bjr help svp résoudre a) 3x²-x-2=0 b) x²+5=0 b) -x²+5x-1

Bonjour

a) 3x²-x-2

calculons delta = b²-4ac

delta = (-1)² - 4(3)(-2) = 1 + 24 = 25

delta = 25 et V(25) = 5

x1 = (1+5) / 6 = 6/6 =1

x1 = 1

x2 = (1-5) / 6 = -4/6 = -2/3

x2 = -2/3

S = {-2/3;1}

b) x²+5=0

 x² = -5 impossible car le carré d'un nombre n'est jamais négatif

 S =  ∅

c) -x²+5x-1 ≤ 0

Calculons delta = b²-4ac

delta = 5² - 4(-1)(-1) = 25 - 4 = 21

delta = 21

x1 = (-5-V21) / -2 = (5+V21) / 2

x2 = (-5 + V21) / -2 = (5-V21) / 2

tableau de signe

x            I -oo               (5-V21)/2                (5+V21)/2              +oo              

-x²+5x-1                -             0                +              0               -

S = ]-oo;(5-V21)/2]U[(5+V21)/2;+oo[