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Δ<0
Δ=k²-4(k+3)=k²-4k+3<0
k²-3k-k+3<0
k(k-3)-(k-3)<0
(k-3)(k-1)<0
k∈(1,3)
b)
Δ=k²-4(k+1)<0
k²-4k-4<0
Δ₁=16+16=32
k₁=1/2*(4-4√2)=2-2√2, k₂=2+2√2
k∈<2-2√2, 2+√2>