Miejsca zerowe paraboli y=1/3xkwadrat-x+2/3 to:
a) -1,-2
b) -1,2
c) 1,-2
d)1,2
proszę obliczenia
Δ=b² -4ac =1 - 4*1/3*2/3=1-8/9=1/9
√Δ=1/3
x1=(-b-√Δ)/2a=(1-1/3)/2/3 = 2/3 : 2/3 =1
x2=(-b+√Δ)/2a=(1+1/3)/2/3 = 4/3 * 3/2 = 2
odp d
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Δ=b² -4ac =1 - 4*1/3*2/3=1-8/9=1/9
√Δ=1/3
x1=(-b-√Δ)/2a=(1-1/3)/2/3 = 2/3 : 2/3 =1
x2=(-b+√Δ)/2a=(1+1/3)/2/3 = 4/3 * 3/2 = 2
odp d