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Adeline15
@Adeline15
May 2019
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Bonjour pouvez vous m aider
F et G sont les fonctions définies sur R par f(x) = 2x(x − 1) et g(x) = −3x + 3.
1. Résoudre algébriquement (c’est-à-dire par le calcul) l’équation f(x) = g(x).
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choupp
F(x) = g(x)
2x(x-1) = -3x + 3
2x² - 2x = -3x + 3
2x² + x - 3 = 0
Δ = 1² - 4*2*(-3) = 1 + 24 = 25
L'équation admet deux solutions réelles distinctes :
=
=
=
= 1
Donc
S = {-3/2 ; 1}
1 votes
Thanks 0
choupp
Delta
choupp
(C'est le discriminant du trinôme)
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Bonjour
2 x( x - 1) = - 3 x + 3
2 x² - 2 x = - 3 x + 3
2 x² - 2 x + 3 x - 3 = 0
2 x² + x - 3 = 0
ton delta = - b² + 4 ac
Δ = 1² + 24 = 25
Δ positif donc 2 racines
( - b² - √Δ)/ 2 a =( - 1 - √25) 4 = - 6/4 =
- 3/2
et ( - b² + √Δ) / 2 a = ( - 1 + √25)/4 = 4/4 =
1
1 votes
Thanks 0
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2x(x-1) = -3x + 3
2x² - 2x = -3x + 3
2x² + x - 3 = 0
Δ = 1² - 4*2*(-3) = 1 + 24 = 25
L'équation admet deux solutions réelles distinctes :
=
= = = 1
Donc S = {-3/2 ; 1}
2 x( x - 1) = - 3 x + 3
2 x² - 2 x = - 3 x + 3
2 x² - 2 x + 3 x - 3 = 0
2 x² + x - 3 = 0
ton delta = - b² + 4 ac
Δ = 1² + 24 = 25
Δ positif donc 2 racines
( - b² - √Δ)/ 2 a =( - 1 - √25) 4 = - 6/4 = - 3/2
et ( - b² + √Δ) / 2 a = ( - 1 + √25)/4 = 4/4 = 1