Réponses

2013-09-21T14:20:00+02:00
Bonjour,

On utilise l'identité remarquable a²-b² = (a+b)(a-b) :
\left(x^2+3x-1\right)^2 - \left(x^2-3x+1\right)^2 = 0\\
\left[\left(x^2+3x-1\right)-\left(x^2-3x-1\right)\right]\left[\left(x^2+3x-1\right)+\left(x^2-3x-1\right)\right]\\
\left(x^2+3x-1-x^2+3x+1\right)\left(x^2+3x-1+x^2-3x-1\right) = 0\\
6x\left(2x^2-2\right) = 0\\
On continue à factoriser :
6x\left(2x^2-2\right) = 0\\
12x\left(x^2-1\right) = 0\\
12x\left(x+1\right)\left(x-1\right) = 0

Donc :
12x = 0
x = 0
OU
x+1 = 0
x = -1
OU
x-1 = 0
x = 1

S = \left\{-1 ; 0 ; 1\right\}