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2009-10-24T15:03:31+02:00
(x+3)²-5(x+3)+4=0

Grupujemy wyrazy

(x+3)[(x+3) -5] +4 =0
(x+3)(x-2)+4=0
x²-2x +3x -6 +4 =0
x² +x -2 =0

Delta:

Δ=b² -4ac
Δ = 1 + 8 =9

x1 = (-b-√Δ)/2a = (2-3)/2 = - 1/2 = -0.5
x2 = (-b+√Δ)/2a = (2+3)/2 = 5/2 = 2,5
2009-10-24T15:06:47+02:00
(x+3)²-5(x+3)+4=0
x²+6x+9-5x-15+4= 0
x²+x-2=0

Δ=b²-4*a*c
Δ=1 -4*1*2
Δ= 9 √Δ=3

x₁= -b+√Δ/2a x₂= -b -√Δ/2a
x₁= -1+3/2 x₂= -1-3/2
x₁= 1 x₂=-2

(x-1)(x+2)= 0
czyli
x-1=0 lub x+2= 0
x=1 lub x= -2

Rozwiązaniem jest x=1 lub x= -2