bonjour
f (x) = - x² + 2 x - 3
Δ = ( - 2 )² - 4 ( - 1 * - 3 )= 4 - 12 = - 8
Δ < 0 ⇔pas de solutions
x² + x + 1/4 = 0
Δ = 1 ² - 4 ( 1 * 1/4 ) = 1 - 4/4 = 1 - 1 = 0
Δ = 0 ⇔ 1 seule solution
x ₀ = - 1 /2
2 x² + 2 x - 12 = 0
Δ = 4 - 4 ( 2 * - 12 ) = 4 + 96 = 100
x 1 = ( - 2 - √100) / 4 = ( - 2 - 10 ) / 4 = - 12/4 = - 3
x 2 = ( - 2 + √100) / 4 = ( - 2 + 10 ) /4 = 6/4 = 3 /2
idem pour la suite
