Question

Dla jakich wartości parametru m równanie -x^2+(2m-1)x+m=0 ma dwa pierwiastki które różnią się co najmniej o 4?

Answer 1

-x²+(2m-1)x+m=0     a=-1,  b=2m-1,  c=m      Δ=b²-4ac=(2m-1)²-4·(-1)·m= 4m²-4m+1+4m

                                                                   Δ=4m²+1  √Δ=√(4m²+1)

 

                                 x₂=x₁-4

 

                                (-b-√Δ)/2a=(-b+√Δ)/2a-4   |·2a

                                (-b-√Δ)=(-b+√Δ)-8a

                                -b-√Δ+b-√Δ+8a=0

                                  -2√Δ+8a=0 |:-2

                                     √Δ+4a=0

                                     √(4m²+1)-4=0                  

                                     √(4m²+1)=4

                                     √(4m²+1)=√16

                                       4m²+1=16

                                         4m²=15

                                           m²=15/4

                                           m=± √15/2