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emtro
@emtro
January 2021
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Bonsoir, je dois développer une inéquation que je n'arrive pas du tout ...
La voici: (x+2) au carré > ou égal à (2x-3) au carré
Merci beaucoup d'avance.
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swnn
(x+2)² ≥ (2x-3)²
x²+4x+4 ≥ 4x²-12x+9
x²+4x+4-4x²+12x-9 ≥ 0
-3x²+16x-5 ≥ 0
calcul du discriminant
δ = b²-4ac
δ = (16)²-4*(-3)*(-5)
δ = 256 - 60 = 196
δ > 0 donc l’inéquation admet deux solutions
x1: (-b-√δ)/2a = (-16-√196)/2*(-3) = (-16-14)/(2*(-3)) = -30/-6 = 5
x2: (-b+√δ)/2a = (-16+√196)/2*(-3) = (-16+14)/(2*(-3)) = -2/-6 = 2/6 = 1/3
S = {1/3;5}
2 votes
Thanks 0
emtro
Merci beaucoup !
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x²+4x+4 ≥ 4x²-12x+9
x²+4x+4-4x²+12x-9 ≥ 0
-3x²+16x-5 ≥ 0
calcul du discriminant
δ = b²-4ac
δ = (16)²-4*(-3)*(-5)
δ = 256 - 60 = 196
δ > 0 donc l’inéquation admet deux solutions
x1: (-b-√δ)/2a = (-16-√196)/2*(-3) = (-16-14)/(2*(-3)) = -30/-6 = 5
x2: (-b+√δ)/2a = (-16+√196)/2*(-3) = (-16+14)/(2*(-3)) = -2/-6 = 2/6 = 1/3
S = {1/3;5}