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równanie mieć rozwiązanie zatem
Δ>0
Δ= b²-4ac, gdzie b= 3, a=1, c=k
Δ= 9-4k
9-4k>0
-4k>-9/:(-4)
k<9/4
b)x²-kx+k+1≥0
Δ= k²-4(k+1)
k²-4k-4>0
Δ=16+16=32
√Δ= 4√2
k= 4-4√2/2= 2-2√2
k= 2+2√2
zatem
k∈(-∞, 2-2√2)u(2+2√2,∞)