bonjour
f (x) = ( 3 x + 1 ) ( x - 4 )
f (x) = 3 x² - 12 x + x - 4
f (x) = 3 x² - 11 x - 4
f (x) = 0 pour x = - 1/3 et 4
f (x) = - 4 ⇒ 3 x² - 11 x - 4 = - 4 ⇒ 3 x² - 11 x = 0
x ( 3 x - 11) = 0 ⇔ x = 0 et 11/3
g (x) = - 2 x² - 6
g (x) = 0 ⇔ -2 x² - 6 = 0 ⇒ - 2 x² = 6 ⇔ pas de solution dans R
g (x) = - 6
- 2 x² - 6 = - 6
- 2 x² = - 6 + 6
- 2 x² = 0 ⇔ x = 0
g (x) = - 8
- 2 x² - 6 = - 8
- 2 x² = - 8 + 6
- 2 x² = - 2
x² = 1 ⇔ x = √1 ou - √1
f (x) = g ( x )
( 3 x + 1 ) ( x- 4 ) = - 2 x² - 6
3 x² - 12 x + x - 4 = - 2 x² - 6
3 x² + 2 x² - 11 x = - 6 + 4
5 x² - 11 x = - 2
5 x² - 11 x +2 = 0
Δ = 121 - 40 = 81
x 1 = ( 11 - 9 ) / 10 = 2/10 = - 1/5
x 2 = ( 11 + 9) /10 = 20 /10 = 2
S ( - 1/5 : 2 )
