Explications étape par étape :
a) x² +3x -10 = 0
∆ = b² - 4ac =
Δ = 3² - 4· 1 · (-10) = 9 + 40 = 49
Δ > 0 2 solutions
x₁ = ( -3 - √49 ) / ( 2·1 ) = ( -3 - 7 ) / 2 = -10 / 2 = -5
x₂ = ( -3 + √49 ) / ( 2·1 ) = ( -3 + 7 ) / 2 = 4/2 = 2
b) -x² + 6x - 9 = 0
∆ = b² - 4ac
Δ = 6² - 4 · (-1) · (-9) = 36 - 36 = 0
Une solution unique:
x = -6 / ( 2 · -1 )
⇔ x = -6 / -2
⇔ x = 3
c) 5x - 4x + 1 = 0
∆ = b² - 4ac
Δ = (-4)² - 4· ( 5 · 1 ) = 16 - 20 = -4
Δ < 0 Pas de solution réelle.
d) 2x²+ 12x = 14
⇔ 2x² + 12x - 14 = 0
⇔ x² + 6x - 7 = 0
∆ = b² - 4ac
Δ = 6² - 4 · 1 · (-7) = 36 + 28 = 64
Δ > 0 2 solutions
x₁ = ( - 6 - √64 ) / 2 = ( -6 - 8 ) / 2 = -14/2 = -7
x₂ = ( - 6 + √64 ) / 2 = ( -6 + 8 ) / 2 = 2/2 = 1
e) -3x² = -6x + 3
⇔ -3x² + 6x - 3 = 0
∆ = b² - 4ac
Δ = 6² - 4· (-3) · (-3) = 36 - 36 = 0
Une solution unique:
x = -6 / (2 · -3)
⇔ x = -6 / -6
⇔ x = 1
