Question

己知关于x的一元二次方程mx²-(2m-1)x+(m-2)=0(m>0).

Answer 1

x1+x2=(2m-1)/m

x1x2=(m-2)/m

x1x2-3(x1+x2)+9=5m

(m-2)/m-3(2m-1)/m=5m-9

(m-2-6m+3)/m=5m-9

1-5m=5m²-9m

5m²-4m-1=0

1 -1

5 1

(m-1)(5m+1)=0

m=1 m=-1/5

b²-4ac

=(2m-1)²-4m(m-2)

=4m²-4m+1-4m²+8m

=4m+1>0

m>-1/4

∴m=1 m=-1/5