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Carabou
@Carabou
April 2019
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Bonjour,
Comment résolvez-vous l'équation suivante ?
C(x) = 2x^2 + 232x + 480 = 3000
Sachant que le but de l'exercice est de trouver la valeur de x.
Ca fait des heures que je cherche en tournant en rond, si quelqu'un aurait la réponse, cela m'aiderait beaucoup, merci.
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Thomas1106
2x²+232x+480 = 3000
2x²+232x+480
-3000
= 3000
-3000
2x²+232x-2520 = 0
Δ = b²-4ac
Δ = (232)²-4*2*(-2520)
Δ = 53824+20160
Δ = 73984
√Δ = 272
x₁ =
(-b-√Δ)/2a = (-232-272)/4 = -504/4 =
-126
x₂ =
(-b+√Δ)/2a = (-232+272)/4 = 40/4 =
10
1 votes
Thanks 1
Commentaires
2x² + 232x + 480 = 3 000
2x² + 232x + 480 - 3 000 = 0
2x² + 232x - 2 520 = 0
x² + 116x - 1260 = 0
Δ = b² - ac
Δ = 116² - 4 x 1 x (- 1260)
Δ = 13 456 + 5 040
Δ = 18 496
On a Δ > 0 donc l'équation a deux racines réelles distinctes.
1 votes
Thanks 1
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2x²+232x+480-3000 = 3000-3000
2x²+232x-2520 = 0
Δ = b²-4ac
Δ = (232)²-4*2*(-2520)
Δ = 53824+20160
Δ = 73984
√Δ = 272
x₁ = (-b-√Δ)/2a = (-232-272)/4 = -504/4 = -126
x₂ = (-b+√Δ)/2a = (-232+272)/4 = 40/4 = 10
2x² + 232x + 480 - 3 000 = 0
2x² + 232x - 2 520 = 0
x² + 116x - 1260 = 0
Δ = b² - ac
Δ = 116² - 4 x 1 x (- 1260)
Δ = 13 456 + 5 040
Δ = 18 496
On a Δ > 0 donc l'équation a deux racines réelles distinctes.