Réponse :
Explications étape par étape
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Résoudre :
x^4 - 8x^2 + 12 = 0
On pose : X = x^2
X^2 - 8X + 12 = 0
X1 = (8 - 4)/(2 * 4) = 4/8 = 1/2
X2 = (8 + 4)/(2 * 4) = 12/8 = 3/2
x^2 = 1/2
x = 1/V2 ou x = -1/V2
x^2 = 3/2
x = V3/V2 ou x = -V3/V2
x^4 + x^3 - x - 1 = 0
On voit qu’il y a une racine évidente : x = 1
x^4 + x^3 - x - 1 = (x - 1)(ax^3 + bx^2 + cx + d)
x^4 + x^3 - x - 1 = ax^4 + bx^3 + cx^2 + dx - ax^3 - bx^2 - cx - d
x^4 + x^3 - x - 1 = ax^4 + (b - a)x^3 + (c - b)x^2 + (d - c)x - d
a = 1
b - a = 1 => b = 1 + a = 1 + 1 = 2
c - b = 0 => c = b = 2
d - c = -1 => d = -1 + c = -1 + 2 = 1
-d = -1 => d = 1
(x - 1)(x^3 + 2x^2 + 2x + 1) = 0
x = -1 est une racine évidente de polynôme de degré 3 :
x^3 + 2x^2 + 2x + 1 = (x + 1)(ax^2 + bx + c)
x^3 + 2x^2 + 2x + 1 = ax^3 + bx^2 + cx + ax^2 + bx + c
x^3 + 2x^2 + 2x + 1 = ax^3 + (a + b)x^2 + (b + c)x + c
a = 1
a + b = 2 => b = 2 - a = 2 - 1 = 1
b + c = 2 => c = 2 - b = 2 - 1 = 1
c = 1
(x - 1)(x + 1)(x^2 + x + 1) = 0
les solutions sont : 1 et -1
x^4 - 4x^3 + 4x - 1 = 0
Une racine évidente 1 donc
x^4 - 4x^3 + 4x - 1 = (x - 1)(ax^3 + bx^2 + cx + d)
x^4 - 4x^3 + 4x - 1 = ax^4 + bx^3 + cx^2 + dx - ax^3 - bx^2 - cx - d
x^4 - 4x^3 + 4x - 1 = ax^4 + (b - a)x^3 + (c - b)x^2 + (d - c)x - d
a = 1
b - a = -4 => b = -4 + a = -4 + 1 = -3
c - b = 0 => c = 0 + b = 0 - 3 = -3
d - c = 4 => d = 4 + c = 4 - 3 = 1
-d = -1 => d = 1
(x - 1)(x^3 - 3x^2 - 3x + 1) = 0
Racine évidente : -1
x^3 - 3x^2 - 3x + 1 = (x + 1)(ax^2 + bx + c)
x^3 - 3x^2 - 3x + 1 = ax^3 + bx^2 + cx + ax^2 + bx + c
x^3 - 3x^2 - 3x + 1 = ax^3 + (b + a)x^2 + (b + c)x + c
a = 1
a + b = -3 => b = -3 - a = -3 - 1 = -4
b + c = -3 => c = -3 - b = -3 + 4 = 1
c = 1
(x - 1)(x + 1)(x^2 - 4x + 1) = 0
[tex]\Delta = (-4)^{2} - 4 * 1 * 1 = 16 - 4 = 12[/tex]
[tex]\sqrt{\Delta} = 2V3[/tex]
x1 = (4 - 2V3)/2 = 2 - V3
x2 = (4 + 2V3)/2 = 2 + V3
(x - 1)(x + 1)(x - 2 + V3)(x - 2 - V3) = 0
S = {1 ; -1 ; 2 - V3 ; 2 + V3}
x^4 - 12x + 3 = 0
Pas de racine évidente, pas de solution
