Oblicz dla jakich wartości parametru m równanie
2x² -(m -1)x+m+1=0 ma dwa pierwiastki ujemne .
ze wzorow Viete'a
-b/a<0 i c/a>0 i Δ>0
(m-1)/2<0 i (m+1)/2>0
m<1 i m>-1
m∈(-1;1)
Δ=m²-2m+1-4m²-8m-4=-3m²-8m-3>0
Δ_m=64-36=28
√Δm=2√7
m=(-8-2√7)/(-6)=(3-√7)/6 v m=3+√7>1∉D
Odp. m∈(3-√7; 1)
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ze wzorow Viete'a
-b/a<0 i c/a>0 i Δ>0
(m-1)/2<0 i (m+1)/2>0
m<1 i m>-1
m∈(-1;1)
Δ=m²-2m+1-4m²-8m-4=-3m²-8m-3>0
Δ_m=64-36=28
√Δm=2√7
m=(-8-2√7)/(-6)=(3-√7)/6 v m=3+√7>1∉D
Odp. m∈(3-√7; 1)