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Oplossen

z²-4+5(z-2)=0
(y+2)·(y+3)=6(y+1)

inge
19-8-2003

Antwoord

Beste Inge,

z² -4 + 5(z - 2) = 0
z² -4 + 5z - 10 = 0
z² + 5z - 14 = 0
(z+7)(z-2) = 0
(z+7)=0 Ú (z-2)=0
z=-7 Ú z=2


(y+2)(y+3)=6(y+1)
y² + 3y + 2y + 6 = 6y + 6
y² + 5y + 6 = 6y + 6
y² + 5y - 6y + 6 -6 = 0
y² - y = 0
y(y - 1) = 0
y = 0 Ú (y - 1) = 0
y = 0 Ú y = 1

Davy
19-8-2003