (x) = 3 x² + 3 x - 3 x - 3 + x² - 4 x

f (x) = 4 x² - 4 x - 3

f(x) = - 4

4 x² - 4 x - 3 = - 4

4 x² - 4 x - 3 + 4 = 0

4 x² - 4 x + 1  = 0

( 2 x - 1 )²  = 0  ⇔ x = 1/2

f(x) = 0

4 x² - 4 x - 3 = 0

Δ  ( -4)² - 4 ( 4 * -3) = 16 + 48 = 64 = 8 ²

x 1 = ( 4 + 8) / 8 = 12/8 = 3/2

x 2 = ( 4 - 8 ) / 8 = -4/8 = - 1 /2

f(x) = - 3

4 x² - 4 x - 3 = - 3

4 x² - 4 x - 3 + 3 = 0

4 x² - 4 x = 0

4 x ( x - 1 ) = 0

x = 0 ou 1

f(x) = 12

4 x² - 4 x - 3 = 12

4 x² - 4 x - 3 - 12 = 0

4 x² - 4 x - 15 = 0

Δ = ( -4)² - 4 ( 4 * - 15) = 16 + 240 = 256 = 16 ²

x1 = ( 4 + 16) / 8 = 20/8 = 5/2

x2 = ( 4 - 16) /8 = - 12/8 = - 3/2

fais le dernier