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Leasa
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January 2021
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Svp je comprends pas
Résoudre x^2 -3x+2 < ( x-2)(3x+4)
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Aftershock
x² -3x+2 < (x-2)(3x+4)
x²-3x+2 < 3x²+4x-6x-8
-2x²-x+10 < 0
Δ = b²-4ac
Δ = (-1)²-4*(-2)*10
Δ = 1-(-80)
Δ = 81
x1 = (-b+
√
Δ)/2a
x1= (-(-1)+
√
81)/(2*(-2))
x1 = (1+9)/-4
x1 = -2,5
x2 =
(-b-√Δ)/2a
x2 = (-(-1)-√
81)/(2*(-2))
x2 = (1-9)/(-4)
x2 = 2
Les 2 racines sont donc S = {-2,5 ; 2}
Or, on sait que dans un polynôme du second degré de la forme
ax²+bx+c, si le coefficient a est négatif, alors la courbe est croissante, puis décroissante
Donc,
-2x²-x+10 est inférieur à 0 sur l'intervalle ]-
∞ ; -2,5[ ∪ ]2 ; +∞[
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x²-3x+2 < 3x²+4x-6x-8
-2x²-x+10 < 0
Δ = b²-4ac
Δ = (-1)²-4*(-2)*10
Δ = 1-(-80)
Δ = 81
x1 = (-b+√Δ)/2a
x1= (-(-1)+√81)/(2*(-2))
x1 = (1+9)/-4
x1 = -2,5
x2 = (-b-√Δ)/2a
x2 = (-(-1)-√81)/(2*(-2))
x2 = (1-9)/(-4)
x2 = 2
Les 2 racines sont donc S = {-2,5 ; 2}
Or, on sait que dans un polynôme du second degré de la forme
ax²+bx+c, si le coefficient a est négatif, alors la courbe est croissante, puis décroissante
Donc, -2x²-x+10 est inférieur à 0 sur l'intervalle ]-∞ ; -2,5[ ∪ ]2 ; +∞[