(m-2)x²+(2m-1)x+m-3=0
Δ=(2m-1)²-4(2m-1)(m-3)=4m²-4m+1-4(2m²-6m-m+3)=4m²-4m+1-8m²+24m+4m-12=
-4m²+24m-11>0⇒(-2m+1)(2m-11)>0
-2m+1=0⇔-2m=-1⇔m=1/2
2m-11=0⇔2m=11⇔m=11/2
Fie aceasta ecuatie o functie de gradul 2.
Deci: f(x)⇒-2m+1>0⇔m<-1/2
2m-11>0⇔m>11/2
si: -2m+1<0⇔m>-1/2
2m-11<0⇔m<11/2
De-aici vom extrage carei valori apartine m.
m∈(-1/2;11/2).
Sper ca te-am ajutat.
