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Ranyaelhassan
@Ranyaelhassan
April 2019
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Je dois resoudre cette equation:
(x+1)^2-3^2=0
^2 signifiant au carre
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pierrebiret
Il faut utiliser la troisième identité remarquable
a^2 - b^2 = (a+b)(a-b)
donc
(x+1)^2 - 3^2 = 0
<=> (x+1+3)(x+1-3) = 0
<=> (x+4)(x-2) = 0
<=> x+4 = 0 ou x-2 = 0 (règle du produit nul)
<=> x=-4 ou x=2
donc S = {-4; 2)
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a^2 - b^2 = (a+b)(a-b)
donc
(x+1)^2 - 3^2 = 0
<=> (x+1+3)(x+1-3) = 0
<=> (x+4)(x-2) = 0
<=> x+4 = 0 ou x-2 = 0 (règle du produit nul)
<=> x=-4 ou x=2
donc S = {-4; 2)