Rozwiązanie:
Zadanie 1.
[tex]x^{2}+(m+2)x+3m-2=0\\\Delta=m^{2}+4m+4-4(3m-2)=m^{2}+4m+4-12m+8=m^{2}-8m+12\\\Delta>0\\m^{2}-8m+12>0\\\Delta_{m}=64-4*1*12=16\\m_{1}=\frac{8+4}{2}=6\\m_{2}=\frac{8-4}{2}=2\\m \in (-\infty,2) \cup (6,\infty)\\x_{1}+x_{2}>0\\-(m+2)>0\\m+2<0\\m<-2\\x_{1}x_{2}>0\\3m-2>0\\3m>2\\m>\frac{2}{3} \\[/tex]
Ostatecznie:
[tex]m \in \varnothing[/tex]
Zatem taka wartość parametru [tex]m[/tex] nie istnieje.
Zadanie 2.
[tex]x^{2}+(m+2)x-2m-1\\\Delta=m^{2}+4m+4+4(2m+1)=m^{2}+4m+4+8m+4=m^{2}+12m+8\\\Delta<0\\m^{2}+12m+8<0\\\Delta_{m}=144-4*1*8=112\\m_{1}=\frac{-12-4\sqrt{7} }{2} =-6-2\sqrt{7}\\m_{2}=\frac{-12+4\sqrt{7} }{2}=-6+2\sqrt{7} \\m \in (-6-2\sqrt{7},-6+2\sqrt{7})[/tex]
