sa se determine m apartine lui r astfel incat dreapta de ecuatia y=3x-2 este secanta parabolei de ecuatie y=(m-2)xpatrat -mx+m
Răspunsuri la întrebare
Explicație pas cu pas:
(m-2)*x²-mx+m=3x-2
(m-2)*x²-mx+m=3x-2(m-2)*x²-x(m+3)+m+2=0
(m-2)*x²-mx+m=3x-2(m-2)*x²-x(m+3)+m+2=0delta>0
(m-2)*x²-mx+m=3x-2(m-2)*x²-x(m+3)+m+2=0delta>0delta=(m+3)²-4(m²-4)=m²+6m+9-4m²+16>0
(m-2)*x²-mx+m=3x-2(m-2)*x²-x(m+3)+m+2=0delta>0delta=(m+3)²-4(m²-4)=m²+6m+9-4m²+16>0-3m²+6m+25>0
(m-2)*x²-mx+m=3x-2(m-2)*x²-x(m+3)+m+2=0delta>0delta=(m+3)²-4(m²-4)=m²+6m+9-4m²+16>0-3m²+6m+25>03m²-6m-25<0
(m-2)*x²-mx+m=3x-2(m-2)*x²-x(m+3)+m+2=0delta>0delta=(m+3)²-4(m²-4)=m²+6m+9-4m²+16>0-3m²+6m+25>03m²-6m-25<0delta= 36+12*25=336
(m-2)*x²-mx+m=3x-2(m-2)*x²-x(m+3)+m+2=0delta>0delta=(m+3)²-4(m²-4)=m²+6m+9-4m²+16>0-3m²+6m+25>03m²-6m-25<0delta= 36+12*25=336x=(6+4rad21)/6=(3+2rad2)/3
(m-2)*x²-mx+m=3x-2(m-2)*x²-x(m+3)+m+2=0delta>0delta=(m+3)²-4(m²-4)=m²+6m+9-4m²+16>0-3m²+6m+25>03m²-6m-25<0delta= 36+12*25=336x=(6+4rad21)/6=(3+2rad2)/3sau x=(3-2rad2)/3
(m-2)*x²-mx+m=3x-2(m-2)*x²-x(m+3)+m+2=0delta>0delta=(m+3)²-4(m²-4)=m²+6m+9-4m²+16>0-3m²+6m+25>03m²-6m-25<0delta= 36+12*25=336x=(6+4rad21)/6=(3+2rad2)/3sau x=(3-2rad2)/3deci: x€( (3-2rad2)/3 , (3+2rad2)/3 )
(m-2)*x²-mx+m=3x-2(m-2)*x²-x(m+3)+m+2=0delta>0delta=(m+3)²-4(m²-4)=m²+6m+9-4m²+16>0-3m²+6m+25>03m²-6m-25<0delta= 36+12*25=336x=(6+4rad21)/6=(3+2rad2)/3sau x=(3-2rad2)/3deci: x€( (3-2rad2)/3 , (3+2rad2)/3 )
UN MODERATOR va rog , care e problema de mi s-a atasat asa raspuns??
Baftă!