quadratic equation問題!!!

f(x)= x^2-4mx-(5m^2-6m+1),where m>1/3.

Let a, b be the roots of the equation of the f(x)=0,where a>b

Express a and b in term of m
a+b=4m...(1)
ab=-(5m^2-6m+1)...(2)
from (1)
a=4m-b
sub into (2)
b(4m-b)=-(5m^2-6m+1)
b^2-4mb-(5m^2-6m+1)=0
b^2-4mb+(-5m+1)(m-1)=0
(b-(-5m+1))(b-(m-1))=0
b=1-5m or b=m-1
a=4m-(1-5m) or a=4m-(m-1)
a=9m-1 or a=3m+1



2006-11-21 19:22:10 補充：
SORRYit should beb=5m-1 or b=m-1 (reject)a=4m-(5m-1)a=-m 1

x^2 - 4mx = 5m^2-6m+1
x^2 - 4mx + 4m^2 = 5m^2-6m+1+4m^2
(x - 2m)^2 = (3m-1)^2
(x - 2m)^2 - (3m-1)^2 = 0
(x - 2m + 3m - 1)(x - 2m - 3m + 1) = 0
(x + m - 1)(x - 5m + 1) = 0
x = 1 - m or x = 5m - 1

Since
m > 1/3
6m > 2
5m + m > 1 + 1
5m - 1 > 1 - m

Since a > b,
then a = 5m - 1, b = 1 - m

Here is the hints.
1. a, b is the root, and a>b
=> distinct root
=> discriminant > 0
2. discriminatnt is b^2-4ac
3. b = -4m
a=1
c= -(5m^2-6m+1)

Hope it can help