Problème mathématiques

[William 1]

Soit a et b deux réel tel que:
a² + b² =5 et a³ +b³ =7
Trouver les trois valeur possible de a+b.

[Noemi]

@William-1 Bonsoir (Marque de politesse à ne pas oublier !!)

Utilise le développement de :
$(a+b{)}^2$
$(a+b{)}^3$
et pose $a+b=S$ et $a\times b=P$

résous ensuite le système.

[keyboard]

@Noemi
(a + b)² = a² + 2ab + b²
(a + b)² = a² + b² + 2ab → given that: a² + b² = 5
(a + b)² = 5 + 2ab

(a + b)³ = (a + b)².(a + b)
(a + b)³ = (a² + 2ab + b²).(a + b)
(a + b)³ = a³ + a²b + 2a²b + 2ab² + ab² + b³
(a + b)³ = a³ + b³ + 3a²b + 3ab²
(a + b)³ = (a³ + b³) + 3ab.(a + b)
(a + b)³ - 3ab.(a + b) = a³ + b³ → given that: a³ + b³ = 7
(a + b)³ - 3ab.(a + b) = 7
(a + b)².(a + b) - 3ab.(a + b) = 7 → recall: (a + b)² = 5 + 2ab
(5 + 2ab).(a + b) - 3ab.(a + b) = 7
(a + b).[(5 + 2ab) - 3ab] = 7
(a + b).(5 + 2ab - 3ab) = 7
(a + b).(5 - ab) = 7
(a + b) = 7/(5 - ab) → you square
(a + b)² = 49/(5 - ab)² → recall: (a + b)² = 5 + 2ab
5 + 2ab = 49/(5 - ab)²
(5 + 2ab).(5 - ab)² = 49
(5 + 2ab).(25 - 10ab + a²b²) = 49 → let: ab = x
(5 + 2x).(25 - 10x + x²) = 49
125 - 50x + 5x² + 50x - 20x² + 2x³ = 49
2x³ - 15x² + 76 = 0
2x³ - (19x² - 4x²) + (38x - 38x) + 76 = 0
2x³ - 19x² + 4x² + 38x - 38x + 76 = 0
2x³ - 19x² + 38x + 4x² - 38x + 76 = 0
(2x³ - 19x² + 38x) + (4x² - 38x + 76) = 0
x.(2x² - 19x + 38) + 2.(2x² - 19x + 38) = 0
(x + 2).(2x² - 19x + 38) = 0

First:
(x + 2) = 0
x = - 2 → recall: ab = x
ab = - 2

Second:
2x² - 19x + 38 = 0
x² - (19/2).x + 19 = 0
x² - (19/2).x + (19/4)² - (19/4)² + 19 = 0
x² - (19/2).x + (19/4)² - (361/16) + 19 = 0
[x - (19/4)]² - (361/16) + 19 = 0
[x - (19/4)]² - (57/16) = 0
[x - (19/4)]² = 57/16
x - (19/4) = ± (√57)/4
x = (19/4) ± (√57)/4
x = (19 ± √57)/4 → recall: ab = x
ab = (19 ± √57)/4

3 cases to be study
First case: ab = - 2
Second case: ab = (19 + √57)/4
Third case: ab = (19 - √57)/4

Restart
(a + b)² = 5 + 2ab → where first case: ab = - 2
(a + b)² = 5 - 4
(a + b)² = 1
a + b = ± 1 → where first case: ab = - 2 → b = - 2/a
a - (2/a) = ± 1
a² - 2 = ± a
a² ± a = 2
a² ± a + (1/2)² - (1/2)² = 2
a² ± a + (1/2)² = 2 + (1/2)²
[a ± (1/2)]² = 9/4
a ± (1/2) = ± 3/2
a = ± (1/2) ± (3/2)

First solution: a = (1/2) + (3/2) → a = 2 → b = - 1
Second solution: a = (1/2) - (3/2) → a = - 1 → b = 2
Third solution: a = - (1/2) + (3/2) → a = 1 → b = - 2
Fourth solution: a = - (1/2) - (3/2) → a = - 2 → b = 1

Restart
(a + b)² = 5 + 2ab → where second case: ab = (19 + √57)/4
(a + b)² = 5 + [(19 + √57)/2]
(a + b)² = (10 + 19 + √57)/2
(a + b)² = (29 + √57)/2
a + b = ± √[(29 + √57)/2] → where second case: ab = (19 + √57)/4 → b = (19 + √57)/4a
a + [(19 + √57)/4a] = ± √[(29 + √57)/2]
4a² + (19 + √57) = ± 4a√[(29 + √57)/2]
4a² ± 4a√[(29 + √57)/2] + (19 + √57) = 0
…to be continued