数学题 方程的 急!!!!

若a=0;ax^2+3x-2b=0解为2b/3;带入3x^2-ax+2b=0得b=-3/2
若a!=0;
方程ax^2+3x-2b=0的根为x1,x2;3x^2-ax+2b=0的根为1/x1,1/x2;
而方程ax^2+3x-2b=0中,x1+x2=-3/a,x1*x2=-2b/a;
方程3x^2-ax+2b=0中,1/x1+1/x2=a/3;1/x1*1/x2=2b/3
有上各式:2ab=9;-4b^2=3a;得a=-3;b=-3/2.

设方程ax^2+3x-2b=0的根是p,q

p+q=-3/a.......(1)
pq=-2b/a.......(2)

则方程3x^2-ax+2b=0的根为1/p,1/q

1/p+1/q=a/3......(3)

1/p*1/q=2b/3......(4)

即(1)(3)得：(p+q)/pq=(-3/a)/(-2b/a)=a/3

3/2b=a/3
2ab=9......(5)

(2)(4)得:1/(-2b/a)=2b/3
-a/2b=2b/3

-3a=4b^2.....(6)

(5)(6)得：-3*9/2b=4b^2

-27/8=b^3

b=-3/2

a=9/2b=9/(-3)=-3