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Syssou
@Syssou
January 2021
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Bonjour j'ai besoin d'aide c'est urgent. Merci à ce qui répondrons ;)
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grr12
Il y a deux solutions et non pas un intervalle de solutions , car le discriminant Δ = b² - 4ac = (-3.9)² − 4×1×3.8 = 0.0099999999999998
Δ > 0 alors l'équation x² − 3.9x + 3.8 = 0 admet 2 solutions réelles x1 et x2
SOLUTIONS dans ℜ :
x1 = (-b − √Δ)/2a = (3.9 − √(0.0099999999999998)) / (2) et x2 = (-b + √Δ)/2a = (3.9 + √(0.0099999999999998)) / (2)
x1 = 1.9 et x2 = 2
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Il y a deux solutions et non pas un intervalle de solutions , car le discriminant Δ = b² - 4ac = (-3.9)² − 4×1×3.8 = 0.0099999999999998
Δ > 0 alors l'équation x² − 3.9x + 3.8 = 0 admet 2 solutions réelles x1 et x2
SOLUTIONS dans ℜ :
x1 = (-b − √Δ)/2a = (3.9 − √(0.0099999999999998)) / (2) et x2 = (-b + √Δ)/2a = (3.9 + √(0.0099999999999998)) / (2)
x1 = 1.9 et x2 = 2